How I'd Prove Cantor's Theorem

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[kylecow]

from doing some straightforward point-set topology we can find
(1): If we have sequentially closed sets K_n where K_i < K_{i-1} and K_1 is sequentually compact and each K_i is nonempty, then we can construct a sequence a_i \in K_i, this has a convergent subsequence a_ji->a then this point is a limit of points in K_j for every j so is in every K_j since they are sequentally closed so is in the intersection so the intersection isnt empty

it's not too hard to see that closed intervals are sequentially compact so are sequentially closed by taking a sequence and constructing the sequence of floor terms {a_j: a_k >= a_j for all k>j} if this set is infinite we have a monotone, bounded sequence so it must converge to its supremum and if this set is finite then there is some final a_j, every other a_i must then have an element below it eventually so we can construct a monotone decreasing sequence which must converge to its infimum so closed intervals are sequentially compact so we can apply (1)

[sphakamisozondi]

I'm not a math major, but your videos are fascinating and I can't stop watching

[richardfarrer]

Thoroughly enjoyed this. It's been a few decades since I did this.

I did have to stop to think why Cantor's theorem implies the reals are uncountable though. I think it would have been good to briefly clarify that.

[timothymattnew]

I think I came up with a slightly different proof. If there is no such point c, then the complements of the closed nested sets form an open cover of the interval I1. Since I1 is compact, the cover must contain a finite subcover. But it's easily seen that a finite number of complements of closed nested intervals can't cover the interval I1, yielding a contradiction.

[geoffreyfaust]

Thanks for this one. Kronecker's attacks on Cantor, based on fear of transfinite numbers, is another despicable chapter in math history. This is kind of like what I emailed you guys about earlier, with the Pythagoreans murdering the guy who proved that the square root of 2 is irrational. Another one you can add is Saint Gauss, who obnoxiously dissed a teenage Janos Bolyai to his father, claiming that Gauss was the true inventor of all non-Euclidean geometry, to diminish the young Bolyai as a mathematician (It worked, unfortunately).
The math establishment does now oppose, and always has opposed, innovation that would shake up the powers that be. Look at the dogma that Kurt Godel had to put up with!

[jakobr_]

8:10 The minimum of B won’t necessarily be sup(A) unless we know these intervals get arbitrarily small.

[TheHebrewMathematician]

7:30 You defined sup with inf and inf with sup, that's circular definition, you should have defined sup as the min of all upper bounds and inf as the max of all lower bounds.

[Himanshu_Khichar]

7:17 Isn't the infimum of both (2, 3) and [2, 3) 2? So inf(I)=a whether the interval is open or closed at a.

[intptointp]

I think the little animations and explained notation was beautifully done. Very little left unexplained.

That being said, I’ve been watching math videos on and off for the past year so a lot of notation is familiar to me.

One suggestion that would really make it crystal clear is some examples for the conclusions made.

Maybe a single example during the walkthrough would provide a concrete thing to walk through as you prove the general idea.

[danielchin]

I got lost at 8:16. Why does it follow? Also, I'd love the video to first establish why this conclusion is counter-intuitive.

[Khashayarissi-obyj]

With luck and more power to you.
hoping for more videos.

[bastamtajik]

Indeed when we know that any complete metric space (with at most finite isolated points) is uncountable AND the Cantor's intersection theorem can be implied for a general complete metric space, the main point of the theorem reduces to completeness of the Euclidean line! If the Euclidean line is assigned another metric which is not complete, then the Cantor's intersection theorem fails. So can't we put the blame for this theorem on completeness of one's preferred choice of metric over real line and not Infinity? 10:33

[charithreddy]

at 7:22, Isn't Inf(I) a set ? so how can you compare it with a number?

[timelsen]

Thought you would prove UNCOUNTABLE in this video. Will you in a following, "great exposition as always?"

[theoneandonly-lucf]

I feel like this doesn't need a proof. "After shrinking a set infinitely many times there is something in the set. Yeah of course. That's a no brainer.

[willnewman]

You use inf in your definition of sup, and sup in your definition of inf, which is not good.

Also, I would have liked you to say something about the existence of sups and infs.

Finally, it is not clear to me why this shows that the real numbers are uncountable.

[adityamishra]

this is a fucked up way to prove cantor's theorem, not to mention this ISN'T even the cantor's theorem, in first place.

[sumdumbmick]

how do you open by mispronouncing Georg?

bro... if you're that lazy you have nothing to say that's worth saying.