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If $A$ and $B$ are two set then $B^A$ is the set of all function from $A$ to $B$, that is $$ B^A:=\{f\in\mathcal P(A\times B):f\,\,\text{function}\} $$ Now if $A$ and $B$ are finite it is not hard to show that the cardinal $|B^A|$ of $B^A$ is the cardinal $|B|$ raised to the power of the cardinal of $|A|$, that is $$ |B^A|=|B|^{|A|} $$ So if $B$ was empty then the set $A\times B$ would be empty so that the set $\mathcal P(A\times B)$ is a singleton set containing the empty set $\emptyset$ as element and thus in this case the set $B^A$ is a singleton set containing the empty function as element and this is consistent with respect the identity $$ n^0=1 $$ for any $n>0$. However if $A$ and $B$ was simultaneously empty then due to the same argumentations $B^A$ is a singleton set containing the empty function but this, apparently, is absurd because in ordinary arithmetic the exponentiation $0^0$ is not defined. In any case I know that if $g_m:\Bbb N\times\Bbb N\rightarrow\Bbb N$ is for any $m\in\Bbb N$ the function defined as $$ g_m(x,y):=x\cdot m $$ for any $x,y\in\Bbb N$ then the recursion theorem guarantees the existence of a function $f_m:\Bbb N\rightarrow\Bbb N$ such that $$ f_m(0)=1\\ f_m(n+1)=g_m\big(f_m(n),n\big)\,\,\forall n\in\Bbb N $$ and is not hard to show that this function is exactly such that $$ f_m(n)=m^n $$ for any $n\in\Bbb N$ so that apparently it is possible to define $0^0$ because we did not establish any constraint on the values of $m$.

So could someone explain if the set $\emptyset^\emptyset$ exist, please?

Eric Wofsey
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Antonio Maria Di Mauro
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    You're over-estimating the importance of "ordinary arithmetic" - just going by the definition, it's clear that $\emptyset^\emptyset$ makes sense and is a set with one element, namely the empty function - so $\emptyset^\emptyset={\emptyset}$ (using the standard set-theoretic approach that a function is simply a set of ordered pairs as opposed to e.g. set of ordered pairs + explicit domain + explicit codomain). I'm posting this as a comment rather than an answer since I'm pretty confident this has been asked before. – Noah Schweber Mar 09 '22 at 16:13
  • I took a look to the duplicate questions but unfortunately I did not find anything about what I asked here, sorry. So I ask foreman about your comment: the set $\emptyset^\emptyset$ exists and it has one element and this is consistent with the ZFC theory, right? – Antonio Maria Di Mauro Mar 09 '22 at 16:19
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    Not only consistent with ZFC, it's a direct consequence of the definition which is outright provable in ZFC (or indeed much less). – Noah Schweber Mar 09 '22 at 16:23
  • Okay, only a bit curiosity: if $B$ was empty the $B^A$ has one element but this is not consistent with the identity $$0^n=0$$ for any $n>0$. Perhaps am I wrong? – Antonio Maria Di Mauro Mar 09 '22 at 16:26
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    No, if $B$ is empty then $B^A$ has at most one element. $B^A$ is empty if $B=\emptyset$ and $A\not=\emptyset$. Don't try to generalize from the arithmetic setting, despite the notational overload, just work directly with the definition. – Noah Schweber Mar 09 '22 at 16:28
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    If $B$ is empty and $A$ is empty, then $B^A$ has one element (the empty function). If $B$ is empty and $A$ is non-empty, then $B^A$ has zero elements (since a function must map every element of $A$ - and there are elements of $A$! - to an element of $B$ - and there are no elements of $B$!). This is consistent with elementary arithmetic: $0^0 = 1$, but $0^n = 0$ when $n>0$. – Alex Kruckman Mar 09 '22 at 16:29
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    Tangentially, looking at standard arithmetic it's worth noting that whether $0^0$ is undefined is a matter of convention. (This is relevant to the last sentence of @AlexKruckman's comment above.) – Noah Schweber Mar 09 '22 at 16:31
  • First of all thanks for your upvoted. Only one dubt: the empty set is obviously a relation that is contained any product $A\times B$ so that if $\emptyset^A$ is empty then I argue that the empty relation is not a relation between $A$ and $B$, right? – Antonio Maria Di Mauro Mar 09 '22 at 16:33
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    I do suspect this question may be a duplicate but it is not answered by any of the questions it was linked to, so I'm reopening. The question here is about an apparent contradiction between the fact that $\emptyset^\emptyset$ is defined and $0^0$ is not, not about whether the empty function is a function. – Eric Wofsey Mar 09 '22 at 16:34
  • I discussed functions with empty domains abd codomains in some detail here: https://math.stackexchange.com/a/4236083/25554 – MJD Mar 09 '22 at 18:14
  • See my answer at https://math.stackexchange.com/q/484962 . – Andreas Blass Mar 09 '22 at 18:25

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