University Mathematics — Year 3 · Bachelor Year 3

6General Topology

The Year 2 volume did analysis in metric spaces: distances, balls, sequences. But the fundamental notions — continuity, compactness, connectedness — never mention the numerical value of a distance, only the family of open sets it generates. This chapter takes that family as the primitive object. The gain is not generality for its own sake: quotient constructions (the circle as $\R/\Z$ , projective spaces), products, and the weak-type topologies of functional analysis simply are not metric-first objects. We rebuild continuity, then treat compactness by open covers (proving it equivalent to Year 2’s sequential definition in metric spaces), and connectedness — closing with the theorem that a continuous bijection cannot identify $\R$ with $\R^2$ : topology can distinguish dimensions.

6.1 Topologies, open sets, continuity

Definition 6.1

A topology on a set $X$ is a family $\mathcal T$ of subsets of $X$ — called open sets — such that: $\varnothing, X \in \mathcal T$ ; any union of open sets is open; any finite intersection of open sets is open. The pair $(X, \mathcal T)$ is a topological space. Complements of open sets are closed. A neighborhood of $x$ is a set containing an open set containing $x$ .

Example 6.2

(a) A metric space, with “open” as in Year 2 (unions of open balls): the metric topology; different metrics can give the same topology (equivalent metrics). A space whose topology arises from some metric is metrizable. (b) The discrete topology (all subsets) and the indiscrete topology $\{\varnothing, X\}$ . (c) The cofinite topology on an infinite set: open $=$ empty or cofinite. Not metrizable, as we shall see (Exercise 6.3). (d) On $\R$ , the usual topology; on $\bar\R = \R \cup \{\pm\infty\}$ , the order topology generated by rays — making “ $x_n \to +\infty$ ” an instance of plain convergence.

Definition 6.3

For $A \subseteq X$ : the interior $\mathring A$ is the largest open set inside $A$ (union of all of them); the closure $\bar A$ the smallest closed set containing $A$ ; the boundary $\partial A = \bar A \setminus \mathring A$ . $A$ is dense if $\bar A = X$ . One has $x \in \bar A$ iff every neighborhood of $x$ meets $A$ (if some neighborhood misses $A$ , its open core’s complement is a smaller closed set around $A$ ; conversely).

Definition 6.4

A basis of $\mathcal T$ is a family $\mathcal B \subseteq \mathcal T$ such that every open set is a union of members of $\mathcal B$ (e.g. open balls in a metric space; open intervals in $\R$ ). A family $\mathcal B$ of subsets of $X$ is a basis of some topology iff it covers $X$ and, for $B_1, B_2 \in \mathcal B$ and $x \in B_1 \cap B_2$ , some $B_3 \in \mathcal B$ has $x \in B_3 \subseteq B_1\cap B_2$ — then “unions of members” is a topology, the topology generated by $\mathcal B$ .

Definition 6.5

$f \colon X \to Y$ is continuous if $f^{-1}(V)$ is open for every open $V \subseteq Y$ — equivalently, preimages of closed sets are closed; equivalently, for every $x$ and neighborhood $V$ of $f(x)$ , $f^{-1}(V)$ is a neighborhood of $x$ (continuity at each $x$ ). It suffices to check on a basis of $Y$ . Compositions of continuous maps are continuous. A homeomorphism is a continuous bijection with continuous inverse; topology studies the properties preserved by homeomorphisms.

Proposition 6.6 (Continuity vs closure; gluing)

(a) $f$ is continuous iff $f(\bar A) \subseteq \overline{f(A)}$ for all $A \subseteq X$ . (b) If $X = F_1 \cup F_2$ with $F_1, F_2$ closed and $f\colon X \to Y$ restricts continuously to each $F_i$ , then $f$ is continuous.

Proof. (a) If $f$ continuous: $f^{-1}(\overline{f(A)})$ is closed and contains $A$ , hence contains $\bar A$ . Conversely apply the criterion to $A = f^{-1}(F)$ , $F$ closed: $f(\bar A) \subseteq \overline{f(f^{-1}(F))} \subseteq \bar F = F$ , so $\bar A \subseteq f^{-1}(F) = A$ : preimages of closed sets are closed. (b) For $F \subseteq Y$ closed: $f^{-1}(F) = (f\restriction_{F_1})^{-1}(F) \cup (f\restriction_{F_2})^{-1} (F)$ , a union of two sets closed in $F_1$ , resp. $F_2$ , hence closed in $X$ ( $F_i$ closed: closed-in-closed is closed). ∎

Definition 6.7

$X$ is Hausdorff (or separated) if any two distinct points have disjoint neighborhoods. Metric spaces are Hausdorff (balls of radius $d(x,y)/2$ ). A sequence $(x_n)$ converges to $x$ if every neighborhood of $x$ contains all but finitely many $x_n$ ; in a Hausdorff space, limits are unique (two limits would have disjoint neighborhoods each containing a tail). In a Hausdorff space, points — hence finite sets — are closed.

Remark 6.8

In metric spaces, sequences detect everything: $x \in \bar A$ iff some sequence of $A$ converges to $x$ (take $x_n \in A \cap B(x, 1/n)$ ), and $f$ is continuous iff it is sequentially continuous. In general spaces both equivalences fail; the correct sequence-substitutes (filters, nets) belong to a more advanced course. We will state sequence-based results in metric spaces and cover-based results in general — and prove them equivalent where they are.

6.2 Subspaces, products, quotients

Definition 6.9

Three ways to make new spaces from old:

Subspace: on $A \subseteq X$ , the open sets are the $U \cap A$ , $U$ open in $X$ — the coarsest topology making the inclusion continuous.
Product: on $X \times Y$ (and finite products), the topology with basis the open boxes $U \times V$ ; on an infinite product $\prod_i X_i$ , the basis consists of boxes $\prod U_i$ with $U_i = X_i$ for all but finitely many $i$ — the coarsest topology making every projection continuous.
Quotient: if $\sim$ is an equivalence on $X$ and $\pi\colon X \to X/{\sim}$ the projection, declare $V \subseteq X/{\sim}$ open iff $\pi^{-1}(V)$ is open — the finest topology making $\pi$ continuous.

Proposition 6.10 (Universal properties)

(a) $f \colon Z \to X \times Y$ is continuous iff both components $\pi_X \circ f$ , $\pi_Y \circ f$ are (same for arbitrary products). (b) $g \colon X/{\sim} \to Z$ is continuous iff $g \circ \pi \colon X \to Z$ is.

Proof. (a) Necessity: compositions. Sufficiency: it is enough to check preimages of basis boxes: $f^{-1}(U \times V) = (\pi_Xf)^{-1}(U) \cap (\pi_Yf)^{-1}(V)$ , open (finitely many factors $\neq X_i$ in the infinite case). (b) Necessity: composition. Sufficiency: for $W \subseteq Z$ open, $\pi^{-1}(g^{-1}(W)) = (g\pi)^{-1}(W)$ is open, which by definition of the quotient topology means $g^{-1}(W)$ is open. ∎

Example 6.11

The quotient $\R/\Z$ (identify $x$ and $x + n$ ) is homeomorphic to the circle $S^1 = \{z \in \C : \abs z = 1\}$ : the map $x \mapsto \eu^{2\iu\pi x}$ passes to a continuous bijection $\R/\Z \to S^1$ (Proposition 6.10(b)); its inverse is continuous by the compactness argument of Corollary 6.14 below (Exercise 6.5 details everything, including why $\R/\Z$ is Hausdorff and compact). Likewise $[0,1]$ with endpoints glued is $S^1$ , the square with opposite sides glued is the torus, and gluing is finally a theorem, not a picture.

6.3 Compactness

Definition 6.12

An open cover of $X$ is a family $(U_i)_{i\in I}$ of open sets with $\bigcup U_i = X$ . $X$ is compact if it is Hausdorff and every open cover admits a finite subcover. Equivalently (taking complements): every family of closed sets with the finite intersection property (all finite subfamilies have nonempty intersection) has nonempty total intersection.

Theorem 6.13 (First properties)

Let $X$ be compact.

A closed subset of $X$ is compact; a compact subset of a Hausdorff space is closed.
A continuous image of a compact space in a Hausdorff space is compact. In particular a continuous $f\colon X \to \R$ is bounded and attains its bounds.
A decreasing sequence of nonempty closed subsets of $X$ has nonempty intersection.

Proof. (1) Let $F \subseteq X$ closed and $(U_i)$ an open cover of $F$ (by opens of $X$ ): adding $X \setminus F$ gives an open cover of $X$ ; a finite subcover, minus $X \setminus F$ , covers $F$ . Subspace Hausdorff is clear. Conversely let $K \subseteq Y$ compact, $Y$ Hausdorff, and $y \notin K$ : for each $x \in K$ pick disjoint open $U_x \ni x$ , $V_x \ni y$ ; finitely many $U_x$ cover $K$ , and the intersection of the corresponding $V_x$ is a neighborhood of $y$ disjoint from $K$ ’s cover, hence from $K$ : the complement of $K$ is open.

(2) If $(V_j)$ covers $f(X)$ , then $(f^{-1}(V_j))$ covers $X$ ; a finite subcover downstairs comes from the finite subcover upstairs. The image is Hausdorff as a subspace. For $f$ real: $f(X)$ is compact in $\R$ , hence closed and bounded (cover by $(-n, n)$ for boundedness; closed by (1)), and a closed bounded set contains its supremum.

(3) If $\bigcap F_n = \varnothing$ , the opens $X \setminus F_n$ cover $X$ ; finitely many suffice, so some $F_{n_0} = \varnothing$ (the sequence decreases) — contradiction. ∎

Corollary 6.14

A continuous bijection from a compact space to a Hausdorff space is a homeomorphism.

Proof. The inverse is continuous iff direct images of closed sets are closed; a closed $F$ is compact (Theorem 6.13(1)), its image is compact (2), hence closed (1) in the Hausdorff target. ∎

Theorem 6.15 (Finite products)

A finite product of compact spaces is compact.

