# Heine theorem without subsequences

I wanted to prove the following without the need of subsequences : let $$f$$ be a continuous fonction on a closed interval of $$\mathbb{R}$$ then it is uniformly continuous.

I found this proof but can't understand it

Let us assume every increasing sequence on [0;1] converges (if anyone can help me proving this) and the IVT

then let us create a sequence $$u_n$$

$$\begin{cases}u_0=0 \\ u_{n+1}=1 \text{ if } \forall x>x_k, |f(x)-f(x_k)|<\varepsilon \text{ and we stop here} \\ \text{otherwise } u_{n+1}=\min(x) \text{ s.t } |f(x)-f(x_k)|=\varepsilon\\ \end{cases}$$

if this sequence is finite, let $$\eta=\max(x_{n+1}-x_n)$$ then $$\forall (x,y),\forall \varepsilon >0$$ $$|x-y|\leq \eta \implies |f(x)-f(y)|<2\varepsilon$$

Otherwise we would have a contradiction.

Can anyone help me understanding

• It looks like $x_k$ is $u_k$... Jan 6, 2019 at 15:48
• Have you looked at en.wikipedia.org/wiki/Heine%E2%80%93Cantor_theorem ? Same argument of your setup. Jan 6, 2019 at 15:50
• I cant understand it thus I ask here Jan 7, 2019 at 19:30