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.