§13 ★ Some Topological Concepts in Metric Spaces

Enrichment section

This section is optional, but Chapter 3’s enrichment sections §21–§22 build on it.

So far we have used the order of $\mathbb{R}$ heavily, for instance in $\limsup$ and $\liminf$. But most of our analysis could have been based on distance alone, which makes it easy to work more generally. Analysis on $k$-dimensional space $\mathbb{R}^k$ is important, and $\mathbb{R}^k$ has no useful natural order when $k > 1$.

Metric spaces

13.1 Definition.

Let $S$ be a set and $d$ a function defined for all pairs $(x, y)$ of elements of $S$ satisfying

• D1. $d(x, x) = 0$ for all $x \in S$, and $d(x, y) > 0$ for distinct $x, y \in S$.
• D2. $d(x, y) = d(y, x)$ for all $x, y \in S$.
• D3. $d(x, z) \le d(x, y) + d(y, z)$ for all $x, y, z \in S$.

Such a $d$ is a distance function or metric on $S$. A metric space is a set $S$ together with a metric; properly speaking it is the pair $(S, d)$, since one set can carry several metrics (Exercise 13.1).

Example 1.

$\operatorname{dist}(a, b) = |a - b|$ is a metric on $\mathbb{R}$; Corollary 3.6 gives D3. For any metric, D3 is called the triangle inequality.

Example 2.

The set of all $k$-tuples $x = (x_1, x_2, \dots, x_k)$ of reals is $k$-dimensional Euclidean space $\mathbb{R}^k$. Its most familiar metric gives ordinary distance in the plane or in 3-space:

$$d(x, y) = \left[ \sum_{j=1}^k (x_j - y_j)^2 \right]^{1/2} \qquad\text{for } x, y \in \mathbb{R}^k.$$

D1 and D2 are obvious; the triangle inequality D3 is not (it follows from the Cauchy–Schwarz inequality).

Try it: balls in three metrics

Drag the centre $x_0$ and the pink point. The disc, square and diamond are the balls of radius $r$ for $d$, $d_1$ and $d_2$ of Exercise 13.1.

radius r1.50

Keyboard: Tab to the center or the probe, then use the arrow keys to move it.

Euclidean d (disc)d₁ = max |xⱼ − yⱼ| (square)d₂ = Σ |xⱼ − yⱼ| (diamond)

Distance from the black centre x₀ to the pink point:d = 1.253 < r (inside)d₁ = 1.100 < r (inside)d₂ = 1.700 ≥ r (outside)

Always d₁ ≤ d ≤ d₂ ≤ 2·d₁ in the plane, so each ball contains a smaller ball of each other type. That is why all three metrics give the same open sets and the same convergent sequences.

13.2 Definition.

A sequence $(s_n)$ in a metric space $(S, d)$ converges to $s \in S$ if $\lim_{n\to\infty} d(s_n, s) = 0$. It is a Cauchy sequence if for each $\varepsilon > 0$ there is $N$ such that $m, n > N$ implies $d(s_m, s_n) < \varepsilon$. $(S, d)$ is complete if every Cauchy sequence in $S$ converges to some element of $S$.

Used in §10

“Complete” now seems to have two meanings, but both express that the space has no gaps. Theorem 10.11 says $(\mathbb{R}, \operatorname{dist})$ is a complete metric space, and its proof used the Completeness Axiom 4.4. We could instead have taken metric completeness as the axiom and proved the least upper bound property. We did not, because least upper bounds seem more fundamental than Cauchy sequences.

When subscripts are needed for coordinates, we write $(x^{(n)})$ for a sequence in $\mathbb{R}^k$, with $x^{(n)} = (x_1^{(n)}, x_2^{(n)}, \dots, x_k^{(n)})$. The metric on $\mathbb{R}^k$ is always that of Example 2 unless stated otherwise.

13.3 Lemma.

A sequence $(x^{(n)})$ in $\mathbb{R}^k$ converges if and only if each coordinate sequence $(x_j^{(n)})$ converges in $\mathbb{R}$, for $j = 1, \dots, k$. It is Cauchy in $\mathbb{R}^k$ if and only if each $(x_j^{(n)})$ is Cauchy in $\mathbb{R}$.

Used in Theorem 13.4, Theorem 13.5, Exercise 13.2

Proof

The first assertion is Exercise 13.2(b). For the second, note that for $x, y \in \mathbb{R}^k$ and each $j$,

$$|x_j - y_j| \le d(x, y) \le \sqrt{k}\, \max\{|x_j - y_j| : j = 1, \dots, k\}. \tag{1}$$

If $(x^{(n)})$ is Cauchy, the first inequality in (1) shows each $(x_j^{(n)})$ is Cauchy. Conversely, if each $(x_j^{(n)})$ is Cauchy and $\varepsilon > 0$, choose $N_j$ with $|x_j^{(m)} - x_j^{(n)}| < \frac{\varepsilon}{\sqrt k}$ for $m, n > N_j$. With $N = \max\{N_1, \dots, N_k\}$, the second inequality in (1) gives $d(x^{(m)}, x^{(n)}) < \varepsilon$ for $m, n > N$.

