Show $A$ is compact subset of a metric space $(X,\mathscr T, d)$ only if for all $x \in X$, $d(x,A)=d(x,a)$ for some $a \in A$.

Question

Topology by James Munkres Exer27.2b

This exercise has been previously asked about, and while some of the proofs are similar or the same as mine, they're outlines left for the reader.

I have 3 proofs which I guess are either wrong or inelegant. Please point out and explain any errors in them. I put the propositions cited at the end.

The outlines of my proofs are as as follows:

1. Outline of Pf 1: The set $\{d(x,a)\}_{a \in A}$ is a compact subspace of $\mathbb R$ and thus is closed in $\mathbb R$ and thus has a minimum in $\mathbb R$.

2. Outline of Pf 2: The set $\{d(x,a)\}_{a \in A}$ is closed in $\mathbb R$, deduced without proving it is a compact subspace of $\mathbb R$ and then thus, has a minimum in $\mathbb R$.

3. Outline of Pf 3: The set $\{d(x,a)\}_{a \in A}$ is a compact subspace of $\mathbb R$ and thus has a minimum in $\mathbb R$ without proving it is closed in $\mathbb R$.

The proofs are as follows:

---

(moved to answers)

---

Cor 17.7

Exer 18.11

Exer 20.3a

Thm 26.3

Thm 26.5

Exer 26.4

Exer 26.6

Heine Borel Theorem (Thm 27.3)

Extreme Value Theorem (Thm 27.4)

Answer 1

Pf 2:

By Exer 20.3(a) and Exer 18.11, for any $x \in X$, the map given by $d': A \to \mathbb R$ s.t. $d'(x,a)=d(x,a)$ is continuous in the variable $a$.

Because $A$ is compact and $\mathbb R$ is Hausdorff, we have by Exer 26.6 that $d'$ is a closed map. Thus, its image, $\{d(x,a)\}_{a \in A}$, is closed in $\mathbb R$.

Therefore $\{d(x,a)\}_{a \in A}$ contains its limit points by Cor 17.7 including its infimum, $d(x,A)$, as a minimum, namely, $d(x,a^{*})$ for some $a^{*} \in A.$

Thanks to freakish: $\{d(x,a)\}_{a \in A}$ is bounded below by $0$ and is closed, and therefore has a minimum, namely its infimum.

QED

Answer 2

Pf 1:

By Exer 20.3(a) and Exer 18.11, for any $x \in X$, the map given by $d': A \to \mathbb R$ s.t. $d'(x,a)=d(x,a)$ is continuous in the variable $a$.

Because $A$ is compact, we have that by Thm 26.5, $d'$s image, $\{d(x,a)\}_{a \in A}$, is compact subspace of $\mathbb R$ and thus, is closed by any of the following:

• Heine Borel Theorem (Thm 27.3)

• Thm 26.3

• Exer 26.4

Therefore $\{d(x,a)\}_{a \in A}$ contains its limit points by Cor 17.7 including its infimum, $d(x,A)$, as a minimum, namely, $d(x,a^{*})$ for some $a^{*} \in A.$

Thanks to freakish: $\{d(x,a)\}_{a \in A}$ is bounded below by $0$ and is closed, and therefore has a minimum, namely its infimum.

QED

Answer 3

Pf 3:

By Exer 20.3(a) and Exer 18.11, for any $x \in X$, the map given by $d': A \to \mathbb R$ s.t. $d'(x,a)=d(x,a)$ is continuous in the variable $a$.

Because $A$ is compact, we have that by Thm 26.5, $d'$s image, $\{d(x,a)\}_{a \in A}$, is compact subspace of $\mathbb R$.

Therefore, by the (argument that proves the) Extreme Value Theorem (Thm 27.4), $\{d(x,a)\}_{a \in A}$ has a mininum $d(x,a^{*})$ for some $a^{*} \in A$, and this minimum is its equal to its infimum, $d(x,A)$.

Thanks to freakish: $\{d(x,a)\}_{a \in A}$ is bounded below by $0$ and is closed, and therefore has a minimum, namely its infimum.

QED