cardinality of division by zero [closed]

let $x=\frac{1}{0}$

If a set has a cardinality of $x$, would its power set also have a cardinality of $x$? I guess the cardinality of $x$ must be greater than $\aleph_n$ for any finite n. Is there any way to prove/disprove this or is it undecidable?

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No set has size $x$. See here for more. –  JavaMan Feb 10 '12 at 13:11
The expression $\frac{1}{0}$ has no meaning. The only sort of related thing one needs to know is that when $\epsilon$ is tiny positive, then $\frac{1}{\epsilon}$ is very large. –  André Nicolas Feb 10 '12 at 13:12
–  Asaf Karagila Feb 10 '12 at 13:35

$\frac 10$ is not defined at all, so in particular it is not a cardinality of any set.
However, in ordinary set theory, Cantor's theorem tells us that $\mathcal P(X)$ and $X$ can never have the same cardinality for any $X$. So if you want to define $\frac 10$ to mean a cardinality such that $2^{\frac 10}=\frac 10$, you had better do that in a heterodox set theory where Cantor's argument doesn't work. (Such set theories have been proposed, e.g. Quine's New Foundations, but to the best of my knowledge they are not generally considered safe for doing ordinary mathematics in).
$P(\aleph_n)\not=\aleph_n$ for any finite n, but what about $\aleph_{\aleph_n}$? –  Angela Richardson Feb 11 '12 at 6:08
@Angela, I repeat: $\mathcal P(X)$ and $X$ can never have the same cardinality for any set $X$. This is independent on where in the aleph sequence $X$ is, or indeed (in the absence of AC) whether $X$ is equinumerous with an aleph number at all. –  Henning Makholm Feb 11 '12 at 14:25