# cardinality of $E^F$, $E, F = \emptyset$

My textbook (Naive Set Theory) asks the reader to show that $\left| E^F \right|$ = $\left| E \right| ^ \left| F \right|$ for all finite sets. In passing to induction, I noticed that this would imply that $0^0 = 1$, since there exists the trivial empty function (i.e., $\emptyset \times \emptyset = \emptyset$). The definition of exponentiation in $\omega$ says that $(\forall m \in \omega$ $(m^0 = 1))$, so this seems reasonable.

Is this correct, or should this case be discarded? If it is correct, is it correct in $\omega$ only, or everywhere?

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It's correct; for cardinal exponentiation, $0^0 = 1$. –  Arturo Magidin Jun 23 '12 at 3:45
In cardinal exponentiation, it is $1$ by definition, not by convention: $0^0 = |\varnothing^{\varnothing}| = |\{\varnothing\}| = 1$. –  Arturo Magidin Jun 23 '12 at 3:49
@user1296727: $|E|$ and $|F|$ refer to natural numbers viewed as cardinals. Remember that natural numbers are playing the role of ordinals and cardinals in set theory, not their usual arithmetical role as a subset of the real numbers. $0^0$, as cardinal exponentiation, is $1$, and the problem you quote is asking you to verify a fact of cardinal exponentiation, not of the exponentiation function of numbers. –  Arturo Magidin Jun 23 '12 at 4:03
@user1296727: I don't have my copy of Halmos near me, but as I recall, he has not defined the usual arithmetic of natural numbers as embedded inside the reals. His definitions are purely for the natural numbers as elements of $\omega$, i.e., as set-theoretic creatures. The distinction I am making is the same that we make when we say that $x^2+1$ has not roots, and then say that it has roots $i$ and $-i$; in one case we are talking in the context of real numbers, in the other in the context of complex numbers. The question is asked in one context, you are trying to think about it in another. –  Arturo Magidin Jun 23 '12 at 4:29
@user1296727: You seem to find the definition of the exponentiation $0^0=1$ difficult to accept, probably because you are used to the fact that in the context of working with the real numbers (e.g., in Calculus), $0^0$ is usually left undefined (or given the value $1$ only by convention for simplicity in writing out Taylor series/polynomials). My entire point is that that context, where $0^0$ is not defined, is not the context you are working on here. You are working in set theory. Looking at natural numbers as set theoretic creatures, not as real numbers embedded in the calculus. –  Arturo Magidin Jun 23 '12 at 4:36

As I mentioned when dealing with this extensively elsewhere, in set theory there is no reason to treat $0^0$ any different than any other set-theory exponentiation (set of all funtions from $\varnothing$ to $\varnothing$) or cardinal exponentiation (the cardinality of the set of all functions from $\varnothing$ to $\varnothing$). The general definition of exponentiation of sets, of cardinals (and ordinals, for that matter) applies to this special case and gives the value $1$. This value fits well with the corresponding uses within set theory.

There are other contexts, separate from set theory (e.g., the Calculus of limits on the real line) where one might wish to avoid giving a specific value to $0^0$; this is not one of them, and so you should feel absolutely no qualms in applying the general definition of exponentiation for sets/cardinals/natural numbers to that special case, in this context.

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The empty product is equal to $1$. To neglect to multiply by anything, is the same as multiplying by $1$.

Look at the well-known identity: $$e^z = \sum_{n=0}^\infty \frac{z^n}{n!}.$$ If $z=0$, then this is $$e^0 = \frac{0^0}{0!}+0+0+0+\cdots.$$ What should the first term be if $e^0=1$?

$0^0$ is an indeterminate form in the sense that if $f\to0$ and $g\to 0$ as $x\downarrow0$, then the limit of $f^g$ could be any non-negative number, depending on which functions of $x$ are $f$ and $g$. But I think if they are analytic, then the limit will always be $1$.

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–  MJD Jun 23 '12 at 4:07
And so does this site: math.stackexchange.com/questions/11150/zero-to-zero-power –  talmid Jun 23 '12 at 4:13
Actually, I'm a bit unclear on this. Your answer seems to assert that $0^0$ should defined as 1 in the real field (I don't believe this follows from anything other than arbitrary definition), but Arturo's linked answer asserts that this is unnecessary and cumbersome (he says it would interfere with existing theorems and require new special cases). –  Chris Jun 23 '12 at 4:30
I don't think it's merely a convention. I.e. it is in a certain sense logically demonstrable that $0^0=1$. –  Michael Hardy Jun 23 '12 at 6:14
@user1296727: For example, $$\lim_{x\to 0^+}x^x=1,$$ so it is entirely reasonable in that sense to call $0^0=1$. This may not be the logical demonstration to which Michael refers, of course. –  Cameron Buie Jun 23 '12 at 15:50