University Mathematics — Year 1 · Bachelor Year 1

12Topology of the Real Line

Limits keep referring to the same geometric vocabulary: points “close to” a set, sets “without boundary leaks”, intervals in which sequences cannot escape. This chapter fixes that vocabulary — open and closed sets, interior and closure, density — on the real line, and proves the compactness of segments in its sequential form. The same notions, in normed vector spaces, are second-year material; on $\R$ they are within reach and immediately useful for Chapter 13.

12.1 Open sets, closed sets

Definition 12.1 (Neighborhood, open set)

A set $V \subseteq \R$ is a neighborhood of $x \in \R$ when it contains an interval $\intoo{x - r}{x + r}$ for some $r > 0$ . A set $U \subseteq \R$ is open when it is a neighborhood of each of its points:

\forall x \in U,\ \exists r > 0, \quad \intoo{x - r}{x + r} \subseteq U .

Example 12.2

Open intervals are open: for $x \in \intoo{a}{b}$ , take $r = \min(x - a,\, b - x) > 0$ . Half-lines $\intoo{a}{+\infty}$ are open; $\R$ and $\emptyset$ are open (the latter vacuously). $\intcc{0}{1}$ is not open: no interval around $0$ stays inside.

Proposition 12.3 (Stability of open sets)

Any union of open sets is open; a finite intersection of open sets is open. Infinite intersections may fail: $\bigcap_{n \geq 1} \intoo{-\frac 1n}{\frac 1n} = \{0\}$ , not open.

Proof. Union: if $x \in \bigcup_i U_i$ , then $x \in U_{i_0}$ for some $i_0$ , and the interval provided by $U_{i_0}$ sits inside the union. Finite intersection: if $x \in U_1 \cap \dots \cap U_k$ , take $r = \min(r_1, \dots, r_k) > 0$ of the radii provided by each $U_j$ . For the counterexample: any interval around $0$ contains some $\frac{1}{n}$ (Archimedes), hence leaves the intersection. ∎

Example 12.4 (Certifying openness with explicit radii)

Is $U = \{x \in \R : x^2 > 2\}$ open? Yes, and the certificate can be written down: $U = \intoo{-\infty}{-\sqrt2} \cup \intoo{\sqrt2}{+\infty}$ , a union of two open half-lines, open by Proposition 12.3. Alternatively, argue point by point: for $x \in U$ with $x > \sqrt 2$ , take $r = x - \sqrt2 > 0$ : every $y \in \intoo{x - r}{x + r}$ satisfies $y > \sqrt 2$ , hence $y^2 > 2$ ; symmetrically on the left. Both styles matter — the structural one (build from known open sets by unions and finite intersections) scales better, the $\varepsilon$ -style one works when no structure is visible; and Chapter 13 will add a third, the most powerful: $U$ is the preimage of the open $\intoo{2}{+\infty}$ under the continuous $x \mapsto x^2$ .

Definition 12.5 (Closed set)

A set $F \subseteq \R$ is closed when its complement $\R \setminus F$ is open. By de Morgan and Proposition 12.3: any intersection of closed sets is closed, finite unions of closed sets are closed.

Theorem 12.6 (Sequential characterization of closed sets)

$F$ is closed if and only if: for every sequence $(u_n)$ of points of $F$ converging to some $\ell \in \R$ , the limit $\ell$ belongs to $F$ . (“Closed” $=$ “stable under limits”.)

Proof. ( $\Rightarrow$ ) Let $F$ be closed, $u_n \in F$ , $u_n \to \ell$ , and suppose $\ell \notin F$ . The complement is open: some $\intoo{\ell - r}{\ell + r}$ avoids $F$ . But convergence puts $u_n$ in that interval for large $n$ : contradiction with $u_n \in F$ .

( $\Leftarrow$ ) Suppose $F$ is not closed: the complement is not open, so some $x \notin F$ has no interval $\intoo{x - r}{x + r}$ inside the complement; taking $r = \frac{1}{n+1}$ , choose $u_n \in F$ with $\abs{u_n - x} < \frac{1}{n+1}$ . Then $u_n \in F$ , $u_n \to x \notin F$ : the sequential property fails. ∎

Example 12.7 (The sequential test, both ways)

Closed: $F = \Z \cup \bigl\{n + \frac1n : n \geq 2\bigr\}$ . Let $u_k \in F$ with $u_k \to \ell$ . The window $\intcc{\ell - 1}{\ell + 1}$ contains only finitely many points of $F$ (finitely many integers, finitely many $n + \frac1n$ ), and beyond some rank all $u_k$ lie in it: the sequence then takes finitely many values, and, converging, is eventually constant (as in Exercise 12.3): $\ell \in F$ . Closed — even though $F$ contains pairs of points at distance $\frac1n$ , arbitrarily close.

Not closed: $G = \bigl\{\frac1m + \frac1n : m, n \in \N^*\bigr\}$ . The sequence $\frac1n + \frac1n \in G$ tends to $0$ , and $0 \notin G$ (a sum of two positive terms): the sequential test fails, $G$ is not closed. Interestingly, each $\frac1m$ does belong to $\overline G \cap G$ : indeed $\frac1m = \frac{1}{m+1} + \frac{1}{m(m+1)} \in G$ . The closing insight: to prove closedness, control all convergent sequences at once (usually via a local finiteness or a closed-formula argument); to disprove it, one well-chosen escaping sequence suffices — the asymmetry makes the negative direction the easy one, and the counterexamples of this chapter all have this one-line shape.

Example 12.8

Segments $\intcc{a}{b}$ , half-lines $\intco{a}{+\infty}$ , finite sets, $\Z$ (a convergent sequence of integers is eventually constant) are closed. $\intoc{0}{1}$ is neither open (fails at $1$ ) nor closed ( $\frac 1n \to 0 \notin$ the set): most sets are neither. $\R$ and $\emptyset$ are both open and closed — and they are the only such subsets of $\R$ (Exercise 12.9).

Example 12.9 (An open set assembled from infinitely many pieces)

$\R \setminus \Z = \bigcup_{n \in \Z} \intoo{n}{n+1}$ : an infinite union of open intervals, open by Proposition 12.3 — so $\Z$ is closed with no sequential argument needed. Note the division of labor in the stability rules: unions of open sets may be arbitrary (each point only needs its own certificate, supplied by the one set containing it), while intersections must stay finite (certificates must be intersected, and infinitely many radii can shrink to nothing). Exercise 12.10 will show this example is the general shape: every open subset of $\R$ is a countable disjoint union of open intervals.

12.2 Interior, closure, density

Definition 12.10 (Interior, closure, boundary)

Let $A \subseteq \R$ .

A point $x$ is interior to $A$ when $A$ is a neighborhood of $x$ ; the interior $\mathring{A}$ is the set of interior points.
A point $x$ is adherent to $A$ when every neighborhood of $x$ meets $A$ ; the closure $\overline{A}$ is the set of adherent points.
The boundary is $\partial A = \overline A \setminus \mathring A$ .

Then $\mathring A \subseteq A \subseteq \overline A$ .

Proposition 12.11 (Main properties)

$\mathring A$ is the largest open set contained in $A$ ; $A$ is open iff $A = \mathring A$ .
$\overline A$ is the smallest closed set containing $A$ ; $A$ is closed iff $A = \overline A$ .
(Sequential characterization of adherence) $x \in \overline A$ if and only if $x$ is the limit of a sequence of points of $A$ .
Complementation exchanges the notions: $\R \setminus \overline A = \bigl(\R \setminus A\bigr)^{\!\circ}$ .

Proof. (4) $x \notin \overline A$ $\iff$ some neighborhood of $x$ avoids $A$ $\iff$ some interval around $x$ lies in $\R \setminus A$ $\iff$ $x$ is interior to $\R \setminus A$ .

(1) $\mathring A$ is open: if $x \in \mathring A$ , some $\intoo{x-r}{x+r} \subseteq A$ ; every point $y$ of that interval has a smaller interval around it inside it, hence inside $A$ : the whole interval is in $\mathring A$ . Any open $U \subseteq A$ consists of interior points of $A$ , so $U \subseteq \mathring A$ : largest. The characterization of openness follows.