Proof. It suffices to treat $X \times Y$ . Hausdorff is inherited (separate in one coordinate). Let $(W_i)$ be an open cover of $X \times Y$ ; we may assume the $W_i$ are basis boxes $U_i \times V_i$ (refine: each point sits in a box inside some $W_i$ ; a finite subcover of boxes yields one of $W_i$ ’s). Fix $x \in X$ : the slice $\{x\}\times Y \cong Y$ is compact, so finitely many boxes $U_1\times V_1, \dots, U_k\times V_k$ cover it, with $x \in U_j$ for all $j$ ; then $U^x = \bigcap_{j \leq k} U_j$ is an open neighborhood of $x$ with $U^x \times Y$ covered by finitely many boxes (the tube lemma: for $(x', y) \in U^x \times Y$ , $y \in V_j$ for some $j$ since the boxes covered $\{x\}\times Y$ at level $y$ , and $x' \in U^x \subseteq U_j$ , so $(x', y) \in U_j \times V_j$ ). Now finitely many $U^x$ cover the compact $X$ ; the corresponding finite collections of boxes cover $X \times Y$ . ∎

Theorem 6.16 (Compactness in metric spaces)

For a metric space $(X, d)$ , the following are equivalent:

$X$ is compact (Borel–Lebesgue);
every sequence in $X$ has a convergent subsequence (sequential compactness — Year 2’s definition);
$X$ is complete and totally bounded: for every $\varepsilon > 0$ , finitely many balls of radius $\varepsilon$ cover $X$ .

Proof. (1) $\Rightarrow$ (2): Let $(x_n)$ have no convergent subsequence. Then each $x \in X$ has an open ball $B_x$ containing $x_n$ for only finitely many $n$ (otherwise a subsequence converges to $x$ : take radii $1/k$ ). Finitely many $B_x$ cover $X$ , so only finitely many indices $n$ exist: absurd.

(2) $\Rightarrow$ (3): Completeness: a Cauchy sequence with a convergent subsequence converges (Year 2). Total boundedness: if some $\varepsilon$ admits no finite cover, choose inductively $x_{n+1}$ outside $\bigcup_{k\leq n} B(x_k, \varepsilon)$ : the sequence has $d(x_m, x_n) \geq \varepsilon$ for $m \neq n$ , no Cauchy subsequence, no convergent one.

(3) $\Rightarrow$ (1): First, (3) implies (2): given $(x_n)$ , cover $X$ by finitely many balls of radius $1$ : one, $B_1$ , contains a subsequence; cover $X$ by balls of radius $1/2$ : one contains a further subsequence; iterate and diagonalize: the diagonal subsequence is Cauchy (two terms beyond stage $k$ lie in a common ball of radius $2^{-k}$ , up to the usual $2\cdot$ ), hence converges. Now let $(U_i)$ be an open cover and suppose no finite subcover. Lebesgue number argument: for each $n$ , some ball $B(y_n, 2^{-n})$ is not covered by finitely many $U_i$ — indeed, cover $X$ by finitely many balls of radius $2^{-n}$ ; if each were finitely covered, so would $X$ be. By (2), a subsequence $y_{n_k} \to y$ ; pick $i$ with $y \in U_i$ and $r > 0$ with $B(y, r) \subseteq U_i$ . For large $k$ , $B(y_{n_k}, 2^{-n_k}) \subseteq B(y, r) \subseteq U_i$ : covered by one $U_i$ — contradiction. ∎

Corollary 6.17 (Heine–Borel; Heine)

(a) A subset of $\R^n$ is compact iff it is closed and bounded. (b) A continuous map from a compact metric space to a metric space is uniformly continuous.

Proof. (a) Closed and bounded $\Rightarrow$ contained in a cube $[-M,M]^n$ , which is compact: $[-M, M]$ is (sequentially, by Bolzano–Weierstrass — or directly by dichotomy for covers), and Theorem 6.15 handles the product; then apply Theorem 6.13(1). Conversely a compact subset is closed (Theorem 6.13(1)) and bounded (cover by concentric balls).

(b) Let $f \colon X \to Y$ , $\varepsilon > 0$ . The balls $B\bigl(x, \delta_x\bigr)$ with $f\bigl(B(x, 2\delta_x)\bigr)\subseteq B\bigl(f(x), \varepsilon/2\bigr)$ cover $X$ ; extract a finite subcover $B(x_j, \delta_{x_j})$ and set $\delta = \min_j \delta_{x_j}$ . If $d(x, x') < \delta$ : $x \in B(x_j, \delta_{x_j})$ for some $j$ , and then both $x, x' \in B(x_j, 2\delta_{x_j})$ , so $d(f(x), f(x')) < \varepsilon$ . ∎

Definition 6.18

$X$ is locally compact if it is Hausdorff and every point has a compact neighborhood ( $\R^n$ ; open subsets of $\R^n$ ; discrete spaces — but not $\Q$ , see Exercise 6.9). Every locally compact space embeds in a compact one: the one-point compactification $\hat X = X \cup \{\infty\}$ , whose opens are those of $X$ together with the complements (in $\hat X$ ) of compact subsets of $X$ . One checks the axioms directly; $\hat X$ is compact (a cover has a member containing $\infty$ , whose complement is compact, covered by finitely many others) and Hausdorff (separate $x$ from $\infty$ by a compact neighborhood of $x$ and its complement). Example: $\hat\R \cong S^1$ , and $\widehat{\R^n} \cong S^n$ by stereographic projection (Exercise 6.11).

6.4 Connectedness

Definition 6.19

$X$ is connected if it is not the union of two disjoint nonempty open sets — equivalently, its only subsets both open and closed are $\varnothing$ and $X$ ; equivalently, every continuous map $X \to \{0, 1\}$ (discrete) is constant. A subset is connected if it is as a subspace.

Theorem 6.20

The connected subsets of $\R$ are exactly the intervals.
Continuous images of connected sets are connected (whence the intermediate value theorem: a continuous real map on a connected space has an interval as image).
If $(A_i)$ are connected with a common point, $\bigcup A_i$ is connected. If $A$ is connected and $A \subseteq B \subseteq \bar A$ , then $B$ is connected.
Finite products of connected spaces are connected.

Proof. Throughout we use the $\{0,1\}$ -criterion: $f \colon X \to \{0,1\}$ continuous must be constant.

(1) A non-interval $A$ misses some $c$ between $a, b \in A$ : $A = (A \cap (-\infty, c)) \sqcup (A \cap (c, +\infty))$ disconnects. Conversely let $I$ be an interval and $f \colon I \to \{0,1\}$ continuous with $f(a) = 0$ , $f(b) = 1$ , $a < b$ . Let $c = \sup\{x \in [a,b] : f(x) = 0\}$ ; continuity at $c$ forces $f(c) = 0$ (limit of values $0$ : every neighborhood of $c$ meets $\{f = 0\}$ ; $\{f=0\}$ is closed) and then $c < b$ with $f \equiv 1$ on $(c, b]$ , so $f(c) = 1$ by the same closure argument on $\{f = 1\} \ni c$ : contradiction.

(2) A continuous $g \colon f(X) \to \{0,1\}$ yields $g \circ f$ constant, so $g$ is constant on the image.

(3) A continuous $f\colon \bigcup A_i \to \{0,1\}$ is constant on each $A_i$ , with the same value at the common point. For the closure: $f \colon B \to \{0,1\}$ is constant $= c$ on $A$ ; any $x \in B \subseteq \bar A$ is in the closure of $A$ , and $f^{-1} (f(x))$ is a neighborhood of $x$ (preimage of open), which must meet $A$ : $f(x) = c$ .

(4) For $X \times Y$ and $f\colon X\times Y \to \{0,1\}$ : any two points $(x,y), (x',y')$ are joined by the “elbow” $\{x\}\times Y \cup X \times \{y'\}$ , a union of two connected sets (homeomorphic to $Y$ , $X$ ) meeting at $(x, y')$ : by (3) and (2), $f$ agrees on the two points. ∎

Definition 6.21

$X$ is path-connected if any two points are joined by a path (continuous $\gamma \colon [0,1] \to X$ ). Path-connected implies connected: two values of a continuous $f \colon X \to \{0,1\}$ at $x, y$ are values of the constant $f\circ\gamma$ (Theorem 6.20(1)–(2)). Convex subsets of normed spaces are path-connected (segments); so are $\R^n \setminus \{0\}$ for $n \geq 2$ (go around the origin), and $S^n$ for $n \geq 1$ (project paths from $\R^{n+1}\setminus \{0\}$ ).

Example 6.22 (The topologist’s sine curve)

Let $\Gamma = \{(x, \sin\frac1x) : 0 < x \leq 1\}$ and $S = \bar\Gamma = \Gamma \cup (\{0\}\times[-1,1])$ (every point $(0, y)$ , $\abs y \leq 1$ , is a limit of points of $\Gamma$ : solve $\sin\frac1x = y$ near $0$ ). Then $S$ is connected — closure of the connected $\Gamma$ , a continuous image of $(0, 1]$ (Theorem 6.20(3)) — but not path-connected: a path from $(1, \sin 1)$ to $(0,0)$ would have to traverse abscissas $\to 0$ while the ordinate oscillates between $\pm1$ ; Exercise 6.9 makes this rigorous. Connectedness and path-connectedness genuinely differ.

The topologist’s sine curve: the graph of 1x (red) accumulates on the whole segment \0\×[-1,1] (blue). The union is connected but not path-connected: no continuous path can cross the infinitely many oscillations in finite parameter time. — The topologist’s sine curve: the graph of $\sin\frac1x$ (red) accumulates on the whole segment $\{0\}\times[-1,1]$ (blue). The union is connected but not path-connected: no continuous path can cross the infinitely many oscillations in finite parameter time.

