Let $(X, d)$ be a metric space, and $K$ an infinite subset of $X$. Show that $K$ is not compact in $(X, d)$.

From what I understand, a set of compact if every open cover of $K$ can be reduced to a finite subcover. Since $K$ is infinite, how can you construct a finite cover? Since you can't cover an "infinite amount of things" with a "finite amount of things," does this make $K$ not compact?

Or, should I take the "compactness $\implies$ closed and bounded" approach? Are all infinite sets not closed and unbounded (since I know $\mathbb{Z}$ is such in $\mathbb{R}$)?

  • 1
    $\begingroup$ There must be more to it... $[0,1]$ is compact but is still infinite. $\endgroup$
    – Clayton
    Apr 10, 2018 at 22:09
  • $\begingroup$ Should "$K$ an infinite subset" read "$K$ an unbounded subset"? $\endgroup$
    – Rob Arthan
    Apr 10, 2018 at 22:17
  • $\begingroup$ @RobArthan, no, it says "$K$ an infinite subset" on my paper. $\endgroup$
    – Axioms
    Apr 10, 2018 at 22:19
  • $\begingroup$ @Axioms Should "a metric space" read "a metric space with discrete metric"? $\endgroup$
    – Aloizio Macedo
    Apr 10, 2018 at 22:20
  • $\begingroup$ @AloizioMacedo no, the question reads exactly the same from the source. $\endgroup$
    – Axioms
    Apr 10, 2018 at 22:25

1 Answer 1


This is not true. Take $[0,1]\subseteq \mathbf{R}$. This is a compact subset of $\mathbf{R}$, being closed and bounded (Heine-Borel Theorem). However, its cardinality is infinite (in fact uncountable). Also, if you don't believe the statement that a finite cover of an infinite set can be constructed, consider the following construction:

A cover in this context is a collection $\mathcal{U}=\{U_i\}$ of open neighborhoods of the space $X$ so that $X=\bigcup_{U_i\in \mathcal{U}} U_i$. In the case of $[0,1]$, we can construct such an open cover by taking $U_1=[0,\frac{1}{2})$, $U_2=(0,1)$ and $U_3=(\frac{1}{2}, 1]$. These are all open subsets of $[0,1]=X$. Further, we can see that $U_1\cup U_2\cup U_3=X$. Finally, the number of elements of the cover is $3$, and so we have constructed a finite open cover of an uncountable set. A more trivial way to do this is just to take the whole space $X$ itself.


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