(2) In detail. By (4), $\R \setminus \overline A$ is the interior of $\R \setminus A$ , an open set by (1): so $\overline A$ is closed, and it contains $A$ . Minimality: let $F \supseteq A$ be closed. Then $\R \setminus F$ is open and contained in $\R \setminus A$ , so by the maximality in (1),

\R \setminus F \subseteq \bigl(\R \setminus A\bigr)^{\!\circ} = \R \setminus \overline A ,

and taking complements again: $\overline A \subseteq F$ . Thus $\overline A$ is the smallest closed superset. Characterization: if $A = \overline A$ then $A$ is closed (just shown); if $A$ is closed, it is itself a closed superset of $A$ , so minimality forces $\overline A \subseteq A$ , and equality.

(3) If $u_n \in A$ , $u_n \to x$ : every neighborhood of $x$ contains some $u_n \in A$ , so $x \in \overline A$ . Conversely, if $x \in \overline A$ : each interval $\intoo{x - \frac{1}{n+1}}{x + \frac{1}{n+1}}$ meets $A$ at some $u_n$ , and $u_n \to x$ . ∎

Example 12.12

$\overline{\intoo{0}{1}} = \intcc{0}{1}$ ; $\mathring{\intcc{0}{1}} = \intoo{0}{1}$ ; $\partial\intoo{0}{1} = \{0, 1\}$ . For $A = \{\frac 1n : n \in \N^*\}$ : $\overline A = A \cup \{0\}$ , $\mathring A = \emptyset$ , $\partial A = A \cup \{0\}$ . For $\Q$ : by density (Theorem 10.14) every real is adherent to $\Q$ , so $\overline{\Q} = \R$ while $\mathring{\Q} = \emptyset$ (every interval contains irrationals): the boundary of $\Q$ is all of $\R$ .

Example 12.13 (A full anatomy)

Let $A = \intoc{0}{1} \,\cup\, \bigl(\Q \cap \intoo{2}{3}\bigr) \,\cup\, \{4\}$ . We compute the three sets of Definition 12.10, piece by piece.

Interior. A point of $\intoo{0}{1}$ has a whole interval inside $A$ : interior. The point $1$ : every interval around it leaks right of $1$ , where $A$ has nothing until $2$ : not interior. No point of $\Q \cap \intoo{2}{3}$ is interior (every interval contains irrationals, Theorem 10.14); neither is the isolated $4$ . So $\mathring A = \intoo{0}{1}$ .

Closure. Limits of points of $A$ : all of $\intcc{0}{1}$ ( $0 = \lim \frac1n$ with $\frac 1n \in A$ ); all of $\intcc{2}{3}$ (every real there is a limit of rationals of the interval, density again); and $4$ . Nothing else: a point outside $\intcc{0}{1} \cup \intcc{2}{3} \cup \{4\}$ has positive distance to that closed set. So $\overline A = \intcc{0}{1} \cup \intcc{2}{3} \cup \{4\}$ .

Boundary. $\partial A = \overline A \setminus \mathring A = \{0, 1\} \cup \intcc{2}{3} \cup \{4\}$ .

The closing insight: the three operations act locally — each piece of $A$ contributes according to its own nature (a solid interval keeps its inside, a dense-but-porous piece turns entirely into boundary, an isolated point is pure boundary), and a two-line drawing of $A$ predicts every answer before any proof is written.

Remark 12.14 (Common pitfalls in point-set reasoning)

(i) “Not open” does not mean “closed”: most sets are neither ( $\intoc{0}{1}$ ), and two sets are both ( $\emptyset$ , $\R$ ) — open and closed are not opposites but duals through complementation. (ii) Interior and closure do not commute: for $A = \Q$ ,

\overline{\mathring A} = \overline\emptyset = \emptyset \qquad\text{while}\qquad \bigl(\,\overline A\,\bigr)^{\!\circ} = \mathring \R = \R :

the two iterated operators differ as much as sets can. (iii) Infinite unions of closed sets can fail to be closed: $\bigcup_{n\geq1} \intcc{\frac1n}{1} = \intoc{0}{1}$ — the mirror of the intersection counterexample of Proposition 12.3. (iv) Dense does not mean big: $\Q$ is dense, countable, with empty interior, and its complement is dense too; density says “arbitrarily close to everything”, not “almost everything” — the weekend problem’s Cantor set (Problem 12.1) makes the opposite point, a topologically small set that is uncountably big.

Example 12.15 (A closure computed exactly)

Let $G = \bigl\{\frac1m + \frac1n : m, n \in \N^*\bigr\}$ (from Example 12.7). Claim:

\overline G = G \,\cup\, \Bigl\{\frac1m : m \in \N^*\Bigr\} \,\cup\, \{0\} .

( $\supseteq$ ) $\frac1m = \lim_n \bigl(\frac1m + \frac1n\bigr)$ and $0 = \lim_n \frac2n$ : adherent by the sequential characterization. ( $\subseteq$ ) Let $x = \lim_k \bigl( \frac{1}{m_k} + \frac{1}{n_k}\bigr)$ ; order each pair so that $m_k \leq n_k$ . If $(m_k)$ is unbounded, a subsequence has $m_k \to \infty$ , hence $n_k \to \infty$ too and $x = 0$ . Otherwise $(m_k)$ takes finitely many values, one of them, say $m$ , infinitely often; along that subsequence $\frac{1}{n_k} \to x - \frac1m$ : if $(n_k)$ is bounded it takes some value $n$ infinitely often and $x = \frac1m + \frac1n \in G$ ; if not, $x = \frac1m$ . Every case lands in the announced set. The closing insight: computing a closure is a compactness-style case analysis on indices — bounded index means finitely many values (pigeonhole), unbounded index means a limit escapes — and the answer displays the typical two-layer structure of limit points: the set, its first-generation limits, and their limit $0$ .

Definition 12.16 (Density, topological form)

$A$ is dense in $\R$ when $\overline A = \R$ — equivalently, every nonempty open interval meets $A$ ; equivalently (by Proposition 12.11 (3)), every real is a limit of elements of $A$ . Examples: $\Q$ , $\R \setminus \Q$ , the dyadics (Exercise 10.8), dense subgroups (Exercise 10.9).

Example 12.17 (Density is relative)

“Dense” as defined here means dense in $\R$ ; a set can instead be dense in a part of the line only. The dyadics of $\intcc{0}{1}$ , i.e. $D \cap \intcc{0}{1}$ (Exercise 10.8), meet every open interval included in $\intcc{0}{1}$ but of course miss $\intoo{2}{3}$ entirely: they are dense in $\intcc{0}{1}$ , meaning $\overline{D \cap \intcc{0}{1}} = \intcc{0}{1}$ . The general phrase “ $A$ is dense in $B$ ” abbreviates $B \subseteq \overline A$ — always name the ambient set, since the weekend problem’s endpoints are dense in the Cantor set while being nowhere dense in $\R$ : the same set, two truthful and opposite-sounding descriptions.

Example 12.18 (Handling density)

Three quick moves that recur constantly. Enlarging: if $A$ is dense and $A \subseteq B$ , then $B$ is dense (every interval already meets $A$ ). Transporting: if $A$ is dense, so is $\lambda A + \mu$ for $\lambda \neq 0$ — an interval $I$ meets $\lambda A + \mu$ iff the interval $\frac{I - \mu}{\lambda}$ meets $A$ ; thus the odd multiples of $10^{-9}$ , say, are dense. Intersecting fails: two dense sets can miss each other entirely ( $\Q$ and $\R \setminus \Q$ ): density survives unions and affine maps, never intersections.

12.3 Compactness of segments

Theorem 12.19 (Segments are sequentially compact)

Let $a \leq b$ . Every sequence of points of $\intcc{a}{b}$ has a subsequence converging to a point of $\intcc{a}{b}$ .

More generally, the subsets of $\R$ with this property (every sequence has a subsequence converging in the set) are exactly the closed and bounded sets.

Proof. A sequence in $\intcc{a}{b}$ is bounded, so Bolzano–Weierstrass (Theorem 11.16) extracts a convergent subsequence; its limit stays in $\intcc{a}{b}$ because segments are closed (Theorem 12.6).