∎

13.4 Theorem.

Euclidean $k$-space $\mathbb{R}^k$ is complete.

Used in Exercise 13.1

Proof

If $(x^{(n)})$ is Cauchy in $\mathbb{R}^k$, each $(x_j^{(n)})$ is Cauchy in $\mathbb{R}$ by Lemma 13.3, so converges to some $x_j$ by Theorem 10.11. By Lemma 13.3 again, $(x^{(n)})$ converges to $x = (x_1, \dots, x_k)$.

∎

A set $S \subseteq \mathbb{R}^k$ is bounded if there is $M > 0$ with $\max\{|x_j| : j = 1, \dots, k\} \le M$ for all $x \in S$.

13.5 Bolzano–Weierstrass Theorem.

Every bounded sequence in $\mathbb{R}^k$ has a convergent subsequence.

Used in §11

Proof

Each $(x_j^{(n)})$ is bounded in $\mathbb{R}$. By Theorem 11.5 we may pass to a subsequence along which $(x_1^{(n)})$ converges; then to a further subsequence along which $(x_2^{(n)})$ converges, while $(x_1^{(n)})$ still converges by Theorem 11.3. After $k$ steps every coordinate converges, and the resulting subsequence converges in $\mathbb{R}^k$ by Lemma 13.3.

∎

Open and closed sets

13.6 Definition.

Let $(S, d)$ be a metric space and $E \subseteq S$. A point $s_0 \in E$ is interior to $E$ if for some $r > 0$, $\{s \in S : d(s, s_0) < r\} \subseteq E$. $E^\circ$ denotes the set of interior points of $E$. $E$ is open in $S$ if every point of $E$ is interior to $E$, i.e. $E = E^\circ$.

13.7 Discussion.

One can show (Exercise 13.4):

1. $S$ is open in $S$ (trivial).
2. The empty set $\varnothing$ is open in $S$ (trivial).
3. The union of any collection of open sets is open.
4. The intersection of finitely many open sets is open.

Sometimes a useful notion of convergence comes from no metric at all. The right setting is then a topology: a set in which certain subsets are decreed open, subject only to (i)–(iv). The open sets of a metric form a topology. Concepts defined in terms of open sets, such as those in 13.8, 13.11 and 22.1, are therefore called topological, which explains this section’s title.

13.8 Definition.

A subset $E$ of a metric space $(S, d)$ is closed if its complement $S \setminus E$ is open. Because of 13.7(iii), any intersection of closed sets is closed (Exercise 13.5). The closure $E^-$ of $E$ is the intersection of all closed sets containing $E$. The boundary of $E$ is $E^- \setminus E^\circ$; its points are boundary points of $E$.

13.9 Proposition.

Let $E$ be a subset of a metric space $(S, d)$.

1. $E$ is closed if and only if $E = E^-$.
2. $E$ is closed if and only if it contains the limit of every convergent sequence of points in $E$.
3. A point is in $E^-$ if and only if it is the limit of some sequence of points in $E$.
4. A point is in the boundary of $E$ if and only if it belongs to the closure of both $E$ and its complement.

Used in Theorem 13.10, Exercise 13.6, Exercise 13.11, Exercise 13.13, Exercise 13.15

Intuitionopen and closed

A set is open when every point in it has some room to move: a small ball around the point stays inside the set. A set is closed when you cannot escape it by taking limits (Proposition 13.9(b)). These are not opposites. $[0, 1)$ is neither open nor closed, while $\mathbb{R}$ and $\varnothing$ are both. “Not open” does not mean “closed”.

Example 3.

In $\mathbb{R}$, open intervals $(a, b)$ are open and closed intervals $[a, b]$ are closed. The interior of $[a, b]$ is $(a, b)$, and the boundary of both is $\{a, b\}$. Every open set in $\mathbb{R}$ is a union of a disjoint sequence of open intervals (Exercise 13.7). A closed set need not be a disjoint union of closed intervals and points; see Example 5. No interval $(a, b)$ or $[a, b]$ with $a < b$ is a disjoint union of two or more closed intervals with more than one point each (Theorem 21.11).

Example 4.

In $\mathbb{R}^k$, open balls $\{x : d(x, x_0) < r\}$ are open and closed balls $\{x : d(x, x_0) \le r\}$ are closed; the boundary of each is the sphere $\{x : d(x, x_0) = r\}$. In $\mathbb{R}^2$, $\{(x_1, x_2) : x_1 > 0\}$ and $\{(x_1, x_2) : x_1 > 0,\ x_2 > 0\}$ are open; with $\ge$ they become closed. Many sets are neither: $[0, 1)$ in $\mathbb{R}$, or $\{(x_1, x_2) : x_1 > 0,\ x_2 \ge 0\}$ in $\mathbb{R}^2$.

13.10 Theorem.

Let $(F_n)$ be a decreasing sequence $F_1 \supseteq F_2 \supseteq \cdots$ of closed, bounded, nonempty sets in $\mathbb{R}^k$. Then $F = \bigcap_{n=1}^\infty F_n$ is also closed, bounded and nonempty.