Definition 6.23

The connected component of $x \in X$ is the union of all connected subsets containing $x$ — the largest one (Theorem 6.20(3)). Components partition $X$ and are closed (closures of connected sets are connected). $X$ is totally disconnected if all components are singletons ( $\Q$ ; the Cantor set of the weekend problem).

Theorem 6.24

$\R$ is homeomorphic to no $\R^n$ with $n \geq 2$ .

Proof. Suppose $h \colon \R \to \R^n$ is a homeomorphism. Removing a point: $\R \setminus \{0\}$ is homeomorphic to $\R^n \setminus \{h(0)\}$ . But $\R\setminus\{0\}$ is disconnected, while $\R^n\setminus\{\text{point}\}$ is path-connected for $n \geq 2$ (Definition 6.21; translate the point to $0$ ): connectedness is a homeomorphism invariant (Theorem 6.20(2)) — contradiction. (That $\R^2 \not\cong \R^3$ requires finer invariants — algebraic topology; the weekend problem shows the danger: continuous surjections $\R \to \R^2$ do exist.) ∎

Method 6.25

To prove a set is connected: exhibit it as a continuous image, a union of overlapping connected sets, a closure, or a product (Theorem 6.20); for subsets of normed spaces, prove path-connectedness with explicit paths (segments, arcs, elbows). To prove two spaces are not homeomorphic: find a topological invariant that differs — compactness, connectedness, number of components, or components after deleting a well-chosen finite set (the trick of Theorem 6.24: it also proves $[0,1) \not\cong (0,1)$ and $S^1 \not\cong [0,1]$ ).

6.5 Exercises

Exercise 6.1 ★

(a) List all topologies on $\{a, b\}$ , classify them up to homeomorphism, and determine which ones are connected and which are Hausdorff. (b) In $\R$ (usual topology), compute interior, closure and boundary of $\Q$ , of $[0,1) \cup \{2\}$ , and of $\{1/n : n \geq 1\}$ .

Solution

Solution of Exercise 6.1.

(a) Four topologies on $\{a,b\}$ : indiscrete $\{\varnothing, X\}$ ; discrete; the two Sierpiński topologies $\{\varnothing, \{a\}, X\}$ and $\{\varnothing, \{b\}, X\}$ . The last two are homeomorphic (swap $a, b$ ): three classes. Connected: all except the discrete one (only the discrete topology contains a proper nonempty clopen set). Hausdorff: only the discrete one (in the others, $b$ ’s only neighborhood is $X$ , or $a$ ’s).

(b) $\Q$ : interior $\varnothing$ (every interval contains irrationals), closure $\R$ (density), boundary $\R$ . $[0,1) \cup \{2\}$ : interior $(0,1)$ , closure $[0,1]\cup\{2\}$ , boundary $\{0, 1, 2\}$ . $\{1/n\}$ : interior $\varnothing$ , closure $\{1/n\}\cup\{0\}$ , boundary the closure itself.

Exercise 6.2 ★

(a) Show that $f\colon X \to Y$ is continuous iff $f^{-1}(B)$ is open for every $B$ in a fixed basis of $Y$ . (b) Show that $f = \mathbf 1_{[0,\infty)} \colon \R \to \R$ is not continuous but is right-continuous. Verify that the half-open intervals $[a, b)$ form a basis of a topology on the source $\R$ (the Sorgenfrey line), that this topology is strictly finer than the usual one, and that a function $\R \to \R$ (usual target) is continuous from the Sorgenfrey line iff it is right-continuous at every point.

Solution

Solution of Exercise 6.2.

(a) Every open $V \subseteq Y$ is a union $\bigcup B_j$ of basis sets, and $f^{-1}(V) = \bigcup f^{-1}(B_j)$ : if the latter are open, so is the former; the converse is trivial.

(b) $f^{-1}\bigl(\intoo{\frac12}{\frac32}\bigr) = [0, \infty)$ is not open: $f$ is not continuous; right-continuity at each point is clear ( $f$ is locally constant on the right of every point). The sets $[a, b)$ satisfy the basis criterion (Definition 6.4): $[a,b)\cap[c,d) = [\max(a,c), \min(b,d))$ . The Sorgenfrey topology is finer than the usual one, since $(a, b) = \bigcup_{a < c < b}[c, b)$ ; strictly: $[0,1)$ is Sorgenfrey-open, not usual-open. Continuity from the Sorgenfrey line at $x$ : a Sorgenfrey neighborhood of $x$ contains a basis set $[a, b) \ni x$ , hence contains $[x, b)$ — so the continuity condition reads: for every $\varepsilon > 0$ there is $\delta > 0$ with $\abs{f(t) - f(x)} < \varepsilon$ whenever $x \leq t < x + \delta$ . That is exactly right-continuity at $x$ .

Exercise 6.3 ★

On an infinite set with the cofinite topology, show: any two nonempty open sets meet (so the space is not Hausdorff, hence not metrizable); every injective sequence converges to every point. Where does the uniqueness-of-limits proof use Hausdorff?

Solution

Solution of Exercise 6.3.

Two nonempty opens have finite complements, so their intersection has finite complement: nonempty (the set is infinite) — no two points have disjoint neighborhoods: not Hausdorff, hence not metrizable (Definition 6.7). Let $(x_n)$ be injective and $x$ arbitrary: a neighborhood of $x$ contains a cofinite open $U$ ; the finitely many points of $X\setminus U$ are hit by at most finitely many indices (injectivity), so a tail of the sequence lies in $U$ : $x_n \to x$ , for every $x$ . The uniqueness proof needs two disjoint neighborhoods to separate two alleged limits — precisely what fails here.

Exercise 6.4 ★★

(a) Show that the projections $\pi_X, \pi_Y$ of a product are continuous and open (images of opens are open), but not closed in general ( $\{xy = 1\}$ in $\R^2$ ). (b) Show that a sequence in a countable product $\prod_n X_n$ of metric spaces converges iff each coordinate converges, and that $d(x, y) = \sum_n 2^{-n}\min(1, d_n(x_n, y_n))$ metrizes the product topology. (c) On $\R^\N$ , show that the “box topology” (all products of opens are open) is strictly finer: the sequence $x^{(k)} = (1/k, 1/k, \dots)$ converges to $0$ in the product topology but not in the box topology.

Solution

Solution of Exercise 6.4.

(a) Continuity is by construction (Definition 6.9). Openness: an open $W \subseteq X \times Y$ is a union of boxes $U_i\times V_i$ , and $\pi_X(W) = \bigcup U_i$ (nonempty boxes project onto their factors), open. Not closed: $H = \{(x,y) : xy = 1\}$ is closed in $\R^2$ (preimage of $\{1\}$ under the continuous product), but $\pi_X(H) = \R\setminus\{0\}$ is not closed.

(b) ( $\Rightarrow$ ) Projections are continuous. ( $\Leftarrow$ ) Let $x^{(k)} \to x$ componentwise and $W = \prod U_n$ a basis neighborhood of $x$ , with $U_n = X_n$ except for $n \in F$ finite. For each $n \in F$ , $x_n^{(k)} \in U_n$ for $k \geq K_n$ ; for $k \geq \max_F K_n$ , $x^{(k)} \in W$ . The formula $d$ defines a metric (each summand is one, up to the standard verification that $\min(1, d_n)$ is); its balls: $B_d(x, 2^{-N})$ contains the basis box $\{y : d_n(x_n,y_n) < 2^{-N},\ n \leq N'\}$ for suitable $N'$ (the tail $\sum_{n > N'} 2^{-n}$ is small), and conversely every basis box contains a $d$ -ball: the two topologies have the same neighborhoods of each point.

(c) In the product topology, $x^{(k)} \to 0$ by (b). The box-open set $U = \prod_n \intoo{-\tfrac1n}{\tfrac1n}$ contains $0$ , but $x^{(k)} \notin U$ for every $k$ (the $n$ -th coordinate $1/k \geq 1/n$ when $n \geq k$ ): no tail enters $U$ . The box topology is strictly finer and not a product-type topology for convergence purposes.

Exercise 6.5 ★★

The circle, three ways. Show that the following are pairwise homeomorphic, with explicit maps: (i) $S^1 \subseteq \R^2$ ; (ii) $\R/\Z$ (quotient topology); (iii) $[0,1]/(0 \sim 1)$ . (For (ii): show $\R/\Z$ is Hausdorff — lift two classes to representatives at distance $\leq \frac12$ — and that $\pi([0,1])$ is everything, so the quotient is compact; then use Corollary 6.14.)

Solution

Solution of Exercise 6.5.

$\varphi \colon \R \to S^1$ , $x \mapsto \eu^{2\iu\pi x}$ , is continuous, surjective, and constant on classes mod $\Z$ : it induces a continuous bijection $\bar\varphi \colon \R/\Z \to S^1$ (Proposition 6.10(b)). The projection $\pi$ is open: for $U \subseteq \R$ open, $\pi^{-1}(\pi(U)) = \bigcup_{n}(U + n)$ is open, so $\pi(U)$ is open. Hausdorff: let $\bar x \neq \bar y$ ; choose representatives with $\delta = \min_{n}\abs{x - y - n} > 0$ ; the images of the intervals of radius $\delta/3$ around $x$ and $y$ are open (openness of $\pi$ ), contain $\bar x, \bar y$ , and are disjoint (two preimage points would be $< 2\delta/3$ apart mod $\Z$ ). Compact: $\R/\Z = \pi([0,1])$ , a continuous image of a compact space (Theorem 6.13(2)). Now $\bar\varphi$ is a continuous bijection from a compact space to the Hausdorff $S^1$ : a homeomorphism (Corollary 6.14).

For (iii): the composite $[0,1] \hookrightarrow \R \xrightarrow{\pi} \R/\Z$ is continuous, surjective, and identifies exactly $0 \sim 1$ : it induces a continuous bijection $[0,1]/(0\sim1) \to \R/\Z$ from a compact space (continuous image of $[0,1]$ under the quotient projection) to a Hausdorff one: a homeomorphism. Composing: all three spaces are homeomorphic.

