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

On Oct 16, 3:00 pm, magi...@xxxxxxxxxxxxxxxxx (Arturo Magidin) wrote:
In article <1192555944.978947.292...@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,



porky_pig...@xxxxxxxxxxx <porky_pig...@xxxxxxxxxxx> wrote:
On Oct 16, 1:24 pm, magi...@xxxxxxxxxxxxxxxxx (Arturo Magidin) wrote:
In article <1192555182.902626.103...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,


Yes, thanks, I've realized I have reversed the domain and codomain
right after I've posted this message. Should keep example in mind:
{0,1}^A is a set of all mappings from A to {0,1} so it is a power set
of A. Thanks again.

There is a natural correspondence between {0,1}^A and the power set of
A, via characteristic functions; but it might be incorrect to say that
{0,1}^A ->is<- the power set of A. For example, if A = {x}, with x
different from 0 and 1, then

{0,1}^A = { {(x,0)}, {(x,1)} } (using the usual definition of
function as sets of ordered pairs)

whereas P(A) = { {}, {x} }.

No element of {0,1}^A is the emptyset, so P(A) is not equal to
{0,1}^A. Of course, there is an obvious correspondence between them,
though, as I mentioned.


Yes, I sort of cut the corners. In fact, the problem I'm working on is
exactly that: given the definition of natural correspondence between
the power set of A and {0,1}^A, prove that that correspondence is
bijection. So it is of course incorrect to say that the indicator
function I_B \subset {0,1}^A *is* the same (in general) as B \subset
A, but they are clearly related.
Thanks again for all the feedback.

.