- Problem Statement
- Using the Bolzano-Weierstrass theorem (for every bounded sequence, there exists a convergent subsequence), prove that a positive derivative implies a growing function.

- Relevant Equations
- The Bolzano-Weierstrass theorem: Every bounded sequence ##c_n## has a convergent subsequence ##c_{n_k}##.

I'm stuck on a proof involving the Bolzano-Weierstrass theorem. Consider the following statement:

$$f'(x)>0 \ \text{on} \ [a,b] \implies \forall x_1,x_2\in[a,b], \ f(x_1)<f(x_2) \ \text{for} \ x_1<x_2 $$ i.e. a positive derivative over an interval implies that the function is growing over the interval. I've been instructed to proceed with a proof by contradiction, yet am slightly confused over how one would start such a proof. I would assume ##f'(x)>0## over a given interval but then also assume the logical negotiation of what that implies, correct? In other words, that the function is then decreasing over the interval and show that this leads to a contradiction. I can also somewhat see how one would go about proving it... ##\exists \ x_3 \in[x_1,x_2]## ...but don't see how the Bolzano Weierstrass theorem comes into play.

$$f'(x)>0 \ \text{on} \ [a,b] \implies \forall x_1,x_2\in[a,b], \ f(x_1)<f(x_2) \ \text{for} \ x_1<x_2 $$ i.e. a positive derivative over an interval implies that the function is growing over the interval. I've been instructed to proceed with a proof by contradiction, yet am slightly confused over how one would start such a proof. I would assume ##f'(x)>0## over a given interval but then also assume the logical negotiation of what that implies, correct? In other words, that the function is then decreasing over the interval and show that this leads to a contradiction. I can also somewhat see how one would go about proving it... ##\exists \ x_3 \in[x_1,x_2]## ...but don't see how the Bolzano Weierstrass theorem comes into play.