Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 2, Problem 10

Problem

Let $X$ be an infinite set. For $p \in X$ and $q \in X$, define

$$ d(p,q) = \begin{cases} 1 \;\; (\text{if $p \neq q$}) \\ 0 \;\; (\text{if $p = q$}). \end{cases} $$

Prove that this is a metric. Which subsets of the resulting metric space are open? Which are closed? Which are compact?

Answer

(1) We first show that $d$ is a metric.

(a) By the definition, $d(p,q)>0$ if $p \neq q$ and $d(p,p)=0$.

(b) If $p \neq q$ then $d(p,q)=1=d(q,p)$. If $p = q$ then $d(p,q) = 0 = d(q,p)$.

(c) If $p \neq q$ then $d(p,q) = 1 \leq d(p,r) + d(r,q)$ because one of $d(p,r)$ and $d(r,q)$ is 1. If $p = q$ then $d(p,q) = 0 \leq d(p,r) + d(r,q)$ for any $r \in X$.

(2) For any $A \subset X$, $N_{1/2}(p) = \{ p \} \subset A$ for all $p \in A$, which means every subset of $X$ is open.

(3) Every subset of $X$ has an open complement, and therefore is closed.

(4) Finite set is obviously compact. For infinite set $A \subset X$, an open cover $\{N_{1/2}(x)\}_{x \in A}$ of $A$ does not have a finite subcover, which means $A$ is not compact. Therefore, only finite subsets of $X$ are compact.

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