Does $k+\aleph_0=\mathfrak{c}$ imply $k=\mathfrak{c}$ without the Axiom of Choice? I'm currently reading a little deeper into the Axiom of Choice, and I'm pleasantly surprised to find it makes the arithmetic of infinite cardinals seem easy. With AC follows the Absorption Law of Cardinal Arithmetic, which states that for $\kappa$ and $\lambda$ cardinal numbers, the larger infinite and the smaller nonzero, then $\kappa+\lambda=\kappa\cdot\lambda=\max(\kappa,\lambda)$.
I was playing around with the equation $k+\aleph_0=\mathfrak{c}$ for some cardinal $k$. From the above, it follows that $\mathfrak{c}=k+\aleph_0=\max(k,\aleph_0)$, which implies $k=\mathfrak{c}$.
I'm curious, can we still show $k=\mathfrak{c}$ without the Axiom of Choice? Is it maybe possible to bound $\mathfrak{c}-\aleph_0$ above and below by $\mathfrak{c}$? But then I'm not quite sure such algebraic manipulations even mean anything, or work like that here. Certainly normal arithmetic does not! Thanks.
 A: There's another way apart from joriki's answer that I think is a bit easier and is usually considered the "standard" way to show what you are asking:
First of all we have that $\mathfrak{c}\times\mathfrak{c}=\mathfrak{c}$. To see this using subsets of natural numbers (for example) observe that given $X,Y\in\mathcal{P}(\omega)$ the function $(X,Y)\mapsto\{2n : n\in X\}\cup\{ 2n+1 : n\in Y\}$ is an injection.
Thus we need to show that given a countable set $X\subset\mathbb{R}\times\mathbb{R}$ we have that $|\mathbb{R}\times\mathbb{R}\setminus X|=\mathfrak{c}$. Take the projection of $X$, $\pi(X)$. $\pi(X)$ is at most countable. This doesn't use the axiom of choice since $X$ is well orderable (because it's countable). Therefore, there exists $x\in\mathbb{R}$ such that $x\notin\pi(X)$. Then the set $\{x\}\times\mathbb{R}$ (that has cardinality $\mathfrak{c}$) is disjoint from $X$ and thus a subset of $\mathbb{R}\times\mathbb{R}\setminus X$. Apply Cantor-Bernstein theorem and you are done.
A: There is a general argument without choice: Suppose ${\mathfrak m}+{\mathfrak m}={\mathfrak m}$, and ${\mathfrak m}+{\mathfrak n}=2^{\mathfrak m}$. Then ${\mathfrak n}=2^{\mathfrak m}.\,$ This gives the result. 

The argument is part of a nice result
  of Specker showing that if CH holds
  for both a cardinal ${\mathfrak m}$
  and its power set $2^{\mathfrak m}$,
  then $2^{\mathfrak m}$ is
  well-orderable. This shows that GCH
  implies choice, and that the proof is
  "local". It is still open whether CH
  for ${\mathfrak m}$ implies that
  ${\mathfrak m}$ is well-orderable.

Anyway, here is the proof of the statement above: Note first that $2^{\mathfrak m}\cdot 2^{\mathfrak m}=2^{{\mathfrak m}+{\mathfrak m}}=2^{\mathfrak m}={\mathfrak m}+{\mathfrak n}$. 
Let $X$ and $Y$ be disjoint sets with $|X|={\mathfrak m}$, $|Y|={\mathfrak n}$, and fix a bijection $f:{\mathcal P}(X)\times{\mathcal P}(X)\to X\cup Y$.
Note that there must be an $A\subseteq X$ such that the preimage $f^{-1}(X)$ misses the fiber $\{A\}\times {\mathcal P}(X)$. Otherwise, the map that to $a\in X$ assigns the unique $A\subseteq X$ such that $f^{-1}(a)$ is in $\{A\}\times {\mathcal P}(X)$ is onto, against Cantor's theorem.
But then, for any such $A$, letting $g(B)=f(A,B)$ gives us an injection of ${\mathcal P}(X)$ into $Y$, i.e., $2^{\mathfrak m}\le {\mathfrak n}$. Since the reverse inclusion also holds, we are done by Schroeder-Bernstein.
(Note the similarity to Apostolos's and Joriki's answers.)
The original reference for Specker's result is Ernst Specker, "Verallgemeinerte Kontinuumshypothese und Auswahlaxiom", Archiv der Mathematik 5 (1954), 332–337. A modern presentation is in Akihiro Kanamori, David Pincus, "Does GCH imply AC locally?", in "Paul Erdős and his mathematics, II (Budapest, 1999)", Bolyai Soc. Math. Stud., 11, János Bolyai Math. Soc., Budapest, (2002), 413–426.

Note that assuming that ${\mathfrak m}$ is infinite is not enough for the
  result. For example, it is consistent
  that there are infinite Dedekind
  finite sets $X$ such that ${\mathcal P}(X)$ is also Dedekind finite. To be
  Dedekind finite means that any proper
  subset is strictly smaller. But if
  $2^{\mathfrak m}$ is Dedekind finite
  and $2^{\mathfrak m}={\mathfrak n}+{\mathfrak l}$ for nonzero
  cardinals ${\mathfrak n},{\mathfrak l}$, then we must have 
  ${\mathfrak n},{\mathfrak l}<2^{\mathfrak m}$.

A: I think that without AC you can show that $a\ge\aleph_0$ implies $\aleph_0+a = a$.
First, you can find a bijection $f$ between $\mathbb N\times\{1\}\cup\mathbb N\times\{2\}$ and $\mathbb N$. (You can write down an explicit formula for it, no choice needed.)
Now, you assume that there is an injection $g: \mathbb N \to A$, where $|A|=a$.
Then you can define $h: \mathbb N\times\{1\}\cup A\times\{2\} \to A$ by
$h(n,1)=g(f(n,1))$
$h(g(n),2)=g(f(n,2))$
$h(x,2)=x$ if $x\in A \setminus g[\mathbb N]$.
EDIT:
The basic idea is just that if you have a bijective copy of the set $\mathbb N$ in the set $A$, you can make two copies of $\mathbb N$ from it. I just tried to write it down formally, so that we can check whether AC is used somewhere.
EDIT: As Apostolos correctly pointed out, this does not answer the original question yet. (I kind of misread the question.) It remains to show that $k+\aleph_0=\mathfrak c$ $\Rightarrow$ $\aleph_0\le k$. This can be down similarly as in joriki's answer - which show that it is much more elegant (more to the point) than my attempt.
A: Given a set $K$ with $\lvert K \rvert=k$ and a bijection $f:\mathbb{R}\to K\cup \mathbb{N}$ (assuming without loss of generality that $K$ is disjoint from $\mathbb{N}$), pick some $x\in\mathbb{R}$ such that $f$ maps no element of $x+\mathbb{N}\subset\mathbb{R}$ to $\mathbb{N}$ (this is just one choice and it's possible since only countably many elements are mapped to $\mathbb{N}$), and define a bijection $g:\mathbb{R}\to K$ by mapping $f^{-1}(\mathbb{N})$ to $f(x+2\mathbb{N})$, mapping $x+\mathbb{N}$ to $f(x+1+2\mathbb{N})$, and mapping $\mathbb{R}\setminus\left((x+\mathbb{N})\cup f^{-1}(\mathbb{N})\right)$ as in $f$. This bijection establishes without choice that $k=\lvert K \rvert=\lvert\mathbb{R}\rvert=\mathfrak{c}$.
