---
title: "Topology of Metric Spaces"
book: "University Mathematics — Year 2"
subject: math
language: en
chapter: 4
exercises: 12
source: https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces
---

# Chapter 4 — Topology of Metric Spaces

The topology of the real line (Year 1 volume) generalizes, almost without changing a word, to any set equipped with a distance. The gain is enormous: sequences of functions, matrices, curves — all become points of [metric spaces](#def-b2-metric-def), and the three pillars proved here — [completeness](#def-b2-metric-complete) with the Banach fixed point theorem, [compactness](#def-b2-metric-compact), [connectedness](#def-b2-metric-connected) — apply to them uniformly. This chapter is the backbone of the whole analysis half of the book.

## 4.1 Metric spaces

**Definition 4.1.**

A *metric space* is a set $X$ with a map $d
\colon X \times X \to \R_+$ such that, for all $x, y, z$:

$$
d(x,y) = 0 \iff x = y,
\qquad
d(x,y) = d(y,x),
\qquad
d(x,z) \leq d(x,y) + d(y,z).
$$

Balls: $B(a, r) = \{x : d(a,x) < r\}$ ([open](#def-b2-metric-topology)), $\overline B(a,r) =
\{x : d(a,x) \leq r\}$ (closed). A subset $A \subseteq X$ becomes a metric space with the induced distance.

**Example 4.2.**

$\R$ with $\abs{x - y}$; $\R^n$ with any of

$$
d_1(x,y) = \sum_i \abs{x_i - y_i},
\quad
d_2(x,y) = \Bigl(\sum_i (x_i - y_i)^2\Bigr)^{1/2},
\quad
d_\infty(x,y) = \max_i \abs{x_i - y_i};
$$

the set $C(\intcc{a}{b})$ of [continuous](#def-b2-metric-continuity) functions with the *sup distance* $d_\infty(f, g) = \sup_{\intcc{a}{b}} \abs{f -
g}$ (finite: $f - g$ is bounded); any set with the *discrete* distance ($d(x,y) = 1$ for $x \neq y$). Distances coming from norms are the subject of [Chapter 5](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#ch-b2-nvs).

**Definition 4.3 (Topology of a metric space).**

$U \subseteq X$ is *open* when every point of $U$ is the center of a ball contained in $U$; $F$ is *closed* when its complement is open. Neighborhoods, interior, closure, density, boundary are defined exactly as on the real line (Year 1 volume), with balls replacing intervals, and the statements proved there — unions/intersections of open sets, characterizations of interior and closure, closure as smallest closed superset — carry over with the same proofs. Open balls are open, closed balls are closed (triangle inequality).

**Example 4.4 (Interior, closure, boundary on one set).**

In $\R$, let $A = \intoc{0}{1} \cup \{2\}$. Interior: $\intoo{0}{1}$ — around any $x \in \intoo01$ a small ball stays in $A$; around $1$, every ball $\intoo{1-r}{1+r}$ leaks out of $A$ on the right, so $1$ is not interior; and the isolated $2$ is not interior either. Closure: $\intcc{0}{1} \cup \{2\}$ (the point $0$ is a limit of $A$, nothing else is added). Boundary (closure minus interior): $\{0, 1, 2\}$. Note the asymmetries worth remembering: an endpoint can belong to a set without being interior ($1$), can be adherent without belonging ($0$), and an isolated point is its own boundary ($2$). The same bookkeeping runs verbatim in any [metric space](#def-b2-metric-def), with balls in place of intervals.

**Definition 4.5 (Limits, continuity).**

$x_n \to x$ in $X$ when $d(x_n, x) \to 0$. A map $f \colon X \to Y$ between [metric spaces](#def-b2-metric-def) is *continuous* at $a$ when

$$
\forall \varepsilon > 0,\ \exists\delta > 0,\quad
d_X(x, a) \leq \delta \implies d_Y\bigl(f(x), f(a)\bigr) \leq
\varepsilon ;
$$

equivalently (same proof as on $\R$), $f(x_n) \to f(a)$ for every sequence $x_n \to a$. $f$ is *Lipschitz* with constant $k$ when $d_Y(f(x), f(y)) \leq k\, d_X(x, y)$ always — then uniformly continuous, hence continuous.

**Theorem 4.6 (Global characterization of continuity).**

$f \colon X \to Y$ is [continuous](#def-b2-metric-continuity) (at every point) if and only if the preimage of every [open](#def-b2-metric-topology) set is [open](#def-b2-metric-topology) — iff the preimage of every closed set is closed.

**Proof.** ($\Rightarrow$) Let $V \subseteq Y$ be [open](#def-b2-metric-topology) and $a \in f^{-1}(V)$: some ball $B(f(a), \varepsilon) \subseteq V$; [continuity](#def-b2-metric-continuity) at $a$ provides $\delta$ with $f(B(a, \delta)) \subseteq B(f(a),
\varepsilon)$, so $B(a, \delta) \subseteq f^{-1}(V)$.

($\Leftarrow$) Given $a$ and $\varepsilon$: $f^{-1}\bigl(B(f(a),
\varepsilon)\bigr)$ is [open](#def-b2-metric-topology) and contains $a$, hence contains a ball $B(a, \delta)$: that is the definition of [continuity](#def-b2-metric-continuity) at $a$. Closed sets: complements ([Proposition 1.1](https://one-course.com/books/math/4/en/chapter/1-sets-and-structures#prop-b2-structures-images)). ∎

## 4.2 Complete spaces

**Definition 4.7.**

A sequence $(x_n)$ is *Cauchy* when $\sup_{p, q \geq N} d(x_p,
x_q) \to 0$ as $N \to \infty$. A [metric space](#def-b2-metric-def) is *complete* when every Cauchy sequence converges. Convergent $\Rightarrow$ Cauchy always; closed subsets of complete spaces are complete, and complete subsets of any space are closed (same proofs as on $\R$: Year 1 volume).

**Example 4.8 (Cauchy without a limit).**

In $X = \Q$ with the usual distance, the decimal truncations of $\sqrt2$,

$$
x_0 = 1,\quad x_1 = 1.4,\quad x_2 = 1.41,\quad x_3 = 1.414,
\quad\dots
$$

satisfy $\abs{x_p - x_q} \leq 10^{-\min(p,q)}$: Cauchy in $\Q$. A limit in $\Q$ would also be the limit in $\R$, namely $\sqrt2 \notin \Q$: no limit exists in $X$. Incompleteness is the presence of such “phantom limits”; [completeness](#def-b2-metric-complete) of $\R$ was engineered in the Year 1 volume precisely to give every Cauchy sequence a home.

**Theorem 4.9.**

$\R^n$ (any of the three distances of [Example 4.2](#ex-b2-metric-examples)) and $\bigl(C(\intcc{a}{b}),
d_\infty\bigr)$ are [complete](#def-b2-metric-complete).

**Proof.** $\R^n$: a Cauchy sequence is Cauchy in each coordinate (each $\abs{x_i - y_i} \leq d(x,y)$ for all three distances), so each coordinate converges ([completeness](#def-b2-metric-complete) of $\R$, Year 1 volume), and coordinatewise convergence implies convergence for $d_\infty$ (finitely many coordinates), hence for all three (the three distances dominate each other within constant factors: $d_\infty
\leq d_2 \leq d_1 \leq n\,d_\infty$).

$C(\intcc{a}{b})$: let $(f_n)$ be $d_\infty$-Cauchy. For each $x$, $(f_n(x))$ is Cauchy in $\R$ ($\abs{f_p(x) - f_q(x)} \leq
d_\infty(f_p, f_q)$): converges to some $f(x)$. Passing to the limit in $\abs{f_p(x) - f_q(x)} \leq \varepsilon$ (valid for $p, q
\geq N_\varepsilon$, all $x$) as $q \to \infty$: $\abs{f_p(x) -
f(x)} \leq \varepsilon$ for all $x$, i.e. $d_\infty(f_p, f) \leq
\varepsilon$: uniform convergence. The limit is [continuous](#def-b2-metric-continuity): given $\varepsilon$, pick $p$ with $\sup\abs{f_p - f} \leq \varepsilon$, then use [continuity](#def-b2-metric-continuity) of $f_p$ at $a$ and the three-term split

$$
\abs{f(x) - f(a)} \leq \abs{f(x) - f_p(x)} + \abs{f_p(x) - f_p(a)}
+ \abs{f_p(a) - f(a)} \leq 3\varepsilon
$$

for $x$ close to $a$. (This “$3\varepsilon$ argument” returns as the uniform-limit theorem of [Chapter 10](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#ch-b2-funcseq).) ∎

**Example 4.10 (Open and closed sets recognized by continuity).**

The global characterization ([Theorem 4.6](#thm-b2-metric-globalcontinuity)) is the everyday tool for topological bookkeeping. In $\R^2$: the set $\{(x, y) : x^2 + y^2
< 1,\ y > x^3\}$ is [open](#def-b2-metric-topology) — it is $g^{-1}(\intoo{-\infty}{1})
\cap h^{-1}(\intoo{0}{+\infty})$ for the [continuous](#def-b2-metric-continuity) $g(x,y) =
x^2 + y^2$ and $h(x, y) = y - x^3$, an intersection of two [open](#def-b2-metric-topology) preimages. In $\bigl(C(\intcc01), d_\infty\bigr)$: the set of functions with $f(0) = f(1)$ and $\int_0^1 f = 0$ is closed — the preimage of $\{(0,0)\}$ under the [continuous](#def-b2-metric-continuity) map $f \mapsto
\bigl(f(0) - f(1),\ \int_0^1 f\bigr)$ into $\R^2$ (each coordinate is $1$-Lipschitz, as in [Exercise 4.3](#exo-b2-metric-3)). The method never draws a picture: exhibit a [continuous](#def-b2-metric-continuity) map, read the set as a preimage, quote the theorem.

**Example 4.11 (A closed set defined by infinitely many conditions).**

In $\bigl(C(\intcc01), d_\infty\bigr)$, the set

$$
L = \{f : \abs{f(x) - f(y)} \leq \abs{x - y}
\ \text{for all } x, y\}
$$

of $1$-Lipschitz functions is closed, although it is cut out by uncountably many conditions: for each fixed pair $(x, y)$, the map $f \mapsto \abs{f(x) - f(y)} - \abs{x - y}$ is [continuous](#def-b2-metric-continuity) (evaluations are $1$-Lipschitz), so each single condition defines a closed set, and $L$ is the *intersection* of this family — an arbitrary intersection of closed sets is closed. The same template certifies closedness for monotone functions, convex functions, functions bounded by a fixed $g$: uniform limits inherit every property expressible as a family of closed pointwise constraints. What uniform limits do *not* automatically inherit — differentiability, for one — is exactly what [Chapter 10](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#ch-b2-funcseq) must labour for.

