I Want to Play a Game: Proving [0,1] is Uncountable
HTML-код
- Опубликовано: 12 окт 2024
- The interval [0,1] is uncountable, and there are many ways to prove that. Cantor's diagonal argument is a classic example. But there's another way: proof by playing a game. This video is an explanation of a proof by contradiction after assuming that the infinite set [0,1] is countable and playing a game on that interval. All we need is basic calculus for the limits of sequences!
Original based on Grossman & Turett: arxiv.org/pdf/...
Calculus Problems playlist: • Calculus Problems
Subscribe to see more new math videos!
Music: C418 - Smooth Fall
Direct proof🚫
Proof by contradiction🚫
Proof by leaving it as an exercise for the student 🚫
Proof by game✅
Nice. Just want to point out that this game won't conclude rational numbers are uncountable since the monotone convergence theorem does not apply to rational numbers (the limit of a sequence of rational numbers can be irrational).
This is essentially Cantor's _first_ uncountability proof, with a spin (the context of a game) on it! Nice!
I appreciate this proof a lot because the completeness property of R is really highlighted in this argument. I don't know any proofs about the uncountability of R which doesn't rely on completeness in some fashion. The Diagonal Argument, on the other hand, really sweeps completeness under the rug by expecting the reader/viewer to accept that decimal notation works the way we expect it to (which relies on completeness!).
Thr proof has to rely on completeness because thats essentially the definition of the real numbers, if not you could do it in Q
heyy welcome back really missed your uploads
it'll be nice if you continue your Number theory vids bc they're phenomenal
keep it up :))
Brilliant stuff. I've never seen anything other than Cantor's argument, and naively didn't suspect there were any other approachable proofs of the fact, and how wrong I was. It's definitely not quite as elegant as Cantor, but it also doesn't rely on any expansions in any specific base, but of course still uses completeness. I have to say I don't have an intuitive feel for the proof idea like I do with Cantor, but the presentation was still very good.
I really enjoyed this! Great!!!! Keep it up :-) Can we expect more stuff like this on our way?
I have a few more videos coming in this batch on topics in calculus, topology, and Galois theory.
@@MuPrimeMath looking forward to it :-) I find your work not only helpful but also quite enjoyable. Keep 'em coming.
Got a little off tracked but still thereabouts. I can see where you are getting at.
1) The game never ends.
2) No number picked by any one player is ever the same as a number picked by the other player.
3) After each player has had a single turn, the latest as also the largest number picked by player A is always less than the latest as also the smallest number picked by player B.
4) Assuming the set of all X[i] is a counting of every number between 0 and 1 "IN PROPER ORDER", hence contains every number between 0 and 1 every number that player A picks and every number that player B picks must also be in the set of all X[i].
If the first number player A picks happens to be the first number X[1], in the set of all X[i] he has no choice but to pick all the numbers of the set of all X[i], in that order, which means player A must eventually pick some number which player B also picks because of 4), which can never happen because of 2) and 3) so there must be some number picked by player B which is not one of the numbers in the set of all X[i], which contradicts the assumption that the set of all X[i] counts every number between 0 and 1.
This could be a winning argument and constitute a valid proof by contradiction I suppose, but I can't be certain, cause the order imposed by a counting operating on a set need not necessarily be the same as any inherent order already present in the elements of the set. The key condition/phrase here is, "IN PROPER ORDER". Under the not unreasonable assumption that the set of all X[i] is a listing of the elements between 0 and 1, "IN PROPER ORDER" , which means for all i, X[i] < X[i + 1], it certainly does.
Even though a set is an unordered collection of elements, but the operation of counting always imposes an order on the elements of the set, since it always enumerates, lists the elements of a finite or an infinite set in some particular order without leaving any element out, and including, counting every element exactly once.
Nice. But don't we need R to be uncountable in the first place to use this theoreom about sequence limit?
We only need completeness of R for this theorem to hold. And completeness can be defined, essentially, as the property that every bounded, monotonically increasing sequence has a limit, among several other equivalent definitions in the realm of ordered fields. So the proof in the video is basically "complete implies uncountable".
Yes. Isn't it completnes equvilent to being uncountable in the case of R being the "complete" version of Q?
@@bartekabuz855 I don't think it is equivalent, but it seems nontrivial to give an example of an uncountable ordered field that isn't complete. Maybe use a Hamel basis somehow...
