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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?

Arturo Magidin
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Chris
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    It's correct; for cardinal exponentiation, $0^0 = 1$. – Arturo Magidin Jun 23 '12 at 03:45
  • What about the exponentiation of natural numbers themselves? – Chris Jun 23 '12 at 03:47
  • (I never saw this with Dedekind cuts, so I assume it doesn't hold in $\mathbf{R}$.) – Chris Jun 23 '12 at 03:47
  • For arithmetical exponentiation, which includes natural numbers qua natural numbers(as opposed to natural numbers as cardinals or ordinals), integers, rationals, and reals, $0^0$ is generally undefined. But this is cardinal exponentiation. – Arturo Magidin Jun 23 '12 at 03:48
  • So it is unity by convention? – Chris Jun 23 '12 at 03:48
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    In cardinal exponentiation, it is $1$ by definition, not by convention: $0^0 = |\varnothing^{\varnothing}| = |{\varnothing}| = 1$. – Arturo Magidin Jun 23 '12 at 03:49
  • $0^0$ is undefined in general, and in each case one defines it as convenient, but in this case $0^0 = 1$ precisely because there's only one function $\emptyset \to \emptyset$. See the very related http://math.stackexchange.com/questions/11150/zero-to-zero-power. – talmid Jun 23 '12 at 03:51
  • (I am mostly just checking if the theorem Halmos states does actually imply that $0^0 = 1$ in the arithmetic of the natural numbers. Does it?) – Chris Jun 23 '12 at 03:52
  • @user1296727: Halmos is not talking about arithmetical exponentiation at all, he's talking about cardinal exponentiation only. When you use the vertical bars, you are automatically talking about cardinals. Writing $|E^F|=|E|^{|F|}$ means that he is talking about cardinals. – Arturo Magidin Jun 23 '12 at 03:56
  • But $\left| E \right|$ and $\left| F \right|$ refer to natural numbers - which can be bijected with $E$ and $F$ - so the fact of cardinal arithmetic seems to imply it in ... arithmetical ... arithmetic. – Chris Jun 23 '12 at 03:59
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    @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 04:03
  • @Arturo How could the naturals be a subset of the reals? Can't there be at best an isomorphism between them and certain elements of the real field? (i.e., these are different from Dedekind cuts, or any other formulation) – Chris Jun 23 '12 at 04:14
  • @Arturo (But I apologize; I haven't read about ordinal arithmetic, and the problem set was introduced without mention of the word "cardinal". So I do not understand the distinction you are making.) – Chris Jun 23 '12 at 04:16
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    @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 04:29
  • @Arturo I'm not sure that I understand you. I was working (explicitly) in $\omega$, whose elements are different from the "corresponding" real-field "natural numbers". I understand that much. I don't understand how the numbers I was working with when discussing cardinality, like $\left E \right$ and $\left F \right$, were different from ordinary elements of $\omega$. (If you can't explain that to me in a chat-box, that's okay.) – Chris Jun 23 '12 at 04:34
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    @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 04:36
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    @user1296727: So that pause you feel in wanting to "skip" $0^0$ (omit it) is out of place; you are not looking at the natural numbers as part of the real number system with functions, and limits, and all those things that make a definition of $0^0$ problematical. You are looking at them only in Set Theory. And the addition and multiplication of natural numbers that you are looking at are the addition and multiplication of natural numbers as cardinals. It just so happens that they will agree with the usual arithmetic one in most instance, but not in all. – Arturo Magidin Jun 23 '12 at 04:38
  • @user1296727: Exponentiation has one of those few exceptions: $0^0$ makes perfect sense in the context of cardinal arithmetic, and there is absolutely no problem with defining it as $1$ and making it fit in the general scheme. And so you should not have any qualms about it. Because, the context where defining $0^0$ as something else is problematical is not the context you are in. – Arturo Magidin Jun 23 '12 at 04:40
  • I wasn't looking at them in a calculus context. I was looking at them as elements of the minimal successor set, with their derived properties. I noticed - as you said - that the cardinal number of $\emptyset^\emptyset$ had to be 1; I also noticed (from the theorem I was supposed to prove!) that this implied $0^0 = 1$ for the natural numbers. I just wanted to know if that definition was indeed upheld in the arithmetic of $\omega$, or if I should discard it as a special case. – Chris Jun 23 '12 at 04:40
  • @user1296727: Finally, having defined cardinal exponentiation for natural numbers, the exercise is asking to you prove that if $E$ can be bijected with $n$, and $F$ can be bijected with $m$, $n,m\in\omega$ (so that $|E|=n$ and $|F|=m$), then $E^F$ can be bijected with $n^m$, the cardinal result of that operation. – Arturo Magidin Jun 23 '12 at 04:41
  • @user1296727: And why would you ever stop to consider whether it was useful or not? Because you are familiar with a context in which $0^0$ is treated with care and as a special case. But that context is not the context you are in, which is why, yes, that's how you define it and it is not treated as a special case at all. It fits perfectly with the standard definition. – Arturo Magidin Jun 23 '12 at 04:42
  • @Arturo Thank you sincerely for your patience. (If you would convert any of these comments to a brief answer, I would accept it.) – Chris Jun 23 '12 at 04:43

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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.

Community
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Arturo Magidin
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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 $1$.

Michael Hardy
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  • Wikipedia has an extensive discussion of this. – MJD Jun 23 '12 at 04:07
  • And so does this site: http://math.stackexchange.com/questions/11150/zero-to-zero-power – talmid Jun 23 '12 at 04:13
  • 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 04:30
  • 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 06:14
  • @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