Thread: some play on empty sets

1. some play on empty sets

Hello

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

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

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

2. Re: some play on empty sets

Originally Posted by issacnewton
I just have some general question regarding functions. Consider the sets A and B. Now if $A\neq \varnothing\mbox{ and }B=\varnothing$, then the definition of the function $f:A\to B$ fails but there is still a function $g:B\to A$, even though in both cases $A\times B=\varnothing$ . So if we define $^AB$ as the set of functions from A to B, then if $B=\varnothing\mbox{ and } A\neq \varnothing$, then
$^AB=\varnothing\mbox{ and } ^BA=\{\varnothing\}$
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 $X~\&~Y$ is a set and $Y^X$ is the set of all functions $X\to Y$ then the quote
" $(i) Y^{\emptyset}$ has exactly one element, namely $\emptyset$, whether $Y$ is empty or not, and (ii) if $X$ is not empty, then $\emptyset^X$ is empty."

3. Re: some play on empty sets

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

4. Re: some play on empty sets

Originally Posted by issacnewton
Plato, thats what I am saying too, am I not ?
I think so. But that notation is confusing.

5. Re: some play on empty sets

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