Understanding the equivalence of Compactness to the Bolzano Weierstrass property in Metric spaces

convergence-divergencemetric-spaces

I have seen two type of definition of compactness:

Compactness (Topological) : Every open cover has finite sub cover

Also in a metric space: Sequential compactness/Bolzano Weierstrass/ Every sequence has a convergent subsequence

The concept as a whole is somewhat overwhelming to me, but the main point I am puzzled is, how exactly does this possibility of a sequence having a convergent subsequence or not depend on the coverability on a set level? Could someone give me a big picture intuition on this?

Reading the proof made me get lost in the details :c

Here are some resources which could be helpful for understanding compactness:

  1. This historical arxiv entry
  2. Michael Hardy's Answer is helpful in understanding why one should care about compactness in the first place

Best Answer

I understand that you don't want a proof (which can be found in many places and in at least a few variants) but you'd like some intuition.

Remember that the equivalence doesn't hold in any topological space, although it holds in metric spaces.

I like to think of compact spaces that compact spaces are small. We see this smallness in both of the definitions.

  • If we consider any sequence then since the space is somehow limited, it has to have an accumulation point so there are infinitely many points around some point. If the space if large then we have more space and we can avoid limit points.
  • Any open cover has a finite subcover. The space is so small then if we assign to any point its small neighbourhood, we can find a finitely many points such that their neighbourhoods cover the whole space.

To explain the equivalence a bit more let's start with the simpler implication, that compactness implies sequential compactness.

Fact. The sequence $(x_n)$ has a convergent subsequence is equivalent to the fact that the sequence has an accumulation point $x$, i.e. for any neighbourhood $U$ of $x$ there are infinitely many elements of the sequence that lies in $U$.

Proof(sketch). If $x$ is a limit of a subsequence then in any neighbourhood of $x$ there are almost all elements of the sequence, so infinitely many elements of the initial sequence lie in it. (Generally: Limits of sequences are accumulation points)

Conversely, assume that $x$ is an accumulation point of $(x_n)$. One can construct inductively an increasing sequence of indices $k_n\in\Bbb N$. Having defined $k_1<k_2<\ldots<k_n$ we define $k_{n+1}$ such that $x_{k_{n+1}}\in B(x,1/(n+1))$ and $k_{n+1}>k_n$ (it's possible since in $B(x,1/(n+1))$ there are infinitely many elements of the sequence. We see that $x_{k_n}\to x$. $\square$

Assume our sequence doesn't have an accumulation point. Therefore no points from $X$ are accumulation points, so each $x\in X$ has a n-hood $U_x$ such that only finitely many elements of the sequence belongs to $U_x$. The sets $U_x$ cover $X$ so we can find a finitely many of them that cover the whole space. We can use the pigeonhole principle to show that this leads to a contradiction. Namely, we covered the whole space with finitely many sets, each of them containing only finitely many elements of our infinite sequence.

The opposite implication is more difficult and is based on some properties of the metric spaces. You can use the fact that sequentially compact spaces are countably compact and Lindelof, which implies compactness.

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