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I am stuck on an elementary proof on the cardinality of sets on the following point:

Given two possibly empty sets, $A$ and $B$, I need to prove the existence of any function $f:A\rightarrow B$. Is it possible? Perhaps using AC? I'm thinking you must have at least a non-empty $B$ (all elements of $A$ mapped to the same point). Any help would be appreciated.

Dan Christensen
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With the issue of possibly empty sets $A, B$, you might want to read up a bit on the existence of the Empty function.

In mathematics, an empty function is a function whose domain is the empty set. For each set A, there is exactly one such empty function

$$f_A: \varnothing \rightarrow A.$$

The graph of an empty function is a subset of the Cartesian product ∅ × A. Since the product is empty the only such subset is the empty set ∅. The empty subset is a valid graph since for every x in the domain ∅ there is a unique y in the codomain A such that (x,y) ∈ ∅ × A. This statement is an example of a vacuous truth since there is no x in the domain.

For perhaps a better explanation than that provided by Wikipedia, read this earlier post:

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amWhy
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  • That makes sense. What about $f:A\rightarrow \varnothing$? That doesn't seem possible for non-empty $A$. – Dan Christensen May 16 '13 at 04:30
  • That's a good question. Regarding your "choice" question, you might want to visit Axiom of Choice – amWhy May 16 '13 at 04:34
  • You are absolutely correct, @Dan. If $A$ is non-empty, then there exists no function $A\to\varnothing.$ This corresponds to the fact that $0^{\mathfrak a}=0$ for any non-zero cardinal $\mathfrak a$. – Cameron Buie May 16 '13 at 04:52
  • @amWhy: nice guidance! =1 – Amzoti May 16 '13 at 05:18
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    Dan, you are absolutely right that for $f: A \to B$, we need only stipulate that $B$ is not empty if A is not empty to guarantee the existence of a function from $f$. – amWhy May 16 '13 at 21:53