Re: What is the meaning of the expression E^F, where both E and F are sets?

On Oct 16, 12:19 pm, "porky_pig...@xxxxxxxxxxx" <porky_pig...@my-
deja.com> wrote:
In both Halmos' Naive Set Theory and Suppes' Axiomatic Set Theory, the
expression E^F is introduced without any comments. The same text by
Suppes (if you skip a few chapters) refers to E^F as a set of all
functions f: E -> F.

In the canon of set theory, 0 = {}, 1 = {0}, 2 = {0,1}, 3 = {0,1,2},
etc. so that A^2 would be the set { (a0,a1): a0, a1 in A } which is
equivalent to the set of functions from 2 to A, noting the
correspondence a(0) = a0, a(1) = a1. Similarly, A^3 consists of
ordered triples.

This generalizes. If, instead of numeric parameters, the functions
have non-numeric parameters, you still want to say that the members of
A^X are ordered X-tuples. If the size of the set X is the number n,
then this is equivalent to A^n, but more direct.

Therefore, the exponential A^X consists of all ordered X-tuples drawn
from A which is equivalent to the set of all functions mapping X to A.

An alternate notation is (X->A). The laws of exponents A^{X x Y} = A^X
x A^Y establishes an equivalence between the respective sets. In the
alternate notation this becomes (X x Y) -> A = (X -> A) x (Y -> A).
This is also a logical tautology, if ()x() is interpreted as
conjunction and ()->() as implication. For the empty set, you have A^0
= 1 which is rendered as (0->A) = 1. This, too, is a tautology if 0 is
interpreted as "false" and 1 as "true".

For compound exponents, you have the equivalence (A^X)^Y = A^(Y x X},
which becomes (Y->(X->A)) = (Y x X) -> A. This, too, is a tautology in
logic, when interpreted as a logical statement.

Thus, there is a deep-seated link between the if-then connective in
logic and the exponential operator.

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