**Theorem 4.12 (Banach fixed point theorem).**

Let $X$ be a nonempty [complete metric space](#def-b2-metric-complete) and $f \colon X \to X$ a *contraction*: [Lipschitz](#def-b2-metric-continuity) with constant $k < 1$. Then $f$ has a unique fixed point $\ell$, and every orbit $x_{n+1} = f(x_n)$ converges to $\ell$, with

$$
d(x_n, \ell) \leq \frac{k^n}{1 - k}\, d(x_1, x_0) .
$$

**Proof.** Uniqueness: two fixed points are at distance $\leq k$ times itself. Existence: $d(x_{n+1}, x_n) \leq k^n d(x_1, x_0)$ by induction, so for $q > p$,

$$
d(x_q, x_p) \leq \sum_{j=p}^{q-1} d(x_{j+1}, x_j)
\leq d(x_1, x_0) \sum_{j \geq p} k^j
= \frac{k^p}{1-k}\, d(x_1, x_0) \xrightarrow[p\to\infty]{} 0 :
$$

Cauchy, hence convergent to some $\ell$; [continuity](#def-b2-metric-continuity) of $f$ passes $x_{n+1} = f(x_n)$ to the limit: $\ell = f(\ell)$. The error bound is the displayed estimate with $q \to \infty$. ∎

**Example 4.13 (An integral equation).**

On $X = C(\intcc{0}{1})$ ([complete](#def-b2-metric-complete), [Theorem 4.9](#thm-b2-metric-rncomplete)), consider $T(f)(x) = 1 + \frac12
\int_0^x f(t)\,\dd t$. For $f, g \in X$:

$$
\abs{T(f)(x) - T(g)(x)} \leq \frac12 \int_0^x \abs{f - g} \leq
\frac12\, d_\infty(f, g),
$$

so $T$ is a $\frac12$-contraction: it has a unique [continuous](#def-b2-metric-continuity) fixed point — the solution of $f' = \frac f2$, $f(0) = 1$, namely $\eu^{x/2}$. This scheme, industrialized, becomes the Cauchy–Lipschitz theorem of [Chapter 16](https://one-course.com/books/math/4/en/chapter/16-differential-equations#ch-b2-diffeq).

**Example 4.14 (A numerical fixed point: x=cos⁡xx = \cos xx=cosx).**

On the [complete](#def-b2-metric-complete) $X = \intcc{0}{1}$, the map $f = \cos$ sends $X$ into $\intcc{\cos 1}{1} \subseteq X$ and is a contraction: by the mean value inequality,

$$
\abs{\cos x - \cos y} \leq \bigl(\sup_{\intcc01}\abs{\sin}\bigr)
\abs{x - y} = (\sin 1)\abs{x - y},
\qquad \sin 1 \approx 0.841 < 1 .
$$

Banach: a unique solution of $x = \cos x$ in $\intcc01$ (hence in $\R$: any real fixed point lies in $\intcc{-1}{1}$, then in $\intcc{\cos 1}{1}$ after one application), and the iteration $x_{n+1} = \cos x_n$ converges to it from any start: $x_\infty
\approx 0.739085$, the famous number obtained by hammering the cosine key of a calculator. The error bound predicts $(\sin1)^n/(1 - \sin1)$ decay — about one digit per $13$ presses; the a posteriori bound of this chapter’s weekend problem (question 14) certifies each step on the fly.

## 4.3 Compactness

**Definition 4.15.**

A [metric space](#def-b2-metric-def) $X$ is *compact* when every sequence in $X$ has a subsequence converging *in* $X$ (the Bolzano–Weierstrass property). A subset is compact when it is so with the induced distance.

**Theorem 4.16 (First properties).**

1. A [compact](#def-b2-metric-compact) subset is closed and bounded; a closed subset of a [compact](#def-b2-metric-compact) space is [compact](#def-b2-metric-compact) .
2. In $\R^n$ , the converse holds: [compact](#def-b2-metric-compact) $\iff$ closed and bounded.
3. A [continuous](#def-b2-metric-continuity) image of a [compact](#def-b2-metric-compact) space is [compact](#def-b2-metric-compact) ; a [continuous](#def-b2-metric-continuity) real function on a nonempty [compact](#def-b2-metric-compact) space is bounded and attains its bounds.
4. (Heine) A [continuous](#def-b2-metric-continuity) map on a [compact](#def-b2-metric-compact) space is uniformly continuous.
5. Products: if $X, Y$ are [compact](#def-b2-metric-compact) , so is $X \times Y$ (with $d\bigl((x,y),(x',y')\bigr) = d(x,x') + d(y,y')$ ).

**Proof.** (1) Same arguments as on the line (Year 1 volume): an escaping-to- infinity or converging-outside sequence has no subsequence converging inside; for the second claim, extract in the ambient [compact](#def-b2-metric-compact) and use closedness.

(2) Bounded sequences in $\R^n$ have componentwise convergent subsequences: extract on the first coordinate (Bolzano–Weierstrass on $\R$), then, from that subsequence, on the second, and so on ($n$ successive extractions); closedness keeps the limit inside.

(3) Given $(f(x_n))$, extract $x_{\varphi(n)} \to x \in X$; [continuity](#def-b2-metric-continuity) gives $f(x_{\varphi(n)}) \to f(x) \in f(X)$. Real case: [compactness](#def-b2-metric-compact) of $f(X) \subseteq \R$ makes it closed and bounded, and $\sup f(X) \in f(X)$ (the sup of a set is adherent to it, and $f(X)$ is closed).

(4) The Year 1 proof transfers verbatim; here it is, in metric dress. Suppose $f \colon X \to Y$ [continuous](#def-b2-metric-continuity) on the [compact](#def-b2-metric-compact) $X$ but not uniformly continuous: some $\varepsilon > 0$ admits, for every $n$, points with

$$
d_X(x_n, y_n) \leq \frac{1}{n+1}
\qquad\text{and}\qquad
d_Y\bigl(f(x_n), f(y_n)\bigr) > \varepsilon .
$$

Extract $x_{\varphi(n)} \to a \in X$; then $y_{\varphi(n)} \to
a$ too (the mutual distances tend to $0$). [Continuity](#def-b2-metric-continuity) at $a$ sends both image sequences to $f(a)$, so $d_Y\bigl(f(x_{\varphi(n)}), f(y_{\varphi(n)})\bigr) \to 0$ — contradicting the uniform gap $> \varepsilon$. [Compactness](#def-b2-metric-compact) supplied exactly one thing: the cluster point $a$ at which to apply plain [continuity](#def-b2-metric-continuity).

(5) Extract on the $X$-coordinates, then again on the $Y$-coordinates. ∎

**Example 4.17 (Heine’s theorem, with and without compactness).**

On $\intcc{0}{1}$, the function $x \mapsto x^2$ is uniformly continuous — Heine says so with no computation, but the direct estimate is instructive:

$$
\abs{x^2 - y^2} = \abs{x + y}\,\abs{x - y} \leq 2\abs{x - y},
$$

so $\delta = \varepsilon/2$ works *for every point at once*. On $\R$ the same function is not uniformly continuous: with $x_n = n$ and $y_n = n + \frac1n$, the gap $\abs{x_n - y_n}
= \frac1n \to 0$ while $\abs{x_n^2 - y_n^2} = 2 + \frac1{n^2}
\geq 2$: no single $\delta$ serves $\varepsilon = 1$. The mechanism is visible: the local [Lipschitz](#def-b2-metric-continuity) constant $\abs{x + y}$ is bounded on a [compact](#def-b2-metric-compact) and unbounded on $\R$ — Heine’s theorem is exactly the statement that [compactness](#def-b2-metric-compact) caps such local constants uniformly.

**Method 4.18 (Proving that a set is compact).**

Three routes, in order of frequency. (1) *Ambient recognition:* in $\R^n$ (or any finite-dimensional normed space, [Chapter 5](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#ch-b2-nvs)), verify closed — typically as a preimage, [Example 4.10](#ex-b2-metric-recognize) — and bounded. (2) *Inheritance:* a closed subset of a known [compact](#def-b2-metric-compact) space is [compact](#def-b2-metric-compact); a finite union or a product of [compacts](#def-b2-metric-compact) is [compact](#def-b2-metric-compact); a [continuous](#def-b2-metric-continuity) image of a [compact](#def-b2-metric-compact) is [compact](#def-b2-metric-compact). (3) *Bare hands:* extract a convergent subsequence from an arbitrary sequence — usually by successive extractions coordinate by coordinate. To prove *non*-compactness, one witness suffices: a sequence with no convergent subsequence, most often points at mutual distance $\geq \varepsilon$.

**Example 4.19 (Distances between sets: compactness earns its keep).**

Let $K$ be [compact](#def-b2-metric-compact), $F$ closed, $K \cap F = \emptyset$ in a [metric space](#def-b2-metric-def). Then

$$
d(K, F) = \inf\,\{d(x, y) : x \in K,\ y \in F\} > 0 :
$$

the function $x \mapsto d(x, F)$ is [continuous](#def-b2-metric-continuity) ([Exercise 4.11](#exo-b2-metric-11)) and positive on $K$ ($d(x, F) = 0$ would put $x \in \overline F = F$), so it attains a positive minimum on the [compact](#def-b2-metric-compact) $K$ ([Theorem 4.16](#thm-b2-metric-compactprops) (3)). [Compactness](#def-b2-metric-compact) is not decorative: for two *closed* sets the infimum can vanish without being attained — in $\R^2$, the hyperbola $F_1 = \{xy
= 1\}$ and the axis $F_2 = \{y = 0\}$ are disjoint closed sets with $d(F_1, F_2) = 0$ (the points $(n, \frac1n)$ approach the axis). Escape to infinity is exactly what [compactness](#def-b2-metric-compact) forbids.

**Theorem 4.20 (Borel–Lebesgue).**

A [metric space](#def-b2-metric-def) $X$ is [compact](#def-b2-metric-compact) if and only if every cover of $X$ by [open](#def-b2-metric-topology) sets has a *finite* subcover.

**Proof.** ($\Leftarrow$) Suppose $(x_n)$ has no convergent subsequence. We claim every $x \in X$ has a ball $B(x, r_x)$ containing $x_n$ for only finitely many indices $n$: otherwise, every ball $B(x, \frac1{k+1})$ would contain infinitely many terms, and choosing indices

$$
\varphi(0) < \varphi(1) < \varphi(2) < \cdots
\quad\text{with}\quad
x_{\varphi(k)} \in B\Bigl(x, \frac{1}{k+1}\Bigr)
$$

(possible at each step precisely because infinitely many candidates remain) would build a subsequence converging to $x$. The balls $B(x, r_x)$ cover $X$; if finitely many of them covered $X$, the index set $\N$ would be a finite union of finite sets: absurd.

