Topological Definition of Compactness implies Metric Definition for R

Question

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 $R^1$)

Thanks!

Answer 1

(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.