Exercise 6.6 ★★

Let $X$ be a metric space. (a) Show that a finite union of compact subsets is compact, and that an arbitrary intersection is. (b) If $K$ is compact, $F$ closed, $K \cap F = \varnothing$ , show $d(K, F) = \inf\{d(x,y) : x \in K, y \in F\} > 0$ ; give a counterexample with two disjoint closed sets. (c) Show that $X$ is compact iff every continuous $f \colon X \to \R$ is bounded. (If some sequence has no convergent subsequence, build an unbounded continuous function supported near its terms; or use $x \mapsto d(x, \cdot)$ -type functions — one clean route: if $(x_n)$ has no cluster point, the set $\{x_n\}$ is closed and discrete, and $f(x_n) = n$ extends continuously by Tietze-free means: $f(x) = \sum_n n\,\max\bigl(0, 1 - \frac{d(x, x_n)}{r_n}\bigr)$ for small enough $r_n$ .)

Solution

Solution of Exercise 6.6.

(a) A cover of $K_1 \cup \dots \cup K_m$ restricts to a cover of each $K_j$ : finitely many opens per piece suffice. An intersection $\bigcap K_i$ is closed in the compact $K_{i_0}$ (compacts are closed in the ambient metric space), hence compact.

(b) $x \mapsto d(x, F)$ is continuous ( $1$ -Lipschitz) and strictly positive on $K$ ( $d(x, F) = 0$ means $x \in \bar F = F$ ); on the compact $K$ it attains a minimum $m > 0$ : $d(K, F) \geq m$ . Counterexample without compactness: $F_1 = \{n : n \geq 2\}$ and $F_2 = \{n + \frac1n : n \geq 2\}$ are closed, disjoint, at distance $\inf_n \frac1n = 0$ .

(c) If $X$ is compact, every continuous $f$ is bounded (Theorem 6.13(2)). Conversely, if $X$ is not compact, take $(x_n)$ with no convergent subsequence (Theorem 6.16); passing to a subsequence we may assume the $x_n$ pairwise distinct, and no point of $X$ is a cluster point of the sequence, so

r_n = \min\Bigl(2^{-n},\ \tfrac13\,d\bigl(x_n, \{x_m : m \neq n\}\bigr)\Bigr) > 0,

and the balls $B(x_n, r_n)$ are pairwise disjoint (a common point of the $n$ -th and $m$ -th would give $d(x_n, x_m) < r_n + r_m \leq \frac23 d(x_n, x_m)$ ). Define

f(x) = \sum_{n} n\,\max\Bigl(0,\ 1 - \frac{d(x, x_n)}{r_n}\Bigr):

at each $x$ at most one summand is nonzero, and $f(x_n) = n$ . Continuity at $x$ : let $y_k \to x$ ; each value $f(y_k)$ is either $0$ or a single bump value $\mathrm{bump}_{n_k}(y_k)$ . If some index $n$ occurs infinitely often, along that subsequence $f(y_k) = \mathrm{bump}_n(y_k) \to \mathrm{bump}_n(x)$ , which equals $f(x)$ (if $x \in B(x_n, r_n)$ , the whole tail lies in this open ball and no other index occurs; if $x \notin B(x_n, r_n)$ then $\mathrm{bump}_n(x) = 0 = f(x)$ , since $x$ , adherent to the $n$ -th ball, lies in no other open ball). If $n_k \to \infty$ : $d(y_k, x_{n_k}) < r_{n_k} \leq 2^{-n_k} \to 0$ , so $x_{n_k} \to x$ , making $x$ a cluster point of $(x_n)$ — excluded; so this case concerns finitely many $k$ only. In every case $f(y_k) \to f(x)$ : $f$ is continuous, and unbounded.

Exercise 6.7 ★★

(Lebesgue number) Let $(U_i)$ be an open cover of a compact metric space $X$ . Show there is $\delta > 0$ such that every subset of diameter $< \delta$ lies in a single $U_i$ . (Otherwise pick $A_n$ of diameter $< 1/n$ in no $U_i$ , and a cluster point of chosen $x_n \in A_n$ .) Deduce Heine’s theorem (Corollary 6.17(b)) again.

Solution

Solution of Exercise 6.7.

Suppose no $\delta$ works: for each $n$ there is $A_n$ of diameter $< 1/n$ contained in no single $U_i$ ; pick $x_n \in A_n$ . By compactness (Theorem 6.16), a subsequence $x_{n_k} \to x$ ; pick $i$ and $r > 0$ with $B(x, r) \subseteq U_i$ . For large $k$ : $d(x_{n_k}, x) < r/2$ and $1/n_k < r/2$ , so $A_{n_k} \subseteq B(x_{n_k}, 1/n_k) \subseteq B(x, r) \subseteq U_i$ — contradiction. Heine: given $\varepsilon$ , cover $Y$ by balls $B(y, \varepsilon/2)$ ; the preimages form an open cover of $X$ ; let $\delta$ be a Lebesgue number: if $d(x, x') < \delta$ , the pair $\{x, x'\}$ has diameter $< \delta$ , lies in one preimage, and $d(f(x), f(x')) < \varepsilon$ .

Exercise 6.8 ★★

(a) Show that $GL_n(\R)$ is an open, dense subset of $M_n(\R)$ , and that it is disconnected: the sign of the determinant separates it into (at least) two pieces. Show on the other hand that $GL_n(\C)$ is path-connected. (For $A, B \in GL_n(\C)$ : $\lambda \mapsto \det\bigl((1 - \lambda)A + \lambda B\bigr)$ is a polynomial, not identically zero, so it has finitely many roots in $\C$ : pick a path of $\lambda$ ’s in $\C$ from $0$ to $1$ avoiding them.) (b) Prove density: $A + \varepsilon I$ is invertible for small $\varepsilon > 0$ .

Solution

Solution of Exercise 6.8.

(a) $GL_n(\R) = \det^{-1}(\R\setminus\{0\})$ is open ( $\det$ is polynomial, continuous). Disconnected: $\det$ maps it onto $\R\setminus\{0\}$ , and a connected space has connected continuous images (Theorem 6.20(2)); $\R\setminus\{0\}$ is not an interval. $GL_n(\C)$ : for $A, B$ invertible, $p(\lambda) = \det\bigl((1-\lambda)A + \lambda B\bigr)$ is a polynomial in $\lambda$ with $p(0), p(1) \neq 0$ , hence not identically zero: it has finitely many roots in $\C$ . The plane minus finitely many points is path-connected (avoid the points by going around), so there is a path $\lambda(t)$ from $0$ to $1$ with $p(\lambda(t)) \neq 0$ : $t \mapsto (1-\lambda(t))A + \lambda(t)B$ is a path in $GL_n(\C)$ .

(b) $\det(A + \varepsilon I) = \chi_A(-\varepsilon)\cdot(\pm1)$ vanishes for at most $n$ values of $\varepsilon$ : invertible matrices $A + \varepsilon I \to A$ as $\varepsilon \to 0$ : density.

Exercise 6.9 ★★

(a) Show that the connected components of $\Q$ are the singletons, and that $\Q$ is not locally compact (a compact neighborhood of $0$ in $\Q$ would contain $[-r, r]\cap\Q$ , whose closure in $\Q$ is not compact: cut at an irrational). (b) Complete Example 6.22: no path joins $(0,0)$ to $\Gamma$ . (If $\gamma = (\gamma_1, \gamma_2)$ is such a path, let $t^* = \sup\{t : \gamma_1(t) = 0\}$ ; for $t > t^*$ the point moves on the graph; choose $t_k \downarrow t^*$ with $\gamma_1(t_k) \to 0$ hitting abscissas where $\sin$ is alternately $\pm1$ — the intermediate value theorem supplies them — contradicting continuity of $\gamma_2$ at $t^*$ .)

Solution

Solution of Exercise 6.9.

(a) Let $A \subseteq \Q$ contain $p < q$ and pick an irrational $\alpha \in (p, q)$ : $A = (A \cap (-\infty, \alpha)) \sqcup (A\cap(\alpha, +\infty))$ splits $A$ into two nonempty relatively open pieces: $A$ is disconnected. Components are singletons. Local compactness fails: a compact neighborhood $V$ of $0$ in $\Q$ contains $[-r, r]\cap\Q$ for some $r > 0$ , which is closed in $V$ , hence compact; but a sequence of rationals in $[-r,r]$ converging (in $\R$ ) to an irrational has no subsequence converging in $\Q$ : contradiction with Theorem 6.16.

(b) Let $\gamma = (\gamma_1, \gamma_2) \colon [0,1] \to S$ be a path with $\gamma(0) = (0,0)$ and $\gamma(1) \in \Gamma$ . The set $\{t : \gamma_1(t) = 0\}$ is closed and does not contain $1$ ; let $t^*$ be its supremum, so $\gamma_1(t^*) = 0$ and $\gamma_1 > 0$ on $(t^*, 1]$ . By continuity of $\gamma_2$ at $t^*$ , choose $\delta$ with $\abs{\gamma_2(t) - \gamma_2(t^*)} < \tfrac12$ for $t \in [t^*, t^* + \delta]$ . Fix $t_0 = t^* + \delta$ (may assume $\leq 1$ ): $\gamma_1(t_0) > 0$ , and $\gamma_1$ takes every value of $(0, \gamma_1(t_0)]$ on $[t^*, t_0]$ (intermediate value theorem, Theorem 6.20). Choose $k$ large with $a_k = \frac1{\pi/2 + 2k\pi} < \gamma_1(t_0)$ and $b_k = \frac1{3\pi/2 + 2k\pi} < \gamma_1(t_0)$ : there are $s, s' \in (t^*, t_0]$ with $\gamma_1(s) = a_k$ , $\gamma_1(s') = b_k$ ; since the points lie on the graph, $\gamma_2(s) = \sin(1/a_k) = 1$ and $\gamma_2(s') = -1$ . Both cannot be within $\frac12$ of $\gamma_2(t^*)$ : contradiction. $S$ is connected but not path-connected.