Used in §4, Proposition 13.13

Proof

Closed and bounded are clear; nonemptiness needs proof. Choose $x_n \in F_n$ for each $n$. By Bolzano–Weierstrass 13.5, a subsequence $(x_{n_m})$ converges to some $x_0 \in \mathbb{R}^k$. To show $x_0 \in F_{n_0}$ for a fixed $n_0$: if $m \ge n_0$ then $n_m \ge n_0$, so $x_{n_m} \in F_{n_m} \subseteq F_{n_0}$. Thus $(x_{n_m})_{m \ge n_0}$ lies in $F_{n_0}$ and converges to $x_0$, so $x_0 \in F_{n_0}$ by Proposition 13.9(b).

∎

Example 5.

The Cantor set is a famous nonempty closed subset of $\mathbb{R}$: $F = \bigcap_{n=1}^\infty F_n$, where $F_1 = [0, 1]$ and each $F_{n+1}$ is obtained from $F_n$ by removing the open middle third of each of its intervals. The total length of $F_n$ is $\left(\frac23\right)^{n-1} \to 0$. Yet $F$ is so large that it cannot be listed as a sequence: it is uncountable. Its interior is empty, so $F$ equals its boundary.

Show F₁ … Fₙ, n =4

F4 is 8 closed intervals of length 3⁻³, total length (2/3)³ ≈ 0.2963. The total length tends to 0, yet the intersection is uncountable.

Intuitionhow much is left of the Cantor set?

The removed intervals have total length $\frac13 + \frac29 + \frac4{27} + \cdots = 1$, the whole length of $[0, 1]$, and yet uncountably many points survive. One way to see the survivors: they are exactly the numbers in $[0, 1]$ with a base-$3$ expansion that uses only the digits $0$ and $2$. Length and number of points measure very different things.

Compactness

13.11 Definition.

A family $\mathcal{U}$ of open sets is an open cover of $E$ if every point of $E$ lies in at least one set of $\mathcal{U}$, i.e. $E \subseteq \bigcup\{U : U \in \mathcal{U}\}$. A subcover is a subfamily that still covers $E$. A cover or subcover is finite if it has finitely many sets (the sets themselves may be infinite). $E$ is compact if every open cover of $E$ has a finite subcover of $E$.

This abstract definition is very important in advanced analysis. In $\mathbb{R}^k$, compact sets have a simple description.

Intuitioncompactness turns local into global

Compactness lets you pass from “near each point, something good happens” to “finitely many such neighbourhoods cover everything”, and finitely many things can be handled one at a time (take a $\max$ or a $\min$). Example 6 below is a typical use. In $\mathbb{R}^k$ there is an equivalent form in terms of sequences: $E$ is compact exactly when every sequence in $E$ has a subsequence converging to a point of $E$. For a closed, bounded set this follows from Bolzano–Weierstrass 13.5 and Proposition 13.9(b). If $E$ is unbounded, a sequence running off to infinity has no convergent subsequence. If $E$ is not closed, a sequence converging to a point outside $E$ has no subsequence with limit in $E$.

13.12 Heine–Borel Theorem.

A subset $E$ of $\mathbb{R}^k$ is compact if and only if it is closed and bounded.

Proof

Suppose $E$ is compact. Let $U_m = \{x \in \mathbb{R}^k : \max_j |x_j| < m\}$. The family $\{U_m\}$ covers $\mathbb{R}^k$, hence $E$, so finitely many $U_m$ cover $E$. If $U_{m_0}$ is the largest of them, $E \subseteq U_{m_0}$, so $E$ is bounded.

To show $E$ is closed, take $x_0 \in \mathbb{R}^k \setminus E$ and let $V_m = \{x : d(x, x_0) > \frac1m\}$. Each $V_m$ is open and $\bigcup_m V_m = \mathbb{R}^k \setminus \{x_0\} \supseteq E$. Finitely many $V_m$ cover $E$, so $E \subseteq V_{m_0}$ for some $m_0$. Then $\{x : d(x, x_0) < \frac1{m_0}\} \subseteq \mathbb{R}^k \setminus E$, so $x_0$ is interior to the complement. Hence $\mathbb{R}^k \setminus E$ is open and $E$ is closed.

Conversely, suppose $E$ is closed and bounded. Then $E \subseteq F = \{x : |x_j| \le m \text{ for all } j\}$ for some $m$. By Exercise 13.12 it suffices to prove $F$ compact, which is Proposition 13.13.

∎

Common pitfallHeine–Borel is special to ℝᵏ

“Closed and bounded” is not the definition of compact, and in other metric spaces it is not enough. In $\mathbb{Q}$ with the usual distance, $E = \{r \in \mathbb{Q} : 0 \le r \le 2\}$ is closed and bounded but not compact. The sets $U_n = \{r \in \mathbb{Q} : |r - \sqrt2| > \frac1n\}$ are open in $\mathbb{Q}$ and together cover $E$, because $\sqrt2$ is not rational. But any finite collection of them is contained in a single $U_N$, which misses the rationals in $E$ within $\frac1N$ of $\sqrt2$. The proof of Theorem 13.12 relies on the completeness of $\mathbb{R}^k$, through Theorem 13.10.