General case. (Closed bounded $\Rightarrow$ compact): let $F$ be closed and bounded, and $(u_n)$ a sequence in $F$ . Boundedness of $F$ bounds the sequence, so Bolzano–Weierstrass extracts $u_{\varphi(n)} \to \ell$ ; and $\ell \in F$ because $F$ is closed and the subsequence is a convergent sequence of points of $F$ (Theorem 12.6): the two hypotheses are consumed one each, boundedness for existence of the limit, closedness for its membership. (Compact $\Rightarrow$ closed and bounded): if $F$ is unbounded, pick $u_n \in F$ with $\abs{u_n} \geq n$ ; every subsequence is unbounded, hence divergent (Proposition 11.4): no convergent subsequence at all. If $F$ is not closed, take $u_n \in F$ with $u_n \to \ell \notin F$ (Theorem 12.6): every subsequence converges to $\ell \notin F$ , so no subsequence converges in $F$ . ∎

Example 12.20 (Nested compact sets)

A first workout for the theorem. Let $K_0 \supseteq K_1 \supseteq K_2 \supseteq \dots$ be nonempty compact (closed bounded) subsets of $\R$ . Then $\bigcap_n K_n \neq \emptyset$ . Indeed, pick $x_n \in K_n$ for each $n$ : the sequence lives in the compact $K_0$ , so a subsequence $x_{\varphi(n)}$ converges to some $x$ (Theorem 12.19). For every fixed $m$ , the terms $x_{\varphi(n)}$ with $\varphi(n) \geq m$ all lie in the closed set $K_m$ , so the limit $x$ lies in $K_m$ (Theorem 12.6); as $m$ was arbitrary, $x \in \bigcap_n K_n$ . A useful companion: if an open set $U$ contains $\bigcap_n K_n$ , then $U \supseteq K_n$ for some $n$ — apply the same argument to points $x_n \in K_n \setminus U$ ; the limit $x$ would lie in $\bigcap K_n \subseteq U$ , yet $U$ open forces $x_{\varphi(n)} \in U$ eventually, a contradiction. Both statements fail without compactness: $\bigcap_n \intoo{0}{\frac 1n} = \emptyset$ and $\bigcap_n \intco{n}{+\infty} = \emptyset$ . The closing insight: compactness converts an infinite chain of non-emptiness assertions into a single limit point — it is the tool that survives passage to the infinite intersection, and the weekend problem (Problem 12.1) will lean on it twice.

The first four stages of the middle-thirds construction: each segment of C_n loses its open middle third, leaving the 2n+1 segments of C_n+1, of total length (2/3)n+1. The Cantor set C = _n C_n — the subject of the weekend problem — is the nonempty compact residue guaranteed by the nested-compacts argument above: length zero, yet uncountably many points survive. — The first four stages of the middle-thirds construction: each segment of $C_n$ loses its open middle third, leaving the $2^{n+1}$ segments of $C_{n+1}$ , of total length $(\frac23)^{n+1}$ . The Cantor set $C = \bigcap_n C_n$ — the subject of the weekend problem Problem 12.1 — is the nonempty compact residue guaranteed by the nested-compacts argument above: length zero, yet uncountably many points survive.

Remark 12.21

This is the engine behind the extreme value theorem (Chapter 13) and Heine’s theorem on uniform continuity. The name “compact” will acquire its general (covering) definition in the second year; on $\R$ , sequential compactness is all we need, and “compact $=$ closed $+$ bounded” is the statement to remember.

Example 12.22 (Boundaries under unions)

Always $\partial(A \cup B) \subseteq \partial A \cup \partial B$ : a point of $\partial(A \cup B)$ has every neighborhood meeting $A \cup B$ (hence $A$ or $B$ , infinitely often one of them) and meeting the complement of $A \cup B$ , which lies in both complements — a short check then places the point in $\partial A$ or $\partial B$ . The inclusion can be spectacularly strict: with $A = \Q$ and $B = \R\setminus\Q$ ,

\partial(A \cup B) = \partial \R = \emptyset , \qquad \partial A \cup \partial B = \R \cup \R = \R :

two ragged sets can glue into a seamless one, their boundaries annihilating each other. The closing insight: interiors and closures behave monotonically under unions and intersections, but boundaries do not — treat $\partial$ as a derived quantity ( $\overline A \setminus \mathring A$ ), never as an operator with algebra of its own.

Remark 12.23 (Perspectives inside this volume)

The vocabulary built here is consumed twice more in this book. In Chapter 13, every theorem is a topology statement in disguise: the intermediate value theorem says continuous maps preserve the interval property, the extreme value theorem says they preserve compactness — and the proofs call Theorems 12.6 and 12.19 by name. In Chapter 25, the same definitions are re-read in $\R^2$ with disks in place of intervals: open sets, closures and compactness transfer word for word, and the two-variable extreme value theorem again rides on Bolzano–Weierstrass (extract on each coordinate). The one notion that does not generalize painlessly is the interval itself — in the plane, connectedness replaces convexity, a story that begins with Exercise 12.9’s “only $\emptyset$ and $\R$ are open and closed”.

12.4 Exercises

Exercise 12.1 ★

For each set, say whether it is open, closed, both, or neither (with justification): $\intoo{0}{1} \cup \intoo{2}{3}$ ; $\;\intco{0}{1}$ ; $\;\{0\} \cup \intcc{1}{2}$ ; $\;\R \setminus \Z$ ; $\;\Q \cap \intoo{0}{1}$ .

Solution

Solution of Exercise 12.1.

$\intoo{0}{1} \cup \intoo{2}{3}$ : open (union of open sets), not closed ( $\frac 1n \to 0$ outside).

$\intco{0}{1}$ : neither. Not open (no interval around $0$ inside); not closed ( $1 - \frac1n \to 1 \notin$ set).

$\{0\} \cup \intcc{1}{2}$ : closed (finite union of closed sets), not open (fails at $0$ ).

$\R \setminus \Z$ : open ( $\Z$ is closed), not closed: the sequence $\bigl(\frac 1n\bigr)$ lies in it, but its limit $0$ belongs to $\Z$ , i.e. escapes the set.

$\Q \cap \intoo{0}{1}$ : neither. Not open: every interval around a rational contains irrationals. Not closed: it contains sequences tending to the irrational $\frac{\sqrt 2}{2}$ (density).

Exercise 12.2 ★

Determine $\mathring A$ , $\overline A$ and $\partial A$ for: $A = \intoc{0}{1} \cup \{2\}$ ; $\;A = \R \setminus \Q$ ; $\;A = \bigl\{\frac{(-1)^n n}{n+1} : n \in \N\bigr\}$ .

Solution

Solution of Exercise 12.2.

$A = \intoc{0}{1} \cup \{2\}$ : $\mathring A = \intoo{0}{1}$ , $\overline A = \intcc{0}{1} \cup \{2\}$ , $\partial A = \{0, 1, 2\}$ .

$A = \R \setminus \Q$ : $\mathring A = \emptyset$ (every interval contains rationals), $\overline A = \R$ (density of the irrationals), $\partial A = \R$ .

$A = \bigl\{\frac{(-1)^n n}{n+1}\bigr\}$ : the even terms tend to $1$ , the odd terms to $-1$ , neither belongs to $A$ . $\mathring A = \emptyset$ (isolated points), $\overline A = A \cup \{-1, 1\}$ , $\partial A = \overline A$ .

Exercise 12.3 ★

Prove that a finite set is closed, first via complements, then via the sequential characterization.

Solution

Solution of Exercise 12.3.

Complements. $F = \{a_1 < a_2 < \dots < a_k\}$ : the complement is the union of the open intervals $\intoo{-\infty}{a_1}$ , $\intoo{a_i}{a_{i+1}}$ , $\intoo{a_k}{+\infty}$ — open by Proposition 12.3.

Sequences. Let $u_n \in F$ , $u_n \to \ell$ . With $\varepsilon = \min\{\abs{a_i - a_j} : i \neq j\}/2 > 0$ (or any $\varepsilon$ if $F$ is a singleton): beyond some rank, all terms are within $\varepsilon$ of $\ell$ , hence within $2\varepsilon$ of each other, which forces them to be one single $a_i$ from that rank on; then $\ell = a_i \in F$ .

