PWrong wrote:There are easier ways to do this. Just try to take the limit of x^y as (x,y) approaches (0,0). You get different answers on different paths. This proves, by definition, that the limit does not exist.
Yes but the point was not whether the limit exists but if it is definable.
Originally I wanted to shed some light on the defineability issue by the long explanations, but it seems it was not very successful. So I give a small survey on definability.
Can we define anything and have to wait until some contradictions are found?
No, but everytime we make a definition we have to show that it is well defined.
Thatswhy I distinguish between natural, rational and real numbers. Though they are embeddable (in that order) not necessarily the operations on it can always be extended to the next number system (though addition and multiplication do).
The natural numbers N (I regard them including 0 here) are defined by the peano axioms, that means I define by induction. And we can arbitrarily choose our induction start (without having to deal with definability). So for example:
m^0=1 for every m
m^(n+1)=m*m^n
But we could also have defined the induction start as
m^0=1 for m>0 and 0^0=0
without changing the other values of x^y.
Every other induction start of 0^0 would change some other values of x^y.
Then we define the integers by natural number paired with sign (except for 0).
Then we define the rational numbers Q by pairs of integer numbers (or signed pair of natural numbers) and the congruence relation a/b=c/d iff a*d=c*b. If I define for example (a/b)^k=a^k/b^k then for well-definition I have to show that it yields the same result for every c/d=a/b (proof left to the reader). Though usually one assumes a/b to be canceled and then only shows that (na)/(nb) yields the same result. If we would admit zero as denominator then 0/0 = a/b for all a,b because 0*b=0=a*0 and so Q would be trivial (and especially no extension of N).
Then we define the real numbers R as Cauchy sequences of rational numbers (the fine thing about Cauchy sequences is, that you dont need to know the limit) with the congruence relation that the difference of both Cauchy sequence tends to zero. So when we define something for real numbers for well-definition we have to prove that the result is again a Cauchy sequence and that for each congruent Cauchy sequences as arguments the results must be again congruent Cauchy sequences.
We can define ^ on RxN by induction (as above taking the multiplication on RxR). And have again that 0^0 can only be defined as 0 or 1 if the operation on the other values of RxN should be left unchanged.
We can define ^ on RxN by Cauchy sequences (taking ^ defined on QxN). Then 0^0 can be defined as 1 and 1 is the only value that it can take.
Extension of ^ to R<sub>+</sub>xR (where R<sub>+</sub> are the positive reals without 0) is a bit more complicated. We mainly do it by defining first ^ on R<sub>+</sub>xQ and then via Cauchy sequences of the exponents. It was already shown that the definition cannot extended to 0^0:=1 because it leads to different results for certain congruent Cauchy sequences of exponents. On the other hand 0^0=0 is that way definable if we extend the definition set to R<sub>+</sub>xR U {0}xR<sub>+</sub>. So the 0^0=0 is the only possible value in this defintion.
That are the (strict) results what regards definability. And I at least had expected from you, Wendy, that you (1.) confirm or deny the restriction on RxN (especially you only use natural numbered exponents in your considerations) and if you deny 1 then (2.) how you integrate the (multiple mentioned) undefinability arguments for RxR into your conviction.
Of course if there are any other definitions of ^ on the reals (which of course have to be equal to exp(y*ln(x)) on R<sub>+</sub>xR) there might come out different results what regards the value 0^0. So any ideas are welcome.
*nudge*
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