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.


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.