Infinite metric space with discrete metric is not compact

Question

Consider the metric space $(X, d)$, where $d$ is the discrete metric; let $K$ be an infinite subset of $X$. Show that $K$ is not compact in $(X, d)$.

To show $K$ is not compact, I've done this:

Since $K$ is an infinite subset of $X$, it follows $K$ is an infinite discrete metric space. Consider $G = $ {{$k$} | $k \in K$}, which is an open cover of $K$. Clearly, $G$ has no finite subcover. Thus, $K$ is not compact.

How does my proof look? Do I need to explicitly show why $G$ has no finite subcover? Thank you.

Answer 1

Alternatively, you can prove this with sequences too.

Since your metric space is infinite, you can extract a countable set $\{a_{1}, \dots, a_{n}, \dots\}$ (with distinct points), thus $$ d(a_{i}, a_{j}) = 1 $$ for all $a_{i}, a_{j}$ (with $i \neq j$). This means that this sequence can't have any Cauchy subsequence.

An additional note - to someone who doesn't have much experience with compactness or the discrete metric - try to think of why this proof fails in a finite set. (Hint: can you extract a constant subsequence from every sequence with only finitely many distinct values?)

Answer 2

Any $H\subset G$ is equal to $\{\{x\}: x\in S_H\}$ for some $S_H\subset X.$ And $H$ is finite iff $S_H$ is finite. So if $H$ is a finite subset of $G$ then $\cup H=S_H \ne X$ because $S_H$ is finite while $X$ is infinite.