Exercise 6.10 ★★★

The middle-thirds Cantor set $C = \bigcap_n C_n$ , where $C_0 = [0,1]$ and $C_{n+1}$ removes the open middle third of each interval of $C_n$ . (a) Show $C = \bigl\{\sum_{n\geq1} a_n3^{-n} : a_n \in \{0,2\}\bigr\}$ , and that $C$ is compact, with empty interior, and has no isolated point (perfect). (b) Show that $C$ is totally disconnected. (c) Show that $x \mapsto (a_n(x))_n$ is a homeomorphism $C \to \{0,2\}^\N$ (product topology on the discrete two-point space); deduce that $C$ is uncountable, and that $C \times C \cong C$ .

Solution

Solution of Exercise 6.10.

(a) $C_n$ consists exactly of the $x$ admitting a ternary expansion with digits in $\{0, 2\}$ up to rank $n$ (induction: removing middle thirds removes first digit $1$ , etc.; endpoints have two expansions, one avoiding $1$ s), so $C = \bigcap C_n$ is the set of sums $\sum a_n3^{-n}$ , $a_n \in \{0,2\}$ . Compact: each $C_n$ is a finite union of closed intervals; $C$ is closed in $[0,1]$ . Empty interior: $C \subseteq C_n$ , a union of intervals of length $3^{-n}$ ; an interior interval of length $\ell > 0$ would fit in one of them for all $n$ . Perfect: given $x \in C$ and $m$ , flip the digit $a_m$ : the new point is in $C$ , distinct, within $2\cdot3^{-m}$ of $x$ .

(b) If $x \neq y \in C$ , their expansions first differ at some rank $k$ ; between them lies a removed middle-third interval (the gap at rank $k$ separating digit $0$ from digit $2$ ), providing a point $c \notin C$ , $x < c < y$ (say): $C = (C \cap (-\infty, c)) \sqcup (C \cap (c, \infty))$ splits any subset containing both points. Components are singletons.

(c) The digit map $\beta\colon x \mapsto (a_n(x))$ is a bijection onto $\{0,2\}^\N$ (existence and uniqueness of $\{0,2\}$ -expansions: distinct digit sequences give points at distance $\geq 3^{-k} > 0$ where they first differ, as in Problem 6.1, question 1). It is continuous: $\abs{x - y} < 3^{-m}$ forces agreement of the first $m$ digits (same computation), so $\beta$ maps small balls into basis boxes. A continuous bijection from the compact $C$ to the Hausdorff product is a homeomorphism (Corollary 6.14; the product is Hausdorff: separate at a differing coordinate). Uncountability: Cantor’s diagonal on $\{0,2\}^\N$ . Finally $\{0,2\}^\N \times \{0,2\}^\N \cong \{0,2\}^\N$ by interleaving digits (a homeomorphism: componentwise continuity both ways), so $C \times C \cong C$ .

Exercise 6.11 ★★★

Stereographic projection: from the north pole $N = (0, \dots, 0, 1)$ of $S^n \subseteq \R^{n+1}$ , the map $\sigma(x) = \frac{(x_1, \dots, x_n)}{1 - x_{n+1}}$ is a homeomorphism $S^n \setminus \{N\} \to \R^n$ (give the inverse explicitly). Deduce that $S^n$ is homeomorphic to the one-point compactification $\widehat{\R^n}$ , and that removing any point of $S^n$ leaves a space homeomorphic to $\R^n$ .

Solution

Solution of Exercise 6.11.

For $y \in \R^n$ , set

\tau(y) = \Bigl(\frac{2y}{\norm y^2 + 1},\ \frac{\norm y^2 - 1}{\norm y^2 + 1}\Bigr) \in \R^n\times\R = \R^{n+1}.

One checks $\norm{\tau(y)} = 1$ , $\tau(y) \neq N$ , and $\sigma(\tau(y)) = y$ , $\tau(\sigma(x)) = x$ : $\sigma$ and $\tau$ are mutually inverse, both continuous (rational formulas with nonvanishing denominators): $S^n\setminus\{N\} \cong \R^n$ . Extend to $\hat\sigma \colon S^n \to \widehat{\R^n}$ by $N \mapsto \infty$ : a bijection, continuous at every point of $S^n\setminus\{N\}$ , and at $N$ : a basis neighborhood of $\infty$ is $\widehat{\R^n}\setminus K$ , $K$ compact, $K \subseteq \bar B(0, R)$ ; since $\norm{\sigma(x)}^2 = \frac{1 + x_{n+1}}{1 - x_{n+1}} \to \infty$ as $x_{n+1} \to 1$ , the set $\{x \in S^n : x_{n+1} > 1 - \eta\}$ maps outside $\bar B(0,R)$ for small $\eta$ : continuity. A continuous bijection from the compact $S^n$ to the Hausdorff $\widehat{\R^n}$ is a homeomorphism. Removing another point $p$ : a rotation of $S^n$ maps $p$ to $N$ (rotations are homeomorphisms), reducing to the computed case: $S^n \setminus \{p\} \cong \R^n$ .

Exercise 6.12 ★★★

(The topologist’s sine curve) Let

T = \bigl\{(x, \sin\tfrac1x) : 0 < x \leq 1\bigr\} \cup \bigl(\{0\}\times\intcc{-1}1\bigr) \subseteq \R^2 .

(a) Show that $T$ is compact, and that it is the closure of the graph part. (b) Show that $T$ is connected (the graph is connected as a continuous image; its closure remains connected). (c) Show that $T$ is not path-connected: no continuous path joins $(0,0)$ to $(1, \sin1)$ . (If $\gamma = (\gamma_1, \gamma_2)$ is such a path, let $t^* = \sup\{t : \gamma_1(t) = 0\}$ ; just after $t^*$ , $\gamma_1$ takes all small positive values (intermediate value theorem), so $\gamma_2$ oscillates between $\pm1$ on every interval $(t^*, t^* + \delta)$ — contradict continuity at $t^*$ .) (d) Conclude that path-connectedness is strictly stronger than connectedness, and show that no such example can be open in $\R^n$ : an open connected subset of $\R^n$ is path-connected (the set of points joinable to a base point by a path is open and closed in the domain).

Solution

Solution of Exercise 6.12.

(a) $T$ is bounded, and closed: a limit of points of $T$ with abscissas $\to x_0 > 0$ stays on the (locally closed) graph by continuity of $\sin\frac1x$ on $\intoc01$ ; a limit with abscissas $\to 0$ has ordinate in $\intcc{-1}1$ , hence lies in the segment. Compact by Heine–Borel (Corollary 6.17). Closure of the graph $\Gamma$ : every point $(0, y)$ , $y \in \intcc{-1}1$ , is a limit of graph points — solve $\sin\frac1x = y$ near $0$ (the function sweeps $\intcc{-1}1$ on each interval $[\frac1{(k+1)\pi}, \frac1{k\pi}]$ ): $\bar\Gamma = T$ .

(b) $\Gamma$ is the continuous image of the connected $\intoc01$ under $x \mapsto (x, \sin\frac1x)$ : connected; and the closure of a connected set is connected (Theorem 6.20): $T = \bar\Gamma$ is connected.

(c) Suppose $\gamma\colon\intcc01\to T$ is continuous with $\gamma(0) = (0,0)$ , $\gamma(1) = (1, \sin1)$ , and let $t^* = \sup\{t : \gamma_1(t) = 0\}$ : by continuity $\gamma_1(t^*) = 0$ and $\gamma_1 > 0$ on $\intoc{t^*}1$ . For every $\delta > 0$ , the interval $\intoc{t^*}{t^*+\delta}$ has $\gamma_1$ taking all values in some $\intoc0\eta$ (intermediate value theorem, $\gamma_1(t^* + \delta) > 0$ ); in particular it contains abscissas of the form $\frac1{\pi/2 + 2k\pi}$ and $\frac1{-\pi/2 + 2k\pi}$ for arbitrarily large $k$ , at which $\gamma_2 = \pm1$ . So on every right-neighborhood of $t^*$ , $\gamma_2$ takes both values $1$ and $-1$ : $\gamma_2$ has no limit at $t^{*+}$ , contradicting continuity. No path exists.

(d) $T$ is connected but not path-connected: the two notions differ. For open $U \subseteq \R^n$ connected and $a \in U$ , let $V$ be the set of points of $U$ joinable to $a$ by a path in $U$ . $V$ is open: around $x \in V$ a ball $B(x, r) \subseteq U$ is starlike, and concatenating the path to $x$ with a segment reaches every point of the ball. $V$ is closed in $U$ : if $x \in U \setminus V$ , the same ball argument shows $B(x, r) \cap V = \varnothing$ (a point of $V$ in the ball would join $a$ to $x$ ). Nonempty ( $a \in V$ ), open and closed in the connected $U$ : $V = U$ . The sine curve evades this by being closed with empty interior: its “bad” point $(0,0)$ has no ball inside $T$ to bridge the oscillations.

6.6 Problem: the Cantor set and a space-filling curve

Problem 6.1

Weekend problem — Peano curves exist, and why they are not homeomorphisms

In 1890 Peano stunned analysis with a continuous surjection $[0,1] \to [0,1]^2$ : a curve filling a square. We build one with bare hands from the Cantor set $C$ of Exercise 6.10 (whose results may be used freely), and then prove that no such map can be injective: squares are not curves. Throughout, elements of $C$ are written $x = \sum_{n \geq 1} \frac{2b_n(x)}{3^n}$ with digits $b_n(x) \in \{0, 1\}$ .

Part I — Reading off digits continuously.

