Very insightful comments and diagrams by the special guest!

The special guest did a great job

8:34 The next generation is prepared for the meme 😂

@________6295

3 жыл бұрын

lol

Hello there young man! Your Dad is an amazing mathematician!!

This is known as the Heine-Borel theorem

Love the guest. It would be great to see more of them:)

It is VERY IMPORTANT to note here that in general, not every closed and bounded set is compact. For instance, consider the set E = {x in l2 : ||x||

@tomatrix7525

3 жыл бұрын

Wouldn’t {0,.......0,.....} be a convergent subsequence? Recall, a subsequence of an, say ank, nk a subset of Natural No., is a subsequence if ank is a sunset of an for all nk you have selected from N, thus we can choose a sunsequence ank from your given sequence, where nk is all Natural No. =/ n, since nth term is 1 you said. This subsequence is just a trail of 0s, and thus converges to 0.

@vance4994

2 жыл бұрын

why are you watching this series if you already know all this?

@thesecondderivative8967

7 ай бұрын

​​​​@@tomatrix7525 No. For a sequence to converge, the distance between adjacent terms has to decrease. But the distance between adjacent terms is √2.... therefore even though (0, 0,...) seems like it is the limit, you can find a ball of radius less than √2 centred at any point as far into the sequence as possible and it will not touch any other point in the sequence. It is jumping dimensions to escape converging.

@alegal695

2 ай бұрын

@RandomBurfness define "l2"

Lovely ending here. New postdoc ?

You are not easily distracted :)

Very clear explanation, thank you professor for this great vedio!

Just a tiny correction, the subsequence a_n_k inside K converging to x does not mean that x is a _limit point_ of K, but only that x ∈ K-bar (the closure of K), which is enough of course. 😊 I myself had forgotten this quirk in the definition of “limit points” that exist in analysis textbooks, until recently I looked it up.

beautiful explanation, and good job explaining that your proof was heine borel and not that compact implies closed and bounded for like other spaces. all people look for in these videos are the reals anyway. and let yo kid continue drawing.

Thank you so much, Dr. Michael Penn

I believe there's a small subtlety in the if direction. You have to prove that the limit point x belongs to K. Compacity implies that there is a subsequence of an that converges to some point y in K. However, since an converges to x, this point y cannot be different. Thus x is in K. That's my suggestion. Love your videos and channel, keep up with the good work

@tomatrix7525

3 жыл бұрын

Let me dare to say you’re wrong. Compactess (Not sure about compacity!) is exactly as he defines it. A limit pt. of course may not be in K, but we know we can find a sequence that converges to it by definition. All subsequences of a convergent sequence converge and converge to the same value of the parent sequence. Thus, all subsequences will converge to x, including the trivial subsequence (duh!) But by definition of compactness, which we are supposing is true in the case of the set K in the forward direction, there exists atleast one subsequence of the parent sequence that converges to some L in K, but all subsequences of our convergent sequence converge to x, x is in K. Thus, unless we have some missunderstanding or I am mistaken, he has proven x is in K.

Nice clear proof!

At 8:00, you draw the implication: (lim k->infinity a_(n_k) = x) => (x is limit point of K) But if, say, a_(n_k) = (x, x, ... , x, ...) and x is an isolated point of K, then you can see that this implication is false. Of course it's still true that x E K, because if it didn't it would be a limit point! Namely: Suppose (not (x E K)) => a_(n_k) =/= x for all k => x is a limit point of K => (b/c K is closed) x E K =>

Thank you so much

pausing at 4:58, I'll point out a small nuance. Michael seems to have left out a vacuous case: he says to suppose x is a limit point of K. It seems like an odd jump to assume that K *has* limit points -- i.e., what if K is a set of isolated points? The answer is that if K has no limit points, then it vacuously contains all of its limit points, and thus is closed. So he just jumps right in and assumes that K has limit points. He would have to prove that K is bounded in the vacuous case as well, so to properly do this proof it'd be more practical to show that K is bounded first, then show it's closed.

5:21 small detail. the contradiction is in the supposition that K is compact.

Hi, Dr. Penn, I have question about the definition. Is the definition an iff statement or it just a one-direction?

How do you know every point of K is a limit point?

*Who else is here while the video's still 18 minutes long? ✋* *Jokes aside, great video!*

Don't we also have to show a_nk =/= x to say x is a limit point of K? An easy way to fix it is to just consider the cases when you do in fact have a_nk =x for some nk, but then x is in K.

Solve for all x such that x+1 is a perfect square and 2x+1 is also a perfect square

5:20

make videos on other topics as well except real analysis

I hope you test her on all that! ;)

Also, the content of the video ends at 8:36, but the video ends at 17:55. Why is that?

@MichaelPennMath

3 жыл бұрын

I think something went weird when I exported the file. I'll trim it in youtube now.

@foreachepsilon

3 жыл бұрын

Jose Franco It was a good place to stop.

thank you sir.nice appear of your mathematicien son

Who is the little curly haired guy in the video? 😁

@rogerkearns8094

3 жыл бұрын

For myself, I felt that I could perceive a familial likeness to the lecturer.

I love the kid! And a very nice set of proofs. 😁

Yeah, who is that kid?

Just like Euler.

A delightful place to stop !

Its not only mouthful. Its brainfulll .

first

@ahmedyacoubchach

3 жыл бұрын

no

@________6295

3 жыл бұрын

@@ahmedyacoubchach yes

@MsOops12

3 жыл бұрын

@@ahmedyacoubchach ok

hhhhhh who is this cute kid