# Topological Proof of Bolzano-Weierstrass Without Constructing Sequences

Published on 7 November 2014

The Bolzano-Weierstrass theorem is a common theorem taught in introductory real analysis and topology courses.

All of the proofs that I’ve read for the Bolzano-Weierstrass theorem involve constructing sequences. In my introductory topology course, I wanted to see if I could write a simple proof of the theorem without having to explicitly construct any sequences.

(It’s worth noting that the proof for compact $$\implies$$ sequentially compact requires constructing sequences. If we take this as given, then we get to avoid sequences.)

## Theorem (Bolzano-Weierstrass)

Any bounded sequence in $$\mathbb{R}^n$$ has a convergent subsequence.

## Proof

We will make use of the following propositions, which I will assume to be previously proved. The proofs of these are in a standard topology textbook (I’m reading from Munkres, 2nd edition).

I’ll redefine various notions of compactness below, since there exists some ambiguity in the different forms of compactness.

Definition. A topological space is sequentially compact if every infinite sequence has a convergent subsequence.

Proposition 1. The interval $$[a, b] \subset \mathbb{R}$$ is compact.

Proposition 2. The product of compact spaces is compact.

In fact, this proposition true in both the infinite case (Tychonoff’s theorem) and the finite case.

Proposition 3. (Munkres, Theorem 28.2) $$X$$ is compact $$\implies$$ $$X$$ is sequentially compact.

We now prove Bolzano-Weierstrass.

Let $$\{ x_i \}$$ be a bounded sequence in $$\mathbb{R}^n$$. We need to show that $$\{ x_i \}$$ has a convergent subsequence.

Since the sequence is bounded in $$\mathbb{R}^n$$, by definition of bounded, every element of the sequence is contained in some closed interval of $$\mathbb{R}^n$$:

$A = [m, M]^n$

where $$m, M$$ denote the “minimum” and “maximum” of our bounded interval.

Note that $$[m, M]$$ is compact (Proposition 1).

It follows that $$A$$ is compact – it is the product of compact spaces (Proposition 2).

Because $$A$$ is compact, $$A$$ is sequentially compact (Proposition 3).

In a sequentially compact space, any sequence of points has a convergent subsequence. Note that $$\{ x_i \}$$ is one such sequence contained in the sequentially compact space $$A$$, and it follows that $$\{ x_i \}$$ must have a convergent subsequence.