Compactness in Metric Spaces: Examples and Counterexamples

$d$ is a metric: (M1) holds by definition; (M2) because the condition $x \ne y$ is symmetric; and (M3) because $d(x,z)$ is $0$ or $1$, and when it is $1$ one has $x \ne z$, so $y$ differs from at least one of $x$ and $z$ and $d(x,y) + d(y,z) \ge 1$.

For $0 < r \le 1$, $d(x,y) < r \le 1$ forces $d(x,y) = 0$ and $y = x$, so $B(x,r) = \{x\}$; for $r > 1$ every $y$ has $d(x,y) \le 1 < r$, so $B(x,r) = S$. Hence every $A \subseteq S$ is open, each $x \in A$ having $B(x,1) = \{x\} \subseteq A$, and every subset is closed as well, its complement being open: claim 1.

Claim 2: let $(x_k)$ be Cauchy and take $K$ with $d(x_m,x_n) < 1$ for $m,n \ge K$; then $d(x_m,x_K) = 0$, that is $x_m = x_K$, for every $m \ge K$, so $d(x_k, x_K) = 0 < \varepsilon$ for $k \ge K$ and every rational $\varepsilon > 0$, and $x_k \to x_K \in S$.

For claim 3, suppose $S$ is finite. If $S = \emptyset$ then the empty subfamily of any family covers it; otherwise list $S = \{s_0, \dots, s_p\}$, let $\mathcal{U}$ be a family of open sets with union $S$, and note that for each $i \le p$ the set $\{U \in \mathcal{U} : s_i \in U\}$ is nonempty, so finite choice supplies $U_0, \dots, U_p \in \mathcal{U}$ with $s_i \in U_i$; their union contains every $s_i$ and hence is $S$. So $(S,d)$ is compact.

If $(S,d)$ is compact it is totally bounded.

If $(S,d)$ is totally bounded, take a finite $(1/2)$-net $F$; then $S = \bigcup_{y \in F} B(y,1/2) = \bigcup_{y \in F} \{y\} = F$ by step 2.1, so $S$ is finite.

Steps 3.1, 4.1 and 5.1 close the cycle finite $\Rightarrow$ compact $\Rightarrow$ totally bounded $\Rightarrow$ finite, so the three conditions of claim 3 are equivalent.

Take a natural $m \ge 1$ with $1/m < \varepsilon/(2M\iota(n))$, which exists because $2M\iota(n) > 0$, and put $h := 2M/m > 0$, so that $\iota(n) h = 2M\iota(n)/m < \varepsilon$.

Let $G$ be the set of points of $\mathbb{R}^n$ each of whose coordinates is $-M + jh$ for some natural $j \le m$; every such point lies in $Q$, since $0 \le jh \le mh = 2M$ gives $-M \le -M + jh \le M$.

$G$ is finite: writing a natural $t < (m+1)^n$ in base $m+1$ gives digits $t_0, \dots, t_{n-1}$, each a natural $\le m$, and $t \mapsto$ the point with $k$-th coordinate $-M + t_k h$ is a function from the natural number $(m+1)^n$ onto $G$, so $G$ is listed by that function.

Let $y \in Q$ and $k < n$; the set of naturals $j \le m$ with $y_k \le -M + jh$ is nonempty, containing $m$ because $y_k \le M = -M + mh$, so it has a least element $j_k$.

Then $|y_k - (-M + j_k h)| \le h$: if $j_k = 0$ then $y_k \le -M$ and also $y_k \ge -M$, so the difference is $0$; and if $j_k \ge 1$ then minimality gives $y_k > -M + (j_k-1)h$, so $-h < y_k - (-M+j_k h) \le 0$.

Writing $g$ for the point of $G$ with $k$-th coordinate $-M + j_k h$, step 4.1 gives $d_\infty(y,g) \le h$, hence $d_2(y,g) \le \iota(n)h < \varepsilon$ by step 1.1, so $y$ lies in the ball of radius $\varepsilon$ about $g$ in the subspace $Q$.

So $G$ is a finite $\varepsilon$-net for $Q$; as $\varepsilon > 0$ was arbitrary, $Q$ is totally bounded.

Let $(U_i)_{i \in I}$ be a family of open subsets of $X$ with $K \subseteq \bigcup_{i \in I} U_i$.

Since $p \in K$ there is $i^{\ast} \in I$ with $p \in U_{i^{\ast}}$, and openness gives a real $r > 0$ with $B(p,r) \subseteq U_{i^{\ast}}$.

Taking a positive rational below $r$, for instance $1/N$ with $N \ge 1$ a natural and $1/N < r$, convergence supplies $K_0 \in \mathbb{N}$ with $d(x_k,p) < 1/N < r$ for every $k \ge K_0$, so $x_k \in B(p,r) \subseteq U_{i^{\ast}}$ for every $k \ge K_0$.

For each $j < K_0$ the set $\{\, i \in I : x_j \in U_i \,\}$ is nonempty, since $x_j \in K$; finite choice applied to $j \mapsto$ that set gives indices $i_0, \dots, i_{K_0-1} \in I$ with $x_j \in U_{i_j}$ for every $j < K_0$, this list being empty when $K_0 = 0$.