The set $F$ above is a $k$-cell: $F = [a_1, b_1] \times [a_2, b_2] \times \cdots \times [a_k, b_k]$, a $k$-dimensional box. A 2-cell is a closed rectangle. Its diameter is

$$\delta = \left[\sum_{j=1}^k (b_j - a_j)^2\right]^{1/2} = \sup\{d(x, y) : x, y \in F\}.$$

Cutting each $[a_j, b_j]$ at its midpoint splits $F$ into $2^k$ $k$-cells, each of diameter $\frac\delta2$.

13.13 Proposition.

Every $k$-cell $F$ in $\mathbb{R}^k$ is compact.

Used in §11, Theorem 13.12

Proof

Suppose an open cover $\mathcal{U}$ of $F$ has no finite subcover, and let $\delta$ be the diameter of $F$. At least one of the $2^k$ half-size cells, call it $F_1$, cannot be covered by finitely many sets of $\mathcal{U}$. Repeating, we get $k$-cells with

$$F_1 \supseteq F_2 \supseteq F_3 \supseteq \cdots; \tag{1}$$$$F_n \text{ has diameter } \delta \cdot 2^{-n}; \tag{2}$$$$F_n \text{ cannot be covered by finitely many sets from } \mathcal{U}. \tag{3}$$

By Theorem 13.10, $\bigcap F_n$ contains a point $x_0$, which lies in some $U_0 \in \mathcal{U}$. Since $U_0$ is open, $\{x : d(x, x_0) < r\} \subseteq U_0$ for some $r > 0$. Then $F_n \subseteq U_0$ as soon as $\delta \cdot 2^{-n} < r$, contradicting (3) dramatically.

∎

Since $\mathbb{R} = \mathbb{R}^1$, all of this applies to $\mathbb{R}$.

Example 6.

Let $E$ be a nonempty subset of a metric space $(S, d)$ and define $d(E, x) = \inf\{d(y, x) : y \in E\}$. Then $|d(E, x_1) - d(E, x_2)| \le d(x_1, x_2)$. We show: if $E$ is compact and $E \subseteq U$ with $U$ open, then for some $\delta > 0$,

$$\{x \in S : d(E, x) < \delta\} \subseteq U. \tag{1}$$

For each $x \in E$ there is $r_x > 0$ with

$$\{y \in S : d(y, x) < r_x\} \subseteq U. \tag{2}$$

The balls $\{y : d(y, x) < r_x/2\}$ cover $E$, so finitely many do: there are $x_1, \dots, x_n \in E$ with $E \subseteq \bigcup_{k=1}^n \{y : d(y, x_k) < \frac{r_k}{2}\}$, writing $r_k = r_{x_k}$. Let $\delta = \frac12\min\{r_1, \dots, r_n\}$. If $d(E, x) < \delta$, pick $y \in E$ with $d(y, x) < \delta$, and $k$ with $d(y, x_k) < \frac{r_k}{2}$. Then

$$d(x, x_k) \le d(x, y) + d(y, x_k) < \delta + \frac{r_k}2 \le \frac{r_k}2 + \frac{r_k}2 = r_k,$$

so $x \in U$ by (2). This proves (1).

Reading alongside Abbott

Abbott’s Chapter 3 covers this material on the real line only: the Cantor set opens the chapter (Section 3.1), then open and closed sets (Section 3.2) and compact sets (Section 3.3). Abbott defines compactness by the sequence property above and proves the open-cover version as part of the Heine–Borel theorem, the reverse of Ross’s order. General metric spaces appear in Abbott only in Section 8.2.

Check your understanding

Take the quiz (3 questions)

Which formula defines a metric on $\mathbb{R}^2$?

$d(x, y) = \max\{|x_1 - y_1|, |x_2 - y_2|\}$$d(x, y) = |x_1 - y_1|$$d(x, y) = (x_1 - y_1) + (x_2 - y_2)$

Which subset of $\mathbb{R}$ is compact?

$[0, 1]$$(0, 1)$$[0, \infty)$

Which subset of $\mathbb{R}$ is closed but not bounded?

$[0, \infty)$$(0, 1]$$\{\frac1n : n \in \mathbb{N}\}$

Exercises

Exercise 13.1

For $x, y \in \mathbb{R}^k$ let $d_1(x, y) = \max\{|x_j - y_j| : j = 1, \dots, k\}$ and $d_2(x, y) = \sum_{j=1}^k |x_j - y_j|$.

1. Show $d_1$ and $d_2$ are metrics for $\mathbb{R}^k$.
2. Show $d_1$ and $d_2$ are complete metrics on $\mathbb{R}^k$.

Hint

(b) $d_1(x, y) \le d(x, y) \le d_2(x, y) \le k\, d_1(x, y)$, so all three metrics have the same Cauchy sequences and the same convergent sequences.

