# Cardinal arithmetic: $b\ge x>1$ and $b^2=b$ implies $x^b=2^b$

Let $x$ and $b$ represent cardinals. Assume that $b\geq x > 1$ and $b^2=b$. Prove that $x^b=2^b$. Thanks!

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Hint: $2^b=2^{b^2}=(2^b)^b\ge b^b$. –  George Lowther Feb 11 '11 at 23:36
Cantor theorem.. Thanks –  user6163 Feb 11 '11 at 23:45
Actually, Cantor's theorem is $2^b > b$. We just need the much weaker $2^b\ge b$. –  George Lowther Feb 11 '11 at 23:50
In this course we cannot rely on the fact that for all infinte cardinals b*b=b –  user6163 Feb 11 '11 at 23:51
Quite right too. That would be assuming the axiom of choice, which is not needed here. –  George Lowther Feb 11 '11 at 23:54

Since $b^2=b$ we know that $b$ is an infinite cardinal. Then we have $2^b \leq x^b \leq b^b$ since $2\leq x$ and $x\leq b$. But $2^b=b^b$ for infinite cardinals.
It was pointed out that we should show the claimed $2^b=b^b$ in the above proof. So let's give a proof of that. First, we note that as $b\geq2$ then we have $2^b\leq b^b$, so we just need to show $b^b\leq 2^b$. We know (left as exercise) that $2^b=|P(b)|$ and $b^b=|\{f:b\to b\}|$. So it will suffice to construct an injection from $\{f:b\to b\}$ (the set of all functions from $b$ to $b$) into $P(b)$ (the powerset of $b$). Now, we know that $b=|b\times b|$ (another exercise), so it is good enough to inject $\{f:b\to b\}$ into $P(b\times b)$. But any function $f:b\to b$ corresponds to the set $\{\langle\alpha,f(\alpha)\rangle\ :\ \alpha\in b\}\in P(b\times b)$. Note that distinct functions will map to distinct elements of $P(b\times b)$. Thus $b^b = |\{f:b\to b\}| \leq |P(b\times b)|=2^b$.
But just stating $2^b=b^b$ is close to simply stating the full result as a fact. –  George Lowther Feb 11 '11 at 23:37