($\Rightarrow$) Two steps. *Lebesgue number:* for an [open](#def-b2-metric-topology) cover $(U_i)$ of a [compact](#def-b2-metric-compact) $X$, there is $\rho > 0$ such that every ball of radius $\rho$ lies in some $U_i$. Otherwise, for each $n$ pick $x_n$ with $B(x_n, \frac{1}{n+1})$ in no $U_i$; extract $x_{\varphi(n)} \to x \in U_{i_0} \supseteq B(x, r)$; for large $n$, $B(x_{\varphi(n)}, \frac{1}{\varphi(n)+1}) \subseteq B(x, r)
\subseteq U_{i_0}$: contradiction. *Total boundedness:* for every $\rho > 0$, finitely many balls of radius $\rho$ cover $X$. Otherwise pick inductively $x_{n+1}$ outside $B(x_0, \rho) \cup
\dots \cup B(x_n, \rho)$: the sequence has pairwise distances $\geq
\rho$, so no Cauchy — hence no convergent — subsequence: contradiction. Combining: cover $X$ by finitely many balls of radius $\rho$ (the Lebesgue number), each inside some $U_i$: a finite subcover. ∎

**Example 4.21 (An ε\varepsilonε-net, counted).**

Total boundedness (from the proof of [Theorem 4.20](#thm-b2-metric-borellebesgue)) is very concrete on $\intcc{0}{1}$: for $\varepsilon > 0$, the $\lceil
\frac{1}{2\varepsilon}\rceil$ balls centered at $\varepsilon,
3\varepsilon, 5\varepsilon, \dots$ of radius $\varepsilon$ cover it — about $\frac1{2\varepsilon}$ balls, and no cover can do with fewer than $\frac{1}{2\varepsilon}$ of them (each ball covers length at most $2\varepsilon$). In $\intcc01^2$ the count squares to order $\varepsilon^{-2}$: covering numbers grow like $\varepsilon^{-d}$ in dimension $d$ — a quantitative face of [compactness](#def-b2-metric-compact), and the reason the infinite-dimensional unit balls of [Chapter 5](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#ch-b2-nvs) (where no finite $\frac13$-net exists at all) cannot be [compact](#def-b2-metric-compact).

**Example 4.22 (Reading compactness on covers).**

The [half-open](#def-b2-metric-topology) interval $\intoc{0}{1}$ is covered by the [open](#def-b2-metric-topology) sets $U_n = \intoo{\frac1n}{2}$, $n \geq 1$; any finite subfamily has a largest index $N$ and misses $\intoc{0}{\frac1N}$: no finite subcover, so $\intoc{0}{1}$ is not [compact](#def-b2-metric-compact) — which the sequential definition sees through $x_n = \frac1n$, whose limit $0$ escapes. On the other hand, adding the single point $0$ repairs both diagnoses at once: on $\intcc{0}{1}$ every such cover must contain a set containing $0$, which swallows a whole initial segment, and finitely many sets finish the rest. The two languages of [Theorem 4.20](#thm-b2-metric-borellebesgue) always fail or succeed together — covers detect escape exactly where sequences do.

**Remark 4.23 (Perspectives within this volume).**

This chapter is the volume’s load-bearing wall; watch where each pillar carries weight. *[Completeness](#def-b2-metric-complete)*: the Cauchy criterion becomes the convergence test for series in Banach spaces ([Chapter 7](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#ch-b2-series)), uniform convergence in [Chapter 10](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#ch-b2-funcseq) is exactly convergence in the [complete](#def-b2-metric-complete) $\bigl(C, d_\infty\bigr)$, and Cauchy–Lipschitz ([Chapter 16](https://one-course.com/books/math/4/en/chapter/16-differential-equations#ch-b2-diffeq)) is the Banach fixed point theorem wearing an integral equation. *[Compactness](#def-b2-metric-compact)*: it proves the equivalence of norms ([Chapter 5](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#ch-b2-nvs)), the attainment of extrema for the optimization of [Chapter 15](https://one-course.com/books/math/4/en/chapter/15-differential-calculus#ch-b2-diffcalc), and the existence of best approximations ([Chapter 5](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#ch-b2-nvs)’s weekend problem). *[Connectedness](#def-b2-metric-connected)*: it globalizes local statements — uniqueness of solutions of differential equations, the intermediate value theorem on curves ([Chapter 18](https://one-course.com/books/math/4/en/chapter/18-curves#ch-b2-curves)), and the two components of $GL_n(\R)$ that orientation theory ([Chapter 20](https://one-course.com/books/math/4/en/chapter/20-line-integrals-and-multiple-integrals#ch-b2-multint)) will keep apart.

**Remark 4.24 (Common pitfalls).**

(i) “Closed and bounded implies [compact](#def-b2-metric-compact)” is a theorem about $\R^n$, not about [metric spaces](#def-b2-metric-def): an infinite set with the discrete metric is closed and bounded in itself yet not [compact](#def-b2-metric-compact) ([Exercise 4.4](#exo-b2-metric-4)), and the closed unit ball of $C(\intcc01)$ fails too ([Chapter 5](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#ch-b2-nvs)). (ii) [Completeness](#def-b2-metric-complete) is a property of the *distance*, not of the topology: $\R$ with $d(x,y) = \abs{\arctan x - \arctan y}$ has the usual convergent sequences but is incomplete ([Exercise 4.1](#exo-b2-metric-1)). (iii) A [continuous](#def-b2-metric-continuity) bijection need not be a homeomorphism — the circle parametrization of [Exercise 4.7](#exo-b2-metric-7); [compactness](#def-b2-metric-compact) of the source repairs it. (iv) Banach’s theorem needs $k < 1$ *uniformly*: the condition $d(f(x), f(y)) < d(x,y)$ alone guarantees nothing on a [non-compact](#def-b2-metric-compact) space ([Exercise 4.5](#exo-b2-metric-5)). (v) [Connected](#def-b2-metric-connected) does not imply [path-connected](#def-b2-metric-connected) in general — but for the [open](#def-b2-metric-topology) subsets of normed spaces met in this book, the two agree ([Chapter 5](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#ch-b2-nvs)).

**Remark 4.25 (Where this chapter is used).**

Everywhere in the analysis half. [Completeness](#def-b2-metric-complete) of $C(\intcc{a}{b})$ powers the convergence theorems of [Chapter 10](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#ch-b2-funcseq) and the Cauchy–Lipschitz theory of [Chapter 16](https://one-course.com/books/math/4/en/chapter/16-differential-equations#ch-b2-diffeq) (this chapter’s weekend problem proves the local Picard–Lindelöf theorem already); [compactness](#def-b2-metric-compact) gives the equivalence of norms in finite dimension ([Chapter 5](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#ch-b2-nvs)) and the existence of extrema in [Chapter 15](https://one-course.com/books/math/4/en/chapter/15-differential-calculus#ch-b2-diffcalc); [connectedness](#def-b2-metric-connected) underlies the intermediate value arguments of [Chapter 8](https://one-course.com/books/math/4/en/chapter/8-functions-of-a-real-variable#ch-b2-realfun) and the uniqueness globalization for differential equations. In the Year 3 volume, [compactness](#def-b2-metric-compact) in function spaces (the Arzelà–Ascoli theorem) and the Baire category theorem ([Exercise 4.12](#exo-b2-metric-12) here) become tools of daily use.

![The first stages of the Cantor set (): each level deletes the open middle third of every segment. The intersection C = _n C_n is compact, has empty interior and length zero, yet is equipotent to ℝ — and it returns as a fixed point of a contraction on sets in this chapter’s weekend problem (question 22).](https://one-course.com/images/onecourse/chapters/math-4/b2-metric/fig-a9e39f8f0d70.svg)

*The first stages of the Cantor set ([Exercise 4.8](#exo-b2-metric-8)): each level deletes the [open](#def-b2-metric-topology) middle third of every segment. The intersection $C = \bigcap_n C_n$ is [compact](#def-b2-metric-compact), has empty interior and length zero, yet is [equipotent](https://one-course.com/books/math/4/en/chapter/1-sets-and-structures#def-b2-structures-countable) to $\R$ — and it returns as a *fixed point* of a contraction on sets in this chapter’s weekend problem (question 22).*

## 4.4 Connectedness

**Definition 4.26.**

$X$ is *connected* when it admits no partition into two nonempty [open](#def-b2-metric-topology) subsets — equivalently, when its only subsets both [open](#def-b2-metric-topology) and closed are $\emptyset$ and $X$. $X$ is *path-connected* when any two points are joined by a [continuous](#def-b2-metric-continuity) map $\gamma \colon \intcc{0}{1} \to X$.

**Theorem 4.27.**

1. The [connected](#def-b2-metric-connected) subsets of $\R$ are exactly the intervals.
2. A [continuous](#def-b2-metric-continuity) image of a [connected space](#def-b2-metric-connected) is [connected](#def-b2-metric-connected) — whence the general intermediate value theorem: a [continuous](#def-b2-metric-continuity) real function on a [connected space](#def-b2-metric-connected) takes every value between any two of its values.
3. [Path-connected](#def-b2-metric-connected) $\Rightarrow$ [connected](#def-b2-metric-connected) . (The converse fails in general; it holds for [open](#def-b2-metric-topology) subsets of normed spaces, [Chapter 5](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#ch-b2-nvs) .)

**Proof.** (1) A non-interval $A$ misses some $z$ between two of its points: $A = (A \cap \intoo{-\infty}{z}) \cup (A \cap \intoo{z}{+\infty})$ splits it into two nonempty [open](#def-b2-metric-topology) (in $A$) pieces. Conversely, let $I$ be an interval and $I = U \cup V$ a partition into nonempty relatively [open](#def-b2-metric-topology) sets; pick $a \in U$, $b \in V$, say $a < b$, and set $s = \sup\,(U \cap \intcc{a}{b})$, a point of $\intcc{a}{b}
\subseteq I$. If $s \in U$: then $s \neq b$, and relative openness of $U$ puts a whole interval around $s$ (intersected with $I$) inside $U$ — so points of $U \cap \intcc{a}{b}$ exceed $s$, contradicting the supremum. If $s \in V$: relative openness of $V$ puts an interval $\intoo{s - r}{s + r} \cap I$ inside $V$; but the supremum is adherent to $U \cap \intcc{a}{b}$, which must meet that interval — contradiction with $U \cap V = \emptyset$. (This is the Year 1 clopen argument for $\R$, run inside $I$.)

(2) If $f(X) = U' \cup V'$ splits into nonempty relatively [open](#def-b2-metric-topology) sets, then $X = f^{-1}(U') \cup f^{-1}(V')$ splits $X$ ([Theorem 4.6](#thm-b2-metric-globalcontinuity)). IVT: $f(X) \subseteq \R$ is [connected](#def-b2-metric-connected), hence an interval by (1).

(3) Suppose $X = U \cup V$, both nonempty [open](#def-b2-metric-topology), and join $a \in U$ to $b \in V$ by a path $\gamma$: then $\gamma^{-1}(U),
\gamma^{-1}(V)$ split $\intcc{0}{1}$, contradicting (1). ∎

**Example 4.28.**

$GL_n(\R)$ is not [connected](#def-b2-metric-connected): $\det$ is [continuous](#def-b2-metric-continuity) (a polynomial in the entries) onto $\R^*$, which is not [connected](#def-b2-metric-connected); the preimages of $\R_+^*$ and $\R_-^*$ split $GL_n(\R)$. (Each piece is in fact [path-connected](#def-b2-metric-connected) — a pleasant exercise beyond our needs.) By contrast $GL_n(\C)$ *is* [path-connected](#def-b2-metric-connected): [Exercise 4.10](#exo-b2-metric-10).

