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First of all I want to say I know there are tons of proofs around there for this theorem, so I'm not asking for a proof. I tried to prove it using my intuition of why it must hold but I'm missing the justification of one fact which I believe it must be true.

Let's say $f$ is a continuous real function defined in a compact set $X \subset \mathbb{R}$, I know it must attain a maximum and a minimum somewhere in the domain, so there is a maximum distance $h$ between $f(x)$ and $f(y)$ for any $x,y \in X$. Now, if we choose $\epsilon>h$ then any $\delta$ would do the job, and if $\epsilon \leq h$, we define $p=\min \{|x-y| : x,y \in X \land |f(x)-f(y)| \geq \epsilon \}$ and we take $\delta= p (\epsilon /h)$, and we would have $\delta \leq p$ and because of that $|f(x)-f(y)|< \epsilon$ whenever $|x-y|<\delta$ (otherwise $p$ wouldn't be the minimum $|x-y|$ satisfying $|f(x)-f(y)|\geq \epsilon$).

This argument seems right to me (is it?) but I still need to justify that $p$ is well defined. I know that if there is indeed a minimum value of $|x-y|$ for which $|f(x)-f(y)|\geq \epsilon$ then it is non-zero and intuitively the reason I see of why it can't approach zero is because in that case the function would blow up somewhere in the domain, i.e. there would be some vertical asymptote and hence the function wouldn't be continuous. I'm asking fore some advice on how to formalize (in the simplest way possible) this reasoning.

Thanks in advance

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    you can justify it using contniuty of some functions in two variables.2017-01-02
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    @JorgeFernándezHidalgo I'm still thinking on your comment but I don't realize what two variables functionI should use2017-01-02

1 Answers 1

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Let me try to formalize it.

The problem is just to prove that $p$ is well defined, we shall do this as follows:

First consider the set $X\times X$, it is compact.

Now consider the two variable function $g(x,y)= |x-y|$ and the function $h(x,y)=|f(x)-f(y)|$. Both of these functions are continuous.

It follows that the set $A=h^{-1}([\epsilon, \infty)$ is a closed set, and closed subsets of compact sets are compact.

So $A$ is compact, notice that $p$ is simply $\min(g(A))$ which exists because $g$ is a real valued continuous function and we are taking the domain to be a compact set.

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    Very very elegant, thanks!2017-01-02