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In set theory, if $A$ and $B$ are sets, then their Cartesian product is defined to be $A\times B$ such that:

$\forall x,y: [(x,y)\in A\times B \iff x\in A \land y\in B]$

Exponentiation (as repeated Cartesian products) is defined as

$A^n = \underbrace{A \times A \times \cdots \times A}_\text{n times}$

where $A$ is a set and $n\in N$.

Exponentiation can also be thought of as repeated multiplication of natural numbers such that

$a^2 = a\times a$

$a^3 = a\times a\times a$

$\cdots$

$a^n = \underbrace{a \times a \times \cdots \times a}_\text{n times}$

where $a, n \in N$.

More formally, we have the following requirements of exponentiation (as repeated multiplication on $N$):

  1. $\forall x\in N: x^2 = x\cdot x$

  2. $\forall x,y\in N: x^{y+1}=x^y\cdot x$

It can be formally proven (see link) that there are infinitely many binary functions on $N$ that satisfy these two requirements. Fortunately, they differ only in the value assigned to $0^0$, which can be any natural number whatsoever. For each of these functions, $x^0=1$ for $x\neq 0$. For these reasons, I think $0^0$ should probably left undefined, as has been an accepted practice for nearly two centuries (since Cauchy, 1820).

From (1) and (2), we can also derive the usual Laws of Exponents provided we avoid any explicit reference to a value for $0^0$. From (1) and (2), we can, for example, derive:

$\forall x,y,z\in N: [x\neq 0 \implies x^y\cdot x^z = x^{y+z}]$

(For a formal development of these ideas, see "Oh, the Ambiguity!" at my math blog.)

My question: While there may be similarities, how is the leap made from repeated Cartesian products to repeated multiplication on the natural numbers (in order to justify $0^0=1$)? Or are they two fundamentally different things?

Added: Are (1) and (2) above insufficient as requirements for repeated multiplication on $N$? If so, what other requirements might reasonably be added to enable us to derive $0^0=1$ (or any other particular value)?

Note: This question is a follow-up to my answer at Why is $ A^0 = \{ \emptyset \}$?

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It seems that the OP doesn't really care about an answer, but rather wants validation to their point. I'm voting to close, questions on this site are not a stage for mathematical propaganda. –  Asaf Karagila Dec 13 '13 at 22:49
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@AsafKaragila I am not making any wild assertions here. The notion that $0^0$ should be left undefined is really quite mainstream. And I have provided over 1,800 lines of formal proof to support this historical position. "Propaganda?" Come now! I just want to know how we go from a definition of exponentiation based on repeated Cartesian products that is used to justify $0^0=1$ to one based on repeated multiplication. –  Dan Christensen Dec 14 '13 at 5:21
    
I originally upvoted the question, because I though it would be a good place to solve some fundamental misunderstading on the matter of defining operations, basic arithmetics and similar. Now, it's not the case, but I can't downvote. As AsafKaragila tells above, and @EmanuelePaolini says in his comments, the OP is not willing to understand why he's wrong, so there was no point in asking. –  rewritten Dec 22 '13 at 16:41
    
If you want to prove things like $4\subset 33$ or $0^0=1$, it seems you will need something other than Peano's axioms as a starting point. Can we not agree on that? –  Dan Christensen Dec 23 '13 at 22:18
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3 Answers 3

Set theory

In set theory, finite Cartesian product of sets is initially defined as the set of tuples with items from the factors, where the tuple $(x_1, x_2,\dots x_n)$ is the set $Pair(x_1, (x_2,\dots x_n))$, where again the $Pair$ function can be defined in different ways, the most common is $Pair(x, y)=\{x,\{x,y\}\}$.

This definition doesn't extend to infinite products, and to empty products. So a new definition can be done, which can be proved equivalent where the first is defined, and that extends naturally to more cases. This definition uses functions from cardinals to the union of the factors, so if $\mathcal{F} = \langle X_i: i<\kappa\rangle$ for a certain cardinal $\kappa$, then the product can be defined as $\prod\mathcal{F} = \{f:\kappa\to\bigcup\mathcal{F}\,\text{such that}\,f(i)\in X_i\}$.

With this definition, finite products can be identified with those defined above, and one can see that the empty product is defined too, in fact for $\kappa=0$, the union of the family is empty and as such, there is only one function $f:\kappa\to\bigcup\mathcal{F}$, the empty function, which voidly satisfies the condition for being in $\prod\mathcal{F}$ as there is no $X_i$ at all.

Also, if any of the $X_i$ is empty, then the product is empty, as no function $f:\kappa\to\bigcup\mathcal{F}$ can satisfy $f(i)\in X_i$ for that empty $X_i$

This definition of infinite product leads to de the definition of exponentiation (in sets) as the family may be constantly a set $A$, so $\bigcup\mathcal{F}=A$. In that case the definition is much simpler, as the $f(i)\in X_i$ is automatically satisfied (when possible): $A^\kappa = \{f:\kappa\to A\}$, which contains just the empty function $\emptyset$ when $\kappa=0$, and which is empty if $\kappa>0$ and $A=\emptyset$. These are not definitions, these are consequences.