Solution

1. Write $a_j = x_j - y_j$, $b_j = y_j - z_j$, so $x_j - z_j = a_j + b_j$. D1 and D2: both $d_1$ and $d_2$ are $\ge 0$, vanish exactly when every $|x_j - y_j| = 0$, i.e. $x = y$, and are symmetric in $x, y$. D3: for each $j$, $|x_j - z_j| \le |x_j - y_j| + |y_j - z_j| \le d_1(x,y) + d_1(y,z)$, and taking the maximum over $j$ gives $d_1(x,z) \le d_1(x,y) + d_1(y,z)$. Summing the coordinate inequalities over $j$ gives $d_2(x,z) \le d_2(x,y) + d_2(y,z)$.
2. Let $d$ be the Euclidean metric. For $a \in \mathbb{R}^k$ we have $\max|a_j| \le \bigl(\sum a_j^2\bigr)^{1/2} \le \sum|a_j| \le k\max|a_j|$ (the middle step because $\sum a_j^2 \le (\sum|a_j|)^2$). Hence $d_1 \le d \le d_2 \le k\,d_1 .$ So a sequence converges to $x$ (or is Cauchy) for one of $d_1, d, d_2$ if and only if it does for the others, since each distance is bounded by a constant multiple of any other. Now $\mathbb{R}^k$ is complete for $d$ (Theorem 13.4): a $d_1$- or $d_2$-Cauchy sequence is $d$-Cauchy, so it converges for $d$ to some $x$, and therefore converges to $x$ for $d_1$ and $d_2$ as well.

Tags: StandardOpen, closed, compact

Used in Definition 13.1

Exercise 13.2

1. Prove (1) in Lemma 13.3.
2. Prove the first assertion in Lemma 13.3.

Solution

1. Fix $j$. Since $(x_j - y_j)^2$ is one term of the sum defining $d(x,y)^2$, we get $|x_j - y_j| \le d(x,y)$. Also each term satisfies $(x_i - y_i)^2 \le m^2$ where $m = \max_i|x_i - y_i|$, so $d(x,y)^2 \le km^2$, i.e. $d(x,y) \le \sqrt k\, m$.
2. Suppose $x^{(n)} \to x$ in $\mathbb{R}^k$. By (1), $|x_j^{(n)} - x_j| \le d(x^{(n)}, x) \to 0$, so each coordinate sequence converges (to $x_j$). Conversely, suppose each $(x_j^{(n)})$ converges, say to $x_j$, and let $x = (x_1, \dots, x_k)$. Given $\varepsilon > 0$ choose $N_j$ with $|x_j^{(n)} - x_j| < \varepsilon/\sqrt k$ for $n > N_j$, and let $N = \max_j N_j$. For $n > N$, (1) gives $d(x^{(n)}, x) \le \sqrt k \max_j|x_j^{(n)} - x_j| < \varepsilon$. So $x^{(n)} \to x$.

Tags: StandardOpen, closed, compact

Used in Lemma 13.3

Exercise 13.3

Let $B$ be the set of all bounded sequences $x = (x_1, x_2, \dots)$ and define $d(x, y) = \sup\{|x_j - y_j| : j = 1, 2, \dots\}$.

1. Show $d$ is a metric for $B$.
2. Does $d^*(x, y) = \sum_{j=1}^\infty |x_j - y_j|$ define a metric for $B$?

Solution

(b) No: for $x = (1, 1, 1, \dots)$ and $y = (0, 0, 0, \dots)$ the sum is $+\infty$.

Tags: StandardOpen, closed, compact

Used in Exercise 13.15

Exercise 13.4

Prove (iii) and (iv) in Discussion 13.7.

Hint

(iv): if $s_0$ lies in $U_1 \cap \cdots \cap U_n$ with balls of radii $r_1, \dots, r_n$ inside each, use $r = \min r_i$. Explain why this fails for infinitely many sets, e.g. $\bigcap_n (-\frac1n, \frac1n) = \{0\}$.

Solution

Let $\{U_\alpha\}$ be open sets.

(iii) If $s_0 \in \bigcup_\alpha U_\alpha$, then $s_0 \in U_\alpha$ for some $\alpha$, and some ball about $s_0$ lies in $U_\alpha \subseteq \bigcup U_\alpha$. So the union is open.

(iv) Let $s_0 \in U_1 \cap \cdots \cap U_n$ with $U_i$ open. For each $i$ there is $r_i > 0$ with $B(s_0, r_i) \subseteq U_i$. Let $r = \min\{r_1, \dots, r_n\} > 0$. Then $B(s_0, r) \subseteq U_i$ for every $i$, so the intersection is open. For infinitely many sets the minimum of the radii can be $0$: each $(-\frac1n, \frac1n)$ is open, but $\bigcap_n(-\frac1n, \frac1n) = \{0\}$ is not.

Tags: StandardOpen, closed, compact

Used in Remark 13.7

Exercise 13.5

1. Verify one of De Morgan’s laws: $\bigcap\{S \setminus U : U \in \mathcal{U}\} = S \setminus \bigcup\{U : U \in \mathcal{U}\}$.
2. Show that the intersection of any collection of closed sets is closed.