Exercise 12.4 ★

Prove that for any $A, B \subseteq \R$ : $\overline{A \cup B} = \overline A \cup \overline B$ . Show by example that $\overline{A \cap B}$ can differ from $\overline A \cap \overline B$ .

Solution

Solution of Exercise 12.4.

$\subseteq$ : $\overline A \cup \overline B$ is closed (finite union) and contains $A \cup B$ , so it contains the smallest closed superset $\overline{A \cup B}$ . $\supseteq$ : $A \subseteq A \cup B$ gives $\overline A \subseteq \overline{A \cup B}$ (closure is monotone: adherent points of $A$ are adherent to the larger set), and likewise for $B$ .

Counterexample for intersections: $A = \intoo{0}{1}$ , $B = \intoo{1}{2}$ : $\overline{A \cap B} = \overline\emptyset = \emptyset$ but $\overline A \cap \overline B = \{1\}$ .

Exercise 12.5 ★★

Let $u_n \to \ell$ in $\R$ . Prove that the set $\{u_n : n \in \N\} \cup \{\ell\}$ is closed (hence compact if we add that it is bounded — which it is).

Solution

Solution of Exercise 12.5.

Use the sequential characterization (Theorem 12.6). Let $S = \{u_n\} \cup \{\ell\}$ and $(v_k)$ a sequence in $S$ with $v_k \to m$ ; show $m \in S$ . Two cases. If some value $v \in S$ is taken by $(v_k)$ infinitely often, then a constant subsequence gives $m = v \in S$ . Otherwise every value is taken finitely often; in particular, for each $n$ , the term $u_n$ appears finitely often, and $\ell$ too. Then for every $N$ , the indices $k$ with $v_k \in \{u_0, \dots, u_N, \ell\}$ are finitely many: the remaining $v_k$ are terms $u_n$ with $n > N$ . Given $\varepsilon > 0$ , choose $N$ with $\abs{u_n - \ell} \leq \varepsilon$ for $n > N$ : all but finitely many $v_k$ satisfy $\abs{v_k - \ell} \leq \varepsilon$ . Hence $v_k \to \ell$ , so $m = \ell \in S$ .

Exercise 12.6 ★★

Let $U$ be open and $A$ arbitrary. Prove that $U + A = \{u + a\}$ is open. Deduce that the sum of an open set and any set is open, and contrast: exhibit two closed sets whose sum is not closed. (Try $\Z$ and $\sqrt 2\,\Z$ , with Exercise 10.9.)

Solution

Solution of Exercise 12.6.

$U + A = \bigcup_{a \in A} (U + a)$ , and each translate $U + a$ is open (translating the interval certificates). A union of open sets is open (Proposition 12.3).

Closed sets: $\Z$ and $\sqrt 2\,\Z$ are closed (as $\alpha\Z$ : convergent sequences are eventually constant, cf. Example 12.8). Their sum $\Z + \sqrt 2\,\Z$ is dense in $\R$ (Exercise 10.9) but is not $\R$ (it is countable, or simply: $\frac{\sqrt 2}{2} \notin \Z + \sqrt2\Z$ , else $\sqrt 2$ would be rational — writing $\frac{\sqrt2}{2} = m + n\sqrt 2$ forces $(2n - 1)\sqrt 2 = -2m$ , so $\sqrt 2 \in \Q$ unless $n = \frac12$ , impossible). A dense proper subset is not closed: its closure is $\R \neq$ itself.

Exercise 12.7 ★★

A point $x \in A$ is isolated in $A$ when some neighborhood of $x$ meets $A$ only at $x$ . Prove that every point of $\Z$ is isolated in $\Z$ , that $A = \{\frac1n\}$ has all its points isolated yet $\overline A \neq A$ , and that a set whose points are all isolated has empty interior.

Solution

Solution of Exercise 12.7.

$\Z$ : the neighborhood $\intoo{n - \frac12}{n + \frac12}$ of $n$ meets $\Z$ only at $n$ .

$A = \{\frac1n : n \in \N^*\}$ : around $\frac 1n$ , the interval of radius $\frac{1}{n} - \frac{1}{n+1} = \frac{1}{n(n+1)}$ (halved, say) isolates it from its neighbors — all points isolated. Yet $0 \in \overline A \setminus A$ : isolated points do not prevent adherent outsiders.

If all points of $A$ are isolated: no point of $A$ is interior, since an interior point has a whole interval of $A$ -neighbors around it (an interval is infinite), contradicting isolation. So $\mathring A = \emptyset$ .

Exercise 12.8 ★★

Prove that the closure of a bounded set is bounded, and that $\sup \overline A = \sup A$ for $A$ nonempty bounded above. Deduce that $\sup A \in \overline A$ : the supremum is always adherent.

Solution

Solution of Exercise 12.8.

If $A \subseteq \intcc{-M}{M}$ , the closed set $\intcc{-M}{M}$ contains $A$ , hence contains $\overline A$ (smallest closed superset): $\overline A$ is bounded.

Let $s = \sup A$ (finite). Since $A \subseteq \overline A$ , $\sup \overline A \geq s$ . Conversely, $\overline A \subseteq \intoc{-\infty}{s}$ : the half-line is closed and contains $A$ ; so every element of $\overline A$ is $\leq s$ , giving $\sup\overline A \leq s$ . Equality.

$s \in \overline A$ : by the $\varepsilon$ -characterization (Proposition 10.4), every interval $\intoo{s - \varepsilon}{s + \varepsilon}$ contains an element of $A$ : $s$ is adherent.

Exercise 12.9 ★★★

Prove that the only subsets of $\R$ that are both open and closed are $\emptyset$ and $\R$ . Hint: suppose $A$ is open, closed, with $A \neq \emptyset$ and $\R \setminus A \neq \emptyset$ ; pick $a \in A$ , $b \notin A$ , say $a < b$ , and consider $s = \sup\,(A \cap \intcc{a}{b})$ ; decide whether $s$ can belong to $A$ or to its complement.

Solution

Solution of Exercise 12.9.

Suppose $A$ is open and closed, with $a \in A$ and $b \in \R \setminus A$ ; without loss of generality $a < b$ . The set $B = A \cap \intcc{a}{b}$ is nonempty ( $a$ ), bounded: let $s = \sup B$ . By Exercise 12.8, $s \in \overline B \subseteq \overline A = A$ ( $A$ closed). Note $s \leq b$ , and since $b \notin A$ : $s < b$ . Now $A$ is open: some interval $\intoo{s - r}{s + r}$ lies in $A$ , and we may take $r < b - s$ . Then $s + \frac{r}{2}$ belongs to $A \cap \intcc{a}{b} = B$ and exceeds $s$ — contradicting $s = \sup B$ . Hence no such pair $(a, b)$ exists: one of $A$ , $\R \setminus A$ is empty.

Exercise 12.10 ★★★

(Structure of open sets) Let $U \subseteq \R$ be open, nonempty. For $x \in U$ , let $I_x$ be the union of all open intervals containing $x$ and contained in $U$ . Prove that $I_x$ is an open interval, that two sets $I_x$ , $I_y$ are equal or disjoint, and that $U$ is a union of countably many pairwise disjoint open intervals (pick a rational in each).

Solution

Solution of Exercise 12.10.

$I_x$ is a union of open intervals all containing $x$ : it is open, and it is an interval, being convex — if $u < z < v$ with $u, v \in I_x$ , then $u$ and $v$ lie in open subintervals $J_u \ni x$ , $J_v \ni x$ of $U$ , and $J_u \cup J_v$ is an interval (both contain $x$ ) inside $U$ containing $z$ ; so $z \in I_x$ (Proposition 10.19).

If $I_x \cap I_y \neq \emptyset$ : $I_x \cup I_y$ is then an open interval (convex: two overlapping intervals) contained in $U$ containing $x$ and $y$ , so $I_x \cup I_y \subseteq I_x$ and $\subseteq I_y$ by maximality of each: $I_x = I_y$ .