Now, to Peano.

Peano's axioms are (rephrased):

  1. $0$ is a natural number.
  2. $S$ is an injective function, whose image excludes $0$.
  3. Induction on first-order sentences.

The fact that every number $a\neq 0$ is a successor can be proved by induction.

On this base, one can recursively define addition: $$ a+0=a \\ a+S(b)=S(a+b) $$ and multiplication $$ a\cdot 0=0 \\ a\cdot S(b)=(a\cdot b) + a $$

Up to here, addition and multiplication correspond to cardinal addition and cardinal multiplication in the standard VN model, where natural numbers are the finite ordinals.

You can also define integer exponentiation. You can define in any means that you want, but the simplest definition, which parallels the definitions above, is $$ a^0=1 \\ a^{S(b)}=(a^b)\cdot a $$ which clearly indicates that $0^0=1$ and $0^n=0$ for any $n>0$.

Again, this definition maps naturally to cardinal exponentiation.

Now, to the real numbers.

One of the reasons many people tries to say that $0^0$ should not be $1$, is because the limit $\lim_{x\to 0^+}x^\delta = 0$ for any positive $\delta$ so "by continuity" it seems natural to define $0^0=0$, but then $\lim_{x\to 0^+}x^x = 1$ and in general $x^y$ has no defined limit for $(x,y)\to(0,0)$.

But.

Real numbers axioms (the theory of real-closed fields, which basically says that you can do addition, multiplication, there is a strict order and all positives have square roots) do not extend Peano's axioms. In fact the RCF theory is complete. And Peano is not complete, as proven by Gödel. What actually proved by Gödel is that any theory strong enough to contain all arithmetic, is not complete.

So it turns out (quite surprisingly) that we can't think about real numbers and hope to define things on the natural numbers on that base.

Epilogue

What's left is that we can define things, and until these defintions show some inconsistency, we can safely keep these definitions.

Note: This answer has been expanded a lot because the OP required it. In fact I think it deserved such an expansion, so thank you Dan for insisting.

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Doesn't it lead only to the conclusion that, for exponentiation as repeated Cartesian products, $\emptyset ^ 0=1$? –  Dan Christensen Dec 12 '13 at 15:06
    
Is it based on the fact that in the language of ZF, we define $\emptyset = 0$? –  Dan Christensen Dec 12 '13 at 15:08
    
It is based on the Von Neumann representation of natural number as finite ordinals, where $n = \{x\in\mathscr{O}:x<n\}$, and yes, in particular, $0 = \emptyset$. –  rewritten Dec 12 '13 at 15:10
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@DanChristensen No. Why should it be? An error, even made by Cauchy, is still an error. –  egreg Dec 12 '13 at 15:21
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Despite the apparent defacement of the question, I would vote it up again. I wonder, @rewritten, have you managed to see my answer and the comments that ensued it on this page? –  Asaf Karagila Dec 22 '13 at 16:12
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If $x$ is the cardinality of finite set $A$, then $x^n$ is the cardinality of the Cartesian power product $A^n$. As longer as either $A\ne\emptyset$ or $n\ne0$.

So: What happens to $\emptyset^0$. What's its cardinality. If you make the exponent a variable: $\emptyset^n$ for $n\ne0$ then $\emptyset^n=\emptyset$, whose cardinality is $0$.

On the other hand if you make the base a variable, for $A\ne\emptyset$ then $A^0=\{()\}$ whose cardinality is $1$. (You can choose to either identify or not the zeroupla $()$ with $\emptyset$.)

So at least you have two conflicting approaches to $\emptyset^0$ much as you have two conflicting approaches to $0^0$. It is posible to claim $()\in\emptyset^0$ (it is false that a member of the zeroupla does not belong to the empty set) or $()\notin\emptyset^0$ (as nothing belong to the empty set claiming that all members of the zeroupla belong to the empty set is nonsense even if the zeroupla has no members).

