Real Analysis | Compact set of real numbers.

We provide a definition of a (sequentially) compact subset of the real numbers and prove a classic theorem that says this definition is equivalent to the set being closed and bounded.

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[TheNiTeMaR]

Very insightful comments and diagrams by the special guest!

[noeticresearch]

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

[tomatrix]

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

[goodplacetostop]

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

[benniepieters]

This is known as the Heine-Borel theorem

[RandomBurfness]

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|| <= 1}. Then E is bounded because every x in E has norm less than or equal to 1, and it's closed because for every point x in E, for every ε > 0 we can clearly find an y in E such that ||x-y|| < ε and x != y. Another way of seeing this is that the function f : l2 -> R which sends x to ||x|| is a continuous function, and E = f^{-1}([0, 1]) and since [0, 1] is closed in R, E is closed since f is continuous and the inverse image of a closed set under a continuous function is closed. But E is not compact because the sequence xn = en where en = (0, ...., 0, 1, 0, ...) where the 1 is in the nth position, admits no convergent subsequence.

[nnaammuuss]

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.

[reelandry]

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.

[wtt]

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

[karlmarxsteingoldberg-kike]

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

[adam-does]

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
=><=

Therefore, x E K.

[erik]

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.

[Queiroz]

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

[PowerhouseCell]

*Who else is here while the video's still 18 minutes long? ✋*

*Jokes aside, great video!*

[sujalsagtani]

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

[sitienlieng]

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

[punditgi]

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

[wernerhartl]

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

[ashwinvishwakarma]

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.

[MsOops]

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