So $U$ is the disjoint union of the distinct sets $I_x$ (each $x \in U$ lies in its own $I_x$ ). Countability: each nonempty open interval $I$ of the family contains a rational $q_I$ (Theorem 10.14), and distinct disjoint intervals get distinct rationals: the family injects into $\Q$ , which is countable (it is indexed by pairs of integers). Hence at most countably many intervals.

Exercise 12.11 ★★

A point $x \in \R$ is an accumulation point of $A$ when every neighborhood of $x$ meets $A \setminus \{x\}$ ; their set is the derived set $A'$ . Prove that $\overline A = A \cup A'$ , and that $A$ is closed if and only if $A' \subseteq A$ . Determine $A'$ for $A = \{\frac 1n : n \in \N^*\}$ , for $A = \Z$ , and for $A = \Q$ .

Solution

Solution of Exercise 12.11.

$\overline A = A \cup A'$ . ( $\supseteq$ ) $A \subseteq \overline A$ always; and if $x \in A'$ , every neighborhood of $x$ meets $A \setminus \{x\} \subseteq A$ , so $x$ is adherent. ( $\subseteq$ ) Let $x \in \overline A$ . If $x \in A$ , done. If $x \notin A$ , every neighborhood of $x$ meets $A = A \setminus \{x\}$ : $x \in A'$ .

Consequently $A$ closed $\iff$ $A = \overline A = A \cup A'$ $\iff$ $A' \subseteq A$ .

$A = \{\frac 1n\}$ : $0$ is an accumulation point ( $\frac 1n \to 0$ , terms $\neq 0$ ); each $\frac 1n$ is isolated (Exercise 12.7), so not in $A'$ ; and a point $x \notin A \cup \{0\}$ has a whole interval avoiding $A$ (between the two neighbors of $x$ in $A \cup \{0\}$ , or beyond $1$ ). Hence $A' = \{0\}$ .

$\Z' = \emptyset$ : every integer is isolated, every non-integer has a neighborhood inside $\R \setminus \Z$ .

$\Q' = \R$ : every interval around any real contains infinitely many rationals (Theorem 10.14), in particular one different from the center.

Exercise 12.12 ★★★

For $A \subseteq \R$ nonempty, define $d_A(x) = \inf\{\abs{x - a} : a \in A\}$ . Prove:

$\abs{d_A(x) - d_A(y)} \leq \abs{x - y}$ for all $x, y$ ( $d_A$ is $1$ -Lipschitz);
$d_A(x) = 0$ if and only if $x \in \overline A$ ; in particular, if $F$ is closed and $x \notin F$ , then $d_F(x) > 0$ ;
for every $\varepsilon > 0$ , the set $V_\varepsilon = \{x : d_F(x) < \varepsilon\}$ is open, contains $F$ , and $\bigcap_{\varepsilon > 0} V_\varepsilon = F$ for $F$ closed: every closed set is a countable intersection of open sets.

Solution

Solution of Exercise 12.12.

For every $a \in A$ : $\abs{x - a} \leq \abs{x - y} + \abs{y - a}$ , so $d_A(x) \leq \abs{x - y} + \abs{y - a}$ ; taking the infimum over $a$ : $d_A(x) \leq \abs{x - y} + d_A(y)$ . Exchanging $x$ and $y$ gives the other inequality: $\abs{d_A(x) - d_A(y)} \leq \abs{x - y}$ .
$d_A(x) = 0$ $\iff$ for every $\varepsilon > 0$ there is $a \in A$ with $\abs{x - a} < \varepsilon$ $\iff$ every interval around $x$ meets $A$ $\iff$ $x \in \overline A$ . If $F$ is closed and $x \notin F = \overline F$ , then $d_F(x) \neq 0$ , i.e. $d_F(x) > 0$ .
If $d_F(x) < \varepsilon$ , set $r = \varepsilon - d_F(x) > 0$ : for $\abs{y - x} < r$ , part (1) gives $d_F(y) \leq d_F(x) + \abs{x - y} < \varepsilon$ : the interval $\intoo{x - r}{x + r}$ lies in $V_\varepsilon$ , which is therefore open; it contains $F$ since $d_F = 0$ there. Finally $x \in \bigcap_{\varepsilon>0} V_\varepsilon$ $\iff$ $d_F(x) < \varepsilon$ for all $\varepsilon$ $\iff$ $d_F(x) = 0$ $\iff$ $x \in \overline F = F$ . Since $\bigcap_{\varepsilon > 0} V_\varepsilon = \bigcap_{n \geq 1} V_{1/n}$ , every closed set is a countable intersection of open sets.

12.5 Problem: The Cantor set, small and enormous at once

Problem 12.1

Weekend problem — the Cantor middle-thirds set: length zero, uncountable, perfect, and $C + C = \intcc{0}{2}$

Remove from $\intcc{0}{1}$ its open middle third, then the middle third of each remaining segment, and repeat forever: what survives is the Cantor set $C$ , the fundamental counterexample factory of analysis. This problem constructs it, reads it through the base- $3$ machinery of Problem 10.1, and establishes its paradoxical portrait: total length zero, yet uncountable; empty interior, yet no isolated point; totally disconnected, yet $C + C$ fills the whole segment $\intcc{0}{2}$ . Formally: $C_0 = \intcc{0}{1}$ , and $C_{n+1}$ is obtained from $C_n$ by deleting the open middle third of each segment of $C_n$ ; finally $C = \bigcap_{n \geq 0} C_n$ . Throughout, a ternary code of $x \in \intcc{0}{1}$ is any digit string $(d_k)_{k\geq1}$ with $d_k \in \{0, 1, 2\}$ whose value $\sup_n \sum_{k=1}^n d_k 3^{-k}$ equals $x$ — improper codes (eventually $2$ ) are allowed; by Problem 10.1 (questions 9–11), every $x \in \intcc{0}{1}$ has one or two codes, two exactly when $x = m/3^N \in \intoo{0}{1}$ .

Part I — The construction.

Describe $C_1$ and $C_2$ explicitly as unions of segments, and prove by induction: $C_n$ is a disjoint union of $2^n$ closed segments, each of length $3^{-n}$ .
Show that $C$ is closed, bounded — hence compact (Theorem 12.19) — nonempty, and that every endpoint of every segment of every $C_n$ belongs to $C$ .
The total length of $C_n$ is $\bigl(\frac23\bigr)^n$ . Deduce that for every $\varepsilon > 0$ , the set $C$ can be covered by finitely many segments of total length $\leq \varepsilon$ : the Cantor set has length zero.
Show that an interval contained in $C$ has length $\leq 3^{-n}$ for every $n$ , hence is a singleton or empty: $\mathring C = \emptyset$ . Being closed with empty interior, $C$ is nowhere dense.

Part II — The ternary code.

Prove the self-similarity recursion

$C_{n+1} = \tfrac13 C_n \,\cup\, \bigl(\tfrac23 + \tfrac13 C_n\bigr), \qquad\text{hence}\qquad C = \tfrac13 C \,\cup\, \bigl(\tfrac23 + \tfrac13 C\bigr),$

the two pieces being disjoint: $C$ is two copies of itself at scale $\frac13$ .
Prove by induction on $n$ : $x \in C_n$ if and only if $x$ has a ternary code whose first $n$ digits lie in $\{0, 2\}$ . Deduce, using the fact that $x$ has at most two codes: $x \in C$ if and only if $x$ has a code with no digit equal to $1$ (a $1$ -free code).
Codes in action: give $1$ -free codes for $0$ , $1$ , $\frac13$ , $\frac23$ ; show $\frac14 = (0.\overline{02})_3$ and $\frac34 = (0.\overline{20})_3$ , so both belong to $C$ ; and check that $\frac14$ is not an endpoint of any $C_n$ (endpoints have the form $m/3^n$ ).
Show that each $x \in C$ has exactly one $1$ -free code (when $x$ has two codes, prove that exactly one of the pair contains the digit $1$ ). Conclude: the value map is a bijection from $\{0,2\}$ -strings onto $C$ .
(Diagonal) Let $k \mapsto x_k$ be any map $\N^* \to C$ . Build a $\{0, 2\}$ -string differing at index $k$ from the code of $x_k$ , and conclude that $C$ is uncountable — while, by contrast, question 3 says it is metrically negligible.

Part III — Topological portrait.

