Can anyone help me with making the logical progression for the equivalence of these two definitions for compactness?

Topological Definition: A topological space $X$ is said to be compact if for each open covering $\{U_\alpha\}$, $\alpha \in I$, there is a finite subcovering $\{U_\beta\}$, $\beta \in J$.

(This is the definition I am used to, at least)

Now, in the book "Elementary Real and Complex Analysis" by Shilov, he gives the definition of compactness for a metric space as so:

A metric space $M$ is said to be compact if every sequence in $M$ has a limit point that is also in $M$.

I found the metric definition to be a bit odd because it is so similar to the definition of completeness.

I am assuming that these two definitions are logically equivalent, but I am not sure why. My first assumption is that Shilov is taking advantage of the Hausdorff property of the real number line, and that any compact subset of a Hausdorff space is closed. From this, by Bolzano-Weierstrass thm, we also know that any closed interval (which could be considered the closure of a neighborhood around the point $\frac{a+b}{2}$, where the interval is $[a,b]$ ) has a limit point lying within our interval. This is somewhat similar to the definition Shilov gives for locally compact:

A set is said to be locally compact if every point of $M$ has a neighborhood whose closure is compact.

Long story short: How can we deduce these definitions for metric compactness from the topological definitions? (Also we are assuming the standard Euclidean metric, or just $d(x,y) = |x-y|$ if we are in $R^1$)


  • $\begingroup$ This distinction is sometimes called "sequential compactness." The Wikipedia page covers it: en.wikipedia.org/wiki/Sequentially_compact_space $\endgroup$ – Thomas Andrews Sep 5 '15 at 23:42
  • $\begingroup$ Good call. I see there's a term called "Limit point compact" which has a virtually identical definition to the one given. In general, though, does the metric definition imply the topological definition? Or vice versa? It says that compactness and sequential compactness are not equivalent for a general topological space, but are for metric spaces. $\endgroup$ – Rellek Sep 5 '15 at 23:48

(Partial answer.)

This alternative definition is often called "sequential compactness." It is equivalent to topological compactness for metric spaces, but not more generally.

Topological Compactness implies Sequential Assume that $a_1,a_2,a_3,\dots,a_n,\dots$ has no sub-sequence which converges in the metric space $X$.

Then, for each $a\in X$, there is an $\epsilon_a$ so that $|a-a_i|>\epsilon$ for all $i$, except when $a=a_i$. Then $X=\bigcup N_{\epsilon_a}(a)$. But there is no finite sub-cover, since the only $N_{\epsilon_a}(a)$ containing $a_i$ is $a=a_i$. So infinitely many of these open sets are needed to cover $X$.

Basically, a sequence without a convergent sub-sequence must yield an open cover without a finite sub-cover.

Sequential Compactness implies Topological

[Still working on this, sorry.]

It's pretty easy to show that a countable open cover must have a finite sub-cover. If $U_1,U_2,\dots$ is an open cover without a finite sub-cover, then choose $x_n\not\in \bigcup_{i=1}^{n} U_i$ for each $n$. Since $X\setminus U_i$ is closed, then a limit point of $x_1,\dots,$ must be in $\bigcap_{i=1}^{\infty} (X\setminus U_i)$. But $\bigcap (X\setminus U_i)=X\setminus (\bigcup U_i)= X\setminus X=\emptyset$.

Not sure how to extend this to uncountable open covers. It seems like if you were able to show that you could combine enough of the $U_i$ to make the resulting cover countable, you are done, but I'm missing the trick to do that.

I think it is that sequentially compact implies there is a countable basis.

  • $\begingroup$ Thanks. Also, since the article you linked says that sequential compactness $\neq$ compactness for a general topological space, perhaps it isn't possible to prove in the opposite direction? At least not without the real number/metric space axioms. $\endgroup$ – Rellek Sep 5 '15 at 23:52
  • $\begingroup$ It says it is eequivalent for metric spaces, which is what I'm trying to prove. $\endgroup$ – Thomas Andrews Sep 5 '15 at 23:54
  • $\begingroup$ Pretty slick for the Sequential Compactness $\rightarrow$ Topological. Does density of the rationals not allow the countable basis we need? $\endgroup$ – Rellek Sep 6 '15 at 0:01
  • $\begingroup$ Not in the general metric space - you need a subset of the points and a subset of the radiuses in a metric space. We don't have immediately a countably dense subset of the points of $X$ in general. It's enough for $X\subseteq R$, however. @Rellek $\endgroup$ – Thomas Andrews Sep 6 '15 at 0:03
  • $\begingroup$ Sorry that is my bad I am way too used to defining metrics only on the reals... $\endgroup$ – Rellek Sep 6 '15 at 0:06

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