Is it true without the axiom of choice that every infinite cardinal is larger than every finite cardinal

Question

It seems I'm going to continue a sequence of dumb set-theoretic questions (here's the first one I asked yeserday: Simple properties of cardinal numbers). Sorry again if it's a duplicate.

I'm interested if $n\leq\mathfrak{m}$ where $n$ is a finite and $\mathfrak{m}$ is an infinite cardinal, and can it be proved without $AC$? My attempts were the following. Let say we want to prove $2\leq\mathfrak{m}$. Consider a set $M$ with cardinality $\mathfrak{m}$. It follows (?) from the definition of infinite sets that:
\begin{align}
\exists(a\in M)\exists(b\in M):a\neq b
\end{align}
Thus there is a subset of $M$ with cardinality $2$. The natural way to generalize these thoughts is to use the induction. Thus after observing $0$-case we're supposing there is a subset $X$ of $M$ set with cardinality $n$. Then we conclude $X\neq M$, and finally:
\begin{align}
\exists(a\in M):\overline{\overline{\{a\}\cup X}}=n+1
\end{align}
The problem is I'm not sure if this pseudo-proof is valid. This raises the question again…

Answer 1

The proof is fine. You're appealing to induction. You would only need to choice if you want to claim that from this proof follows that every infinite set has a countably infinite subset.

There are two key facts here:

1. An infinite set is not empty, since the empty set is finite.
2. Removing finitely many elements from an infinite set still gives us an infinite set, since the union of two finite sets is finite.

That means that if we managed to somehow find a subset of size $n$, then there is a subset of size $n+1$. And by induction that means there are subsets of every finite size.