I don't know in general sence but intuitivly the complete version of Q would have to be uncountable. I will give it some formal thought
I don't see the issue. R's defining feature is it's completeness, and this applies to Cantor's proof, too, doesn't it, since decimal (or other base) expansions assume completeness, right? Of course intuitively completeness seems like it is the same as being a "continuum" and thus uncountable, but the whole point is to prove that.
Nicely done, as always.
What's your stand on the Continuum Hypothesis? Personally, I favor it, but that's likely largely influenced by my early exposure to the book _One, Two, Three ... Infinity_ by Gamow, which pretty much assumed it. Currently, my thinking is that, since Q is dense in R, how could you fit something in between? (Yes, argument by lack of imagination isn't very compelling.)
I wouldn't say that the continuum hypothesis is universally true or false. Rather, its decidability, and its truth or falsity where decidable, is a property of a given axiomatic system. But I am not educated on this topic, so there may be more to say.
@@MuPrimeMath I'm a Platonist, verging on Pythagorean, so I believe it "actually" is either true or false in the "real" world. But then, I've always tended to be quantized in my viewpoints. If you're in the middle of the road, you're likely to get run over.😂
Congratulations on finishing Caltech. It took me a total of nine years and two more schools to get a degree.
Look up the Cantor set. It isn't dense anywhere in R, yet it is uncountable
@@MuPrimeMathdoes ZFC imply CH?
No, CH is independent of ZFC.
On the line of length 1: you could say 1/0 points of length 0 or uncountably infinite points of length 1 by your proof. By definition, 1/0 * 0 =1, and by the axiom a=b and b=c implies a=c so 1/0 is the same as uncountable infinite. Other deductions like limits or -1! show it is also unsigned infinity, like points at infinity on a perspective graph. Another thing to define 1/0 is that if n/0=m/0 then math breaks after *0 and to be closed, n/0/0 is different from n/0 so also implies that you can’t substitute 0 for 0 if they aren’t the same version of 0. To define the default 0, 1-1=0*1=0=true 0, other zeroes are like 2(1-1) so 2*0 so could be phrased like this: a is not congruent to b even if a=b when a is c*0 and b is d*0 unless c=d and c is congruent to d, where if numbers are congruent then they can be substituted for each other. For examples, 0/0 is not congruent to 1/0, and 1/-(0) is not congruent to 1/0. The remaining issue is n^(k/0) or 0th root of (n^k) which can be defined with a new number system like this: NAN(x)^0=x
And recursion like this: NAN(NAN(x))=NAN^2(x) and that to the 0 is NAN(x) most common use of 1/0 is slope of vertical lines, so on the step function has 1/0 and 0 slope via average rate of change so just not sure the slope of x=n since seems like it can be m/0 but maybe perspective geometry could solve that. Indeterminate means forms of math where congruence was not used to get that form. 0^0=0!=1 And that’s a good place to stop.
but now you have to use AC
I'm not too educated on all this, but why on Earth is AC so controversial, to the point where people are very careful about noting when it's being used?
@@fahrenheit2101 It gives rise to a lot of paradoxes like the banach tarski paradox which says that it allows you to split a ball into two balls that are identical with the original one.
@@melf5057 I'm aware of Banach-Tarski. I just don't see the fuss over it. There are plenty of counter-intuitive results even without AC. This one seems fine.
Indeed I never struggled with the idea that you can put a continuum in bijection with 2 copies of itself in a 1d case.
@@fahrenheit2101 I mean, it certainly baffles me. All the operations one does should maintain volume but somehow it gets doubled. But other controversial examples would be predicting ANY real function's value or the prisoner dilemma.
@@melf5057 That's not volume, though is it? You're not duplicating the ball, you're creating a bijection between the uncountable infinitude of points on its surface, and 2 copies of that. Points and the continuum have always been a bit counterintuitive anyways.
I'm aware of the prisoner dilemma but didn't think AC was at all relevant, nor even that the dilemma was truly problematic?
As for predicting any function's value, I simply do not know what you mean.
Though I do know the adage:
"The axiom of choice is obviously true, the well-ordering principle obviously false, and who can tell about Zorn's Lemma"
The joke therein of course being that the 3 statements are equivalent.
I'm sure there are many valid reasons to be wary of AC, I've just never been massively perturbed by Banach-Tarski.
Congratulations, you’ve proven you can’t count to infinity just that we have a structure for counting towards it infinitely.
Do you think that some people actually think that we can count to infinity?
@@synaestheziac They say it’s countable 🤣