**Example 4.29 (A fixed point from connectedness alone).**

Every [continuous](#def-b2-metric-continuity) $f \colon \intcc01 \to \intcc01$ has a fixed point — no contraction hypothesis, no iteration. Consider $g(x) = f(x) - x$, [continuous](#def-b2-metric-continuity) on the [connected](#def-b2-metric-connected) $\intcc01$:

$$
g(0) = f(0) \geq 0,
\qquad
g(1) = f(1) - 1 \leq 0 ,
$$

and the intermediate value theorem ([Theorem 4.27](#thm-b2-metric-connectedness) (2)) delivers a zero of $g$, i.e. a fixed point of $f$. Contrast with Banach ([Theorem 4.12](#thm-b2-metric-banach)): here existence is topological and free, but uniqueness and the algorithm are lost — $f =
\mathrm{id}$ has every point fixed, and iteration of a non-contracting $f$ may cycle forever. The two fixed point theorems of this chapter answer different questions with different currencies.

**Example 4.30 (R\RR and R2\R^2R2 are not homeomorphic).**

[Connectedness](#def-b2-metric-connected) is a topological fingerprint. Suppose $h \colon
\R^2 \to \R$ were a homeomorphism (a [continuous](#def-b2-metric-continuity) bijection with [continuous](#def-b2-metric-continuity) inverse). Remove one point $a \in \R^2$: the restriction $h \colon \R^2\setminus\{a\} \to
\R\setminus\{h(a)\}$ is still a homeomorphism. But $\R^2$ minus a point is [path-connected](#def-b2-metric-connected) — join any two points by a segment, detouring along a second segment through an auxiliary point if $a$ blocks the direct one — hence [connected](#def-b2-metric-connected) ([Theorem 4.27](#thm-b2-metric-connectedness) (3)); while $\R$ minus a point splits into two nonempty [open](#def-b2-metric-topology) half-lines: not [connected](#def-b2-metric-connected). [Connectedness](#def-b2-metric-connected) is preserved by [continuous](#def-b2-metric-continuity) maps: contradiction. The plane and the line are genuinely different *as topological spaces* — a fact that cardinality alone ([Exercise 1.3](https://one-course.com/books/math/4/en/chapter/1-sets-and-structures#exo-b2-structures-3)-style bijections do exist!) is too coarse to see.

## 4.5 Exercises

**Exercise 4.1 ★.**

On $\R$, check that $\delta(x, y) = \min(1, \abs{x - y})$ and $d(x,y) = \abs{\arctan x - \arctan y}$ are distances. Which sequences converge for each? Is $(\R, d)$ [complete](#def-b2-metric-complete)?

**Solution of Exercise 4.1.**

$\delta$: symmetry and separation are clear; triangle inequality: $\min(1, u + v) \leq \min(1,u) + \min(1,v)$ for $u, v \geq 0$ (if either min is $1$, the right side is $\geq 1$; else it is $u + v$). $d$: it is the pullback of $\abs{\cdot}$ by the injective $\arctan$: the three axioms transfer.

Convergence: for $\delta$, $\delta(x_n, x) \to 0 \iff \abs{x_n - x}
\to 0$ (for small values the two distances agree): same convergent sequences as usual. For $d$: $d(x_n, x) \to 0 \iff \arctan x_n \to
\arctan x \iff x_n \to x$ ([continuity](#def-b2-metric-continuity) and strict monotonicity of $\arctan$ and of its inverse on the relevant ranges): again the usual convergence.

$(\R, d)$ is *not* [complete](#def-b2-metric-complete): $x_n = n$ satisfies $d(x_p, x_q)
= \abs{\arctan p - \arctan q} \to 0$ (both tend to $\frac\pi2$): Cauchy; but $(x_n)$ does not converge for $d$ (its $d$-limit would be an ordinary limit). [Completeness](#def-b2-metric-complete) is a property of the *distance*, not just of the convergent sequences.

**Exercise 4.2 ★.**

In a [metric space](#def-b2-metric-def), prove that a convergent sequence is Cauchy and bounded, and that a Cauchy sequence with a convergent subsequence converges. Deduce again that [compact metric spaces](#def-b2-metric-compact) are [complete](#def-b2-metric-complete).

**Solution of Exercise 4.2.**

Convergent $\Rightarrow$ Cauchy: $d(x_p, x_q) \leq d(x_p, \ell) +
d(\ell, x_q)$. Bounded: beyond $N$, $d(x_n, \ell) \leq 1$; the finitely many first terms are within some radius too.

Cauchy $+$ convergent subsequence $x_{\varphi(n)} \to \ell$: given $\varepsilon$, for large $n$, $d(x_n, \ell) \leq d(x_n,
x_{\varphi(n)}) + d(x_{\varphi(n)}, \ell) \leq 2\varepsilon$ (the first term by Cauchy, since $\varphi(n) \geq n$).

[Compact](#def-b2-metric-compact) $\Rightarrow$ [complete](#def-b2-metric-complete): a Cauchy sequence has a convergent subsequence ([compactness](#def-b2-metric-compact)), hence converges.

**Exercise 4.3 ★.**

In $\bigl(C(\intcc{0}{1}), d_\infty\bigr)$, compute the distance between $f(x) = x$ and $g(x) = x^2$; describe the closed ball $\overline B(0, 1)$; and prove that the set $\{f : f(0) = 0\}$ is closed while $\{f : f(0) > 0\}$ is [open](#def-b2-metric-topology).

**Solution of Exercise 4.3.**

$d_\infty(f, g) = \sup_{\intcc{0}{1}} \abs{x - x^2} = \frac14$ (maximum of $x - x^2$ at $x = \frac12$).

$\overline B(0, 1) = \{f : \sup\abs f \leq 1\}$: the [continuous](#def-b2-metric-continuity) functions with values in $\intcc{-1}{1}$.

$\{f : f(0) = 0\}$ is the preimage of $\{0\}$ under the *evaluation* $f \mapsto f(0)$, which is $1$-Lipschitz ($\abs{f(0) - g(0)} \leq d_\infty(f,g)$), hence [continuous](#def-b2-metric-continuity): the set is closed ([Theorem 4.6](#thm-b2-metric-globalcontinuity)). Likewise $\{f :
f(0) > 0\}$ is the preimage of the [open](#def-b2-metric-topology) $\intoo{0}{+\infty}$: [open](#def-b2-metric-topology).

**Exercise 4.4 ★★.**

Prove that the discrete [metric space](#def-b2-metric-def) $X$ (any set) is [complete](#def-b2-metric-complete), and that it is [compact](#def-b2-metric-compact) if and only if $X$ is finite. Which subsets are [connected](#def-b2-metric-connected)?

**Solution of Exercise 4.4.**

*Complete:* a Cauchy sequence with $\varepsilon = \frac12$ is eventually constant, hence convergent.

*[Compact](#def-b2-metric-compact) iff finite:* if $X$ is finite, any sequence takes some value infinitely often (constant subsequence). If $X$ is infinite, a sequence of pairwise distinct points has all mutual distances $1$: no Cauchy subsequence, so no convergent one.

*[Connected](#def-b2-metric-connected) subsets:* the singletons (and $\emptyset$). Any $A$ with two points $x \neq y$ splits as $\{x\} \cup (A
\setminus\{x\})$, both [open](#def-b2-metric-topology) in $A$ (every subset of a discrete space is [open](#def-b2-metric-topology) — balls of radius $\frac12$ are singletons).

**Exercise 4.5 ★★.**

Let $X$ be [compact](#def-b2-metric-compact) and $f \colon X \to X$ with

$$
d\bigl(f(x), f(y)\bigr) < d(x, y) \quad \text{for all } x \neq y .
$$

Prove that $f$ has a unique fixed point *(minimize $x \mapsto
d(x, f(x))$)*, and give an example on $X = \intco{1}{+\infty}$ (not [compact](#def-b2-metric-compact)) with no fixed point.

**Solution of Exercise 4.5.**

The function $g(x) = d(x, f(x))$ is [continuous](#def-b2-metric-continuity) on the [compact](#def-b2-metric-compact) $X$ ($\abs{g(x) - g(y)} \leq 2d(x,y)$ by two triangle inequalities), so it attains its minimum at some $a$ ([Theorem 4.16](#thm-b2-metric-compactprops)). If $f(a) \neq a$:

$$
g\bigl(f(a)\bigr) = d\bigl(f(a), f(f(a))\bigr) < d(a, f(a)) = g(a),
$$

contradicting minimality. So $f(a) = a$; uniqueness as usual (two fixed points $a \neq b$ give $d(a,b) = d(f(a), f(b)) < d(a,b)$).

[Non-compact](#def-b2-metric-compact) example: $f(x) = x + \frac1x$ on $\intco{1}{+\infty}$: $\abs{f(x) - f(y)} = \abs{x - y}\,\abs{1 - \frac{1}{xy}} < \abs{x -
y}$ for $x \neq y$ (as $xy > 1$), yet $f(x) > x$ everywhere.

**Exercise 4.6 ★★.**

(Nested [compacts](#def-b2-metric-compact)) Let $(K_n)$ be a decreasing sequence of nonempty [compact](#def-b2-metric-compact) subsets of a [metric space](#def-b2-metric-def). Prove that $\bigcap_n K_n \neq
\emptyset$ *(pick $x_n \in K_n$ and extract)*. Show by example that nonempty nested *closed* sets in $\R$ can have empty intersection.

**Solution of Exercise 4.6.**

Pick $x_n \in K_n$. All terms from rank $n$ on lie in $K_n$; in particular the whole sequence lies in the [compact](#def-b2-metric-compact) $K_0$: extract $x_{\varphi(k)} \to \ell$. For each fixed $n$, the terms $x_{\varphi(k)}$ with $\varphi(k) \geq n$ lie in the *closed* $K_n$, so the limit $\ell \in K_n$. Hence $\ell \in \bigcap K_n$.

Closed counterexample: $F_n = \intco{n}{+\infty}$ in $\R$: nested, closed, nonempty, empty intersection.

**Exercise 4.7 ★★.**

Let $K$ be [compact](#def-b2-metric-compact) and $f \colon K \to Y$ [continuous](#def-b2-metric-continuity) and bijective. Prove that $f^{-1}$ is [continuous](#def-b2-metric-continuity) *(use closed sets: [Theorem 4.6](#thm-b2-metric-globalcontinuity) and [Theorem 4.16](#thm-b2-metric-compactprops))*. Give a counterexample without [compactness](#def-b2-metric-compact) ($\gamma(t) = (\cos t, \sin t)$ on $\intco{0}{2\pi}$).

**Solution of Exercise 4.7.**

[Continuity](#def-b2-metric-continuity) of $f^{-1}$ means: images $f(F)$ of closed sets $F
\subseteq K$ are closed (preimages under $f^{-1}$ are images under $f$). A closed $F$ in the [compact](#def-b2-metric-compact) $K$ is [compact](#def-b2-metric-compact) ([Theorem 4.16](#thm-b2-metric-compactprops) (1)); its [continuous](#def-b2-metric-continuity) image $f(F)$ is [compact](#def-b2-metric-compact), hence closed. So $f^{-1}$ is [continuous](#def-b2-metric-continuity): $f$ is a homeomorphism.

Counterexample: $\gamma(t) = (\cos t, \sin t)$ from $\intco{0}{2\pi}$ (not [compact](#def-b2-metric-compact)) onto the unit circle is a [continuous](#def-b2-metric-continuity) bijection, but $\gamma^{-1}$ is discontinuous at $(1,0)$: points on the circle just below the axis have parameters near $2\pi$, not near $0$.

