 Because the natural numbers are a proper subset of the real numbers, then in some sense the real numbers are more infinite than the natural numbers. And this leads to the following question, might there be even more infinite sets? Well, let's consider. The following results should be obvious, well, no, easy to... no, no, that's not true. Approvable. So, let A be a finite set, then the cardinality of the power set of A is strictly greater than the cardinality of A. And then let A be an infinite set. Then the power set of A is also an infinite set. Since we've seen that some infinities can be larger than others, might the power set of A be more infinite than A itself? And the answer is, yes. And this leads to what's known as Cantor's theorem. Let A be an infinite set, then the cardinality of the power set of A is strictly greater than the cardinality of A. To prove this, we'll show that the power set of natural numbers has a greater cardinality than the set of natural numbers itself. The proof offers us insight into the general case. First, note that the cardinality of the set of natural numbers must be greater than or equal to the cardinality of the set of natural numbers. So let's suppose that the cardinality of the power set of natural numbers is the same as the cardinality of the set of natural numbers. Then there is a one-to-one correspondence between the set of natural numbers and the power set of natural numbers. Now, when we prove the uncountability of the real numbers, we relied on being able to construct a real number different from every real number on the list. So can we construct a subset different from every subset on our list? Well, let's try it. So let's consider any x among the natural numbers and because there's a one-to-one correspondence between the set of natural numbers and the power set, then every natural number corresponds to some set, for example. And as we did with the proof of the uncountability of the real numbers, we don't really care what these sets are. So we might proceed as before and try to produce a set by choosing an element not in each set. But the thing to remember is that since this is the set of all subsets, one of these subsets is the set of natural numbers itself. And that means that we won't be able to choose a number that's not in this subset. Well, that didn't work. So what else can we do? Well, remember, try something easy. So rather than trying to find an element not in the set, let's focus on one specific element. How about the index itself? And so in the subset we're constructing, we can choose to include the index or not. Now, since we want our constructed set to be different from all the sets on our list, let's include the index if it's not in the set and then exclude the index if it is. We'll express this idea formally as follows. Let f of x be the set corresponding to the natural number x. For example, 0 corresponds to this set, so f of 0 would be this set, 1 corresponds to this set, so f of 1 would be this set, and so on. We'll define our set c to be the set of elements x where x is not in the set that corresponds to it. And so if our set has these lists, we see that 0 is not in the set that corresponds to it, and so c will include 0. 1 is in the set that corresponds to it, so 1 is not in our set, 2 is not in the set that corresponds to it, so 2 is in our set, and so on. We claim that c itself is not on our list of subsets. And we can prove that as follows. Suppose f of x is c, that is to say c is a subset that corresponds to a natural number, then x must correspond to c. And so the question is, where is x? Well, there's only two real choices. If x is an element of our set c, then by the definition of our set c, x is not an element of the set that corresponds to x. But the set that corresponds to x is c, so x is not in c, and that's a contradiction. Now, in an ordinary proof by contradiction, we would stop at this point and say that since x in c leads to a contradiction, x can't be in c. But what makes set theorists a little bit strange, even by mathematician standards, is that they ask the next question, well, suppose x is not in c, then what happens? If x is not in c, then x is not an element of the set that corresponds to it. So by our definition of what c is, if x is not an element of the set that corresponds to it, then x must be in c, and this is another contradiction. Now, if you're not totally confused at this point, you haven't really understood what we've just done here. So here's a recap. This set c is a subset of natural numbers. And if we assume we can list all these subsets of natural numbers in this way, then c has to be someplace on the list. But if c is on the list, then the index corresponding to c can't be in c, and it can't not be in c. Well, how do we get out of this double contradiction? Remember that we actually started with the assumption that some natural number corresponded to our subset c. And this led to those contradictions, and so this is the assumption that must be false. And here's the punchline. While we constructed our proof for the power set of natural numbers, nothing in the proof relied on the set being countable. In particular, let there be any infinite set A. If the cardinality of A is the cardinality of the power set of A, then there's a one-to-one correspondence between an element of A and A element of the power set of A. So we can still construct our set c that consists of the elements that are not in their corresponding subset. And as before, x not in c means that x is in c, which is a contradiction, while x in c means that x is not in c, which leads to another contradiction. Now, since not all infinites are the same size, it helps to have some notation. And so we'll define the cardinality of the set of natural numbers to be this symbol, which is the Hebrew letter aleph, and subscript of zero. So we can either call this aleph zero or aleph null. Then aleph one is the next larger infinity. And we know that the set of real numbers is more infinite than the set of natural numbers. And so the question is, could the cardinality of the set of real numbers be this next larger infinity aleph one? And this leads to what's known as the continuum hypothesis. Since aleph one is the next larger infinity, then we know that aleph one has to be smaller than the cardinality of the set of real numbers. The continuum hypothesis claims that aleph one is the cardinality of the set of real numbers. Proving the continuum hypothesis was one of the problems posed by David Hilbert in 1900. At a 1963, Paul Cohen proved the continuum hypothesis independent of the other axioms of set theory. In other words, it could be assumed true and no contradictions would arise, but also it could be assumed false and no contradictions would arise.