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?
On
$C$ 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?
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?