Solution

1. For $x \in S$: $x \in \bigcap_{U}(S \setminus U)$ iff $x \notin U$ for every $U \in \mathcal{U}$ iff $x \notin \bigcup_{U}U$ iff $x \in S \setminus \bigcup_U U$.
2. Let $F_\alpha$ be closed subsets of $S$ and put $U_\alpha = S \setminus F_\alpha$, which is open. By (a), $\bigcap_\alpha F_\alpha = \bigcap_\alpha(S \setminus U_\alpha) = S \setminus \bigcup_\alpha U_\alpha$. The union $\bigcup_\alpha U_\alpha$ is open by 13.7(iii), so the intersection is closed.

Tags: StandardOpen, closed, compact

Used in Definition 13.8, Exercise 13.6

Exercise 13.6

Prove Proposition 13.9.

Solution

First, an open ball $B(s,r) = \{t : d(t,s) < r\}$ is open: if $t \in B(s,r)$ and $\rho = r - d(s,t) > 0$, then $d(u,t) < \rho$ implies $d(u,s) < r$ by D3, so $B(t,\rho) \subseteq B(s,r)$.

A characterization. $s \in E^-$ iff every ball $B(s,r)$ meets $E$. Indeed, if some ball $B(s,r)$ misses $E$, then $S \setminus B(s,r)$ is a closed set containing $E$ but not $s$, so $s \notin E^-$. Conversely, if $s \notin E^-$, then $S \setminus E^-$ is open (the closure is an intersection of closed sets, hence closed by Exercise 13.5), so a ball about $s$ lies in $S \setminus E^-$ and misses $E$.

1. If $E$ is closed, it is one of the closed sets whose intersection defines $E^-$, and it is contained in each of them, so $E^- = E$. Conversely, if $E = E^-$ then $E$ is an intersection of closed sets, hence closed.
2. Suppose $E$ is closed, $s_n \in E$ and $s_n \to s$. If $s \notin E$, then $S \setminus E$ is open and contains a ball $B(s,r)$, but $s_n \in B(s,r)$ for large $n$, contradicting $s_n \in E$. So $s \in E$. Conversely, suppose $E$ contains the limits of all its convergent sequences, and let $s \in S \setminus E$. If $s$ were not interior to $S \setminus E$, each ball $B(s, \frac1n)$ would contain some $s_n \in E$; then $s_n \to s$ and so $s \in E$, a contradiction. Hence $S \setminus E$ is open.
3. If $s \in E^-$, every ball $B(s, \frac1n)$ meets $E$ by the characterization, so choose $s_n \in E$ in it; then $s_n \to s$. Conversely, if $s_n \in E$ and $s_n \to s$, every ball about $s$ contains some $s_n \in E$, so $s \in E^-$ by the characterization.
4. By the characterization, $s \notin E^\circ$ iff every ball about $s$ contains a point not in $E$ iff $s \in (S \setminus E)^-$. Hence $s \in E^- \setminus E^\circ$ iff $s \in E^-$ and $s \in (S\setminus E)^-$.

Tags: StandardOpen, closed, compact

Used in Exercise 13.8

Exercise 13.7

Show that every open set in $\mathbb{R}$ is the disjoint union of a finite or infinite sequence of open intervals.

Hint

For $x$ in the open set $U$, take the largest open interval in $U$ containing $x$. Distinct such intervals are disjoint, and each contains a rational, so there are at most countably many.

Solution

Let $U \subseteq \mathbb{R}$ be open. For $x \in U$ let $I_x$ be the union of all open intervals $J$ with $x \in J \subseteq U$. Such $J$ exist since $U$ is open, and $I_x$ is an open interval (a union of intervals sharing the point $x$ is an interval, and it is open as a union of open sets). It contains $x$, lies in $U$, and by construction contains every open interval with these two properties.

If $I_x \cap I_y \ne \varnothing$, then $I_x \cup I_y$ is an interval (the two share a point), open, contained in $U$, and contains $x$; so $I_x \cup I_y \subseteq I_x$, and likewise $\subseteq I_y$. Therefore $I_x = I_y$. So any two of the intervals $I_x$ are equal or disjoint, and their union is $U$ (each $x$ lies in $I_x$).

Each distinct $I_x$ contains a rational number, and disjoint intervals contain different rationals. Choosing one rational per interval gives a one-to-one map from the family of distinct intervals into $\mathbb{Q}$, which is countable. So the family is finite or countably infinite, and $U$ is the disjoint union of a finite or infinite sequence of open intervals.