Show that if $x, y \in C$ satisfy $\abs{x - y} < 3^{-m}$ , then $b_n(x) = b_n(y)$ for all $n \leq m$ . (If the first differing digit is $b_k$ , then $\abs{x - y} \geq \frac{2}{3^k} - \sum_{j>k}\frac2{3^j} = \frac1{3^k}$ .)
Deduce that each digit function $b_n \colon C \to \{0, 1\}$ is continuous, and re-derive the homeomorphism $C \cong \{0,1\}^\N$ of Exercise 6.10(c).

Part II — A continuous surjection $C \to [0,1]^2$ .

Define $g \colon C \to [0,1]$ by $g(x) = \sum_{n\geq1} b_n(x)\,2^{-n}$ (read the Cantor digits as binary). Show that $g$ is continuous (use question 1) and surjective. Is it injective?
Define $\Phi \colon C \to [0,1]^2$ by

$\Phi(x) = \Bigl(\ \sum_{n\geq1} b_{2n-1}(x)\,2^{-n},\ \sum_{n\geq1} b_{2n}(x)\,2^{-n}\ \Bigr)$

(odd digits give the abscissa, even digits the ordinate). Show that $\Phi$ is continuous and surjective.

Part III — Filling the gaps: the Peano curve.

The complement $[0,1] \setminus C$ is a countable union of disjoint open intervals $(u, v)$ (the removed middle thirds) whose endpoints lie in $C$ . Define $f \colon [0,1] \to [0,1]^2$ by $f = \Phi$ on $C$ , extended affinely on each gap:

$f(t) = \frac{v - t}{v - u}\,\Phi(u) + \frac{t - u}{v - u}\,\Phi(v), \qquad t \in (u, v).$

Show that $f$ is well defined and surjective onto $[0,1]^2$ .
Show that $f$ is continuous at every point of $[0,1] \setminus C$ (locally affine), and at every point of $C$ : given $\varepsilon$ , pick $m$ with $2^{-m} < \varepsilon/2$ and use question 1 to control $\Phi$ on $C$ near $x$ , then check that the affine interpolation cannot escape: on a gap $(u,v) \subseteq (x - 3^{-m}, x + 3^{-m})$ , the values $f(t)$ lie on the segment $[\Phi(u), \Phi(v)]$ , both ends close to $\Phi(x)$ . Conclude: $f$ is a continuous surjection $[0,1] \to [0,1]^2$ .
Deduce continuous surjections $[0,1] \to [0,1]^n$ for every $n \geq 2$ , and $\R \to \R^2$ .

Part IV — But never injective.

Show that a continuous injection $f \colon [0,1] \to [0,1]^2$ would be a homeomorphism onto its image (Corollary 6.14).
Show that $[0,1]$ and $[0,1]^2$ are not homeomorphic: remove a well-chosen point and compare connectedness (Method 6.25).
Conclude: a continuous surjection $[0,1] \to [0,1]^2$ can never be injective — an injective one would make $[0,1]^2 = f([0,1])$ homeomorphic to $[0,1]$ by question 8, contradicting question 9. Where exactly did compactness of $[0,1]$ enter the argument?
(Culmination) Assemble the moral: there is a continuous surjection but no continuous bijection $[0,1] \to [0,1]^2$ . What does this say about “dimension” as a topological notion? Formulate precisely one theorem proved in this problem and one plausible statement that remains beyond our tools (invariance of domain).

Part V — The arithmetic of the Cantor set.

Prove that $C + C = [0, 2]$ : given $t \in [0, 1]$ , write $t = \sum_n d_n3^{-n}$ with digits $d_n \in \{0, 1, 2\}$ and split each digit as $d_n = a_n + b_n$ with $a_n, b_n \in \{0, 1\}$ ; conclude that $\frac C2 + \frac C2 = [0, 1]$ (which subset of $[0,1]$ is $\frac C2$ , in terms of ternary digits?), then rescale. No carries are ever needed — say why this is the crux.
Interpret geometrically: the “Cantor dust” $C \times C \subseteq \R^2$ , of empty interior, casts a full shadow on the diagonal: the projection $(x, y) \mapsto x + y$ maps it onto $[0, 2]$ . Also record the self-similarity $C = \frac C3 \cup \bigl(\frac C3 + \frac23\bigr)$ , the equation behind every picture of $C$ .
Show that digit interleaving defines a homeomorphism $C \to C \times C$ , and deduce $C \cong C^n$ for every $n$ — the Cantor set is its own square, cube, … Which familiar spaces share this property?
Show that $\frac14 \in C$ (compute its ternary expansion: $\frac14 = \sum_{k\geq1} 2\cdot3^{-2k}$ ) although $\frac14$ is an endpoint of no removed interval; deduce — counting endpoints — that endpoints form a countable, proper subset of $C$ .
Show that the binary-reading map $g$ of question 3 is at most $2$ -to- $1$ , and describe exactly which points of $[0,1]$ have two preimages. ( $g$ collapses $C$ onto $[0,1]$ by gluing countably many pairs: the combinatorial shadow of the Cantor staircase met again in Chapter 9.)

Part VI — Perfect compact sets: Cantor everywhere. A nonempty compact metric space $P$ is perfect if it has no isolated point.

Let $P$ be perfect compact and let $x \in P$ , $r > 0$ . Show that the ball $B(x, r)$ contains two points $y_0 \neq y_1$ of $P$ , and hence two disjoint closed balls $\bar B(y_0, \rho)$ , $\bar B(y_1, \rho)$ inside $B(x, r)$ , centered at points of $P$ , of radius $\rho \leq r/4$ as small as desired. Explain why perfection (no isolated points) is exactly what allows this splitting to be repeated inside each of the two new balls.
Iterate: build closed sets $F_w \neq \varnothing$ indexed by finite binary words $w$ , with $F_{w0}, F_{w1} \subseteq F_w$ disjoint and $\operatorname{diam}F_w \leq 2^{-\abs w} \operatorname{diam}P$ . Show that for every infinite word $\varepsilon \in \{0,1\}^\N$ the intersection $\bigcap_mF_{\varepsilon_1\cdots\varepsilon_m}$ is a single point $h(\varepsilon)$ (finite intersection property of the compact $P$ ).
Show that $h\colon\{0,1\}^\N \to P$ is injective and continuous, and conclude: every perfect compact metric space is uncountable — in fact of cardinality at least that of $\{0,1\}^\N$ . Recover: $[0,1]$ and $C$ are uncountable.
Show that $h$ is a homeomorphism onto its image (continuous injection from a compact, Corollary 6.14): every perfect compact metric space contains a homeomorphic copy of the Cantor set. The Cantor set is not an oddity but the universal germ of compact perfection.
Deduce that every countable compact metric space has an isolated point, and exhibit one where the isolated points are dense but not everything: $K = \{0\} \cup \{\frac1n : n \geq 1\}$ .
(Cantor–Bendixson for $\R$ ) Let $F \subseteq \R$ be closed. Call $x$ a condensation point of $F$ if every neighborhood of $x$ meets $F$ uncountably. Show that the condensation points of an uncountable closed $F$ form a nonempty perfect closed set $P$ , and $F \setminus P$ is countable (cover the non-condensation points by countably many rational intervals meeting $F$ countably). Conclude: every closed subset of $\R$ is countable or has the cardinality of the continuum — the continuum hypothesis holds for closed sets.

Part VII — Codas: how regular, and how far from perfect.

(Hölder regularity of the curve) Set $\alpha = \frac{\ln 2}{2\ln 3} \approx 0.3155$ , and note the identity $3^\alpha = \sqrt2$ . Using question 1, show that

$\norm{\Phi(x) - \Phi(y)}_\infty \leq 2\,\abs{x - y}^\alpha \qquad (x, y \in C),$

then propagate the estimate through the affine gaps: show that the Peano curve $f$ of question 6 satisfies $\norm{f(t) - f(s)}_\infty \leq 6\abs{t - s}^\alpha$ on all of $[0,1]$ (treat a pair in the same gap by interpolation, then a general pair by passing through the extreme points of $C \cap [t, s]$ ).
(The exponent $\frac12$ is a wall) Show that no surjection $F \colon [0,1] \to [0,1]^2$ can be $\alpha$ -Hölder with $\alpha > \frac12$ : cut $[0,1]$ into $N$ intervals, bound the diameters of their images, and count the points of the grid $\bigl(\frac iM, \frac jM\bigr)$ , $0 \leq i, j \leq M$ , that a set of diameter $< \frac1M$ can contain; choose $M$ of order $N^\alpha$ and let $N \to \infty$ . Locate our curve ( $\alpha \approx 0.3155$ ) relative to the wall, and record without proof that Hilbert’s curve achieves the critical exponent $\frac12$ .
(Derived sets: measuring imperfection) For $F$ closed in $\R$ , let $F'$ be the set of limit points of $F$ (a closed subset), and iterate: $F^{(0)} = F$ , $F^{(i+1)} = (F^{(i)})'$ . Verify that

$K = \{0\} \cup \Bigl\{\frac1m\Bigr\}_{m \geq 1} \cup \Bigl\{\frac1m + \frac1n : m \geq 1,\ n > m^2\Bigr\}$

is a countable compact set with $K' = \{0\} \cup \{\frac1m\}$ , $K'' = \{0\}$ , $K''' = \varnothing$ (check that the $m$ -th cluster lives in the interval $(\frac1m, \frac1{m-1})$ , so the clusters do not interleave), and that its isolated points are dense in $K$ , as question 21 predicts. Describe the induction producing, for every $j$ , a countable compact $K_j$ with $K_j^{(j)} = \{0\}$ and $K_j^{(j+1)} = \varnothing$ , and contrast with the perfect sets of question 22, for which the derivation never moves: finite rank is the exact opposite of perfection.

Solution

Solution of Problem 6.1.

1. Suppose the expansions of $x, y \in C$ first differ at rank $k \leq m$ , say $b_k(x) = 1$ , $b_k(y) = 0$ . Then

x - y \;\geq\; \frac{2}{3^k} - \sum_{j > k}\frac{2}{3^j} = \frac{2}{3^k} - \frac{1}{3^k} = \frac1{3^k} \geq \frac1{3^m}:

contrapositive: $\abs{x - y} < 3^{-m}$ forces agreement up to rank $m$ .

