Equality of Cardinality of Cartesian Sets

Question

I'm having a hard time thinking about this problem.

My question: Let A be a set. Define C to be the collection of all functions f: {0,1} --> A. Prove that |A x A| = |C| by constructing a bijection F: A x A --> C.

I'm assuming A x A has the same cardinality as A itself. Also, to create a bijection from A x A --> C, I think I need to prove |A x A|≤|C| and |A x A|≥|C| through Cantor Schroder-Bernstein Theorem.

Can someone please tell me how to solve this?

"I'm assuming A x A has the same cardinality as A itself. " -- You shouldn't. This doesn't hold, for example, for finite sets. — PrincessEev, Nov 29 '19 at 05:56
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"I'm assuming A x A has the same cardinality as A itself. " Well, no. if there are $CARDINALITY$ $a_\alpha \in A$ how many $(a_\alpha, a_\beta) \in A\times A$ are there? — fleablood, Nov 29 '19 at 06:07
@EeveeTrainer Oh okay. But it does hold for infinite sets like N right? Like, |N x N| = |N|? — GetRekt5000, Nov 29 '19 at 06:07
"I think I need to prove |A x A|≤|C| and |A x A|≥|C| through Cantor Schroder-Bernstein Theorem. " Far easier to just find the bijection itself. A typical element of $A\times B$ would be $(a,b)$ and a typical function of $C$ would be a $f:{0,1} \to A$ via $f(0) = a$ and $f(1) = b$. Can you see what would be the intended bijection? — fleablood, Nov 29 '19 at 06:10
For infinite sets ... yes and no. It (somewhat) depends on whether you assume the axiom of choice (see an elaboration here). If you don't mind assuming the axiom of choice, though, yes: for an infinite set $S$, $|S \times S|=|S|$. — PrincessEev, Nov 29 '19 at 06:12
Countable Infinite? Yes, but that's because $|\mathbb N|^2 = |\mathbb N|$. It's not an easy thing to prove $|\mathbb R \times \mathbb R| = |\mathbb R|$. ... And there is nothing that says $A$ is infinite. — fleablood, Nov 29 '19 at 06:14
Hint: If $f \in C$ then $f(0)\in A$ and $f(1) \in A$ and $(f(0), f(1)) \in A\times A$. — fleablood, Nov 29 '19 at 06:26

score 1 · Answer 1 · answered Nov 29 '19 at 06:24

$C$ is a set of functions from $\{0,1\} \to A$. A function $f\in C$ would be defined by simply spelling out what the two values of $f(0) =a$ and $f(1)=b$ are.

$A\times A$ is the set of all ordered pairs $(a,b)$ where $a,b \in A$.

Can you construct a bijection.

Consider $A =\{a,b,c\}$.

Then $A\times B = \{(a,a),(a,b),(a,c), (b,a), (b,b),(b,c), (c,a), (c,b), (c,c)\}$

ANd $C = \{\{f(0)=a; f(1)=a\}, \{g(0)=a; g(1)=b\}, \{h(0)=a; h(1)=c\},etc.\}$.

Can you see the bijection between them?

score 0 · Answer 2 · answered Nov 29 '19 at 06:18

Intuitively, the statement is true, because there are $|A|^2$ functions from $\{0,1\}$ to $A$-you have $|A|$ choices for $f(0)$ and again for $f(1)$. This is independent of whether $A$ is finite or infinite. The easiest approach is not through C-S-B but to directly construct the isomorphism. If I give you an $f$, what element of $C$ is it natural to pair with it?

Equality of Cardinality of Cartesian Sets

2 Answers2