I'm having this discussion with some friends, and it seems we all don't really understand this topic. Now, I understand that Cantor's diagonal argument is supposed to prove that there are "bigger infinities" than others. Now, honestly, I only think that there is something I must not understand here. I don't really know what it is here, but here I watched a video on YouTube by Numberphile titled "Infinity is bigger than you think". I'd post the link but I'm not sure I'm allowed to in this forum.
ANYWAY, here I'll try to describe what I don't understand: it seems to me that the argument goes "no matter what your list is, with this method I can write a number that is not on that list". But then again, at the same time, the entire topic is about infinite lists. If you had an infinite list of numbers, it would clearly contain every number, right? I mean, if you had a list that was truly INFINITE, then you simply couldn't find a number that is not on the list! For example:
. . . 0.12345 0.12346 0.12347 0.12348 0.12349 . . .
The first thing i already don't understand is this: do you have to add a digit to every number on the list for every number that you add to the list? Because I mean, if I had 6 numbers on the list above instead of five, how could you possibly exectute this diagonal-method thing when there are only five digits after the 0.? And the second thing I don't understand: if you had a truly INFINITE list, how could you find a number that is not on that list? It seems to me that this only works with finite lists, doesn't it? Because a truly infinite list would contain all numbers. The guy in the video claims that you could always make another number that's not on that list, but on an infinite list, what number could there possibly be that doesn't show up on the list?
Guys, I am not even really sure what it is that I don't understand, I just think that I must not be understanding something. If you can somehow see what my thinking error is, please let me know. And try to answer my questions. Thanks a bunch.
For your specific questions:
(This is the question in the title as of the time I write this.) It proves that the set of real numbers is strictly larger than the set of positive integers. In other words, there are more real numbers than there are positive integers. (There are various other equivalent ways of stating it but that's the way I prefer and have most often seen.) One might think "well, obviously that's the case" but nothing is really that obvious when dealing with infinity. For example, the set of positive integers $\{1, 2, 3, 4, 5, \dots\}$ and the set of positive, even integers $\{2, 4, 6, 8, 10, \dots\}$ have exactly the same size, even though one might intuitively expect the latter to be exactly half the size of the former. (To show why these have the same size is relatively straightforward but too much of a deviation from your main question IMO.)
Yes, this is necessary in order for the proof to work. Of course in practice we can't continue this process forever, but the idea of the proof is that we can indefinitely do this in theory. And this is legitimate because the list is supposed to be a list of real numbers. The set of real numbers contains infinitely many numbers whose decimal expansions continue indefinitely, such as rational numbers like $1/11 = 0.09090909...$ and irrational numbers like $\sqrt{2} = 1.41421536...$. So there is certainly no shortage of numbers like this to include in your list (because there are infinitely many such numbers).
You seem to think that "infinite list" is the same thing as "list containing everything possible" but this is not true. It's entirely possible to have an infinite list of numbers that does not contain every possible number.
To more easily understand this, consider the positive integers:
$$ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, \dots$$
I think we can agree that this list is infinite, because there are infinitely many positive integers. (To find the next integer, just add $1$ to the current integer.)
Now, what about positive, even integers? That list would look like this:
$$ 2, 4, 6, 8, 10, 12, \dots $$
I think we can also agree that this list is infinite, because there are infinitely many positive, even integers. (To find the next even integer, just add $2$ to the current integer.)
So, the second list is an infinite list of positive integers (specifically, the even ones) but it does not contain all of the positive integers.
Similarly, in Cantor's diagonal argument the point is to construct an infinite list of real numbers, where each real number is "labeled" with $1, 2, 3, 4, 5, \dots$. That is, you pair up each real number with a positive integer in an attempt to show that there are what's called a "countable" or "countably infinite" amount of real numbers. (Infinite sets whose sizes are equal to the size of the positive integers are called countable because we can essentially "count" the elements of the set by pairing each element in the set with exactly one positive integer.) But then the diagonal argument shows how to construct a number that can't possibly be on the list. Therefore the list can't possibly contain every real number (which, again, doesn't contradict the fact that the list is infinite), which means the amount of real numbers is greater than the amount of positive integers.
Re: your comment
No. The diagonal argument explicitly shows you which number can't be on the list. "Infinite list of numbers" and "contains all possible numbers" are not the same thing but you seem to still be thinking they are. See my positive integer vs. positive, even integer example in my original post above. Did you perhaps understand that but you're just not convinced how that logic extends to real numbers in general and not just integers? Here's another example with more general numbers.
Consider the set of all real numbers whose first digit after the decimal is $1.$ This list contains numbers such as the following: \begin{align*} -&9.148754829...\\ &0.1326544685...\\ &12.1389749827429857...\\ &0.1111111111...\\ &31.123123123123...\\ &0.1487648348999... \end{align*} Now suppose we remove from this list all numbers that are strictly larger than $1$ or less than $0$. Then our new list would actually be a subset of the old list. And the first, third, and fifth numbers I listed above would not be on this list. Do you agree that we are removing an infinite number of numbers from the list? We are. But do you really think that the new list is finite? It isn't, because even between $0$ and $1$ there are infinitely many different numbers whose first decimal digit is $1$. \begin{align*} &0.1326544685...\\ &0.1111111111...\\ &0.1487648348999...\\ &0.1\\ &0.197865413897...\\ &0.125 \end{align*} This is an infinite list of real numbers that doesn't contain every real number. It doesn't even contain every real number whose first decimal digit is $1$, but it's still an infinite list.
This train of thought is actually at the center of why some people reject Cantor's argument but the mathematical community by and large accepts it. It actually doesn't depend on the list ending, because not all real numbers have decimal expansions that end.
The fact that the argument or process can never be finished doesn't make it invalid or incorrect. By that same token, one could say that there aren't infinitely many integers because we can't write all of them down. Would you agree with that statement? Personally I wouldn't, because we can still describe the set of all integers and exhibit a pattern that tells us how to find the "next" integer. And it's common knowledge that to find the "next" (as in, next one to the right on the number line) integer we just add $1$ to the "current" integer. But some people reject this notion on the grounds that this is not a finite process. This is something I've only recently learned about - on this site, actually. It's called ultrafinitism.