TITLE: For a set $A$ what is $A^0$?
QUESTION [5 upvotes]: $A^0 = \varnothing$ seems wrong because then $A^1 \times A^0 = A \times \varnothing = \varnothing \neq A^{1 + 0}$. A singleton set seems more sensible, but is there a "canonical" singleton set to use? (Ie, maybe just the set $1 = \{0\} = \{\varnothing\}$ ?)

REPLY [13 votes]: In general, the set $A^{B}$ is defined to be the set of all functions
from $B$ to $A$. 
$$
A^{B}:=\{f \mid f:B\to A\}
$$
In your case, $B=\emptyset$, and there is only one function $f:\emptyset\to A$
which is the empty function (recall a function is a special set of
pairs).
Hence, $A^{0}=\{\emptyset\}$