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Let $A$ and $B$ be sets. Suppose that $\#A<\#B$. That is, $B$ has strictly greater cardinality than $A$.

Is it necessarily true that $\#(2^A)<\#(2^B)$? In particular, is this question decidable in ZFC without the (generalized) continuum hypothesis?

I know that $\#(2^A)=\#(2^B)\Longrightarrow\#A=\#B$ is not decidable in ZFC, but this seems to be a different question (though I'm not 100% certain). It is also true (in ZFC) that $\#A<\#B$ implies $\#(2^A)\leq\#(2^B)$. The question is whether the “weak inequality” can be strengthened.

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1 Answers1

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Is the same question. Cardinals are always comparable (with AC) and $\#(2^A)=\#(2^B)\Longrightarrow\#A=\#B$ is equivalent to $\#A\ne\#B\Longrightarrow\#(2^A)\ne\#(2^B)$.

Martín-Blas Pérez Pinilla
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