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).

@PhotonEcho

3 жыл бұрын

Rudin was just kind enough to leave this version as an exercise.

@get2113

3 жыл бұрын

I think Bartle is better but to each own.

@hydraslair4723

3 жыл бұрын

e n t o u r a g e

@gustavocardenas6489

3 жыл бұрын

Baby Rudin ch4 ex 10

Can't wait to see integrals in this course

1:30 that's better

@pyy6433

3 жыл бұрын

Thanks

@natepolidoro4565

3 жыл бұрын

ty bro

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

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

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.

😮 I like your lecture

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

@get2113

3 жыл бұрын

I totally agree.

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.

@armansimonyan5772

2 ай бұрын

By Bolzano Weierstrass the subsequential limit exists if the sequence belongs to some compact set A and belongs to the same set A

Thank you so much for these videos!

8:43

@pbj4184

3 жыл бұрын

And that's a good place-

@chessematics

3 жыл бұрын

@@pbj4184 the stop stopped the TO STOP

🔥🔥🔥

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.

@chuckaway6580

3 жыл бұрын

You could, but then the proof won't generalize to general metric spaces, as compact = closed and bounded doesn't apply to all metric spaces (in fact, not even all Banach spaces).

@ruanramon1

3 жыл бұрын

Maybe theclosed part is used when he says in 4:51 that the subsequence converge to a point in the compact. Think in a bounded open set, it is possible a convergence to a limit point not contained in the open

@chuckaway6580

3 жыл бұрын

@@ruanramon1 Yes, you need closed to say that it converges to a point in the set, but it's not clear to me that we actually need that in this case. Seems like the proof still works if it converges to a point outside the set, unless you don't assume continuity there, which I guess is possible.

@tomkerruish2982

3 жыл бұрын

No. Consider tan x on (-pi/2, pi/2). Choosing x sufficiently close to an endpoint will invalidate any chosen (positive) delta for a given (positive) epsilon.

@chuckaway6580

3 жыл бұрын

@@tomkerruish2982 There it is. I think I follow, but to build your point out more for others: Consider tan on (-pi/2 , pi/2) U (pi/2, 3pi/2). This is clearly a bounded set on which tan is continuous. Take x_n -> pi/2 from the left and y_n -> pi/2 from the right. Then |x_n - y_n| -> 0, but |f(x_n) - f(y_n)| -> infinity. We need for the limit of the convergent subsequences of x_n and y_n to be in the set or the proof falls apart. Kind of subtle, though. It would be easy to write a somewhat convincing fake proof that leaves off closure. The fake proof becomes a real proof if you assume that the function is continuous on some closed set containing the bounded set.