**Exercise 4.8 ★★.**

The *Cantor set* $C$ is obtained from $\intcc{0}{1}$ by repeatedly deleting [open](#def-b2-metric-topology) middle thirds. Prove that $C$ is [compact](#def-b2-metric-compact), has empty interior, and is infinite — indeed [equipotent](https://one-course.com/books/math/4/en/chapter/1-sets-and-structures#def-b2-structures-countable) to $\{0,1\}^{\N}$ *(ternary expansions with digits $0,2$; [Exercise 1.3](https://one-course.com/books/math/4/en/chapter/1-sets-and-structures#exo-b2-structures-3))*.

**Solution of Exercise 4.8.**

$C = \bigcap_n C_n$ where each $C_n$ (union of $2^n$ closed intervals of length $3^{-n}$) is closed: $C$ is closed and bounded in $\R$, hence [compact](#def-b2-metric-compact) ([Theorem 4.16](#thm-b2-metric-compactprops) (2)).

Empty interior: $C$ contains no interval of length $> 3^{-n}$ (it sits inside $C_n$, whose components have that length), for every $n$.

Cardinality: the points of $C$ are exactly the reals $\sum_{n\geq1}
a_n 3^{-n}$ with digits $a_n \in \{0, 2\}$ (at each stage, the deleted middle third removes the digit $1$); the map $(a_n) \mapsto
\sum a_n 3^{-n}$ is a bijection from $\{0,2\}^{\N^*}$ onto $C$ (injectivity as in [Exercise 1.3](https://one-course.com/books/math/4/en/chapter/1-sets-and-structures#exo-b2-structures-3)). So $C$ is [equipotent](https://one-course.com/books/math/4/en/chapter/1-sets-and-structures#def-b2-structures-countable) to $\{0,1\}^{\N}$: uncountable, though of “length zero”.

**Exercise 4.9 ★★★.**

Let $X$ be a *[compact](#def-b2-metric-compact)* [metric space](#def-b2-metric-def) and $f \colon X \to X$ an isometry: $d(f(x), f(y)) = d(x, y)$. Prove that $f$ is surjective. *Hint: if $a \notin f(X)$, then $\varepsilon = d(a, f(X)) > 0$ (why?); study the orbit $a, f(a), f^2(a), \dots$ and show its points are pairwise $\geq \varepsilon$ apart — contradiction with [compactness](#def-b2-metric-compact).*

**Solution of Exercise 4.9.**

Suppose $a \notin f(X)$. The image $f(X)$ is [compact](#def-b2-metric-compact) ([continuous](#def-b2-metric-continuity) image), hence closed; so

$$
\varepsilon = d\bigl(a, f(X)\bigr)
= \inf_{y \in f(X)} d(a, y) > 0
$$

(the infimum of a [continuous](#def-b2-metric-continuity) function on a [compact](#def-b2-metric-compact) set is attained; if it were $0$, $a$ would be adherent to the closed $f(X)$, hence in it).

Consider the orbit $x_n = f^n(a)$ ($x_0 = a$). For $p < q$:

$$
d(x_p, x_q) = d\bigl(f^p(a), f^p(f^{q-p}(a))\bigr)
= d\bigl(a, f^{q-p}(a)\bigr) \geq \varepsilon,
$$

(isometry iterated $p$ times; and $f^{q-p}(a) \in f(X)$ since $q - p
\geq 1$). A sequence with mutual distances $\geq \varepsilon$ has no convergent subsequence — contradicting [compactness](#def-b2-metric-compact). Hence $f(X) =
X$.

**Exercise 4.10 ★★★.**

Prove that $GL_n(\C)$ is [path-connected](#def-b2-metric-connected). *Hint: given $A, B$ invertible, consider $p(z) = \det\bigl((1 - z)A + zB\bigr)$ for $z
\in \C$: a polynomial in $z$, not identically zero, so it has finitely many roots; join $0$ to $1$ in $\C$ by a path avoiding them.*

**Solution of Exercise 4.10.**

Let $A, B \in GL_n(\C)$ and $p(z) = \det\bigl((1-z)A + zB\bigr)$: a polynomial in $z$ (each entry is affine in $z$; the [determinant](https://one-course.com/books/math/4/en/chapter/2-linear-algebra#def-b2-linalg-det) is a polynomial in the entries). $p(0) = \det A \neq 0$: $p$ is not identically zero, so it has finitely many roots $z_1, \dots, z_m$ (none equal to $0$ or $1$: $p(1) = \det B \neq 0$). The plane $\C$ minus finitely many points is [path-connected](#def-b2-metric-connected): a path from $0$ to $1$ avoiding the $z_i$ exists (take a broken line through a point far from all roots, or a circular arc; only finitely many obstacles). Along such a path $\gamma$, $t \mapsto (1 - \gamma(t))A +
\gamma(t)B$ is a [continuous](#def-b2-metric-continuity) path *inside* $GL_n(\C)$ from $A$ to $B$ (the [determinant](https://one-course.com/books/math/4/en/chapter/2-linear-algebra#def-b2-linalg-det) never vanishes on it). Hence $GL_n(\C)$ is [path-connected](#def-b2-metric-connected) — unlike its real cousin ([Example 4.28](#ex-b2-metric-glnr)): the complex plane has room to walk around obstacles.

**Exercise 4.11 ★★.**

For $\emptyset \neq A \subseteq X$, set $d(x, A) = \inf_{a \in A}
d(x, a)$. Prove that $x \mapsto d(x, A)$ is $1$-Lipschitz, that $d(x, A) = 0$ iff $x \in \overline A$, and that for disjoint nonempty *closed* sets $A, B$ the function

$$
\varphi(x) = \frac{d(x, A)}{d(x, A) + d(x, B)}
$$

is well defined, [continuous](#def-b2-metric-continuity), equal to $0$ exactly on $A$ and to $1$ exactly on $B$ — a [continuous](#def-b2-metric-continuity) “switch” separating any two disjoint closed sets.

**Solution of Exercise 4.11.**

*[Lipschitz](#def-b2-metric-continuity):* for $a \in A$, $d(x, a) \leq d(x, y) + d(y,
a)$; take the infimum over $a$: $d(x, A) \leq d(x, y) + d(y,
A)$, and swap $x, y$: $\abs{d(x,A) - d(y,A)} \leq d(x,y)$.

*Vanishing:* $d(x, A) = 0$ iff there are $a_n \in A$ with $d(x, a_n) \to 0$ iff $x$ is a limit of points of $A$ iff $x \in
\overline A$.

*The switch:* for disjoint closed $A, B$: the denominator $d(x,A) + d(x,B)$ never vanishes (it would force $x \in \overline
A \cap \overline B = A \cap B = \emptyset$), so $\varphi$ is well defined, and [continuous](#def-b2-metric-continuity) as a quotient of [continuous](#def-b2-metric-continuity) functions with nonvanishing denominator. $\varphi(x) = 0$ iff $d(x, A) = 0$ iff $x \in A$; $\varphi(x) = 1$ iff $d(x, B) = 0$ iff $x \in B$; and $0 \leq \varphi \leq 1$ everywhere.

**Exercise 4.12 ★★★.**

(Baire) Let $X$ be a [complete metric space](#def-b2-metric-complete) and $(U_n)_{n\geq1}$ a sequence of dense [open](#def-b2-metric-topology) subsets. Prove that $\bigcap_n U_n$ is dense in $X$ *(inside any ball, build nested closed balls $\overline B(x_n, r_n) \subseteq U_n$ with $r_n \to 0$ and use [completeness](#def-b2-metric-complete))*. Deduce that $\R$ is not a [countable](https://one-course.com/books/math/4/en/chapter/1-sets-and-structures#def-b2-structures-countable) union of closed sets with empty interior, and recover — again — that $\R$ is uncountable.

**Solution of Exercise 4.12.**

Let $B(x_0, r_0)$ be any ball; we find a point of $\bigcap U_n$ in it. Since $U_1$ is dense and [open](#def-b2-metric-topology), $U_1 \cap B(x_0, r_0)$ is nonempty and [open](#def-b2-metric-topology): it contains a closed ball $\overline B(x_1,
r_1)$ with $0 < r_1 \leq \frac{r_0}2$ (shrink the radius). Inductively, $U_{n+1} \cap B(x_n, r_n)$ is nonempty [open](#def-b2-metric-topology): pick $\overline B(x_{n+1}, r_{n+1}) \subseteq U_{n+1} \cap B(x_n,
r_n)$ with $r_{n+1} \leq \frac{r_n}2$. For $p, q \geq n$, both $x_p, x_q$ lie in $B(x_n, r_n)$ with $r_n \leq 2^{-n}r_0$: the sequence is Cauchy, and converges to some $\ell$ by [completeness](#def-b2-metric-complete). For each $n$, the tail of the sequence lies in the *closed* ball $\overline B(x_{n+1}, r_{n+1}) \subseteq U_{n+1}
\cap B(x_0, r_0)$, so $\ell \in U_{n+1}$ for every $n$ and $\ell
\in \overline B(x_1, r_1) \subseteq B(x_0, r_0)$. Hence $\bigcap_n U_n$ meets every ball: dense.

*Application:* if $\R = \bigcup_n F_n$ with $F_n$ closed with empty interior, then $U_n = \R \setminus F_n$ are dense ($\overline{U_n} = \R$ iff $F_n$ has empty interior) and [open](#def-b2-metric-topology), and Baire gives a point in $\bigcap U_n = \R \setminus \bigcup
F_n$: contradiction. In particular $\R \neq \bigcup_{x \in D}
\{x\}$ for a [countable](https://one-course.com/books/math/4/en/chapter/1-sets-and-structures#def-b2-structures-countable) $D$ (singletons are closed with empty interior): $\R$ is uncountable — [Theorem 1.9](https://one-course.com/books/math/4/en/chapter/1-sets-and-structures#thm-b2-structures-cantor) by another route.

## 4.6 Problem: Picard Iteration

[Completeness](#def-b2-metric-complete) plus contraction is a solving machine: feed it an equation written as a fixed point problem, and it returns existence, uniqueness, an algorithm, and error bars. This weekend problem runs the machine at full power on the equation $y' = f(t,
y)$: we prove the local *Picard–Lindelöf theorem* (the nonlinear heart of [Chapter 16](https://one-course.com/books/math/4/en/chapter/16-differential-equations#ch-b2-diffeq)’s Cauchy–Lipschitz theory), watch every hypothesis earn its keep through counterexamples, and collect purely metric dividends — [continuous](#def-b2-metric-continuity) dependence on the data, Kepler’s equation, and the self-similarity of the Cantor set.

