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Is it true that if two sets have the same cardinality, their power sets have the same cardinality? If so, how to prove it?

user107952
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    Yes. Try mapping each subset of X to a subset of Y via the injection between X and Y. – Michael Jul 18 '19 at 21:12
  • For any set with cardinality $k$, its power set has cardinality $2^k$. – herb steinberg Jul 18 '19 at 21:31
  • @herbsteinberg and "how to prove it"? – fleablood Jul 18 '19 at 21:40
  • This is a different question! The general idea is that for each subset, every element is either in or out. This leads to $2^k$ subsets of a set with $k$ elements. – herb steinberg Jul 18 '19 at 21:49
  • @herbsteinberg depending on your set theory (if AC is not necessarily assumed) two sets $X$ and $Y$ may have the same cardinality (i.e. there is a bijection between them) without either of them having cardinality $\kappa$, for any $\kappa$ (that is, there need not be a bijection between $X$ and any well-ordered set). It also depends on your terminology, usually "cardinality $\kappa$ " implies that $\kappa$ is a cardinal, i.e. an initial ordinal, though one may also define cardinality $|X|$ as the class of all sets that have the same cardinality as $X$ (without reference to well-orders). – Mirko Jul 19 '19 at 04:52
  • @Mirko Your statement is too deep for my relatively limited exposure to cardinals. I simply assumed every set has a cardinality, the number of its elements. – herb steinberg Jul 19 '19 at 15:11

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Yes. Let $f: X \to Y$ be a bijection.

Then show that $\hat{f}: \mathscr{P}(X) \to \mathscr{P}(Y)$ given by $$\hat{f}(A) := f[A] (= \{f(x): x \in A\})$$

is a bijection between their power sets.

Henno Brandsma
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  • Where the $(\ )$ in the last line intentional? And is there a difference between $f(A)$ and $f[A]$? – gen-ℤ ready to perish Jul 18 '19 at 22:02
  • @let'shaveabreakdown Yes, $f[A]$ is a standard notation for the image of a subset $A$ under $f$ (between braces because it's a definition you already know but I put it there for completeness). And $\hat{f}$ is a map with inputs subsets, $f$ is not, so in type there is a difference. But if you know the definition of $f[A]$, $\hat{f}(A) = f[A] \in \mathscr{P}(Y)$, for all $A \in \mathscr{P}(X)$, suffices as the definition of the bijection of powersets. – Henno Brandsma Jul 19 '19 at 04:22