TEDxHampshireCollege - Sarah Mullens - The Continuum Hypothesis: A Biography 

Sure. Here's a theory where continuum hypothesis can be proven: "ZFC + continuum hypothesis". (If set theory itself is consistent, then so is the same theory with continuum hypothesis added as an additional axiom.)

Pc I do not understand your question; which criteria are you dealing with? 🤔

Pc To them ∞a + 1 = ∞a

When she says "has more members" or "has fewer members" she means in the sense of cardinalities, not in the sense of superset/subset. E.g. {1,2,3} contains more members than {3,4}, even though {1,2,3} is not a superset of {3,4}. In the sense of cardinalities, the set of all integers contains exactly the same "number" of members as does the set of all integers together with pi.

Stalker.

You could, but then you've got a new list, which still isn't complete (you can repeat the process and construct another number not in the new list). It doesn't matter what list you come up with, it will never be complete.

You don't get it that this is a TEDX talk. They only have 20 min so it is not a lecture platform. It is to introduce a new point of discussion. Decades ago there was a 5 min program every night on BBC called "Thought of the Night" where they introduced a topic and a new angle to view it and let you contemplate it. So she just introduced this subject that hardly anyone knows about with a new angle to view it. You will be amazed that how a single comment can change people's view and life and it happens so often and it is usually very simple. So do not dismiss simplicity, you just did not listen to it with the right mindset.

Are you suggesting that one day a mathematician might figure out how to prove the Continuum Hypothesis?

@Social Mathematics: Well, you can't prove the continuum hypothesis. It is known that the axioms of set theory are insufficient to settle the question one way or the other (at least, as long as set theory itself is consistent). It is true in some models of set theory, and false in others.

This is pretty clearly a talk for laypeople. Also, the story of the history is in fact valuable information. It isn't mathematical information, but it is valuable. Anyway, I hope you've grown up in the last 10 years.

() 〉 "Action" She makes an error: you are true that is the latter she actually proves! 😉

() 〉 "Action" But I am sure you have known it, for 4 years. 😜

She skipped some of the details because people can get lost in the weeds, but the nuances of this argument are important. In math there is a proof technique called "proof by contradiction." Essentially if you're trying to prove that something IS the case, one way you can do this is assume IT IS NOT the case, and see where this takes you (important, you can't assume it IS the case, you can only assume IT IS NOT the case). If you end up with a logical or mathematical contradiction (it doesn't matter where this contradiction comes from btw), you know that whatever you assumed isn't true i.e. the original thing you were trying to prove is true. So if we want to show that the real numbers are not listable (or that you can't find a 1 to 1 correspondence from the natural numbers (1,2,3,4...) to the real numbers, it's important to think of this as an ordered list), using proof by contradiction we assume that they ARE listable and see what happens. The important part of this proof is that you can ALWAYS make a number that is not on the list no mater how long (or even infinitely long) the list is. I can't ALWAYS show that your list is incomplete. This is inconsistent with out assumption and therefore this is a contradiction. Therefore the real numbers aren't countable or listable. Don't worry, many people get confused by this proof and some people don't accept it to this day (though they should, the mathematical community accepts it and it is logically consistent and a good argument). Just because you don't understand something doesn't make it not true.

I CAN always show you that the list is incomplete*** typo, sorry. And I can't edit because mobile ): And also, once you add that new number that was not on the list, to the list, you can do the whole process over again and generate a new number.

+Alex Marsh Thanks Alex. I know about proofs by contradiction. For example I understand the proof that uses this method to show that there are infinitely many primes. However in that proof the steps progress logically. I think that their is a contradiction within this proof (if you understand what I mean). Also this proof seems to assume that mathematical reasoning that applies to finite sets can be extended to infinite sets as well.

+Alex Marsh *typo their s/be there!!

