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

Possible Duplicate:
No set to which all functions belong

I'm having trouble proving that there is no set of all functions. I tried the approach similar to the approach used to show that there is no set of all sets, but was not able to make this work.

First I assumed the set $A$ of all functions, and then constructed the subset $B$ where each member of the subset is not a member of itself. I could then prove that the subset was not a member of $A$. To complete the proof I need to show that $B$ is a function, but $B$ doesn't seem to be a function.

Would like some guidance on the remaining steps, or a better approach to take.

share|cite|improve this question

marked as duplicate by t.b., Asaf Karagila, Zev Chonoles Jul 13 '12 at 5:19

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

up vote 6 down vote accepted

For every set $X$ there exists a (unique) function $\{X\}\to \{1\}$.

Now, if $A$ were the set of all functions, then what would $\bigcup_{f\in A} \operatorname{Dom} f$ be?

share|cite|improve this answer
Beat me to it! :-) – Asaf Karagila Dec 3 '11 at 23:37
This gets particularly annoying if "functions" is replaced by "nets on a particular set". – dfeuer Dec 4 '11 at 4:37

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