5
$\begingroup$

$A^0 = \varnothing$ 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 $1 = \{0\} = \{\varnothing\}$ ?)

$\endgroup$
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. $\endgroup$
    – anon
    Sep 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. $\endgroup$
    – alecbz
    Sep 21, 2013 at 0:23

1 Answer 1

13
$\begingroup$

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, $A^{0}=\{\emptyset\}$

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$
    – alecbz
    Sep 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. $\endgroup$
    – guy
    Sep 21, 2013 at 1:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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