2
$\begingroup$

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

  • 0
    Quite right too. That would be assuming the axiom of choice, which is not needed here.2011-02-11

1 Answers 1

6

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 $b^b = |\{f:b\to b\}| \leq |P(b\times b)|=2^b$.

  • 0
    sure, sorry guys :)2011-02-12