Let $A$ a infinite set, and $a$ the cardinal number of $A$, then $a.a=a$

Question

Let $A$ a infinite set, and $a$ the cardinal number of $A$, then $a.a=a$

My attempt:

We know $a.a=card(A\times A)$ and $card(A)=a$, then we need prove $card(A\times A)=card(A)$

Let $f:A\times A\rightarrow A$ such that $f(x,y)=x$ with $x,y \in A$ this is a bijective function and this implies $card(A\times A)=card(A)$

is correct this?

The function you described is not bijective! Let $x,y \in A$ be distinct. Then $f(x,y) = f(x,x)$ but $(x,y) \neq (x,x)$, so $f$ is not injective. — diracdeltafunk, Nov 04 '19 at 20:14
$f$ is not bijective. Indeed, let $y_1,y_2\in A$ such that $t_1\neq y_2$. Then $f(x,y_1)=f(x,y_2)$ for any $x\in A$. — InsideOut, Nov 04 '19 at 20:15
In fact, stating that $A \times A$ and $A$ have the same cardinality for arbitrary infinite sets is equivalent to the axiom of choice, which is equivalent to Zorn's lemma. — Ben Grossmann, Nov 04 '19 at 20:24

score 3 · Answer 1 · edited Nov 05 '19 at 03:21

Let $X = \{(S,f) : S \subseteq A, f : S \times S \to S \;\text{is a bijection}\}$. Order $X$ by $(S,f) \leq (T,g)$ iff $S \subseteq T$ and $g|_S = f$. Use Zorn's Lemma to prove that $X$ has a maximal element, and use the fact that $\operatorname{Card}(A) + \operatorname{Card}(A) = \operatorname{Card}(A)$ (see your previous post) to show that this maximal element must be of the form $(A,f)$ (so that there exists a bijection $f : A \times A \to A$).

Let $A$ a infinite set, and $a$ the cardinal number of $A$, then $a.a=a$

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