Real Analysis | Uniform continuity and compact sets.

We prove that every function which is continuous on a compact set is uniformly continuous on this set.

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[chuckaway]

This is actually a very sneaky proof and much more elegant than the one you see in baby Rudin. Few outclass baby Rudin when it comes to proof elegance. Makes me suspect this proof is wrong in a way that I haven't noticed yet.

Edit:
I think I figured out why Rudin doesn't go this way with the proof. He tends to prefer epsilonics to sequences in baby Rudin (arguably a flaw with the text, but I think I know why he takes this approach). In particular, he never proves the sequences characterization of uniform continuity. As such, this proof is not really an option for him.

I should add that this proof method really is just better and there's no reason not to prefer it if you've built up the right machinery. It readily generalizes to general metric spaces, and since that's the most general setting where uniform continuity is defined, that's all you need (okay, we could talk about uniform spaces, but let's please not).

[timurpryadilin]

Can't wait to see integrals in this course

[chessematics]

Make a playlist on Number Theory, for you have the best content and many videos on it

[hydraslair]

Michael, any future plans for a series on point-set topology? Would be great to go through it with your intuition about the subject.

[carl]

Small point, I think the modulus at around 6:45 isn’t necessary. Continuity give the limits f(x), f(y), so adding the convergent series (without modulus) has the limit f(x)-f(x)=0, the modulus then comes back in the next line by definition of limit.

[get]

The really needed condition is absolute continuity in order to recover function by integrating its derivative. AC is a little stronger than uniformly

[사기꾼진우야내가죽여]

I think we also need to show that x which is the limit of the two subsequences of Xn and Yn is contained in K using the fact that x is a limit point of K and K is closed.

[daniellin]

Could we change the condition from "K is compact" to "K is bounded"? Because the "closed" part of the definition of the compactness is not used in the proof.