Assemble the record so far: $C$ is compact, uncountable, of length zero, nowhere dense. Which single containment $C \subseteq C_n$ carries each property?
( $C$ is perfect) Let $x \in C$ with $1$ -free code $(d_k)$ . Flipping the digit $d_n$ ( $0 \leftrightarrow 2$ ) produces $x_n \in C$ with $\abs{x_n - x} = 2 \cdot 3^{-n}$ . Conclude that $C$ has no isolated point: every point of $C$ is a limit of other points of $C$ .
Show that the endpoints of question 2 form a countable dense subset of $C$ (truncate the code after $n$ digits and continue with $0$ s; countability as in Exercise 12.10). Conclude: the typical point of $C$ — like $\frac14$ — is not an endpoint: endpoints are a countable skeleton inside an uncountable body.
(Totally disconnected) Let $x < y$ in $C$ . Choose $n$ with $3^{-n} < y - x$ and produce a point $z \in \intoo{x}{y}$ with $z \notin C$ . Conclude that the only nonempty intervals contained in $C$ are singletons.

Part IV — Arithmetic of $C$ .

Show $1 - C = C$ (what does $x \mapsto 1 - x$ do to a $1$ -free code? recall $1 = (0.\overline{2})_3$ ).
(Addition of codes) Show that if $x$ , $x'$ have codes $(a_k)$ , $(b_k)$ , then $x + x' = \lim_n\,(t_n + t'_n)$ where $t_n, t'_n$ are the partial sums. Deduce: every $y \in \intcc{0}{1}$ is the midpoint of two points of $C$ — given a code $(e_k)$ of $y$ , choose digits $a_k, b_k \in \{0, 2\}$ with $\frac{a_k + b_k}{2} = e_k$ .
Conclude $C + C = \intcc{0}{2}$ and, with question 14, $C - C = \intcc{-1}{1}$ . Concrete instance: write $1$ as a sum of the two non-endpoints found in question 7.
Reflect: a set of length zero whose difference set fills $\intcc{-1}{1}$ . Why is there no contradiction between “ $C$ is metrically negligible” and “ $C + C$ has full length”? (One sentence; think about what length controls and what it does not.)

Part V — Members, rational and irrational.

Combine question 8 with the periodicity criterion of Problem 10.1: a point of $C$ is rational if and only if its $1$ -free code is eventually periodic. Run the base- $3$ long division to check $\frac1{13} = (0.\overline{002})_3 \in C$ .
Produce an explicitly irrational member of $C$ : the value of the code with $d_k = 2$ at the triangular positions $k = \frac{j(j+1)}{2}$ and $d_k = 0$ elsewhere. Justify irrationality by the growing-gaps argument of Problem 10.1 (question 20).
(Onto a full segment) Consider $h$ mapping the point of $C$ with $1$ -free code $(d_k)$ to the value of the binary string $\bigl(\frac{d_k}2\bigr)$ , i.e. $h(x) = \sup_n \sum_{k=1}^n \frac{d_k}{2}\,2^{-k}$ . Show that $h$ maps $C$ onto $\intcc{0}{1}$ . So the negligible $C$ surjects onto a segment of full length — a second proof that $C$ is uncountable.
(Self-similar length) Suppose some notion of length $L$ were defined for $C$ and its shrunken copies, respecting scaling ( $L(\lambda A) = \lambda L(A)$ ), translation invariance, and additivity over the disjoint decomposition of question 5. Show that then $L(C) = \frac23\,L(C)$ , forcing $L(C) = 0$ : self-similarity alone already sentences $C$ to length zero.

Part VI — A fat cousin, and the moral.

(Fat Cantor set) Repeat the construction, but at stage $n$ ( $n = 0, 1, 2, \dots$ ) remove from each of the $2^n$ current segments a central open interval of length $4^{-(n+1)}$ only. Show that the segment lengths $l_n$ obey $l_{n+1} = \frac{l_n - 4^{-(n+1)}}{2}$ , $l_n = \frac{2^n + 1}{2\cdot 4^n} > 0$ , that the resulting $K = \bigcap K_n$ is compact with empty interior, and that the total removed length is $\sum_{n\geq0} 2^n 4^{-(n+1)} = \frac12$ . Admitting the (intuitive, Year 3) additivity of length for finite unions of intervals, and using both statements of Example 12.20, show that any finite family of open intervals covering $K$ has total length $\geq \frac12$ : $K$ is nowhere dense but not negligible. Smallness has several inequivalent meanings.
(Distances) Show that for a nonempty closed $F \subseteq \R$ and $x \in \R$ , the infimum $d(x, F)$ is attained (minimizing sequence plus Bolzano–Weierstrass). Then compute

$\max_{y \in \intcc{0}{1}} d(y, C) = \frac16 ,$

attained exactly at the center $y = \frac12$ (a point of a gap created at stage $n$ is within $\frac{3^{-n}}{2}$ of the gap’s endpoints, which lie in $C$ ).
(Every point a subsequential limit) Using questions 12 and 2, produce a single sequence in $C$ whose set of subsequential limits is all of $C$ . (Compare: for a convergent sequence that set is one point — $C$ realizes the opposite extreme among compacts.)
Synthesis, one sentence each: (i) which theorems of this chapter did the construction actually consume (stability of closed sets, compactness, sequential characterizations)? (ii) list the four paradoxical pairings of the portrait (length zero/uncountable, closed/empty interior, perfect/totally disconnected, negligible/ $C+C$ full); (iii) where does $C$ resurface later (the devil’s staircase built on $h$ in the theory of continuity, and the measure theory of the Year 3 volume, where $C$ separates “countable” from “negligible”)?

Solution

Solution of Problem 12.1.

1. $C_1 = \intcc{0}{\frac13} \cup \intcc{\frac23}{1}$ and

C_2 = \intcc{0}{\tfrac19} \cup \intcc{\tfrac29}{\tfrac13} \cup \intcc{\tfrac23}{\tfrac79} \cup \intcc{\tfrac89}{1} .

Induction: if $C_n$ is a disjoint union of $2^n$ closed segments of length $3^{-n}$ , deleting the open middle third of each leaves two closed segments of length $3^{-n-1}$ per parent: $2^{n+1}$ segments, pairwise disjoint (children of distinct parents are separated because the parents were; children of one parent are separated by the removed gap).

2. Each $C_n$ is a finite union of segments, hence closed; $C = \bigcap C_n$ is an intersection of closed sets: closed (Definition 12.5); bounded ( $\subseteq \intcc{0}{1}$ ): compact by Theorem 12.19. Nonempty: $0$ lies in the leftmost segment of every $C_n$ . Let $a$ be an endpoint of a segment $S$ of $C_n$ . For $m \leq n$ , $a \in C_n \subseteq C_m$ . For the later stages: the middle-third deletion never removes an endpoint, and $a$ is again an endpoint of one of the two children of $S$ (the child touching $a$ ); by induction $a \in C_m$ for all $m \geq n$ : $a \in C$ .

3. Total length of $C_n$ : $2^n \cdot 3^{-n} = (\frac23)^n \to 0$ . Given $\varepsilon > 0$ , choose $n$ with $(\frac23)^n \leq \varepsilon$ : then $C \subseteq C_n$ , a union of finitely many segments of total length $\leq \varepsilon$ .

4. Let $I \subseteq C$ be an interval with two distinct points. For every $n$ : $I \subseteq C_n$ , and $I$ , being convex, must lie inside a single segment of $C_n$ (meeting two segments would force $I$ to contain a point of the gap between them, which is outside $C_n$ ). Hence the length of $I$ is $\leq 3^{-n}$ for all $n$ : contradiction. So the only intervals inside $C$ are empty or singletons; in particular no $\intoo{x-r}{x+r}$ fits inside $C$ : $\mathring C = \emptyset$ . As $C$ is closed, $\overline C = C$ has empty interior: $C$ is nowhere dense.

