Is infinite a infinite or finite

Question

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?

Answer 1

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.

Answer 2

No.

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Answer 3

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..