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We are given an infinite set $A$ (countable or uncountable) and a finite set $B$. Let $P = \{ f : B \rightarrow A \} $. Is there a bijection between $P$ and $A$?

I thought of using Schroder-Bernstein theorem. It was easy to show that an injective function exists from $A$ to $P$. But I couldn't find an injective function from $P$ to $A$.

I also thought that if $|B| = b$, then $P$ is basically $A^b$, but still I couldn't think of an injective function from $A^b$ to $A$.

Trying to find a solution to this lead me to ZFC and Axiom of choice, but these topics are strictly excluded from my Math for CS course, so this should be solved without using those. Any other approach to this problem is also welcome.

Shaurya Agarwal
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  • You may find this helpful:https://math.stackexchange.com/questions/1383755/cardinality-of-the-cartesian-product-of-two-equinumerous-infinite-sets – GBA Aug 03 '24 at 13:57
  • All the math for cs courses I am familiar with implicitly assume the axiom of choice, so a naive argument would suffice. – Ethan Bolker Aug 03 '24 at 14:15

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There isn't a nice and simple answer that would sidestep axiom of choice issues for arbitrary sets $A$ here. In fact, the existence of a bijection between $P$ and $A$ for all $A$ is equivalent to the axiom of choice over ZF.

What you can do with elementary methods is to construct such bijections for the case where $A = \mathbb{N}$, $A = 2^*$ and $A = 2^\mathbb{N}$ (which should cover the cases most relevant for a Math for CS course).

Arno
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  • For $n > 2$, the existence of a bijection $A^n \to A$ is equivalent to the existence of a bijection $A^2 \to A$ by the Schroder-Bernstein theorem. One direction (from $A^2$ to $A^n$) is obvious. For the other direction, there is an obvious injection $A \to A^2$ (namely, the diagonal), while the injection $A^2 \to A^n, (a, b) \mapsto (a, b, b, ..., b)$ ($n-1$ $b$s) composed with a bijection $A^n \to A$ gives an injection $A^2 \to A$, from which the Schroder-Bernstein theorem applies. – Geoffrey Trang Aug 03 '24 at 14:25