Yeah, I definitely understand your issues with it. When I first saw Cantor's diagonal proof, it was in a general education undergraduate math class that gave an overview of "academic math," and many of the students and myself felt iffy about the proof. Very similar criticisms that you give. However, once I took Real Analysis, it becomes much more clear why it works. WARNING: Some slightly more advanced math ahead and this is long.... I won't go into tooooo much detail, but it’s really hard to explain this without actually talking about cardinality, bijective functions, and hebrew letters :) You're right, the logic for comparing finite sets IS different than the logic comparing infinite sets. The size of a set is called its cardinality. The cardinality of a finite set is simply how many members it has. However, we get the cardinality of an infinite set by comparing it to another infinite set that we already know the size of. The "smallest" infinite set is the natural numbers (starting at 1, NOT zero, then 2, 3, 4, 5... etc, from now on I will denote the natural numbers as the set N) and we give that cardinality the name Aleph Naught. If a set has the cardinality Aleph Naught, we refer to it as a countable set (or listable if that makes more sense). If we want to compare the size of an infinite set to a set we already know is countable, we have to find a bijective function between the two (for the most part… in certain circumstances just a surjective function will work and in others, just an injective function will work). I know, these terms you probably don’t know. I won’t explicitly define what injective, surjective, and bijective mean, but I will vaguely define what they mean below. You seem like you have a decent math education, so I am sure you understand what a function is: a relation between a set of x's in the domain and a set of y's in the codomain (not technically the range though we won't go there) where every x in the domain is related to one and only one y in the codomain. I'm sure you're used to seeing notation like this f(x)=2x+5, and while this is the most common and useful notation in math, it doesn't have to fit this notation. It could simply be a "list" of commands about how to sort numbers or objects. All that matters is that every x is related to only one y. A bijective function is essentially a function that you CAN find the inverse for. If you remember, only certain functions can have an inverse. The inverse of f(x)=2x is f(x)=(1/2)x. The inverse of f(x)=x^2 is f(x)=sqrt(x), but only for positive x’s. There are essentially two “criteria” a function must meet in order to have an inverse (or be bijective), one of these is that every y is mapped to ONLY one x (the “opposite” of the definition of a function); this means the function is injective. The second is that every y in the range must be mapped to at least one x; this means the function is surjective. Not all injective functions are surjective, and not all surjective functions are injective, however, all bijective functions are both injective and surjective by definition. A bijective function is essentially a way to create a unique ordered list of the range. This is why Cantor’s diagonal proof starts off by assuming that we have a list of the Real Numbers. What we are essentially assuming is that there is a bijective function between N and the Real Numbers. If you don’t like the term list since the set is infinite, just think of it as bijective function from N to the Real Numbers. Alright, so back to infinite sets. So like I said above, if we want to compare the cardinality of a general infinite set to the cardinality of a countable set (for right now N), we have to find a bijective* function between the two sets. If we CAN find this function, then the sets are the same size. If we cannot find a bijective* function between the two sets, then the sets are not the same size. So I told you that Aleph Naught is the smallest infinite set and, for right now, the only set that I’ve told you is of the cardinality Aleph Naught is N (again, the natural numbers). Now, if I want to add a finite set to an infinite, countable set (or find the union of a finite set and N if you’re familiar with unions and intersections), is this new set still countable? YES. While there is a pretty easy proof for this, I’ll show you an easy example. This is NOT a proof since you cannot prove by example, but it at least shows you that this is true. Let’s define set A = N and set B = N with zero added to it or B=N U{0} (since zero is not technically a natural number). So for A and B to have the same cardinality (Aleph Naught), we need to find a bijective* function from A (the domain) to B (the range), and we can. One function that works is f(n)=n-1 where n is in N (though there are many functions that will work). So n=1 in the domain is mapped to f(1)=0 in the range, n=2 is mapped to f(2)=1 in the range, n=3 is mapped to f(3)=2 in the range, etc. As n goes to infinity, every B will be mapped to an element in A. So A and B are the same cardinality and B is countable. Now, is set A (again, the natural numbers) the same cardinality as all the integers or in other words, set N U{0}U -N (if the notation is confusion, this is the set {…-3,-2,-1,0,1,2,3…})? YES! Though this function is a bit harder to express in f(n) notation, it is nevertheless a bijective function. Let set C = the integers or N U{0}U -N. We need to find a bijective* function from A (the domain N) to C (the range, the integers). Here is the function: map n=1 to f(1)=0, map n=2 to f(2)=-1, map n=3 to f(3)=1, map n=4 to f(4)=-2, map n=5 to f(5)=2, etc. In order to express this in the normal f(n) notation, we would need to have a piece wise or “split” function, but nevertheless, this is a bijective function. So the integers ARE countable. This is why I said think of an ordered list in my first comment. Because by relating N to another set, we are essentially saying that I can tell you which number is FIRST (or n=1) on my list, I can tell you which number is SECOND (or n=2) on my list, etc. It really is an ordered list in some regard. The set is “countable” or listable. I will not show you a bijective* function that maps N to the rational numbers {1/1,1/2,1/3,1/4…,2/1,2/2,2/3,2/4…,3/1,3/2,3/3,3/4…} because it is a bit more complicated and “mathy,” however, you can take my word that the rational numbers ARE countable. NOW, we can finally address the real question, are the Real Numbers, denoted as R, (or the union of the rational numbers and the irrational numbers) countable? Well, I have already told you that the rational numbers ARE countable, and I have “showed” you that the union of two countable sets is also countable. So if R is the union of the rational numbers and the irrational numbers, and the rational numbers are countable, what we are essentially asking is “Are the irrational numbers countable?” This answer is NO! You will not find a bijective* function between N and R (which really means you cannot find a bijective function between N and the irrational numbers). The real problem here is that you cannot find a SURJECTIVE function that maps the domain (N) to the range (R), which is what Cantor’s Diagonal proof actually shows. No matter how you relate N to R, I can ALWAYS find a number in R by using the diagonal “if the first digit of the first number is equal to 1, make it 2, if it is anything besides 1, make it 1” technique which means there can NEVER be a function from N to R where every element in R is mapped from at least one element in N. No matter what kinda function you have, I CAN ALWAYS FIND A NUMBER IN THE RANGE THAT IS NOT MAPPED FROM SOMETHING IN THE DOMAIN. This is why Cantor’s Diagonal proof works. I know this was long… Analysis is difficult and rigorous and overly complex. However, that’s why analysis has proven so much. I hoped this helped in some way…