$A^0 = \varnothing$ 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 $1 = \{0\} = \{\varnothing\}$ ?)
$\begingroup$
$\endgroup$
2
-
1$\begingroup$ A function $0\to A$ is a subset of $\,0\times A$. There is only one subset of $\varnothing$, namely $\varnothing$ itself. And it vacuously satisfies the property of being a function ($\forall x\in\varnothing\,\exists!\,y\in A:(x,y)\in 0\times A$). So $A^0=\{\varnothing\}$ which has cardinality $1$, and $A^1\times A^0\cong A^{1+0}$ as a result. $\endgroup$– anonSep 21, 2013 at 0:20
-
$\begingroup$ @anon ah ok, that makes more sense. I intended $A^0$ to be the "0-ary" cartesian product of $A$ with itself, but the interpretation as functions from $0$ to $A$ makes it clearer I guess. $\endgroup$– alecbzSep 21, 2013 at 0:23
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
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, $A^{0}=\{\emptyset\}$
-
$\begingroup$ I believe the intended interpretation (this was being used in a problem) was a "0-ary" cartesian product (it was expressed in the form "$A^n$ where $n \in \mathbb{N}$", including $n = 0$), but yeah, the interpretation as functions from $A$ to $0 = \varnothing$ makes it clearer. $\endgroup$– alecbzSep 21, 2013 at 0:26
-
$\begingroup$ @alecbenzer one could regard $n$ as $\{0,1,...,n-1\}$ so that talking about Cartesian products is captured by this definition. $\endgroup$– guySep 21, 2013 at 1:50