2. By question 1, $b_n$ is constant on $C \cap (x - 3^{-n}, x + 3^{-n})$ : locally constant, hence continuous. The map $x \mapsto (b_n(x))_n \in \{0,1\}^\N$ is continuous (componentwise, Proposition 6.10(a)) and bijective (unique $\{0,2\}$ -digit expansions); from the compact $C$ to a Hausdorff space, it is a homeomorphism (Corollary 6.14).

3. Continuity: if $\abs{x - y} < 3^{-m}$ , the first $m$ digits agree, so $\abs{g(x) - g(y)} \leq \sum_{n > m}2^{-n} = 2^{-m}$ . Surjectivity: every $t \in [0,1]$ has a binary expansion $t = \sum b_n2^{-n}$ , and $t = g\bigl(\sum 2b_n3^{-n}\bigr)$ . Not injective: $g$ identifies the two Cantor points with digits $(0,1,1,1,\dots)$ and $(1,0,0,\dots)$ — both map to $\frac12$ (the dyadic ambiguity $0.0111\ldots_2 = 0.1000\ldots_2$ ).

4. Each component of $\Phi$ is continuous by the same estimate (its digits are a subsequence of the $b_n$ ). Surjectivity: given $(s, t) \in [0,1]^2$ , choose binary digits $(c_n)$ of $s$ and $(d_n)$ of $t$ , and interleave: the point $x \in C$ with $b_{2n-1}(x) = c_n$ , $b_{2n}(x) = d_n$ satisfies $\Phi(x) = (s,t)$ .

5. The gaps are pairwise disjoint with endpoints in $C$ , so the formula defines $f$ unambiguously on $[0,1]$ ; at the endpoints of a gap the affine formula returns $\Phi(u)$ , $\Phi(v)$ : consistent with $f = \Phi$ on $C$ . Surjectivity: already $f(C) = \Phi(C) = [0,1]^2$ .

6. At $t_0 \notin C$ : $t_0$ lies in an open gap on which $f$ is affine: continuous. At $x \in C$ : we first record two estimates.

(i) On $C$ : if $\abs{y - x} \leq 3^{-M}$ , $y \in C$ , then the digits agree up to $M$ , so each component of $\Phi(y) - \Phi(x)$ is at most $\sum_{n > \lfloor M/2\rfloor} 2^{-n} = 2^{-\lfloor M/2\rfloor}$ .

(ii) Across a gap: a gap $(u,v)$ removed at stage $k$ has length $3^{-k}$ , and its endpoints have digits agreeing up to rank $k - 1$ (they differ from rank $k$ on), so $\norm{\Phi(v) - \Phi(u)}_\infty \leq 2^{-\lfloor (k-1)/2\rfloor}$ .

Now let $\abs{t - x} < 3^{-M}$ with $t$ in a gap $(u, v)$ of stage $k$ , and say $u$ is the endpoint on $x$ ’s side, so $\abs{u - x} < 3^{-M}$ and $0 < t - u < 3^{-M}$ . Then

\norm{f(t) - \Phi(x)}_\infty \leq \norm{\Phi(u) - \Phi(x)}_\infty + \frac{t - u}{v - u}\,\norm{\Phi(v) - \Phi(u)}_\infty \leq 2^{-\lfloor M/2\rfloor} + \min\bigl(1,\ 3^{k - M}\bigr)\, 2^{-\lfloor(k-1)/2\rfloor} .

For $k \geq M$ the last term is $\leq 2^{-\lfloor(M-1)/2 \rfloor}$ ; for $k < M$ , since $2^{-\lfloor(k-1)/2\rfloor} \leq \sqrt2\,\cdot 2^{-k/2}$ ,

3^{k-M}\,2^{-\lfloor(k-1)/2\rfloor} \leq \sqrt2\,\Bigl(\frac{3}{\sqrt2}\Bigr)^{k} 3^{-M} \leq \sqrt2\,\Bigl(\frac{3}{\sqrt2}\Bigr)^{M} 3^{-M} = \sqrt2\; 2^{-M/2}

(the middle quantity increases in $k$ ). In all cases $\norm{f(t) - \Phi(x)}_\infty \leq C\,2^{-M/2}$ with an absolute constant $C$ : letting $M \to \infty$ proves continuity at $x$ (points $t \in C$ are covered by (i)). Hence $f$ is a continuous surjection $[0,1] \to [0,1]^2$ — indeed the estimates show it is Hölder of exponent $\log 2/(2\log 3)$ -flavor, but continuity is all we claimed.

7. For $[0,1]^n$ : split the digits of $x \in C$ into $n$ interleaved subsequences and repeat questions 4–6 verbatim. For $\R \to \R^2$ : let $h \colon [0,1] \to [-1,1]^2$ be a continuous surjection (rescale $f$ ). Define $F$ on $[m, m + \frac12]$ ( $m \in \Z$ ) as a copy of $h$ rescaled to fill $[-(m+1), m+1]^2$ for $m \geq 0$ (and symmetrically for $m < 0$ ), and on $[m + \frac12, m + 1]$ as the affine segment joining the endpoint values: $F$ is continuous (gluing on closed pieces, Proposition 6.6(b)) and its image contains $\bigcup_m [-(m+1), m+1]^2 = \R^2$ .

8. $[0,1]$ is compact and $[0,1]^2$ Hausdorff: a continuous injection is a homeomorphism onto its image (Corollary 6.14 applied to the corestriction).

9. Remove $\frac12$ : $[0,1]\setminus\{\frac12\}$ is disconnected. If $h \colon [0,1] \to [0,1]^2$ were a homeomorphism, $[0,1]^2\setminus\{h(\frac12)\}$ would be disconnected too (homeomorphic images of disconnected spaces are disconnected); but the square minus a point is path-connected: join two points by a two-segment path avoiding the puncture. Contradiction: $[0,1] \not\cong [0,1]^2$ .

10. By question 8, an injective continuous surjection $[0,1] \to [0,1]^2$ would be a homeomorphism, contradicting question 9. Compactness entered exactly in Corollary 6.14: it is what makes the continuous bijection’s inverse continuous (images of closed sets are compact, hence closed). Without compactness the conclusion genuinely fails: $[0, 2\pi) \to S^1$ is a continuous bijection that is not a homeomorphism.

11. Proved here: there is a continuous surjection, but no continuous bijection, from $[0,1]$ onto $[0,1]^2$ ; in particular $[0,1] \not\cong [0,1]^2$ . So “dimension” is not preserved by continuous surjections — cardinality and even continuity cannot see it — but it is a topological invariant at the level of homeomorphisms, at least for $1$ versus $2$ dimensions, where connectedness-after-deletion suffices. The general statement — invariance of domain: $\R^m \cong \R^n$ implies $m = n$ , and a continuous injection $\R^n \to \R^n$ is open — is true but requires algebraic topology (homology), beyond this course.

12. In ternary, $\frac C2 = \{\sum_na_n3^{-n} : a_n \in \{0, 1\}\}$ (halving digits $\{0,2\}$ gives digits $\{0,1\}$ ). For $t \in [0,1]$ , pick any ternary expansion $t = \sum_nd_n3^{-n}$ , $d_n \in \{0,1,2\}$ , and split every digit as $d_n = a_n + b_n$ with $a_n, b_n \in \{0,1\}$ ( $0 = 0{+}0$ , $1 = 1{+}0$ , $2 = 1{+}1$ ): then $t \in \frac C2 + \frac C2$ . The crux is that each digit splits within $\{0,1\}$ , so no carry ever propagates and the digits can be treated independently. The reverse inclusion is clear (digit sums stay $\leq 2$ ). Rescaling by $2$ : $C + C = [0,2]$ .

13. The map $(x, y) \mapsto x + y$ sends $C \times C$ onto $C + C = [0,2]$ : the dust, which contains no square (empty interior: Exercise 6.10), projects along the antidiagonal onto a full segment. Self-similarity: the first ternary digit of a Cantor point is $0$ or $2$ , and stripping it gives $C = \frac C3 \cup \bigl(\frac C3 + \frac23\bigr)$ — the fixed-point equation that generates every picture of $C$ .

14. Through $C \cong \{0,1\}^\N$ (question 2), the interleaving $(\varepsilon, \varepsilon') \mapsto (\varepsilon_1, \varepsilon_1', \varepsilon_2, \varepsilon_2', \dots)$ is a bijection $\{0,1\}^\N\times\{0,1\}^\N \to \{0,1\}^\N$ , continuous both ways (each output coordinate depends on a single input coordinate; product topology). Hence $C\times C \cong C$ and inductively $C^n \cong C$ . Familiar spaces do not share this: $\R^2 \not\cong \R$ and $[0,1]^2 \not\cong [0,1]$ (question 11); infinite products do: $([0,1]^\N)^2 \cong [0,1]^\N$ by the same interleaving.

15. $\sum_{k\geq1}2\cdot3^{-2k} = 2\cdot\frac{1/9}{1 - 1/9} = \frac14$ : the ternary expansion of $\frac14$ is $0.\overline{02}$ , with digits in $\{0,2\}$ , so $\frac14 \in C$ ; being neither a finite expansion nor an eventually- $2$ tail, it is an endpoint of no removed middle third. Endpoints form a countable set (two per removed interval, countably many intervals), while $C \cong \{0,1\}^\N$ is uncountable (diagonal argument, or question 19): almost every Cantor point is, like $\frac14$ , invisible in the usual endpoint picture.

16. $g(x) = g(y)$ means that the binary sequences $(b_n(x))$ , $(b_n(y))$ represent the same real. Distinct binary sequences representing the same number occur exactly in the dyadic ambiguity $0.w01^\infty = 0.w10^\infty$ : at most two preimages, and exactly two precisely at the dyadic rationals of $(0,1)$ . So $g$ collapses $C$ onto $[0,1]$ by gluing countably many pairs — the combinatorial skeleton of the Cantor staircase of Chapter 9.

