The Bolzano-Weierstrass theorem is a fundamental result in real analysis and states that each infinite bounded real sequence has a convergent subsequence. Also, it has some applications in data science and economics.

In order to prove the Bolzano-Weierstrass theorem, first we need to prove a result called the monotone subsequence theorem. Note that the monotone subsequence theorem states that each infinite bounded sequence has a monotone subsequence.

#### The monotone subsequence theorem

To prove the monotone subsequence theorem, first we need to define the concept of a peak in real sequences. Let \((x_n)\) be a real sequence. By definition, the \(m\)th term \(x_m\) is a peak if for all \(n \geq m\), we have \(x_m \geq x_n\).

**Result** (the monotone subsequence theorem). Every real sequence has a monotonic subsequence.

The proof is as follows:

Let \((x_n)\) be a real sequence. We distinguish two cases:

- Assume that \((x_n)\) has infinitely many peaks. Imagine the peaks are \((x_{n_i})^{+\infty}_{i=1}\) where \(n_i < n_{i+1}\). So, $$x_{n_1} \geq x_{n_2} \geq \cdots \geq x_{n_k} \geq \cdots$$ is a decreasing subsequence of \((x_n).\)
- Assume that the only peaks of the sequence \((x_n)\) are $$x_{n_1}, x_{n_2}, \dots, x_{n_k},$$ where $$n_1 < n_2 < \cdots < n_k.$$ Let \(s_1 = n_k + 1\). So, \(x_{s_1}\) is not a peak which means that there is an index \(s_2\) such that \(s_1 < s_2\) and \(x_{s_1} < x_{s_2}\). Again since \(x_{s_2}\) is not a peak, there is an index \(s_3\) such that \(s_2 < s_3\) and \(x_{s_2} < x_{s_3}\). If we continue the process, we obtain an increasing subsequence.

Hence in any case, the real sequence \((x_n)\) has a monotonic subsequence, as required.

### The Bolzano-Weierstrass theorem

Now, we are in the position to prove the following result:

**Result** (The Bolzano-Weierstrass theorem). Each infinite bounded real sequence has a convergent subsequence.

For proving the Bolzano-Weierstrass theorem, assume that \((a_n)\) is an infinite bounded real sequence. By the monotone subsequence theorem, \((a_n)\) has a monotonic subsequence. Any subsequence of \((a_n)\) is bounded because the sequence \((a_n)\) is itself bounded. Now, by the monotone convergence theorem, the monotone subsequence of \((a_n)\) is convergent because it is bounded. Thus \((a_n)\) has a convergent subsequence and the proof is complete.

#### An application of the Bolzano-Weierstrass theorem

Finally, we use the Bolzano-Weierstrass theorem to prove the “boundedness theorem”:

**Result**. Let a function \(f : [a,b] \rightarrow \mathbb{R}\) be continuous. Then, \(f\) is bounded, i.e. there is a positive real number \(M\) such that $$\vert f(x) \vert \leq M \text{ for all } x \in [a,b].$$

The proof is as follows:

On the contrary, assume that \(f\) is not bounded. Then, for each positive integer \(n\), there is an \(x_n\) in the closed interval \([a,b]\) such that $$\vert f(x_n) \vert \geq n.$$ By the Bolzano-Weierstrass theorem, the real sequence \((x_n)\) has an infinite convergent subsequence. However, by a result proved in the post on convergent real sequences, any infinite subsequence of \((x_n)\) cannot be bounded, contracting the existence of an infinite convergent subsequence for \((x_n)\). Thus \(f\) is bounded and the proof is complete.

**Remark**. A general form of the Bolzano-Weierstrass theorem has some applications in general equilibrium theory which discusses the mathematical economic theory of price determination and resource allocation.