Thanks for such a great content with love from India

You are a phenomenal teacher Sir ! Thank you for making our life simpler.

Amazing video. Thanks very much.

Nice video excellent method of teaching actually makes the student want to listen being an absent minded one myself I'm surprised

Very nice. I heard this proof before in class, but it got foggy after we used completness. Thank you for clearing it up.

Very nice video sir!!!

go bears!!! im currently learning this in math 104 (introduction to real analysis) at berkeley, and your videos are actually amazing

@drpeyam

2 жыл бұрын

Awwwww 104!!!! That brings me back, I took that class back in summer 2007 at Cal and we used the Pugh book 🙂

@user-iq9pe4ls2j

4 ай бұрын

whose 104 taught the baire category theorem? mine just skipped it.

Very helpful video !

Great . You have to teach my doctor so that he might be like you!!

Ok. Thank you very much.

thank you so much, but I am struggling with approving that the intersection is of second category in R. Maybe If should also prove that the intersection is open?

That Kilmt makes for a great background.

Love from SoCal

Merry christmas from Middle East!!!

Thanku sir ❤️

Nice👍

My analysis final just ended a few days ago. Only if this would've happened just a few days earlier 🤯

It is nice as it starts from scratch and I don't have to Google meaning of the terms

Wait, is a closed ball in a complete metric space guaranteed to be compact? I always forget which results I know for R^n vs more general situations.

@drpeyam

3 жыл бұрын

No it’s not always compact in infinite dimensions, like for bounded sequences with the Sup norm. Check out my non compactness video on it. There is Alaoglu’s theorem though

@martinepstein9826

3 жыл бұрын

@@drpeyam Thanks! I didn't know Alaoglu's theorem. At the end you showed that in a complete metric space a nested sequence of closed balls must have nonempty intersection, and this result has a compactness flavor if you as me.

I'm new to this, but Quick Question: You write that we WTS that (1) B intersects every Un, meaning that (2) there is a y in B that intersects every Un. I understand that (2) => (1) so if we prove (2) we are done. But it doesn't seem to me that the two are equivalent.

Make Way for the Baire!

Category analysis

But where is the bear in all that

Cauchy is kosher.

bro, I don't even see a single category here 🤣 Slightly disappointed

@drpeyam

3 жыл бұрын

Haha