Bolzano-Weierstrass without a ruler

The Bolzano-Weierstrass theorem is often thought of as a critical piece of elementary education in real analysis. While its criticality is not up for debate, at least with me, its position as a theorem of real analysis certainly is! In fact, it is possible to derive the theorem by shear topological methods, and with a little help from a base idea of functional analysis- no metric required.

Ordinarily, the theorem is stated as the result that a sequence is compact (read: sequentially compact) iff it is closed and bounded. This, however, is not an appropriate statement on a generalized topological space which, while closed in many cases, does not offer a solid interpretation of boundedness (in particular, there is no concept of a "ball" in a general topological space, just neighbourhoods- which may be infinite). That means we're going to have to get a little creative. An alternative formulation to the above is the following, as presented in Wald's General Relativity:

Theorem: Balzano-Weierstrass - Let $(X, \mathscr{T})$ be a topological space and let $A \subset X$. If $A$ is compact, then every sequence $\{x_n\}$ of points in $A$ has an accumulation point lying in $A$. Conversely, if $(X, \mathscr{T})$ is second-countable, and every sequence in $A$ has an accumulation point in $A$, then $A$ is compact. Thus, if $(X, \mathscr{T})$ is second-countable, $A$ is compact if and only if every sequence in $A$ has a convergent subsequence whose limit lies in $A$.

For the uninitiated, there's quite a lot to unpack here. There are handy resources online for learning what all of these terms mean, and I encourage the reader to read those if they seek great detail. I will go over some of these definitions in brief. I am going to assume a passing familiarity with topology, and thus a pre-existing understanding of compactness and the properties of topological spaces.

We call a point $y$ an accumulation point or limit point of a sequence $\{x_n\}$ if any neighbourhood of $y$ contains an infinite number of points contained in $\{x_n\}$. As every neighbourhood of the limit itself contains the limit, the limit is automatically an accumulation point. It is not guaranteed in general that every accumulation point $y$ has a subsequence $\{y_n\}$ of $\{x_n\}$ such that $\{y_n\}$ converges to $y$, though it is possible if the space where these sequences lie is first-countable, which we define below.

A topological space $(X, \mathscr{T})$ is called second-countable if every open set in $X$ is expressible as a union of open sets from some countable set of open sets. i.e. where $A$, $\mathcal{O}_\alpha$ are open sets, $$ \forall A \subset X, A = \bigcup\limits_\alpha \mathcal{O}_\alpha$$ In this way, it is easy to see that second-countability implies first-countability, which is the notion that every point in $X$ is contained in at least one of a countable collection of open sets.

With the bookkeeping out of the way, we get to the proof of the theorem.

PROOF: we start in the forward direction: if $A$ is compact, then the sequences $\{x_n\}$ in $A$ have accumulation points in $A$. The definition of compactness of a set is that any open covering of that set must admit a finite subcovering. We attempt a proof by contradicton, in that we start with a comapct set set in which the seqeunces do not have accumulation points in the set. It is our hope that this leads to a regime in which $A$ cannot be compact, thus invalidating the assumption.

For each $a \in A$, we choose an open set $\ \mathcal{U}_a$ containing $a$ such that a sequence $\{x_n\}$ contains a finite amount of points in $\ \mathcal{U}_a$. $\{\mathcal{U}_a | a \in A \}$ is an open cover of $A$, namely the set $\{ \mathcal{U}_{a_1}, \ldots , \mathcal{U}_{a_n} \}$. However, if $\{ \mathcal{U}_{a_1}, \ldots , \mathcal{U}_{a_n} \}$ is an open cover, $\{x_n \} \subset \{ \mathcal{U}_{a_1}, \ldots , \mathcal{U}_{a_n} \}$. Then some $\mathcal{U}_{a_i}\}$ in $\{ \mathcal{U}_{a_1}, \ldots , \mathcal{U}_{a_n} \}$ contains infinitely many points of $\{x_n\}$ as the sequence $\{x_n\}$ is infinite. A contradiction! If there are to be finitely many sets covering $A$, these sets must contain accumulation points.

