Re: Well Ordering the Reals

Tony Orlow says...

>If a set has no end, then asking what the mapping to its last element is

Nobody is asking that. We're asking what element is mapped to the
entire set. If you claim that you have a bijection f from some set
A to P(A), then that means

forall y in P(A), there exists x in A such that f(x) = y

In particular, when y = A. So if you claim to have a bijection,
you need to show that

there exists x in A such that f(x) = A.

>makes no sense. Of course you're going to get a contradiction.

Yes, the contradiction follows from your assumption that you
have a bijection between a set and its power set. Normal people
would say "My assumptions lead to a contradiction. Therefore,
one of my assumptions is wrong."

If you say that it is impossible to come up with an element
x that maps to the entire set, then that means that you *don't*
have a bijection. It's that simple. For any set A, A is an element
of P(A), by definition. If you claim to have a bijection between
A and P(A), then you are claiming to have an element of A that
maps to A, the entire set. If you say that there is no such A,
then you that means you *don't* have a bijection. It's that
simple: YOU DON'T HAVE A BIJECTION!

Sets do not have to have an "end" in order for them to have a bijection
to another set. For example, the function

f(x) = x*2

is a bijection between the set of finite naturals { 0, 1, 2, ... }
and the set of finite even naturals { 0, 2, 4, ... }. You don't
need a "last natural"; all you need is to be able to show:

forall finite even naturals y, there exists a finite natural x such
that y=f(x)

That's all it takes to show a surjection (for f to be a bijection,
you also need that if x is not equal to y, then f(x) is not equal
to f(y)).

Bijections don't have anything to do with there being a "last element".

--
Daryl McCullough
Ithaca, NY

.

Relevant Pages

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