Re: infinity

David R Tribble said:
> David R Tribble wrote:
> >> Not quite. You can define a bijection between *N and P(N), but you
> >> still can't define a bijection between *N and P(*N).
>
> Tony Orlow wrote:
> > For ANY set you have, you can represent the powerset as a set of all binary
> > strings which have the same number of bits as the original set has of
> > elements.
> > If your original set consisted of all bit strings of length x, then there
> > were 2^x such strings, and the powerset is now represented by a set of all
> > bit strings of length 2^x, the power set of that by all bit strings of
> > length 2^(2^x), etc.
>
> Yep.
>
>
> > Now, if your original set is the set of all infinite
> > bit strings, such as the binary representations of all reals in [0,1), or
> > the binary representations of all whole numbers finite and infinite, then
> > the power set of such a set is also a set of all infinite bit strings.
>
> That would mean that any power set, or a power set of a power set,
> is no bigger than a set of infinite bit strings, thus all power sets
> would be essentially the same size. Which is obviously false.
Painfully obvious isn't it? And yet, such a bijection is easily constructed
according to standard approaches.
>
> Given the set of all finite naturals N, its power set P(N) has
> 2^card(N) members; those members can form a bijection to the members
> of *N, but not to the members of N. Thus card(P(N)) = card(*N).
That is assuming you have an infinite set of finite numbers, and thus
infintiely long bit strings denoting each subset of N. If, however, you do not
have an infinite set of finite naturals, then your bijection fails, since each
of those strings is then a finite string, and not one of the infinite strings
in *N.
>
> But *N, the set of finite and infinite naturals, cannot be bijected to
> its power set P(*N). It stands to reason if you think about it; there
> are just simply not enough members in *N, infinite or otherwise, to
> denumerate all of the possible combination subsets of themselves.

If both bitstrings in both sets go on forever, then you have a bijection just
like any other.

> And the same thing is true for any subset of R and its power set,
> such as [0,1) and P([0,1)).
Since the power set is actually bigger, I won't even try to make one in the
reals. My point is that bijections alone don't work.
>
> But feel free to provide a demonstration to the contrary.
Given in last post.
>
>

--
Smiles,

Tony
.

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