Small claim regarding addition of cardinals I want to show that if $\alpha,\beta$ are cardinals such that $\alpha=\alpha+\beta$ and $0<\beta$ then $ \aleph_{0}\leq\alpha$
It should be fairly simple but for some reason I keep getting stuck.
 A: Given any finite cardinal $n$ you can show that $n < n + \beta$ for any non-zero cardinal $\beta$.  This then gives the result.
A: The given arguments assume choice, so that "$\alpha\ge\aleph_0$" is the same as "$\alpha$ is infinite". 
Without choice, we need to say a bit more. Note first (by Bernstein-Schroeder) that if $\alpha=\alpha+\beta$, then $\alpha=\alpha+1$: Since $\alpha\le\alpha+1\le\alpha+\beta=\alpha$, then also $\alpha=\alpha+1$. 
It follows that if $A$ has size $\alpha$ and $*\notin A$, then there is a bijection $f$ between $A\cup\{*\}$ and $A$. Using $f$, it is easy to find a sequence $x_0,x_1,\dots$ of distinct elements of $A$, namely $$f(*), f(f(*)),f(f(f(*))),\dots$$ This sequence witnesses that indeed $\aleph_0\le\alpha$.
A: If $\alpha+\beta=\alpha$ then certainly $\beta\leq\alpha$. If $\alpha$ is finite, $\alpha<\alpha+1\leq\alpha+\beta=\alpha$, which is a contradiction.
It only remains to show that $\alpha<\alpha+1$, but this is the most basic application of the pigeonhole principle.