5. Write $\varphi_0(x) = \frac x3$ and $\varphi_2(x) = \frac{2 + x}{3}$ , increasing affine bijections of $\intcc{0}{1}$ onto $\intcc{0}{\frac13}$ and $\intcc{\frac23}{1}$ . Claim: $C_{n+1} = \varphi_0(C_n) \cup \varphi_2(C_n)$ . For $n = 0$ this is question 1. Induction: an increasing affine map sends the middle third of a segment to the middle third of the image segment, so deleting middle thirds commutes with $\varphi_0$ and $\varphi_2$ ; applying the deletion step to $C_{n+1} = \varphi_0(C_n) \cup \varphi_2(C_n)$ yields $C_{n+2} = \varphi_0(C_{n+1}) \cup \varphi_2(C_{n+1})$ . Intersecting over $n$ : for $x \leq \frac13$ , $x \in C \iff x \in \varphi_0(C_n)$ for all $n$ $\iff 3x \in \bigcap C_n = C$ ; likewise on $\intcc{\frac23}{1}$ ; and no point of $\intoo{\frac13}{\frac23}$ lies in $C_1$ . Hence $C = \varphi_0(C) \cup \varphi_2(C)$ , disjointly.

6. Induction on $n$ ; the case $n = 0$ says every $x \in \intcc{0}{1}$ has a code, which is Problem 10.1 (question 9 for $x < 1$ ; $1 = (0.\overline 2)_3$ ). Suppose the equivalence at rank $n$ . If $x \in C_{n+1}$ : by question 5, $x = \varphi_i(z)$ with $z \in C_n$ and $i \in \{0, 2\}$ ; if $(e_k)$ is a code of $z$ with first $n$ digits $1$ -free, then $(i, e_1, e_2, \dots)$ has partial sums $\frac i3 + \frac13\sum_{k\leq m} e_k 3^{-k} \to \varphi_i(z) = x$ : a code of $x$ with first $n + 1$ digits $1$ -free. Conversely, if $x$ has a code $(d_k)$ with $d_1, \dots, d_{n+1} \in \{0, 2\}$ : the shifted string $(d_2, d_3, \dots)$ has some value $z \in \intcc{0}{1}$ , its first $n$ digits are $1$ -free, and the partial-sum computation read backwards gives $x = \varphi_{d_1}(z)$ ; by induction $z \in C_n$ , so $x \in C_{n+1}$ by question 5. Finally: a fully $1$ -free code puts $x$ in every $C_n$ , hence in $C$ ; conversely if $x \in C$ , then for every $n$ one of the at most two codes of $x$ (Problem 10.1, question 11) has its first $n$ digits $1$ -free; one fixed code must work for arbitrarily large $n$ (pigeonhole between two codes), and a code whose first $n$ digits are $1$ -free for arbitrarily large $n$ is $1$ -free outright.

7. $0 = (0.\overline 0)_3$ , $1 = (0.\overline 2)_3$ , $\frac13 = (0.0\overline{2})_3$ (the improper twin of $(0.1)_3$ ), $\frac23 = (0.2\overline{0})_3$ . Geometric sums:

(0.\overline{02})_3 = \sum_{j\geq1} \frac{2}{9^{\,j}} = \frac{2/9}{1 - 1/9} = \frac14 , \qquad (0.\overline{20})_3 = \sum_{j\geq1} \frac{2}{3\cdot 9^{\,j-1}} = \frac{2/3}{1 - 1/9} = \frac34 ,

both $1$ -free: $\frac14, \frac34 \in C$ . (The infinite sums abbreviate suprema of partial sums, as in Problem 10.1.) Endpoints of segments of $C_n$ are of the form $m/3^n$ (induction: children endpoints are parent endpoints or differ from one by a multiple of $3^{-n-1}$ ). If $\frac14 = \frac{m}{3^n}$ then $3^n = 4m$ , and $4 \nmid 3^n$ : impossible. So $\frac14 \in C$ without ever being an endpoint.

8. Suppose $x$ had two distinct $1$ -free codes. Having two codes at all means (Problem 10.1, question 11, base $3$ ) that $x = m/3^N \in \intoo{0}{1}$ and the two codes are: the terminating one, with last nonzero digit $d_N \in \{1, 2\}$ followed by $0$ s, and its twin, with $d_N - 1$ at position $N$ followed by $2$ s. If $d_N = 1$ the first contains a $1$ ; if $d_N = 2$ the twin carries $d_N - 1 = 1$ . Either way at most one of the pair is $1$ -free: contradiction. So each $x \in C$ has exactly one $1$ -free code (existence by question 6), and distinct $\{0,2\}$ -strings have distinct values. Every $\{0,2\}$ -string has value in $\intcc{0}{1}$ (partial sums $\leq 1$ ) with all prefixes $1$ -free, hence value in every $C_n$ , i.e. in $C$ : the value map is a bijection from $\{0,2\}$ -strings onto $C$ .

9. Let $(d^{(k)})$ be the $1$ -free code of $x_k$ and set $e_k = 2 - d^{(k)}_k \in \{0, 2\}$ : a $\{0,2\}$ -string whose value $y$ lies in $C$ and has $(e_k)$ as its unique $1$ -free code (question 8). For each $k$ the codes of $y$ and $x_k$ differ at position $k$ , so $y \neq x_k$ : no map $\N^* \to C$ is surjective. An uncountable set of length zero: bigness in cardinality, smallness in measure — simultaneously.

10. Compactness: closedness of the infinite intersection plus boundedness (question 2) — the one property not carried by a single containment. Length zero: $C \subseteq C_n$ with total length $(\frac23)^n$ (question 3). Nowhere density: $C \subseteq C_n$ forces intervals inside $C$ to have length $\leq 3^{-n}$ (question 4). Uncountability rides on no containment at all: it needs the full intersection structure, encoded in the bijection of question 8.

11. Flip $d_n$ to $2 - d_n$ : the new string is still a $\{0,2\}$ -string, so its value $x_n$ lies in $C$ ; the partial sums beyond rank $n$ differ by exactly $2\cdot3^{-n}$ , so $\abs{x_n - x} = 2\cdot3^{-n}$ . Thus $x_n \neq x$ and $x_n \to x$ : every point of $C$ is a limit of other points of $C$ — $C$ is perfect, with no isolated point.

12. By the induction of question 5, the segments of $C_n$ are exactly the $\intcc{t}{t + 3^{-n}}$ where $t$ runs over the values of length- $n$ $\{0,2\}$ -strings. Given $x \in C$ with code $(d_k)$ , the truncation $t_n$ (digits $d_1 \dots d_n$ then $0$ s) is therefore a left endpoint, and $0 \leq x - t_n \leq 3^{-n}$ : endpoints are dense in $C$ . They form a subset of $\{m/3^n : m, n\}$ , a set indexed by pairs of integers, hence countable (as for $\Q$ in Exercise 12.10). Since $C$ is uncountable (question 9), all but countably many points of $C$ are not endpoints — $\frac14$ (question 7) is the visible tip of that iceberg.

13. Pick $n$ with $3^{-n} < y - x$ . Both $x, y \in C_n$ , and they cannot lie in the same segment (length $3^{-n} < y - x$ ): the removed gap between their segments provides $z$ with $x < z < y$ and $z \notin C_n \supseteq C$ . Hence any two points of $C$ are separated by the complement: the only convex subsets of $C$ are singletons — $C$ is totally disconnected.

14. If $(d_k)$ is the $1$ -free code of $x$ , the string $(2 - d_k)$ is again a $\{0,2\}$ -string, with partial sums

\sum_{k=1}^{n} (2 - d_k)3^{-k} = (1 - 3^{-n}) - \sum_{k=1}^n d_k 3^{-k} \longrightarrow 1 - x :

so $1 - x \in C$ . Thus $1 - C \subseteq C$ , and applying the map twice gives $1 - C = C$ : the Cantor set is symmetric about $\frac12$ .

15. The partial sums $t_n \to x$ and $t'_n \to x'$ (increasing sequences converge to their supremum, i.e. the value), so $t_n + t'_n \to x + x'$ by Theorem 11.5. Given $y \in \intcc{0}{1}$ with code $(e_k)$ , choose $(a_k, b_k) = (0,0), (0,2), (2,2)$ according as $e_k = 0, 1, 2$ : then $a_k + b_k = 2e_k$ , the strings $(a_k)$ , $(b_k)$ are $\{0,2\}$ -strings with values $x, x' \in C$ , and

x + x' = \lim_n\,(t_n + t'_n) = \lim_n 2\sum_{k=1}^n e_k 3^{-k} = 2y :

every $y \in \intcc{0}{1}$ is the midpoint of two points of $C$ .