Then $K \subseteq U_{i^{\ast}} \cup U_{i_0} \cup \dots \cup U_{i_{K_0-1}}$: the point $p$ and every $x_k$ with $k \ge K_0$ lie in $U_{i^{\ast}}$ by steps 2.1 and 3.1, and every $x_j$ with $j < K_0$ lies in $U_{i_j}$. So finitely many members of the family cover $K$, and $K$ is compact.

For the instance in $\mathbb{R}$, the terms $1/(k+1)$ satisfy $|1/(k+1) - 0| = 1/(k+1) < \varepsilon$ for every $k$ with $k+1 \ge N$, where $N \ge 1$ is a natural with $1/N < \varepsilon$; so $1/(k+1) \to 0$ and $\{0\} \cup \{1/(k+1) : k \in \mathbb{N}\}$ is compact by the general claim.

The map $f : K \to \mathbb{R}$, $f(u) := d(x,\{u\}) = d(x,u)$, is the restriction to $K$ of $u \mapsto d(u,\{x\})$, which is Lipschitz with constant $1$ and hence continuous; $K$ with the restricted metric is a nonempty compact metric space.

So $f$ attains a least value at some $a \in K$: $d(x,a) \le d(x,u)$ for every $u \in K$.

Hence $d(x,a)$ is a lower bound of $\{d(x,u) : u \in K\}$ that belongs to the set, so it is the infimum: $d(x,a) = d(x,K)$, which is claim 1.

For claim 2, the map $g : K \to \mathbb{R}$, $g(u) := d(u,L)$, is continuous by the same argument, so it attains a least value at some $b \in K$.

$g(b) > 0$: otherwise $d(b,L) = 0$ would put $b$ in the closure of $L$, which equals $L$ because $L$ is compact and hence closed, contradicting $K \cap L = \emptyset$.

$d(K,L) = g(b)$: every $u \in K$ and $v \in L$ satisfy $d(u,v) \ge d(u,L) \ge g(b)$, so $g(b)$ is a lower bound of $\{d(u,v) : u \in K, v \in L\}$; and $d(K,L) \le d(b,v)$ for every $v \in L$, so $d(K,L)$ is a lower bound of $\{d(b,v) : v \in L\}$ and therefore $d(K,L) \le d(b,L) = g(b)$.

Combining, $d(K,L) = g(b) > 0$ and the value is attained at $b \in K$: claim 2.

$U$ and $V$ are open in $[0,1]$, being the traces of $(-1,2/3)$ and $(1/3,2)$, and $U \cup V = [0,1]$, because a point of $[0,1]$ is either below $2/3$, and then in $U$, or at least $2/3 > 1/3$, and then in $V$.

Let $A \subseteq [0,1]$ be nonempty with $\operatorname{diam}(A) < 1/3$, and suppose first that every $y \in A$ satisfies $y < 2/3$; then $A \subseteq U$.

Otherwise some $a \in A$ has $a \ge 2/3$, and then every $y \in A$ satisfies $y \ge a - |a - y| \ge a - \operatorname{diam}(A) > 2/3 - 1/3 = 1/3$, so $A \subseteq (1/3,1] = V$.

In both cases $A$ lies in a single member of $\{U,V\}$, so $1/3$ is a Lebesgue number: claim 1.

For claim 2 let $\delta' > 1/3$ be real and put $A := [1/3, 2/3]$, a nonempty subset of $[0,1]$ with $|u-v| \le 1/3$ for $u,v \in A$ and with $|2/3 - 1/3| = 1/3$ attained, so $\operatorname{diam}(A) = 1/3 < \delta'$.

$A \not\subseteq U$ because $2/3 \in A$ and $2/3 \notin [0,2/3)$, and $A \not\subseteq V$ because $1/3 \in A$ and $1/3 \notin (1/3,1]$; so no $\delta' > 1/3$ is a Lebesgue number for this cover, and $1/3$ is the largest one: claim 2.

Each $e_n$ is a bounded function $\mathbb{N} \to \mathbb{R}$, its range being contained in $\{0,1\}$, so $e_n \in \mathcal{B}(\mathbb{N})$; and $d_\infty(e_n,\mathbf{0}) = \sup\{|e_n(s)| : s \in \mathbb{N}\} = 1$, so $e_n \in K$.

For $m \ne n$ the set $\{|e_m(s) - e_n(s)| : s \in \mathbb{N}\}$ is $\{0,1\}$, since the value is $1$ at $s = m$ and at $s = n$ and $0$ elsewhere; hence $d_\infty(e_m,e_n) = 1$.

$K$ is the closed ball $\bar B(\mathbf{0},1)$, hence closed in $\mathcal{B}(\mathbb{N})$, and it is contained in $B(\mathbf{0},2)$, hence bounded.

Suppose $K$ had a finite $1/2$-net $F$; then $F$ is empty or listable as $\{y_0, \dots, y_p\}$, and it is not empty because $K$ contains $e_0$. Define $\varphi : \mathbb{N} \to \sigma(p)$ by letting $\varphi(n)$ be the least $i \le p$ with $d_\infty(e_n, y_i) < 1/2$, which exists because the $1/2$-balls about the $y_i$ cover $K$ and $e_n \in K$.

$\varphi$ is injective: if $\varphi(m) = \varphi(n) = i$ with $m \ne n$, then $d_\infty(e_m,e_n) \le d_\infty(e_m,y_i) + d_\infty(y_i,e_n) < 1/2 + 1/2 = 1$, contradicting step 2.1.