Tags: ChallengeOpen, closed, compact

Exercise 13.8

1. Verify the assertions in the first paragraph of Example 3.
2. Verify the assertions in Example 4.

Solution

1. $(a,b)$ is open: if $x \in (a,b)$, then $r = \min\{x - a, b - x\} > 0$ and $(x - r, x + r) \subseteq (a,b)$. $[a,b]$ is closed: its complement $(-\infty, a) \cup (b, \infty)$ is a union of open intervals of the same kind, hence open. Interior: every point of $(a,b)$ is interior to $[a,b]$; but $a$ is not, since every interval about $a$ contains points less than $a$ (similarly for $b$). So the interior of $[a,b]$ is $(a,b)$. Boundary: $[a,b]$ is closed, so its closure is itself and its boundary is $[a,b] \setminus (a,b) = \{a,b\}$. For $(a,b)$ the interior is $(a,b)$ and the closure is $[a,b]$ (it is closed, and $a, b$ are limits of points $a + \frac{b-a}{n}$, $b - \frac{b-a}{n}$ of $(a,b)$), so the boundary is again $\{a,b\}$.
2. Balls. The open ball $B(x_0,r)$ is open (shown in the solution of Exercise 13.6 using D3). The closed ball $\{d(x,x_0) \le r\}$ is closed: if $d(x,x_0) > r$, put $\rho = d(x,x_0) - r$; for $d(y,x) < \rho$ we get $d(y,x_0) \ge d(x,x_0) - d(x,y) > r$, so the complement is open. Boundary: a point $x$ with $d(x,x_0) = r$ is the limit of $x_0 + (1 - \frac1n)(x - x_0)$, which lie in the open ball, so the closure of the open ball is the closed ball; and $x$ is not interior to the closed ball, since $x + \frac1n(x - x_0)$ has distance $r(1 + \frac1n) > r$ from $x_0$ and tends to $x$. So the interior of the closed ball is the open ball and both sets have boundary $\{d(x,x_0) = r\}$. Quadrants. If $x_1 > 0$ and $d(x,y) < x_1$, then $|y_1 - x_1| \le d(x,y) < x_1$ gives $y_1 > 0$, so $\{x_1 > 0\}$ is open; the same with radius $\min\{x_1, x_2\}$ shows $\{x_1 > 0, x_2 > 0\}$ is open. With $\ge$, the complements $\{x_1 < 0\}$ and $\{x_1 < 0\} \cup \{x_2 < 0\}$ are open by the same argument, so these sets are closed. Neither: $[0,1)$ is not open ($0$ is not interior) and not closed ($1 - \frac1n \to 1 \notin [0,1)$). The set $\{x_1 > 0, x_2 \ge 0\}$ is not open ($(1,0)$ has nearby points $(1,-\varepsilon)$ outside it) and not closed ($(\frac1n, 0)$ lies in it but tends to $(0,0)$, which does not).

Tags: Warm-upOpen, closed, compact

Exercise 13.9

Find the closures of the following sets:

1. $\{\frac1n : n \in \mathbb{N}\}$
2. $\mathbb{Q}$
3. $\{r \in \mathbb{Q} : r^2 < 2\}$

Solution

(a) $\{\frac1n : n \in \mathbb{N}\} \cup \{0\}$ (b) $\mathbb{R}$ (c) $[-\sqrt2, \sqrt2]$

Tags: Warm-upOpen, closed, compactComputation

Exercise 13.10

Show that the interior of each of the following sets is empty:

1. $\{\frac1n : n \in \mathbb{N}\}$
2. $\mathbb{Q}$
3. The Cantor set in Example 5

Hint

(b) Every interval contains an irrational (Exercise 4.12). (c) Any interval in $F$ lies in $F_n$ for all $n$, so its length is at most $3^{-(n-1)}$ for all $n$.

Solution

In each case suppose an interval $(c - r, c + r)$ with $r > 0$ lay in the set.

1. Every interval contains an irrational number (Exercise 4.12), but every element of $\{\frac1n\}$ is rational. So no interval lies in the set.
2. Same: every interval contains an irrational, which is not in $\mathbb{Q}$.
3. If an interval $I$ lies in the Cantor set $F$, then $I \subseteq F_n$ for every $n$. The set $F_n$ is a union of closed intervals of length $3^{-(n-1)}$ separated by removed open gaps. An interval cannot contain points from two different pieces without containing the gap between them, so $I$ lies inside one piece and its length is at most $3^{-(n-1)}$. This holds for every $n$, so the length of $I$ is $0$, a contradiction.

Tags: Warm-upOpen, closed, compact

Exercise 13.11

Let $E \subseteq \mathbb{R}^k$. Show that $E$ is compact if and only if every sequence in $E$ has a subsequence converging to a point in $E$.

Solution

By the Heine–Borel Theorem, $E$ is compact iff it is closed and bounded.

($\Rightarrow$) Let $E$ be compact and $(x^{(n)})$ a sequence in $E$. It is bounded, so by Bolzano–Weierstrass it has a subsequence converging to some $x \in \mathbb{R}^k$. Since $E$ is closed, $x \in E$ by Proposition 13.9(b).

($\Leftarrow$) Assume every sequence in $E$ has a subsequence converging to a point of $E$. Bounded: if not, pick $x^{(n)} \in E$ with $\max_j|x^{(n)}_j| > n$; every subsequence is unbounded, hence not convergent, contradicting the hypothesis. Closed: let $x^{(n)} \in E$ converge to $x$. By hypothesis some subsequence converges to a point of $E$; but every subsequence converges to $x$, so $x \in E$. By Proposition 13.9(b), $E$ is closed. Hence $E$ is compact.

Tags: ChallengeOpen, closed, compactSubsequences

Used in Exercise 13.14

Exercise 13.12

Let $(S, d)$ be any metric space.

1. Show that if $E$ is a closed subset of a compact set $F$, then $E$ is compact.
2. Show that a finite union of compact sets in $S$ is compact.