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I don't understand your last argument. Indeed the $0$-tuple is not an element of $\varnothing$, but then again why is $\varnothing^0=\varnothing$ to begin with? It seems to me that the $0$-tuple is indeed an element of $\varnothing^0$, because it's a $0$-tuple and all the elements appearing in allllll the coordinates of the $0$-tuple are elements of the empty set. –  Asaf Karagila Dec 15 '13 at 15:35
    
The main difference is that there is only one definition of $A^B$ that extends naturally to the case $\emptyset^\emptyset$. That definition identifies $A^B$ with the set of functions from $B$ to $A$ (which if $A$ is a finite set, is naturally identified with a cartesian product, but it's not the cartesian product), and that definition says that $\emptyset^\emptyset=\{\emptyset\}=1$. By the way there is no such thing as a zero-uple. If it existed, then it would be another proof that $\emptyset^0 = A^0$ for any A, as the "product" will be always the zero-uple. –  rewritten Dec 20 '13 at 10:36
    
@rewritten Incorporate the ideas in your first 2 sentences here with your comment above about $0^0$ being "undefined in Peano," and I may be able to officially accept it as an answer. –  Dan Christensen Dec 20 '13 at 18:03
    
Again, it would appear that $0^0=1$ cannot be derived from Peano's Axioms. Additional assumptions would have to be made, e.g. numbers having some kind of internal structure: $0=\emptyset$, $1=\{0\}$, $2=\{0,1\}$, $3=\{0,1,2\}$, etc. –  Dan Christensen Dec 20 '13 at 21:59
    
@DanChristensen I agree. And $4^2=16$ cannot be "derived" from Peano either, just because exponentiation is not part of Peano's axioms. You just define the operations on the base of the axioms, which include $0$, the injective successor function and the first-order induction principle. –  rewritten Dec 21 '13 at 9:02
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With your definition, elements of $A^3=A\times A \times A$ are couples of the form $((x,y),z)$ with $x,y,z \in A$. Notice that from this it follows that $A\times (A\times A) \neq (A\times A)\times A$ and your definition does not apply to the cases $n=1$ and $n=0$.

It seems better to define $A^n$ as the set of $n$-uples of elements of $A$. An $n$-uple is nothing else than a function $I_n\to A$ where $I_n = \{0,1,\dots,n-1\}$. Notice that in the most popular construction of natural numbers (as finite ordinals) one has $I_n = n$. In this case one would have defined $A^n$ for all natural numbers $n$ and one finds that $A^1 = \{(a) \colon a \in A\}$ is naturally isomorphic to $A$, while $A^0 = \{()\}$ has a single element. Here $(a)$ denotes the function $0\mapsto a$ while $()$ denotes the only function $\emptyset\to \emptyset$. Everybody agrees that defining $A^1$ is useful, so this definitions is better than the one you have proposed.

I would call these operations powers because exponentiation is a term usually referring to some sort of natural power i.e. powers with a fixed base and variable exponent. For example: $e^x$ (Euler's number), $e^A$ (matrix exponential), $exp(tv)$ (exponential map).

You define the powers of natural numbers again starting from power $2$ (why?!?). Anyway you assume that $$ \forall x,y\in N: x^{y+1}=x^y\cdot x $$ and here your are assuming that $0^0$ is defined, contrary to your opinion.

Coming to your questions.

There is a strong similarity in set theoretic exponentiation $A^B$ and exponentiation of natural numbers $a^b$ in the fact that $$ |A^B| = |A|^{|B|}. $$ This formula holds true whenever $|A|\neq 0$ or $|B|\neq 0$. If one wants that it holds true also in the case $|A|=0$, $|B|=0$ one should define $0^0=1$.

About added question. You already noticed that the assumptions (1) and (2) are not enough to define the power $a^b$ for every $a,b\in \mathbb N$. To have a power function defined on all natural numbers you could use the following axioms: $$ a^{b+1} = a^b \cdot a \qquad \forall a,b\in\mathbb N\\ a^0 = 1 \qquad \forall a\in \mathbb N $$

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I define the powers of natural numbers starting with powers of $2$ because it seems more intuitive, and more natural. The values for powers of 1 and 0 can be derived from this starting point using $x^{y+1}=x^y \cdot x$. Perhaps there is a better way to express it informally, but, as I have shown elsewhere, there are an infinite number of binary functions on $N$ that satisfy these requirements. And they differ only in the value assign to $0^0$. So, I treat $0^0$ as ambiguous, as an unspecified natural number than can be multiplied by $0$ like any other number to get $0$. –  Dan Christensen Dec 21 '13 at 6:57
    
Correct me if I am wrong, but it would seem to me that you cannot derive $0^0=1$ from Peano's Axioms alone. It seems to me that you would have to use other axioms, and redefine multiplication in terms of Cartesian products, and exponentiation in terms of repeated Cartesian products. –  Dan Christensen Dec 21 '13 at 7:16
    
Peano axioms don't define the power operation. You must explicitly define it. –  Emanuele Paolini Dec 21 '13 at 7:54
    
Peano's axioms define only the successor function. From them, you must prove the existence of unique binary functions for addition and multiplication. You can prove the existence of a unique exponentiation function if you assign a particular value to $0^0$. The trouble, as I see it, is that any value will work as far as satisfying (1) and (2) above. –  Dan Christensen Dec 21 '13 at 15:28
    
@Dan you are commenting your question, not my answer... –  Emanuele Paolini Dec 21 '13 at 17:47
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