# 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

Any closed interval of real numbers has the following property; For any infinite subset of the interval there exists a point in the interval for which we can find an infinite amount of elements of said subset which are as close as we wish. Since you are trying to prove something I assume you will want to prove this statement as well.

To prove this statement think about the following procedure. Let's say we have some infinite subset of a real interval. Now we may cut the interval in half and at least one of the halves must contain an infinite number of members of said subset. We may continue this procedure as often as we wish, or else the infiniteness of the subset would clearly be contradicted.

Now we may look at some sequence of intervals, whose length gets halved with each step, and every element of the sequence contains infinitely many points of the subset.

To finish the proof show that there exists a number which is a member of all the intervals of the sequence and that this numbers has the desired property.( use the least upper bound property of the reals)

To prove your original question assume, for a contradiction, that you have a continious but non uniformly continuous function on a closed real intervall. Using the nonuniformity construct a sequence of points so that the nth member of the sequence has a point which has a distance of 1/n or less and the distance between the projection of the points is larger than a fixed number. Now use the first statemet to produce a point which is infinetly close to infinetly many members of this sequence. And finally use this point to produce a contradiction to the assumed continuity of the function.