![Picard iterates for y' = y, y(0) = 1: each pass through T(y)(t) = 1 + ∈t_0t y adds one Taylor term, and the contraction squeezes the whole sequence uniformly onto \,t.](https://one-course.com/images/onecourse/chapters/math-4/b2-metric/fig-ca7ff92a0207.svg)

*Picard iterates for $y' = y$, $y(0) = 1$: each pass through $T(y)(t) = 1 + \int_0^t y$ adds one Taylor term, and the contraction squeezes the whole sequence uniformly onto $\eu^{\,t}$.*

**Problem 4.1.**

Weekend problem — the Picard–Lindelöf theorem

Throughout, $t_0 \in \R$, $y_0 \in \R$, $a, b > 0$, and $f$ is a [continuous](#def-b2-metric-continuity) function on the rectangle $R = \intcc{t_0 - a}{t_0 +
a} \times \intcc{y_0 - b}{y_0 + b}$, bounded by $M =
\sup_R\,\abs f$, and *$L$-Lipschitz in its second variable*: $\abs{f(t, y) - f(t, z)} \leq L\abs{y - z}$ whenever both points lie in $R$. Set

$$
h = \min\Bigl(a, \frac bM\Bigr) \quad (\text{with } h = a
\text{ if } M = 0), \qquad I = \intcc{t_0 - h}{t_0 + h}.
$$

**Part I — The [complete](#def-b2-metric-complete) stage.**

1. Prove the two statements quoted in [Definition 4.7](#def-b2-metric-complete) : a closed subset of a [complete metric space](#def-b2-metric-complete) is [complete](#def-b2-metric-complete) , and a [complete](#def-b2-metric-complete) subset of any [metric space](#def-b2-metric-def) is closed. Deduce that every closed subset of $\bigl(C(I), d_\infty\bigr)$ is a [complete metric space](#def-b2-metric-complete) .
2. Show that the map $C(I) \to C(I)$ , $y \mapsto \bigl(t  \mapsto \int_{t_0}^{t}y(s)\,\dd s\bigr)$ , is $h$ -Lipschitz for $d_\infty$ .
3. (Fixed points move less than the maps) Let $g \colon X  \to X$ be a $k$-contraction of a [metric space](#def-b2-metric-def) with fixed point $\ell_g$, and $\widetilde g \colon X \to X$ *any* map with a fixed point $\ell_{\widetilde g}$. Prove $$d(\ell_g, \ell_{\widetilde g}) \leq  \frac{d\bigl(g(\ell_{\widetilde g}),  \widetilde g(\ell_{\widetilde g})\bigr)}{1 - k}  \leq \frac{\sup_{x \in X} d\bigl(g(x), \widetilde  g(x)\bigr)}{1 - k}.$$
4. (The iterate trick) Let $X$ be [complete](#def-b2-metric-complete) nonempty and $g  \colon X \to X$ a map — not assumed [continuous](#def-b2-metric-continuity) — such that some iterate $g^m$ is a $k$ -contraction. Prove that $g$ has a unique fixed point $\ell$ and that *every* orbit $x_{n+1} = g(x_n)$ converges to $\ell$ . *(Fixed points of $g$ are fixed points of $g^m$; conversely $g(\ell)$ is a fixed point of $g^m$; split the orbit along residues mod $m$.)*

**Part II — The Picard–Lindelöf theorem.**

5. Show that a function $y \colon I \to \intcc{y_0 - b}{y_0  + b}$ is $C^1$ with $y(t_0) = y_0$ and $y' = f(t, y)$ on $I$ if and only if it is [continuous](#def-b2-metric-continuity) and satisfies the integral equation $$y(t) = y_0 + \int_{t_0}^{t} f\bigl(s, y(s)\bigr)\dd s  \qquad (t \in I).$$
6. Let $X_h = \{y \in C(I) : \abs{y(t) - y_0} \leq b  \text{ on } I\}$ and let $T$ be defined by $T(y)(t) =  y_0 + \int_{t_0}^{t}f(s, y(s))\dd s$ . Show that $X_h$ is a nonempty closed subset of $C(I)$ , hence [complete](#def-b2-metric-complete) , and that $T$ maps $X_h$ into $X_h$ — this is where $h \leq  b/M$ works.
7. Show that $d_\infty\bigl(T(y), T(z)\bigr) \leq  Lh\,d_\infty(y, z)$ on $X_h$ : if $Lh < 1$ , Banach’s theorem already concludes. We remove this smallness condition next.
8. Prove by induction on $n$: $$\abs{T^n(y)(t) - T^n(z)(t)} \leq  \frac{\bigl(L\abs{t - t_0}\bigr)^n}{n!}\,  d_\infty(y, z) \qquad (y, z \in X_h,\ t \in I),$$ so some iterate of $T$ is a contraction. Conclude with question 4 (*the Picard–Lindelöf theorem*): the Cauchy problem $y' = f(t,y)$, $y(t_0) = y_0$ has exactly one solution on $I = \intcc{t_0 - h}{t_0 + h}$ with values in $\intcc{y_0 - b}{y_0 + b}$.
9. Show that the restriction “with values in $\intcc{y_0 -  b}{y_0 + b}$ ” is automatic: any solution of the Cauchy problem defined on $I$ whose graph starts in $R$ stays in $\intcc{y_0 - b}{y_0 + b}$ *(consider the first exit time and bound $\abs{y(t) - y_0}$ by $M\abs{t -  t_0}$)* . Hence uniqueness holds among all solutions on $I$ .
10. Run the machine on $y' = y$ , $y(0) = 1$ , starting from the constant $y^{(0)} \equiv 1$ : compute the Picard iterates $y^{(n)}$ , identify them, and describe the convergence.

**Part III — Every hypothesis earns its keep.**

11. ( [Lipschitz](#def-b2-metric-continuity) fails, uniqueness fails) For $y' =  2\sqrt{\abs y}$ , $y(0) = 0$ : check that $y \equiv 0$ and, for every $c \geq 0$ , the function $y_c(t) = 0$ for $t \leq c$ , $y_c(t) = (t - c)^2$ for $t > c$ , are all $C^1$ solutions on $\R$ . Where exactly does $y \mapsto  2\sqrt{\abs y}$ fail to be [Lipschitz](#def-b2-metric-continuity) ?
12. (Locality is real) For $y' = y^2$ , $y(0) = 1$ : solve explicitly, give the maximal interval of existence, and compute the best $h$ the theorem can certify over all choices of the rectangle ( $a$ large, $b$ free): show $h_{\max} = \sup_{b>0} \frac{b}{(1+b)^2} = \frac14$ , while the true solution lives on $\intoo{-\infty}{1}$ .
13. ( [Completeness](#def-b2-metric-complete) is not decor) On $X = \Q \cap  \intcc{1}{2}$ with the usual distance, let $g(x) =  \frac x2 + \frac1x$ . Show $g(X) \subseteq X$ , that $g$ is a $\frac12$ -contraction *(mean value inequality)* , and that $g$ has no fixed point in $X$ . Which hypothesis of Banach’s theorem fails, and what is the fixed point in the completion?
14. (Error bars) For a $k$ -contraction $g$ on a [complete](#def-b2-metric-complete) space, prove the *a posteriori* estimate $d(x_n,  \ell) \leq \frac{k}{1-k}\,d(x_n, x_{n-1})$ . For Heron’s map $g(x) = \frac x2 + \frac 1x$ on $\intcc{1}{2}$ (fixed point $\sqrt2$ ), starting at $x_0 = \frac32$ : how many steps does the *a priori* bound $\frac{k^n}{1-k}d(x_1, x_0)$ demand for accuracy $10^{-6}$ , and how many steps suffice in reality? (Compute $x_1, x_2, x_3$ and their errors; the contraction bound is honest but pessimistic — Heron converges quadratically.)

**Part IV — [Continuous](#def-b2-metric-continuity) dependence.** In this part $Lh < 1$, so $T$ itself is a contraction on $X_h$ (question 7); write $y[\,y_0\,]$ for the solution with initial value $y_0$.

15. (Dependence on the initial value) Let $z_0$ be another initial value with $\abs{z_0 - y_0}$ small enough that both problems fit in the rectangle. Using question 3, prove $$d_\infty\bigl(y[y_0], y[z_0]\bigr) \leq  \frac{\abs{y_0 - z_0}}{1 - Lh}.$$
16. (Long intervals by chaining) Suppose the solutions exist on a long segment cut into $m$ consecutive pieces on each of which the previous bound applies with $Lh \leq  \frac12$ . Show the deviation grows by a factor at most $2$ per piece, hence $d_\infty \leq 2^m\abs{y_0 - z_0}$ overall — an exponential-in-length bound, the discrete shadow of the $\eu^{L\abs{t - t_0}}$ of Gronwall’s lemma ( [Chapter 16](https://one-course.com/books/math/4/en/chapter/16-differential-equations#ch-b2-diffeq) ).
17. (Dependence on the field) Let $g$ be another field on $R$ , also $L$ -Lipschitz in $y$ , with $\sup_R \abs{f - g}  \leq \varepsilon$ . Prove that the corresponding solutions satisfy $d_\infty \leq  \frac{\varepsilon h}{1 - Lh}$ : modelling error propagates linearly.
18. (Parameters) If a family $f_\lambda$ of fields is $L$ -Lipschitz in $y$ uniformly and $\sup_R\abs{f_\lambda  - f_\mu} \leq C\abs{\lambda - \mu}$ , deduce that $\lambda \mapsto y_\lambda$ is [Lipschitz](#def-b2-metric-continuity) from the parameter space into $\bigl(C(I), d_\infty\bigr)$ .
19. (Systems cost nothing) Explain why Parts I, II and IV hold verbatim for $y$ with values in $\R^n$ (sup distances built on any of the distances of [Example 4.2](#ex-b2-metric-examples) ), then compute all Picard iterates for the system $y' = Ay$ , $y(0) = (c_1,  c_2)$ , $A = \left(\begin{smallmatrix} 0 & 1\\ 0 &  0\end{smallmatrix}\right)$ : show the iteration becomes stationary at the exact solution after one step.

