The proof of “continuous mappings on compact metric spaces are uniformly continuous” is rather convoluted and opaque, as given in most real analysis textbooks (Rudin included). There is a lot of unnecessary bookkeeping, and the treatment is not really motivated. Given below is my own proof of the theorem, which is inspired by the existing proof to an extent I am unaware of. I hope it helps the readers. The motivation for each step is explicitly stated.

A mapping is uniformly continuous if for every , there exists such that for all .

Let be the continuous mapping under consideration. For every , take the inverse .

forms a cover of . Call this cover .

We will do two things with this cover .

Firstly, as is compact, let us choose a finite subcover . We shall call this cover .

Secondly, we form another cover $latex \bigcup B(x,\lambda_x/2)_{x\in X}$. Call it . As is compact. too will have a finite subcover. We shall call it .

Both and consist of balls centred on certain points in . The set of centres of these balls may be distinct. Let the set of centres of be and the corresponding set for be . Seek out points in $latex S\setminus R$, and add them to . Call this new set $latex latex R’=R\cup S\setminus R$. Hence, . For the points in , draw $latex \lambda_x$-balls around them, and add them to . Note that still remains a finite cover of . Call the new finite subcover .

Final bit of notation: let .

Take any two points and such that $latex d(a,b)=\delta$. Clearly $latex a$ is contained within some set of $latex W’$. Let that set of have as centre the point . Note that is also contained within , and hence has a $\lambda_p$ ball around it. We have .

In addition to this, we have . Hence, $latex d(f(p),f(b))\leq\epsilon/2$.

Now $latex d(f(a),f(b))\leq d(f(a),f(p))+d(f(p),f(b))\leq\epsilon/2+\epsilon/2=\epsilon$.

Hence proved.

An explanation for some steps:

The essential motive in this proof is to determine a distance such that for any two points and separated by the distance , we should find that their mappings are separated by . For this we need both points to be in such a ball which maps to an $latex \epsilon/2$-ball.

I will continue this explanation when I get the time.

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