# Is infinite a infinite or finite

Long back I have watched a documentary based on a mathematician named Cantor. According to that documentary, Cantor claimed that infinite does not exists, it is only finite. Is that True?

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Google "Cantor", read a little about his work, and get convinced what a huge nonsense that documentary claimed. –  DonAntonio Sep 26 '12 at 10:48
Quite the opposite: Cantor showed that there's not only one infinity, but infinitely many. –  celtschk Sep 26 '12 at 10:48
@VAR121 If you are talking about "to infinity and beyond", you are quite wrong! –  saadtaame Sep 26 '12 at 11:16
@saadtaame: But then if it is that documentary, OP might be mixing up Cantor with Zeilberger. –  Marc van Leeuwen Sep 26 '12 at 12:58

Here is some elaboration on the comments.

What Cantor showed was more precisely that if $X$ is any set, then the powerset of $X$ (that is, the set consisting of all subsets of $X$, written $\mathcal{P}(X)$) is strictly larger than $X$, even if $X$is infinitely large.

To make sense of this mathematically, we need to define what it means for one set to be strictly larger than another. One way to do this is to instead define what it means for one set to be at most as large as another, and then negate that.

I will for simplicity use the definition that $X\leq Y$ if there exists a surjective map $f: Y\to X$ (usually it is defined using an injective map in the other direction, but at least assuming the axiom of choice, these definitions are equivalent, and it makes the following argument simpler).

Now, to show that $\mathcal{P}(X)$ is strictly larger than $X$, we just need to show that $\mathcal{P}(X)\leq X$ is false.

So how to show that there cannot be a surjective map from $X$ to $\mathcal{P}(X)$? Well, this is where Cantor came up with a nice "trick".

Let $f: X\to\mathcal{P}(X)$ be an map. We wish to find some element in $\mathcal{P}(X)$ which is not in the image of $f$. To do this, we define the subset $Y$ of $X$ as follows: $Y = \{x\in X | x\not\in f(x)\}$. This is certainly a subset of $X$, so it is an element in $\mathcal{P}(X)$, and I now claim that this element is not in the image of $f$.

To see this, assume for the sake of contradiction that $y\in X$ is given such that $f(y) = Y$. Now we know that we must have either $y\in Y$ or $y\not\in Y$. But if $y\in Y$ then by definition of $Y$, we have $y\not\in f(y)$ but since we assumed $f(y) = Y$ this is a contradiction.

On te other hand, if $y\not\in Y$ then again by the definition of $Y$, we must have $y\in f(y)$ (since otherwise, we would have $y\in Y$. But again, since we assumed $f(y) = Y$ this is a contradiction.

All in all, our conclusion is that there cannot be an element $y\in X$ such that $f(y) = Y$, so $f$ is not surjective. But since this was for an arbitrary function $f$ from $X$ to $\mathcal{P}(X)$, there can be no surjective function from $X$ to $\mathcal{P}(X)$.

The previous arguments do not, however, show that there are any infinite sets at all, but if we want to have a theory of mathematics that includes all the natural numbers for example, we need to assume that some infinite set exists. And once we have one infinite set, the above arguments show us that we can always find a strictly larger set than any given one (as long as we are allowed to take powersets, but it would be hard to work without this).

Finally, it should be noted that if you are not used to working with this concept of "size" of sets, then it might seem obvious that we can always get a larger set, by just adding an extra element. But it turns out that once you start looking at infinite sets, it takes some work to make them strictly larger, so for example, taking the cartesian product of two infinite sets does not result in a strictly larger set.

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No.

...required to include 30 characters...

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$$\huge{\text{Absolutely not!}}$$ has just over $30$ characters. :-) –  Brian M. Scott Sep 26 '12 at 10:53
@KevinCarlson Is this a joke? :D –  saadtaame Sep 26 '12 at 11:17
@saadtaame mostly-although it is an accurate answer to the question as posed. –  Kevin Carlson Sep 26 '12 at 11:31

The 'existence' kind of stuffs are rather phylosophical questions. I would say, within mathematics, anything can exist which doesn't lead to logical contradiction.

Well, though it is proved that we cannot know if the set theorety we use (called ZFC) leads to any logical contradiction or not, mathematicians regard objects in ZFC as existing ones..

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you cannot have an infinity as a set of parts in the real world like you do in math.

You cannot have infinity apples. You can have a process where apples are added indefinitely every second heading towards infinity but you'll never reach the amount of infinity apples. Hence, to think of an actual infinity existing in the here and now as being comprised of an infinite set of finites is an idea based on a fallacy. There will never be a point in time where this infinity will exist because such an infinity is not a number but rather, an indefinite process.

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You don't know that. It is possible that there are infinitely many universes, with infinitely many planet Earth, so there are infinitely many apples. (And according to some metaphysical interpretations of modern physical theories, this is a reasonable assumption.) –  Asaf Karagila May 27 '13 at 21:17
But there have been infinite moments of time which elapsed between my writing this comment and your reading it. Is that not in the real world? –  Double AA Oct 20 '13 at 16:46

I didn't see the documentary so can't say for sure, but possibly what it said is that Cantor claimed that INFINITE NUMBERS don't exist. Cantor's vitriolic opposition both to infinitesimals and infinite numbers is a well-known fact; see for example the recent text http://dx.doi.org/10.1007/s10670-012-9370-y In fact, Cantor called infinitesimals "the cholera bacillus of mathematics". His other epithets for infinitesimals include "paper numbers" and "an abomination", see for example http://link.springer.com/article/10.1007%2Fs00407-005-0102-4

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