I am struggling with the idea that all compact subsets of a metric space are closed after reading chapter 2 of Rudin's Principles of Mathematical Analysis.

The reason I am confused is that it seems unreasonable that a compact subset must be closed. If we were to take an open subset $K$ of a compact subset $Y$ in a metric space $X$, where $K$ is open relative to $X$, then every finite subcover of $Y$ would also be a finite subcover of $K$. But then, $K$ would be both compact and open in $X$, which contradicts the theorem that all compact subsets in $X$ are closed.

Could someone please explain to me why my reasoning is incorrect and how it is that compact subsets of metric spaces must be closed?

Thank you,


  • $\begingroup$ In addition to what the answers have pointed out, it seems you are making the popular mistake of thinking that open and closed are opposites. They are not. A set can be open, closed, both, or neither. $\endgroup$ – dfeuer Dec 8 '13 at 7:20

The problem with your reasoning is that $K$ may have an open cover $\mathscr{U}$ that does not arise from an open cover of $Y$, so that compactness of $Y$ doesn’t tell you anything about $\mathscr{U}$.

It’s easiest, I think, to prove the contrapositive: if $A\subseteq X$ is not closed, then $A$ is not compact. Let $\langle X,d\rangle$ be a metric space, and suppose that $A\subseteq X$ is not closed. Then there is a point $p\in(\operatorname{cl}A)\setminus A$. For $n\in\Bbb Z^+$ let $$U_n=\left\{x\in X:d(x,p)>\frac1n\right\}\;,$$ and show that $\{U_n:n\in\Bbb Z^+\}$ is an open cover of $A$ with no finite subcover.

  • $\begingroup$ Thank you for your explanation! I understand it now. $\endgroup$ – Evan Phibbs Dec 8 '13 at 20:00
  • $\begingroup$ @Evan: You’re welcome! $\endgroup$ – Brian M. Scott Dec 8 '13 at 20:06

Let $F$ be our compact set. Then take $x\in X\setminus F$. Because {$x$} and $F$ are compact ,then they have a distance,say $d$.And thus the open disk $D(x,\frac {d}{2})\subset X\setminus F$. So $X\setminus F$ is open and thus $F$ is closed.


Alternate proof if you know that compactness is equivalent to sequential compactness: suppose $Y$ is compact, and $x$ is a limit point of $Y$. Let $(x_n) \subset Y$ with $x_n \to x \in X$ and for all $n$, $x_n \neq x$. $(x_n)$ has a convergent subsequence converging to some $y \in Y$ because $Y$ is compact. But since the whole sequence $(x_n)$ is convergent and converges to $x$, $x=y$. So $x_n \to y \in Y$. $Y$ contains all its limit points. $Y$ is closed.


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