First let us clear up the definitions, since there may be some confusion about them.
We say that a set $A$ is finite, if there is a natural number $n$ and a bijection between $A$ and $\{0,\ldots,n-1\}$. If $A$ is not finite, we say that it is infinite.
We say that a set $A$ is Dedekind-finite, if whenever $f\colon A\to A$ is an injective function, then $f$ is a bijection. If $A$ is not Dedekind-finite, we say that it is Dedekind-infinite.
What can we say immediately?
- Every finite set is Dedekind-finite, and every Dedekind-infinite set is infinite.
- $A$ is Dedekind-infinite if and only if it has a countably infinite subset if and only if there is an injection from $\Bbb N$ into $A$.
- Equivalently, $A$ is Dedekind-finite if and only if it has no countably infinite subset if and only if there is no injection from $\Bbb N$ into $A$.
While in some low-level courses the definitions may be given as synonymous, the equivalence between finite and Dedekind-finite (or infinite and Dedekind-infinite) requires the presence of countable choice to some degree. This was shown originally by Fraenkel in the context of $\sf ZFA$ (where we allow non-set objects in our universe), and later the proof was imitated by Cohen in the context of forcing and symmetric extensions for producing a model of $\sf ZF$ without atoms where this equivalence fails.
Interestingly enough, Dedekind-finiteness can be graded into various level of finiteness, so some sets are more finite than others. For example, it is possible for a Dedekind-finite set to be mapped onto $\Bbb N$, which in some way makes it "less finite" than sets which cannot be mapped onto $\Bbb N$.
Yes, you have a problem there. Specifically, your function returns singletons of elements of $A$, not elements of $A$.
The obvious way to correct this is to take $E=A\cup\mathcal P(A)$ and $G=\{(B,x)\mid x\in B\subseteq A\}$.
Best Answer
I will assume you are familiar with the definitions of cardinality and the ordering of cardinals, even in the absence of choice. If anything is unclear please let me know, and I will see to elaborate if needed.
(This is a very nice exercise, in case you do not see it right away).
When assuming the axiom of choice, $\aleph(\kappa)$ is $|\kappa|^+$ (that is the successor cardinal of $|\kappa|$). we have, if so, that $\aleph(\omega)=\omega_1$.
Without the axiom of choice this gets slightly more complicated, since there are sets which cannot be well ordered. In a way, Hartogs number measures how much of the set can we well order.
Consider, for example, the case of $A$ being an amorphous set (that is, $A$ is infinite, and every $B\subseteq A$ is either finite or $A\setminus B$ is finite). Since $A$ is infinite every finite cardinal can be embedded into it. However $\aleph_0\nleq |A|$, therefore $\aleph(A)=\omega$.
Otherwise there is $f\colon \aleph(A)\to\alpha$ which is a injection, and $g\colon\alpha\to A$ which is an injection. Composition of injective functions is injective, therefore $g\circ f\colon \aleph(A)\to A$ is an injection of $\aleph(A)$ into $A$, which is a contradiction to the definition of $\aleph(A)$.
Before proving this lemma, let us consider this useful corollary:
Proof: Since $\aleph(\kappa)$ cannot be injected into $\kappa$, in particular we have that $\aleph(\kappa)\nleq\kappa$, therefore $\kappa<\aleph(\kappa)$ and so it can be well ordered.
Proof of Lemma: Let $A$ be of cardinality $\kappa$ and $P$ of cardinality $\aleph_\alpha$, without loss of generality we can assume the two sets are disjoint.
Since $|A\times P|=|A\cup P|$, we can also assume there are two disjoint sets $|A'|=|A|$ and $|P'|=|P|$ such that $A\times P=A'\cup P'$.
Since we divide an infinite set into two parts, at least one of the following is bound to occur:
There exists some $a\in A$ such that $\langle a,p\rangle\in A'$ for every $p\in P$, and then we have that $\kappa\ge\aleph_\alpha$ (by the map $p\mapsto\langle a,p\rangle$).
If there is no $a$ as above, then for every $a\in A$ there is some $p\in P$ such that $\langle a,p\rangle\notin A'$. For every $a\in A$ let $p_a\in P$ be the minimal $p\in P$ such that $\langle a,p\rangle\in P'$, and so the injective function $a\mapsto\langle a,p_a\rangle$ is an injective function of $A$ into $P'$, and therefore $\kappa\le\aleph_\alpha$.
Now the main theorem, proved by Tarski.
Proof: Let $\kappa$ be an infinite cardinal. Consider $\kappa+\aleph(\kappa)$. Our assumption gives us:
$$\kappa+\aleph(\kappa)=(\kappa+\aleph(\kappa))^2=\kappa^2+2\kappa\cdot \aleph(\kappa)+\aleph(\kappa)^2\ge\kappa\cdot \aleph(\kappa)\ge\kappa+\aleph(\kappa)$$
Where the last $\ge$ sign is due to the function from $\kappa\cup \aleph(\kappa)$: $$x\mapsto\begin{cases} \langle x,0\rangle & x\in\kappa,x\neq k\\ \langle t,x\rangle & x\in \aleph(\kappa)\\ \langle k,1\rangle & x=k\end{cases}$$ (For fixed $t,k\in\kappa$ and $t\neq k$. Note that we use the fact $\kappa$ is infinite, therefore it has at least two elements).
By the Corollary we have that $\kappa$ can be well ordered.
Pointed out to me on the MathOverflow post of this question, that there was a request to avoid the ordinals. I did think about it, a little bit.
I do remember that studying this proof originally, as well as the proof that "If $\kappa$ is such that $|A|\le\kappa<|\mathcal P(A)|$, then $|A|=\kappa$" implies choice, offers no immediate intuition as for why does the proof works. The use of Hartogs number looks like a magic trick.
However, since the first time I had studied this proof I have gained more than a bite of intuition with regards to the axiom of choice. It seems that this is a good place to slightly give the intuitive explanation for this proof.
I will start by saying that I do not believe that any proof of this theorem will be both concise and reveal the intuition behind it. So even if I were to think really hard and transform this proof into a choice function sort of proof, I doubt it will be any more revealing in its intuition, which is why I have decided to write the following instead -- it seems more profitable to the reader.
It is a rather well known fact that the axiom of choice is equivalent to the assertion that every set can be well ordered. In the absence of choice there are sets that cannot be well ordered. Hartogs number is a way to measure how much out of the set can be well ordered.
The axiom of choice is equivalent, as well, to the assertions that the cardinalities of any two sets can be compared. This is exactly due to the idea of Hartogs number - if we can compare a cardinal with its Hartogs then it can be well ordered.
So essentially what we try to have is "enough" comparability to deduce the well ordering principle. This is exactly what the tricky lemma gives us.
The idea behind the proof of the lemma is that if we break the multiplication into a sum then we can find a way to compare the sets. We can replace the requirement that the second set is well ordered simply by "has a choice function on its power set" (this, however, is equivalent to saying that the set can be well ordered).
From this the theorem follows, proving that we can compare every infinite set with its Hartogs number, therefore implying the set can be well ordered.