TITLE: What is the definition for an infinite set?
QUESTION [8 upvotes]: I'm physicist, not a mathematician, but this time I have following question related to Math:
What is an infinite set per definition?
Of course I know, that a set is infinite, when it is not finite - clear.
But lets say I have infinite set M.
Assume there is a surjection $f: M \rightarrow \mathbb{N}$
So every element of $m \in M$ points to a given $n \in N$.
Such a set M must be, of course infinite, but does the definition suffice for all infinite sets, including non countable ones so I can say:
"Each infinite set has a surjection $f: M \rightarrow \mathbb{N}$" ?
For example a surjective function for the non countable infinite set $\mathbb{R^+_0}$ could be
$x \in [n-1, n) \mapsto n$
But does my definition include all possible infinite sets and is this definition reasonable at all?
Sorry for mathematical incorrectness, but I think one can understand what I mean.

REPLY [2 votes]: This is an interesting question, because we can interpret it as asking if there is any infinite cardinality strictly smaller than $\aleph_0$.
Infinite sets are usually defined in terms of finite sets. A set is finite if there is a one-to-one mapping between it and a set of the form $\boldsymbol n = \{0, 1, ..., n\}$ for some $n\in \mathbb{N}$.
So this interesting question is, is there any set which satisfies the standard definition but does not have a surjection onto $\mathbb{N}$?
Let's try to show that.
Let us suppose we have a set $S$ such that for every $n\in \mathbb N$ there is no one-to-one mapping $f:S\to \boldsymbol n$.
First we note that at the very least, for every $n\in \mathbb N$ we will always be able to find at least a surjective $g_n:S\to \boldsymbol n$.
We can show this by induction. The base case for $n=0$ is trivial. Suppose that there is a surjective $g_n:S\to \boldsymbol n$. Since we assumed that there are no one-to-one mappings between $S$ and $\boldsymbol n$, there must be two elements $a_n,b\in S$ such that $g_n(a) = g_n(b)$. Hence we can define $g_{n+1}(x) = g_n(x)$ if $x\not = a$ and $g_{n+1}(a) = n+1$.
Ok so now we have a surjection $g_n: S \to \boldsymbol n$ for every $n$. Not only that, but we have also defined along the way a series $A = \{a_0, ..., a_n\}$ such that $g_n(a_n)=n$ for every $n$.
Let us define $g:S \to \mathbb N$ such that $g(a_n) = n$ ($g(x) = 0$ if $x \not\in A$.
Hence we have defined a surjection $g:S \to \mathbb N$, and thus we have seen that any infinite set satisfying the classical definition also satisfies that there is a surjection between it and the natural numbers.
PS: I am 78% sure that this construction relies on the axiom of choice, but right now I cant see exactly how.