Hint

(a) Given an open cover of $E$, add the open set $S \setminus E$ to get an open cover of $F$.

Solution

1. Let $\mathcal{U}$ be an open cover of $E$. Since $E$ is closed, $S \setminus E$ is open, and $\mathcal{U} \cup \{S \setminus E\}$ covers $F$ (points of $F$ in $E$ are covered by $\mathcal{U}$, the others lie in $S \setminus E$). By compactness of $F$ a finite subfamily covers $F$. Discarding $S \setminus E$, which contains no point of $E$, leaves a finite subfamily of $\mathcal{U}$ that covers $E$.
2. Let $F_1, \dots, F_n$ be compact and $\mathcal{U}$ an open cover of $F_1 \cup \cdots \cup F_n$. Then $\mathcal{U}$ covers each $F_i$, so some finite $\mathcal{U}_i \subseteq \mathcal{U}$ covers $F_i$. The finite family $\mathcal{U}_1 \cup \cdots \cup \mathcal{U}_n$ covers the union.

Tags: StandardOpen, closed, compact

Used in Theorem 13.12

Exercise 13.13

Let $E$ be a compact nonempty subset of $\mathbb{R}$. Show $\sup E$ and $\inf E$ belong to $E$.

Solution

A compact subset of $\mathbb{R}$ is closed and bounded (Heine–Borel), so $s = \sup E$ is a real number. By Exercise 11.11 there is a sequence in $E$ converging to $s$ (take $s_n = s$ if $s \in E$). Since $E$ is closed, Proposition 13.9(b) gives $s \in E$. For the infimum, apply the same argument to the compact set $-E = \{-x : x \in E\}$, or argue symmetrically.

Tags: StandardOpen, closed, compactSup & inf

Exercise 13.14

Let $E$ be a compact nonempty subset of $\mathbb{R}^k$ and $\delta = \sup\{d(x, y) : x, y \in E\}$. Show $E$ contains points $x_0, y_0$ with $d(x_0, y_0) = \delta$.

Solution

$E$ is bounded, so $\delta < \infty$. For each $n$ choose $x_n, y_n \in E$ with $d(x_n, y_n) > \delta - \frac1n$. By compactness (Exercise 13.11), a subsequence $(x_{n_k})$ converges to some $x_0 \in E$, and then a further subsequence of $(y_{n_k})$ converges to some $y_0 \in E$; relabel so that $x_{n_k} \to x_0$ and $y_{n_k} \to y_0$ along one subsequence. By the triangle inequality $|d(x,y) - d(x',y')| \le d(x,x') + d(y,y')$, so $d(x_{n_k}, y_{n_k}) \to d(x_0, y_0)$. But $\delta - \frac1{n_k} < d(x_{n_k}, y_{n_k}) \le \delta$, so these distances also tend to $\delta$. Hence $d(x_0, y_0) = \delta$.

Tags: StandardOpen, closed, compactSubsequences

Exercise 13.15

Let $(B, d)$ be as in Exercise 13.3, and let $F$ consist of all $x \in B$ with $\sup\{|x_j|\} \le 1$.

1. Show $F$ is closed and bounded. (A set $F$ in a metric space is bounded if $F \subseteq \{s : d(s, s_0) \le r\}$ for some $s_0 \in S$ and $r > 0$.)
2. Show $F$ is not compact.

Hint

(b) For $x \in F$ let $U(x) = \{y \in B : d(y, x) < 1\}$, and consider the cover of $F$ by all $U(x)$. Let $x^{(n)}$ have $x_n^{(n)} = -1$ and $x_j^{(n)} = 1$ for $j \ne n$. Distinct $x^{(n)}$ are at distance $2$, so no $U(x)$ contains two of them.

Solution

1. Bounded: with $0$ the zero sequence, $d(x, 0) = \sup|x_j| \le 1$ for $x \in F$. Closed: let $x^{(n)} \in F$ and $x^{(n)} \to x$ in $B$. For each $j$, $|x_j| \le |x^{(n)}_j| + d(x^{(n)}, x) \le 1 + d(x^{(n)}, x)$, and letting $n \to \infty$ gives $|x_j| \le 1$. So $x \in F$, and $F$ is closed by Proposition 13.9(b).
2. The open balls $U(x)$, $x \in F$, cover $F$ since $x \in U(x)$. Define $x^{(n)} \in F$ by $x^{(n)}_n = -1$ and $x^{(n)}_j = 1$ for $j \neq n$. For $m \ne n$, $d(x^{(n)}, x^{(m)}) \ge |x^{(n)}_n - x^{(m)}_n| = |-1 - 1| = 2$. If some $U(x)$ contained two of them, say $x^{(n)}$ and $x^{(m)}$, then $d(x^{(n)}, x^{(m)}) \le d(x^{(n)},x) + d(x,x^{(m)}) < 2$, a contradiction. So each $U(x)$ contains at most one $x^{(n)}$, and no finite subfamily can cover the infinitely many distinct points $x^{(n)}$. Thus $F$ is closed and bounded but not compact.

Tags: ChallengeOpen, closed, compact