Is it true that $A^A\simeq P(A)$?

Question

Let $A$ be an infinite set and $$A^A=\{f | f:A\to A\,\,\, \text{is a function}\}.$$
Is there any natural bijection between $A^A$ and $P(A)$?

I don't think there could be any bijection because cardinalities are different. $|A^A|=|A|^{|A|}$ but $|P(A)|=2^{|A|}$ — Alberto Andrenucci, Jun 24 '17 at 21:57
@AlbertoAndrenucci: See the comments of this question. I came up with this question from there. — Bumblebee, Jun 24 '17 at 22:00
That doesn't mean their cardinalities are different,@AlbertoAndrenucci For example $|2^{\mathbb N}|=|\mathbb N^{\mathbb N}|$ — Thomas Andrews, Jun 24 '17 at 22:00
The question is about a "natural bijection", or whether a bijection exists? I don't know how to find a natural bijection between $A$ and $A\times A$ for $A$ infinite. Or even between $A$ and $A \sqcup A$. — orangeskid, Jun 25 '17 at 15:59

score 3 · Accepted Answer · answered Jun 24 '17 at 22:04

3

Hint: You need to know for infinite $A$ that $|A\times A|=|A|$, and that for any $A$, $|P(A)^A|=|P(A\times A)|$.

It's obvious that $|P(A)|\leq |A^A|$, so you need to show that $|A^A|\leq |P(A)|$.

answered Jun 24 '17 at 22:04

Thomas Andrews

177,126

In terms of cardinal arithmetic, we might expect that $|(2^A)^A| = |2^{A \times A}|$. A nice bijection between $(2^{A})^A$ and $2^{A \times A}$ is given by "uncurrying". – Ben Grossmann Jun 24 '17 at 22:14

score 2 · Answer 2 · answered Jun 25 '17 at 14:18

2

$|A|^{|A|} \le |\mathscr{P}(A)|^{|A|} = (2^{|A|})^{|A|} = 2^{|A \times A|} = 2^{|A|} \le |A|^{|A|}$, so we have equality.

answered Jun 25 '17 at 14:18

Henno Brandsma

242,131

Wow! very nice one line solution. – Bumblebee Jun 25 '17 at 16:13
1

If you know the rules of cardinal arithmetic, yes – Henno Brandsma Jun 25 '17 at 16:17

score 2 · Answer 3 · answered Jun 25 '17 at 17:18

2

There is a natural injective function $F : A^A \to P(A\times A)$:

$$F(f)=\{ (x,f(x)) | x \in A \}$$ (ie $f \to Graph(f)$).

This shows that $$|A^A| \leq |P(A \times A)|=|P(A)|$$

Also, there is a natural injective function $G : P(A) \to A^A$ given by the characteristic function. Indeed, fix two elements $a,b \in A$ and then define $G(B)(x)=a$ if $x \in B$ and $G(B)(x)=b$ if $x \notin B$.

answered Jun 25 '17 at 17:18

N. S.

132,525

Wait, so if I understand this correctly, we don't need any form of choice anywhere, right? (The other two answers use $|A \times A| = |A|$ for infinite $A$.) – Aryaman Maithani Jul 30 '21 at 10:38
1

@AryamanMaithani The answer also uses that result, namely in the step $|P(A \times A)|=|P(A)|$. – N. S. Jul 30 '21 at 15:20

Is it true that $A^A\simeq P(A)$?

3 Answers3

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