16. Question 15 gives $\intcc{0}{2} = 2\,\intcc{0}{1} \subseteq C + C$ , and $C + C \subseteq \intcc{0}{1} + \intcc{0}{1} = \intcc{0}{2}$ : equality. Then, using $1 - C = C$ :

C - C = C + (C - 1) = (C + C) - 1 = \intcc{-1}{1} .

Concrete instance: $1 = \frac14 + \frac34$ , a sum of two non-endpoint members of $C$ .

17. Length measures how much of the line the set itself occupies; it says nothing about the set of sums, which is the image of the two-parameter family $C \times C$ under $(x, x') \mapsto x + x'$ — the two digit strings are chosen independently, and that freedom is exactly what fills $\intcc{0}{2}$ . No theorem bounds the length of a sumset by the lengths of the summands, and $C$ is the proof that none can.

18. By Problem 10.1 (question 18), $x$ is rational iff its proper expansion is eventually periodic. The $1$ -free code of $x \in C$ is either that proper expansion or the improper twin of a terminating one; a terminating string and its twin (eventually constant $2$ s) are both eventually periodic, so periodicity of the $1$ -free code is equivalent to rationality of $x$ . Long division of $\frac1{13}$ in base $3$ ( $r_0 = 1$ ): $3 = 13\cdot0 + 3$ , $9 = 13\cdot0 + 9$ , $27 = 13\cdot2 + 1$ , and the remainder returns to $1$ : digits $\overline{002}$ , so $\frac1{13} = (0.\overline{002})_3$ , $1$ -free and periodic: a rational member of $C$ . (Check: $\frac{2/27}{1 - 1/27} = \frac{2}{26} = \frac1{13}$ .)

19. The string with $d_k = 2$ at the triangular positions $k = \frac{j(j+1)}{2}$ and $0$ elsewhere is a $\{0,2\}$ -string, so its value $x^*$ belongs to $C$ (question 8). It has infinitely many $2$ s with gaps $j + 1 \to \infty$ between consecutive ones, so it is not eventually periodic (a period $T$ would eventually force $2$ s at gaps $\leq T$ : the growing-gaps argument of Problem 10.1, question 20); by question 18, $x^* \notin \Q$ . And by question 9 plus the countability of $\Q$ , all but countably many members of $C$ are irrational: $x^*$ is the norm, not the exception.

20. Let $y \in \intcc{0}{1}$ : it has a binary code $(c_k)$ with $c_k \in \{0, 1\}$ (Problem 10.1, question 9, base $2$ ; $y = 1$ takes the all- $1$ s string). Then $(2c_k)$ is a $\{0,2\}$ -string, its value $x$ lies in $C$ , and $h(x)$ is the value of $(c_k)$ , namely $y$ : $h$ maps $C$ onto $\intcc{0}{1}$ . If $C$ were the image of a map from $\N^*$ , composing with $h$ would list all of $\intco{0}{1}$ , contradicting the diagonal theorem of Problem 10.1 (question 22): $C$ is uncountable, again. A length-zero set surjecting onto a full segment.

21. By question 5, $C$ is the disjoint union of $\varphi_0(C)$ and $\varphi_2(C)$ , each a translate of the scaled copy $\frac13 C$ . Additivity, scaling and translation invariance give

L(C) = L(\varphi_0(C)) + L(\varphi_2(C)) = \tfrac13 L(C) + \tfrac13 L(C) = \tfrac23\,L(C),

so $\frac13 L(C) = 0$ : $L(C) = 0$ . Self-similarity alone sentences $C$ to length zero — question 3 merely executed the sentence.

22. A segment of length $l_n$ loses a central interval of length $4^{-(n+1)}$ , leaving two segments of length $l_{n+1} = \frac{l_n - 4^{-(n+1)}}{2}$ ; from $l_0 = 1$ , induction confirms $l_n = \frac{2^n + 1}{2\cdot4^n}$ : indeed $\frac12\Bigl(\frac{2^n+1}{2\cdot4^n} - \frac{1}{4^{n+1}}\Bigr) = \frac{2(2^n + 1) - 1}{2\cdot4^{n+1}} = \frac{2^{n+1} + 1}{2\cdot4^{n+1}}$ , and $l_n > 0$ always: the construction never starves. $K = \bigcap K_n$ is closed and bounded, hence compact; an interval inside $K$ lies in one segment of $K_n$ , of length $l_n \to 0$ : empty interior. Removed length: $\sum_{n\geq0} 2^n \cdot 4^{-(n+1)} = \frac14\sum_{n\geq0}\bigl(\frac12\bigr)^n = \frac12$ , and each $K_n$ has total length $2^n l_n = \frac{2^n + 1}{2^{n+1}} > \frac12$ . Now let finitely many open intervals have union $U \supseteq K$ . By the companion statement of Example 12.20, $U \supseteq K_n$ for some $n$ ; admitting additivity of length on finite unions of intervals, the total length of the covering intervals is at least that of $K_n$ , which exceeds $\frac12$ . So $K$ is nowhere dense, yet no cheap cover exists: topological smallness (nowhere dense) and metric smallness (length zero) are genuinely different notions, and $K$ separates them.

23. Attainment: let $d = d(x, F)$ and pick $a_k \in F$ with $\abs{x - a_k} \leq d + \frac1k$ : the $a_k$ are bounded, so Bolzano–Weierstrass (Theorem 11.16) extracts $a_{\varphi(k)} \to a$ , with $a \in F$ ( $F$ closed, Theorem 12.6) and $\abs{x - a} = \lim \abs{x - a_{\varphi(k)}} = d$ . Now the maximum: if $y \in C$ , $d(y, C) = 0$ ; otherwise $y$ lies in a gap removed at some stage $n \geq 1$ , an open interval of length $3^{-n}$ whose two endpoints belong to $C$ (question 2), so $d(y, C) \leq \frac{3^{-n}}{2} \leq \frac16$ , with equality requiring $n = 1$ and $y$ at the center of the gap $\intoo{\frac13}{\frac23}$ , i.e. $y = \frac12$ ; and indeed $d\bigl(\frac12, C\bigr) = \frac16$ since $C \cap \intoo{\frac13}{\frac23} = \emptyset$ and $\frac13, \frac23 \in C$ . Hence $\max_{y\in\intcc{0}{1}} d(y, C) = \frac16$ , attained exactly at $\frac12$ .

24. The endpoints form a countable dense subset of $C$ (question 12): list them as a single sequence $(e_j)_{j\geq1}$ , a sequence in $C$ . Its subsequential limits all lie in $C$ ( $C$ closed). Conversely, fix $x \in C$ : for each $n$ , the segments of the $C_m$ containing $x$ ( $m \geq n$ ) have their endpoints within $3^{-m} \leq 3^{-n}$ of $x$ , so infinitely many distinct endpoints lie within $3^{-n}$ of $x$ ; choose indices $j_1 < j_2 < \dots$ with $\abs{e_{j_n} - x} \leq 3^{-n}$ : a subsequence converging to $x$ . So the set of subsequential limits of $(e_j)$ is exactly $C$ — one sequence clustering at uncountably many points, the opposite extreme from a convergent sequence, whose cluster set is a singleton.

25. (i) The construction consumed: stability of closed sets under arbitrary intersection (existence of $C$ as a closed set), the compactness theorem Theorem 12.19 (questions 2, 22, 23), and the sequential characterizations of closedness and adherence (the nested-compacts argument of Example 12.20 and question 23). (ii) The four pairings: length zero yet uncountable (questions 3, 9); closed yet with empty interior (question 4); perfect — no isolated point — yet totally disconnected (questions 11, 13); negligible yet with $C + C = \intcc{0}{2}$ (question 16). (iii) The surjection $h$ of question 20, made continuous and nondecreasing, becomes the devil’s staircase in the theory of continuous functions; and in the Year 3 volume’s measure theory, $C$ is the standard witness that “negligible” does not mean “countable”, with its fat cousin (question 22) separating “nowhere dense” from “negligible”.