**Part V — Metric dividends and synthesis.**

20. (Perturbation of the identity) Let $X$ be a [complete](#def-b2-metric-complete) normed-space-like stage: take $X = C(I)$ or $\R^n$ . If $\eta \colon X \to X$ is $k$ -Lipschitz with $k < 1$ , prove that $x \mapsto x + \eta(x)$ is a bijection of $X$ whose inverse is $\frac1{1-k}$ -Lipschitz *(for each $y$, apply Banach to $x \mapsto y - \eta(x)$)* . This is the metric heart of the inverse function theorem ( [Chapter 15](https://one-course.com/books/math/4/en/chapter/15-differential-calculus#ch-b2-diffcalc) ).
21. (Kepler’s equation) For $0 \leq e < 1$ and $m \in \R$ , prove that $x = m + e\sin x$ has exactly one solution, that the iteration $x_{n+1} = m + e\sin x_n$ converges to it from any start, and estimate: for $e = \frac12$ , $m = 1$ , how many iterations guarantee an error $\leq  10^{-3}$ by the a priori bound? (The solution is $x  \approx 1.4987$ .)
22. (The Cantor set is a fixed point) Let $S_1(x) = \frac  x3$ and $S_2(x) = \frac x3 + \frac23$ on $\R$ , and let $C$ be the Cantor set of [Exercise 4.8](#exo-b2-metric-8) . Prove $C = S_1(C) \cup S_2(C)$ , and explain in one sentence why no *other* nonempty [compact](#def-b2-metric-compact) set satisfies this equation (the map $A \mapsto S_1(A) \cup S_2(A)$ is a contraction for a distance between [compact](#def-b2-metric-compact) sets — the Hausdorff distance, made honest in the Year 3 volume).
23. ( [Connectedness](#def-b2-metric-connected) globalizes uniqueness) Let $f$ be locally [Lipschitz](#def-b2-metric-continuity) in $y$ on an [open](#def-b2-metric-topology) set, and let $y, z$ be two solutions of $y' = f(t, y)$ on a common interval $J$ with $y(t_0) = z(t_0)$ . Prove $y = z$ on $J$ : show that $\{t \in J : y(t) = z(t)\}$ is nonempty, closed in $J$ , and [open](#def-b2-metric-topology) in $J$ (by local uniqueness), and use the [connectedness](#def-b2-metric-connected) of intervals ( [Theorem 4.27](#thm-b2-metric-connectedness) ).
24. (No smallness for linear equations) For $y' =  \alpha(t)y + \beta(t)$ with $\alpha, \beta$ [continuous](#def-b2-metric-continuity) on a segment $\intcc{A}{B}$ , adapt question 8 to show the factorial bound holds on the *whole* segment, so existence and uniqueness are global there — the scalar case of [Chapter 16](https://one-course.com/books/math/4/en/chapter/16-differential-equations#ch-b2-diffeq) ’s Cauchy–Lipschitz theorem, with no restriction on the length $B - A$ .
25. (Synthesis) One sentence each: what [completeness](#def-b2-metric-complete) contributed; what the contraction contributed; what the iterate trick bought compared with plain Banach; where [connectedness](#def-b2-metric-connected) entered; and which counterexample of Part III guards which hypothesis. Name the summit theorem, and say what replaces contraction when $f$ is merely [continuous](#def-b2-metric-continuity) (Peano’s theorem, via [compactness](#def-b2-metric-compact) in function spaces — the Year 3 volume’s Arzelà–Ascoli).

**Solution of Problem 4.1.**

**1.** Let $F$ be closed in the [complete](#def-b2-metric-complete) $X$ and $(x_n)
\subseteq F$ Cauchy: it converges in $X$ to some $\ell$, and $\ell \in F$ because $F$ is closed (limits of sequences of $F$ stay in $\overline F = F$): $F$ is [complete](#def-b2-metric-complete). Conversely let $A
\subseteq X$ be [complete](#def-b2-metric-complete) and $x \in \overline A$: some sequence of $A$ converges to $x$; it is Cauchy, so it converges *in* $A$; limits are unique, so $x \in A$: $A$ closed. Since $\bigl(C(I), d_\infty\bigr)$ is [complete](#def-b2-metric-complete) ([Theorem 4.9](#thm-b2-metric-rncomplete)), its closed subsets are [complete](#def-b2-metric-complete).

**2.** For $y, z \in C(I)$ and $t \in I$:

$$
\Bigl|\int_{t_0}^{t}y - \int_{t_0}^{t}z\Bigr| \leq
\abs{t - t_0}\,\sup_I\abs{y - z} \leq h\,d_\infty(y, z),
$$

and take the sup over $t$.

**3.** Using the two fixed point equations and the triangle inequality:

$$
d(\ell_g, \ell_{\widetilde g})
= d\bigl(g(\ell_g), \widetilde g(\ell_{\widetilde g})\bigr)
\leq d\bigl(g(\ell_g), g(\ell_{\widetilde g})\bigr) +
d\bigl(g(\ell_{\widetilde g}), \widetilde g(\ell_{\widetilde
g})\bigr)
\leq k\,d(\ell_g, \ell_{\widetilde g}) +
d\bigl(g(\ell_{\widetilde g}), \widetilde g(\ell_{\widetilde
g})\bigr),
$$

and solve for $d(\ell_g, \ell_{\widetilde g})$ (the coefficient $1 - k$ is positive). The second inequality bounds the evaluated gap by the uniform one.

**4.** $g^m$ is a contraction on a nonempty [complete](#def-b2-metric-complete) space: it has a unique fixed point $\ell$ ([Theorem 4.12](#thm-b2-metric-banach)). Then $g^m(g(\ell)) =
g(g^m(\ell)) = g(\ell)$: $g(\ell)$ is a fixed point of $g^m$, so $g(\ell) = \ell$ by uniqueness. Any fixed point of $g$ is one of $g^m$: uniqueness for $g$. Orbits: fix $r \in \{0, \dots,
m-1\}$; the subsequence $(x_{qm + r})_q$ is the $g^m$-orbit started at $x_r$, so it converges to $\ell$ as $q \to \infty$ (Banach again). All $m$ subsequences converge to the same $\ell$, hence $x_n \to \ell$: given $\varepsilon$, each residue class is eventually within $\varepsilon$, and there are finitely many classes.

**5.** If $y$ is [continuous](#def-b2-metric-continuity) with values in $\intcc{y_0 -
b}{y_0 + b}$, the integrand $s \mapsto f(s, y(s))$ is [continuous](#def-b2-metric-continuity) on $I$ (composition), so the right-hand side is $C^1$ with derivative $f(t, y(t))$ (fundamental theorem of calculus, Year 1 volume). If $y$ satisfies the integral equation, it is that $C^1$ function, $y(t_0) = y_0$, and $y' = f(t, y)$. Conversely, integrating $y' = f(s, y(s))$ from $t_0$ to $t$ gives the integral equation.

**6.** $X_h$ contains the constant $y_0$; it is closed as the preimage of $\intcc{0}{b}$ under the [continuous](#def-b2-metric-continuity) map $y
\mapsto d_\infty(y, y_0)$ (distances are $1$-Lipschitz), hence [complete](#def-b2-metric-complete) by question 1. Stability: for $y \in X_h$ and $t \in
I$,

$$
\abs{T(y)(t) - y_0} = \Bigl|\int_{t_0}^{t}f(s, y(s))\dd s\Bigr|
\leq M\abs{t - t_0} \leq Mh \leq b ,
$$

the last step by $h \leq b/M$ (or $M = 0$, trivial). And $T(y)$ is [continuous](#def-b2-metric-continuity) ($C^1$ even, question 5): $T(y) \in X_h$.

**7.** For $t \in I$:

$$
\abs{T(y)(t) - T(z)(t)} \leq \int_{t_0}^{t}\abs{f(s, y(s)) -
f(s, z(s))}\,\abs{\dd s} \leq L\abs{t - t_0}\,d_\infty(y,z)
\leq Lh\,d_\infty(y,z).
$$

If $Lh < 1$: $T$ is a contraction of the nonempty [complete](#def-b2-metric-complete) $X_h$, and Banach gives a unique fixed point — by question 5, the unique solution.

**8.** Induction; the case $n = 1$ is question 7’s middle inequality. Assuming the bound for $n$, for $t \geq t_0$ (the case $t \leq t_0$ is symmetric):

$$
\abs{T^{n+1}(y)(t) - T^{n+1}(z)(t)}
\leq L\int_{t_0}^{t}\abs{T^n(y)(s) - T^n(z)(s)}\dd s
\leq L\int_{t_0}^{t}\frac{L^n(s - t_0)^n}{n!}\dd s\;
d_\infty(y,z),
$$

and the integral evaluates to $\frac{L^n(t -
t_0)^{n+1}}{(n+1)!}$: the bound with $n + 1$. Hence $d_\infty(T^n y, T^n z) \leq \frac{(Lh)^n}{n!}d_\infty(y, z)$, and $\frac{(Lh)^n}{n!} \to 0$ (the exponential series converges): some $T^m$ is a contraction. Question 4 applies on the [complete](#def-b2-metric-complete) $X_h$: $T$ has a unique fixed point, i.e. the Cauchy problem has exactly one solution on $I$ with values in $\intcc{y_0 - b}{y_0 + b}$.

**9.** Let $y$ solve the problem on $I$ and suppose the set $E = \{t \in I, t > t_0 : \abs{y(t) - y_0} > b\}$ is nonempty (the side $t < t_0$ is symmetric); let $\tau = \inf E$. By [continuity](#def-b2-metric-continuity), $\abs{y(s) - y_0} \leq b$ for $s \in
\intcc{t_0}{\tau}$, so the graph lies in $R$ there, the integral equation holds up to $\tau$, and

$$
\abs{y(\tau) - y_0} = \Bigl|\int_{t_0}^{\tau}f\bigl(s,
y(s)\bigr)\dd s\Bigr| \leq M(\tau - t_0) \leq Mh \leq b .
$$

If $\tau < t_0 + h$, points of $E$ arbitrarily close to $\tau$ from the right give, by [continuity](#def-b2-metric-continuity), $\abs{y(\tau) - y_0} \geq
b$, hence $= b$; but then the display forces $M(\tau - t_0) =
Mh$, i.e. $\tau = t_0 + h$: contradiction. So $\tau = t_0 + h$, $E \subseteq \{t_0 + h\}$, and the display (at $\tau = t_0 + h$) gives $\abs{y(t_0 + h) - y_0} \leq b$, contradicting membership in $E$. Hence $E = \emptyset$: every solution on $I$ stays in the band, is a fixed point of $T$ in $X_h$, and uniqueness is unconditional.

**10.** $T(y)(t) = 1 + \int_0^t y$. From $y^{(0)} \equiv
1$:

$$
y^{(1)}(t) = 1 + t,\quad
y^{(2)}(t) = 1 + t + \frac{t^2}2,\quad\dots\quad
y^{(n)}(t) = \sum_{k=0}^{n}\frac{t^k}{k!}
$$

(induction: integrating the partial sum adds the next term). These are the Taylor partial sums of $\eu^{\,t}$; on any bounded $I$ they converge uniformly to $\eu^{\,t}$ (the tail is dominated by the convergent numerical series $\sum h^k/k!$), which is indeed the unique solution.

**11.** $y \equiv 0$ is a solution. For $y_c$: it is $C^1$ (both pieces are, and at $t = c$ the derivatives match: $0$ and $2(t - c) \to 0$), and for $t > c$: $y_c' = 2(t - c) =
2\sqrt{(t-c)^2} = 2\sqrt{\abs{y_c}}$; for $t \leq c$ both sides vanish. So the Cauchy problem $y(0) = 0$ has infinitely many solutions ($c \geq 0$ arbitrary, and $y \equiv 0$). The field $\varphi(y) = 2\sqrt{\abs y}$ is not [Lipschitz](#def-b2-metric-continuity) near $0$: $\frac{\varphi(y) - \varphi(0)}{y - 0} = \frac{2}{\sqrt y} \to
+\infty$ as $y \to 0^+$: no constant $L$ works on any neighbourhood of $0$ — exactly where all the solutions branch.

**12.** Separating variables (or checking directly), the unique local solution is $y(t) = \frac1{1 - t}$, defined on $\intoo{-\infty}{1}$ and blowing up at $t = 1$. For the rectangle $\intcc{-a}{a} \times \intcc{1 - b}{1 + b}$: $M =
\sup y^2 = (1 + b)^2$, so the certified half-width is $h =
\min\bigl(a, \frac{b}{(1+b)^2}\bigr)$. Maximizing $\frac{b}{(1+b)^2}$: derivative zero at $b = 1$, value $\frac14$. So the theorem guarantees life only on $\intcc{-\frac14}{\frac14}$ — correctly less than the true lifespan $1$ forward, and infinitely less backward: the theorem is local by nature, and blow-up shows it cannot be otherwise.

