Re: some play on empty sets

Quote:

Originally Posted by

**issacnewton** I just have some general question regarding functions. Consider the sets A and B. Now if $\displaystyle A\neq \varnothing\mbox{ and }B=\varnothing$, then the definition of the function$\displaystyle f:A\to B$ fails but there is still a function $\displaystyle g:B\to A$, even though in both cases $\displaystyle A\times B=\varnothing$ . So if we define $\displaystyle ^AB$ as the set of functions from A to B, then if $\displaystyle B=\varnothing\mbox{ and } A\neq \varnothing$, then

$\displaystyle ^AB=\varnothing\mbox{ and } ^BA=\{\varnothing\}$

is that right reasoning ? I need this result in some proof.

This question can start an argument. The emptyset is problematic.

That said, I will give a quote directly from Halmos.

If each of $\displaystyle X~\&~Y$ is a set and $\displaystyle Y^X$ is the set of all functions $\displaystyle X\to Y$ then *the quote*

"$\displaystyle (i) Y^{\emptyset}$ has exactly one element, namely $\displaystyle \emptyset$, whether $\displaystyle Y$ is empty or not, and (ii) if $\displaystyle X$ is not empty, then $\displaystyle \emptyset^X$ is empty."

Re: some play on empty sets

Plato, thats what I am saying too, am I not ?

Re: some play on empty sets

Quote:

Originally Posted by

**issacnewton** Plato, thats what I am saying too, am I not ?

I think so. But that notation is confusing.

Re: some play on empty sets

Thats what Daniel (Velleman) uses in his book. Just following him.......