Ok, so that's step one. The second step is considerably more involved. We aim to prove that if $(X,\mathscr{T})$ is 2nd-countable, and if every sequence $\{x_n\}$ in $A$ has an accumulation point in $A$, that $A$ is compact. This will be done directly, though we will invoke an idea from a hitherto-unused part of mathematics.

First- if $(X,\mathscr{T})$ is 2nd-countable, it is first countable and that guarantees that every convergent sequence in $(X,\mathscr{T})$ has convergent subsequences- i.e. every accumulation point $y$ of a sequence $\{x_n\}$ may be written as the limit of a sequence $\{y_n\}$ such that $\{y_n\} \subset \{x_n\}$. Thus for what it's worth, we can say that the set $A$ is closed, as $A = \overline{A}$.

Now we need the concept of "nets", from functional analysis. A "net" is a function taking a "directional set" to a topological space. A directional set is a set with an implicit ordering, like the integers. As an example case, we say $A$ is a set with an index set on the elements of $A$, $\{n,n-1,\ldots,1\}$, such that elements follow the relation according to their index: $a_n \geq a_{n-1} \geq \ldots \geq a_1$. That is, in plain english, the last element is greater than or equal to the second-last, and so-on until you reach the terminal element, if applicable.

Say a net $\{X_{\lambda}\}_{\lambda \in \Lambda}$ has a "cluster point" $x$ if for every neighbourhood $\mathcal{O}$ of $X$ the index set $\{\lambda : X_{\lambda} \in \mathcal{O}\}$ is cofinal in $\Lambda$. $x$ is a cluster point of the net $\{X_{\lambda}\}_{\lambda \in \Lambda}$ iff some subset $\{X_{\gamma}\}_{\gamma \in \Gamma}$ converges to $x$. $x$ is an accumulation point of $A$ iff $\exists$ a net $X \notin \{X_{\lambda}\}_{\lambda \in \Lambda} \subset A$ that converges to $x$. $x \in X$ is an accumulation point of $A$ if $\mathcal{O}(x)$ contains infinitely many limit points. We have a lemma to prove:

Lemma: $A \subset X$ is countably compact iff every infinite subset has a limit point iff every sequence has a cluster point.

PROOF:We start with the last point — let $K$ be an infinite set, and suppose every sequence has a cluster point. Clearly, we may construct a sequence in $K$ such that its cluster point is a limit point in $K$. Now on the other hand, suppose that any infinite subset has an accumulation point. We take some infinite subset $k \in K$ : consider any distinct sequence $\{x_n\}$ with distinct elements. Any of these elements is a cluster point.

Now suppose $X$ is compact. Let $\{x_n\}$ be a sequence in $X$ and say $B_n = \overline{\{x_i\}}_{i=n}^{\infty}$ so that $\exists x \in \cap_{n=1}^{\infty}B_n$. Any neighbourhood of $x$, say $\mathcal{O}$, intersects infinitely many points of $\{x_n\}$ as otherwise $\cap_{n=1}^{\infty}B_n = \emptyset$. On the other hand, suppose every sequence has a cluster point, and let $B_n$ be a sequence of closed sets such that they satisfy the property of finite intersections. We choose any $x_n \in B_1 \cap \ldots \cap B_n$. Any cluster point of $\{x_n\}$, like $x$, is therefore in $\cap_{n=1}^{\infty}B_n$. $\blacksquare$

From our construction, we know that countable compactness implies limit point compactness, and due to first-countability, vice-versa. Due to second-countability, we know that every open covering must be made up of a countable union of open sets from some countable set of open sets- thus countable compactness implies compactness. This only works because our space is countable! Were it not for that fact, one might take some uncountable open covering of $X$ which would admit no finite subcovering. This acknowledgement concludes the proof. $\blacksquare$