17. $x$ is not isolated in $P$ , so $B(x, r/2)$ contains some $y_0 \in P \setminus \{x\}$ ; $y_0$ is not isolated either, so $B(x, r/2)$ contains a second point $y_1 \in P$ . Any $\rho \leq \min\bigl(\frac{d(y_0,y_1)}3, \frac r4\bigr)$ makes $\bar B(y_0, \rho)$ and $\bar B(y_1, \rho)$ disjoint and contained in $B(x, r)$ . Both are centered at points of $P$ , where the same two-point extraction can be repeated: perfection is the inexhaustible supply of nearby points that keeps the recursion alive forever.

18. Build $F_w$ by induction on the word length: $F_\varnothing = P$ , and inside each ball $B(y_w, \rho_w)$ question 17 yields two disjoint closed balls of radius $\leq \min(\rho_w/2,\ 2^{-\abs w-1}\operatorname{diam}P)$ centered at points of $P$ ; set $F_{wi} = \bar B(y_{wi}, \rho_{wi}) \cap P$ : nonempty (its center) and compact. For infinite $\varepsilon$ , the nested nonempty compacts $F_{\varepsilon_1\cdots\varepsilon_m}$ have nonempty intersection (finite intersection property in the compact $P$ ), of diameter $0$ : a single point $h(\varepsilon)$ .

19. Words differing first at rank $m$ send their images into the two disjoint sets $F_{w0}, F_{w1}$ (common prefix $w$ ): $h$ is injective. If two words agree to rank $m$ , both images lie in a set of diameter $\leq 2^{-m}\operatorname{diam}P$ : $h$ is continuous. So the uncountable $\{0,1\}^\N$ injects into $P$ : every perfect compact metric space is uncountable — $[0,1]$ and $C$ among them.

20. $h$ is a continuous injection from the compact $\{0,1\}^\N$ to the Hausdorff (metric) $P$ : Corollary 6.14 upgrades it to a homeomorphism onto its image; and $\{0,1\}^\N \cong C$ (question 2). Every perfect compact metric space contains a copy of the Cantor set: perfection has a universal germ, and it is Cantor’s.

21. A countable compact metric space cannot be perfect (question 19), so it has an isolated point. In $K = \{0\}\cup\{\frac1n : n \geq 1\}$ every point $\frac1n$ is isolated and $0$ is not: isolated points can even be dense without the space being discrete — compactness keeps their limit inside.

22. Let $P$ be the set of condensation points of $F$ and $\mathcal B$ the countable family of open intervals with rational endpoints. Every non-condensation point of $F$ lies in some $I \in \mathcal B$ with $I \cap F$ countable, so $F \setminus P$ is contained in the union of these countably many countable traces: countable. Since $F$ is uncountable, $P \neq \varnothing$ ; $P$ is closed (if $x \notin P$ , the witnessing interval $I$ contains no condensation point at all: $I \cap F$ countable) and $P \subseteq F$ ( $F$ closed: a condensation point is in particular adherent). $P$ is perfect: were $I \cap P = \{x\}$ for some interval $I$ , then $I \cap F \subseteq \{x\} \cup (F \setminus P)$ would be countable, contradicting $x \in P$ . Now the tree construction of questions 17–19 runs verbatim inside $P$ : the pieces $\bar B(y, \rho) \cap P$ are compact (closed bounded subsets of $\R$ ), every center is non-isolated in $P$ , and the argument never used more. Hence $\abs P \geq \abs{\{0,1\}^\N} = \mathfrak c$ , and $\abs F \leq \mathfrak c$ trivially: an uncountable closed $F \subseteq \R$ has cardinality exactly $\mathfrak c$ . Closed sets cannot witness a failure of the continuum hypothesis.

23. Let $x \neq y$ in $C$ and choose $m \geq 0$ with $3^{-(m+1)} \leq \abs{x - y} < 3^{-m}$ . By question 1 the digits $b_n(x) = b_n(y)$ agree for $n \leq m$ ; the first coordinate of $\Phi$ uses $b_1, b_3, \dots$ , the second $b_2, b_4, \dots$ , so in each coordinate the two points share at least $\lfloor m/2\rfloor$ leading binary digits:

\norm{\Phi(x) - \Phi(y)}_\infty \leq \sum_{k > \lfloor m/2\rfloor}2^{-k} = 2^{-\lfloor m/2\rfloor} \leq \sqrt2\cdot2^{-m/2} = \sqrt2\,\bigl(3^{-m}\bigr)^\alpha \leq \sqrt2\,\bigl(3\abs{x-y}\bigr)^\alpha,

using $2^{-m/2} = 3^{-m\alpha}$ (take logarithms); and $\sqrt2\cdot3^\alpha = \sqrt2\cdot\sqrt2 = 2$ : the bound $2\abs{x-y}^\alpha$ on $C$ . Same gap $(u, v)$ : $f$ is affine there, so for $t, s \in [u, v]$ ,

\norm{f(t) - f(s)} = \frac{\abs{t-s}}{v-u}\, \norm{\Phi(v) - \Phi(u)} \leq \frac{\abs{t-s}}{v-u}\,2(v-u)^\alpha \leq 2\abs{t-s}^\alpha,

since $\frac{\abs{t-s}}{v-u} \leq \bigl(\frac{\abs{t-s}} {v-u}\bigr)^\alpha$ when $\abs{t-s} \leq v - u$ . General $t < s$ : if $[t, s]$ meets $C$ , let $u'$ and $v'$ be the smallest and largest points of the compact $C \cap [t, s]$ ; then $t$ lies in a gap (or at a point of $C$ ) whose right endpoint is $u'$ , $s$ in one with left endpoint $v'$ , and

\norm{f(t) - f(s)} \leq 2\abs{t - u'}^\alpha + 2\abs{v' - u'}^\alpha + 2\abs{s - v'}^\alpha \leq 6\abs{t - s}^\alpha ;

if $[t, s]$ misses $C$ , the same-gap case applies. So $f$ is $\alpha$ -Hölder with $\alpha = \frac{\ln2}{2\ln3}$ .

24. Suppose $\norm{F(t) - F(s)} \leq C\abs{t-s}^\alpha$ with $F$ surjective and $\alpha > \frac12$ . Cut $[0,1]$ into $N$ intervals $I_1, \dots, I_N$ of length $\frac1N$ : each image $F(I_j)$ has diameter $\leq CN^{-\alpha}$ . Two distinct points of the grid $G_M = \bigl\{\bigl(\frac iM, \frac jM\bigr)\bigr\}_{0 \leq i, j \leq M}$ are at distance $\geq \frac1M$ , so a set of diameter $< \frac1M$ contains at most one of them. Choose $M = \lfloor N^\alpha/(2C)\rfloor$ (with $N$ large): then $CN^{-\alpha} \leq \frac1{2M} < \frac1M$ , and surjectivity puts every one of the $(M+1)^2$ grid points into some $F(I_j)$ , whence

\frac{N^{2\alpha}}{4C^2} \leq (M + 1)^2 \leq N,

and $N^{2\alpha - 1} \leq 4C^2$ fails for large $N$ when $2\alpha > 1$ : contradiction. Our curve, with $\alpha \approx 0.3155$ , sits below the wall, as it must; Hilbert’s curve (admitted) is $\frac12$ -Hölder, so the critical exponent $\frac12$ is attained — Hölder regularity, unlike injectivity, is a matter of degree, and $\frac12$ is exactly the frontier that dimension $2$ imposes on a map from dimension $1$ .

25. The $m$ -th cluster $\{\frac1m + \frac1n : n > m^2\}$ lies in $\bigl(\frac1m, \frac1m + \frac1{m^2}\bigr] \subseteq \bigl(\frac1m, \frac1{m-1}\bigr)$ for $m \geq 2$ , because $\frac1{m-1} - \frac1m = \frac1{m(m-1)} > \frac1{m^2}$ (and the first cluster lies in $(1, \frac32]$ ): the clusters occupy disjoint intervals. Limit points of $K$ : inside the $m$ -th interval, only $\frac1m$ (the cluster converges to it and is discrete in itself); globally, $0$ (every neighborhood of $0$ contains whole clusters). So $K' = \{0\} \cup \{\frac1m\}$ , then $K'' = \{0\}$ (each $\frac1m$ is isolated in $K'$ ), $K''' = \varnothing$ ; $K$ contains its limit points, hence is closed, bounded, countable, compact, and its isolated points — the cluster points $\frac1m + \frac1n$ — are dense in $K$ : every element of $K'$ is their limit, as question 21’s mechanism predicts. Induction: $K_1 = \{0\}\cup\{\frac1n\}$ has $K_1^{(1)} = \{0\}$ ; given a countable compact $K_j \subseteq [0, 1]$ with $K_j^{(j)} = \{0\}$ and $K_j^{(j+1)} = \varnothing$ , set

K_{j+1} = \{0\} \cup \bigcup_{m \geq 1} \Bigl(\frac1m + \varepsilon_m K_j\Bigr),

with $\varepsilon_m$ small enough that the $m$ -th copy lies in $\bigl(\frac1m, \frac1{m-1}\bigr)$ . Derivation acts copy by copy (the copies live in disjoint open intervals), so $K_{j+1}^{(i)} = \{0\} \cup \bigcup_m\bigl(\frac1m + \varepsilon_mK_j^{(i)}\bigr)$ for $i \leq j$ ; at $i = j$ this reads $\{0\} \cup \{\frac1m\}$ , then $K_{j+1}^{(j+1)} = \{0\}$ and $K_{j+1}^{(j+2)} = \varnothing$ . Every finite rank occurs. A perfect set is the other extreme: $P' = P$ , the derivation never moves — and Cantor–Bendixson (question 22) says precisely that every closed set splits into a perfect core, invisible to derivation, and a countable remainder that derivation eats away.