**13.** $g$ maps $\intcc12$ into itself: $g$ decreases on $\intcc{1}{\sqrt2}$ and increases after (study $g'(x) = \frac12
- \frac1{x^2}$), with $g(1) = g(2) = \frac32$ and minimum $g(\sqrt2) = \sqrt2 > 1$: $g(\intcc12) \subseteq
\intcc{\sqrt2}{\frac32} \subseteq \intcc12$; and $g$ maps rationals to rationals. Contraction: $\abs{g'(x)} = \abs{\frac12
- \frac1{x^2}} \leq \frac12$ on $\intcc12$ ($\frac1{x^2} \in
\intcc{\frac14}{1}$), so the mean value inequality gives $\abs{g(x) - g(y)} \leq \frac12\abs{x - y}$. A fixed point satisfies $\frac x2 = \frac1x$, i.e. $x^2 = 2$: impossible in $\Q$. The failing hypothesis is [completeness](#def-b2-metric-complete) of $X$ ($\Q \cap
\intcc12$ is not [complete](#def-b2-metric-complete)); in the completion $\intcc12$ the fixed point is $\sqrt2$ — Banach’s theorem run on the rationals *creates* the irrational.

**14.** A posteriori: $d(x_n, \ell) \leq d(x_n, x_{n+1}) +
d(x_{n+1}, \ell) \leq k\,d(x_{n-1}, x_n) + k\,d(x_n, \ell)$, whence $d(x_n, \ell) \leq \frac{k}{1-k}d(x_n, x_{n-1})$. Heron from $x_0 = \frac32$: $x_1 = \frac{17}{12}$, $d(x_1, x_0) =
\frac1{12}$, $k = \frac12$: the a priori bound $\frac{k^n}{1-k}d(x_1,x_0) = \frac{2^{-n+1}}{12}$ drops below $10^{-6}$ first at $n = 18$. In reality $x_1 = \frac{17}{12}
\approx 1.41667$ (error $2.5\cdot10^{-3}$), $x_2 =
\frac{577}{408} \approx 1.4142157$ (error $2.1\cdot10^{-6}$), $x_3 \approx 1.41421356237469$ (error $1.6\cdot10^{-12}$): three steps suffice. Each Heron step roughly *squares* the error (quadratic convergence, a Newton phenomenon: [Chapter 8](https://one-course.com/books/math/4/en/chapter/8-functions-of-a-real-variable#ch-b2-realfun)); the contraction estimate, which only halves it, is honest for the worst case but pessimistic here.

**15.** Apply question 3 with $g = T_{y_0}$ (an $Lh$-contraction, $Lh < 1$) and $\widetilde g = T_{z_0}$, whose fixed point is $y[z_0]$. For any $y$,

$$
\abs{T_{y_0}(y)(t) - T_{z_0}(y)(t)} = \abs{y_0 - z_0},
$$

(the integrals are identical), so $\sup_y
d_\infty(T_{y_0}(y), T_{z_0}(y)) = \abs{y_0 - z_0}$, and question 3 gives

$$
d_\infty(y[y_0], y[z_0]) \leq \frac{\abs{y_0 - z_0}}{1 - Lh} .
$$

**16.** On each piece, question 15 applied with the left-endpoint values as initial data bounds the deviation at the right endpoint:

$$
d_\infty \leq \frac{1}{1 - 1/2}\,(\text{left deviation})
= 2\,(\text{left deviation}) .
$$

By induction over the $m$ pieces, the final deviation is at most $2^m\abs{y_0 - z_0}$, and the uniform deviation over the whole segment obeys the same bound (each piece’s sup is controlled at its stage). With pieces of length $h \asymp \frac1{2L}$, the factor is $2^m = 2^{\,\text{length}\cdot 2L}$: exponential in the length of the interval, exactly as Gronwall’s bound $\eu^{L\abs{t-t_0}}$ predicts, with better constants.

**17.** Same scheme: for $y \in X_h$,

$$
\abs{T_f(y)(t) - T_g(y)(t)} \leq \int_{t_0}^t \abs{f(s,y(s)) -
g(s,y(s))}\,\abs{\dd s} \leq \varepsilon h ,
$$

so question 3 (with $T_f$ the contraction, $T_g$ the perturbed map) yields $d_\infty \leq \frac{\varepsilon h}{1 - Lh}$.

**18.** By question 17 applied to $f = f_\lambda$, $g =
f_\mu$: $d_\infty(y_\lambda, y_\mu) \leq \frac{Ch}{1 -
Lh}\abs{\lambda - \mu}$: the solution map is [Lipschitz](#def-b2-metric-continuity) with constant $\frac{Ch}{1-Lh}$.

**19.** Every argument used only: the metric axioms, the [completeness](#def-b2-metric-complete) of the stage, the bound $\abs{\int} \leq
\int\abs{\cdot}$, and the [Lipschitz](#def-b2-metric-continuity) property of $f$ — all available for $\R^n$-valued functions with $d_\infty$ built on any of the equivalent distances of [Example 4.2](#ex-b2-metric-examples) ([complete](#def-b2-metric-complete) by [Theorem 4.9](#thm-b2-metric-rncomplete)). For $y' = Ay$ with the nilpotent $A$: $y^{(0)} \equiv (c_1, c_2)$,

$$
y^{(1)}(t) = (c_1, c_2) + \int_0^t (c_2, 0)\,\dd s
= (c_1 + tc_2,\; c_2),
$$

and $Ay^{(1)}(s) = (c_2, 0)$ again: $y^{(2)} = y^{(1)}$. The iteration is stationary from $n = 1$ on, at the exact solution $y(t) = (c_1 + tc_2, c_2) = \eu^{tA}y(0)$ — nilpotency truncates the exponential series, and Picard notices.

**20.** Fix $y \in X$ and let $g_y(x) = y - \eta(x)$: a $k$-contraction of the [complete](#def-b2-metric-complete) $X$, so there is exactly one $x$ with $x + \eta(x) = y$: the map $\Phi = \mathrm{id} + \eta$ is bijective. [Lipschitz](#def-b2-metric-continuity) inverse: if $\Phi(x) = y$ and $\Phi(x') =
y'$,

$$
d(x, x') \leq d(y, y') + d\bigl(\eta(x), \eta(x')\bigr)
\leq d(y, y') + k\,d(x, x'),
$$

so $d(x, x') \leq \frac{1}{1-k}d(y, y')$. (Distances here come from the norm structure, so $d(a - c, b - c) = d(a, b)$, which the first inequality used.)

**21.** $g(x) = m + e\sin x$ is $e$-Lipschitz on the [complete](#def-b2-metric-complete) $\R$ (mean value inequality, $\abs{g'} = \abs{e\cos x}
\leq e < 1$): Banach gives a unique solution and global convergence of the iteration. For $e = \frac12$, $m = 1$, $x_0
= 1$: $x_1 = 1 + \frac12\sin 1 \approx 1.42074$, $d(x_1, x_0)
\approx 0.4207$, and the a priori bound $\frac{(1/2)^n}{1/2}\cdot 0.4207 \leq 10^{-3}$ first holds at $n = 10$: ten iterations certified (the true value $x \approx
1.4987$ is in fact reached to $10^{-3}$ already around $n =
5$).

**22.** Use the digit description ([Exercise 4.8](#exo-b2-metric-8)): $C$ is the set of sums $\sum_{n\geq1}a_n3^{-n}$, $a_n \in \{0, 2\}$. Then $S_1(C) =
\{x/3 : x \in C\}$ is the subset with $a_1 = 0$, and $S_2(C) =
\{x/3 + 2/3\}$ the subset with $a_1 = 2$: their union, over the free choice of $a_1$, is exactly $C$. Uniqueness in one sentence: on the space of nonempty [compact](#def-b2-metric-compact) subsets of $\intcc01$ metrized by the Hausdorff distance, $A \mapsto
S_1(A) \cup S_2(A)$ is a $\frac13$-contraction of a [complete](#def-b2-metric-complete) space, so Banach allows only one fixed set — the Year 3 volume makes the Hausdorff metric and this argument rigorous.

**23.** Let $Z = \{t \in J : y(t) = z(t)\}$: nonempty ($t_0 \in Z$), closed in $J$ (equalizer of two [continuous](#def-b2-metric-continuity) maps: preimage of $\{0\}$ under $y - z$). [Open](#def-b2-metric-topology): if $t_1 \in Z$, apply the local theorem (question 8) at the point $(t_1, y(t_1))$, in a rectangle where $f$ is [Lipschitz](#def-b2-metric-continuity): on a small interval around $t_1$ both $y$ and $z$ solve the same Cauchy problem, so they coincide there (question 9’s unconditional uniqueness): a neighbourhood of $t_1$ lies in $Z$. A nonempty subset of the interval $J$ that is both [open](#def-b2-metric-topology) and closed in $J$ is all of $J$ (intervals are [connected](#def-b2-metric-connected), [Theorem 4.27](#thm-b2-metric-connectedness)): $y = z$ on $J$.

**24.** Here $f(t, y) = \alpha(t)y + \beta(t)$ is $L$-Lipschitz in $y$ on all of $\intcc AB \times \R$ with $L =
\sup\abs\alpha$ (finite: $\alpha$ [continuous](#def-b2-metric-continuity) on a segment), and no band $\intcc{y_0 - b}{y_0 + b}$ is needed: take $X = C(\intcc
AB)$ entire, on which $T$ is well defined. The induction of question 8 runs verbatim and gives $d_\infty(T^ny, T^nz) \leq
\frac{(L(B - A))^n}{n!}d_\infty(y, z)$: an iterate is a contraction whatever the length $B - A$, and question 4 concludes: one and only one solution on the whole segment. Linearity enters exactly once: it makes the [Lipschitz](#def-b2-metric-continuity) bound global in $y$, removing the rectangle and its $h \leq b/M$.

**25.** *[Completeness](#def-b2-metric-complete)* turned the Cauchy sequence of iterates into an actual solution (questions 1, 6, 8), and its absence let $\sqrt2$ escape from $\Q$ (question 13). *Contraction* gave uniqueness, the algorithm, and the error bars (questions 7, 14). *The iterate trick* removed the smallness condition $Lh < 1$, so the certified interval depends only on $M$, not on $L$ — and made linear equations global (questions 8, 24). *[Connectedness](#def-b2-metric-connected)* promoted local uniqueness to global uniqueness (question 23). The counterexamples: $2\sqrt{\abs y}$ guards [Lipschitz](#def-b2-metric-continuity) (question 11), $y^2$ guards locality (question 12), $\Q$ guards [completeness](#def-b2-metric-complete) (question 13). The summit is the Picard–Lindelöf theorem (question 8); when $f$ is merely [continuous](#def-b2-metric-continuity), existence survives but uniqueness dies, and the proof trades the contraction for [compactness](#def-b2-metric-compact) of function sets — Peano’s theorem via Arzelà–Ascoli, in the Year 3 volume.