Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

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:

share|cite|improve this answer
That makes sense. What about $f:A\rightarrow \varnothing$? That doesn't seem possible for non-empty $A$. – Dan Christensen May 16 '13 at 4:30
That's a good question. Regarding your "choice" question, you might want to visit Axiom of Choice – amWhy May 16 '13 at 4:34
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 4:52
@amWhy: nice guidance! =1 – Amzoti May 16 '13 at 5:18
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.