Re: infinity ...

In article <MPG.1dc0376e59b8e8c98a4e1@xxxxxxxxxxxxxxxxxxxxxxxxx>,
Tony Orlow <aeo6@xxxxxxxxxxx> wrote:

>
> > Tony Orlow wrote:
> > > Not enough infinite binary strings? How many in *N and in P(*N)?
> > > Are you saying that I cannot construct a bijection between the
> > > two on an element-by-element basis which continues infinitely
> > > through the set of binary strings?
> >
> > That's exactly what I'm saying.
> >
> > card(*N) = c, but card(P(*N)) = 2^c, and c < 2^c.
> At what point does the bijection break down? It certainly works for
> all finite cases, does it not? What makes you think it falls apart at
> some point? For which element is there not a corresponding subset?
> For which subset is there no corresponding element? Name either one
> of these, as a counterexample, or explain why you think the bijection
> fails.
> >
> >
> > > Where does it stop? You aren't beginning to see that determining
> > > the infinite value range is crucial to this problem after all,
> > > are you?
> >
> > If you think that approach will work, then show us how.
> >
> > By the way, what is the "range" of a powerset, whose members are
> > sets themselves?
> Sets don't have an inherent measure beyond raw size, and the power
> set is not well ordered as far as subset size goes, so it doesn't
> make sense to talk about a value range on the power set exactly.
> However, one might think of the binary mapping of the power set to
> the set, and say that the power set of an ordered set has an order
> based on that mapping, taking each bitstring as a natural. In this
> case, you might be tempted to say, if the set has N elements, the
> power set has a range of 2^N-1.
> >
> >
> > > What do you want me to try, anyway, and infinite mapping,
> > > element-by-element? A bijection's a bijection, right?
> >
> > Yes, that would be nice. Please show us your bijection.
> f(0) = ...000 = {} f(1) = ...001 = {0} f(2) = ...010 = {1} f(3) =
> ...011 = {0,1} f(4) = ...100 = {2} f(5) = ...101 = {0,2} f(6) =
> ...110 = {1,2} f(7) = ...111 = {0,1,2}
>
> etc. Any questions?
> >
> >
> > > These sets are obviously of the same cardinality, since I have a
> > > bijection which carries to infinity, right?
> >
> > No, their sizes are different cardinalities, which happen to be
> > different infinities. Both sets are infinite, but one set is
> > larger than the other.
> But I have constructed a bijection between the two using an
> intermediate binary representation. What is the specific rule I have
> broken concerning the construction of bijections. If I haven't broken
> any such rule, then is it true that a bijection between two sets
> means that have the same size, or even cardinality?
> >
> >
> > > For every subset there is a unique natural and for every natural
> > > there is a unique subset. If you disagree, please state which of
> > > either has no corresponding element in the other.
> >
> > Like I said, there are not enough naturals to map to every subset
> > of the naturals.
> Which subset, specifically, is left unmapped? If you claim there is
> one, then surely you can name it?
> >
> > This is true whether you've got infinite naturals or not; there are
> > not enough members in N to enumerate all the subsets in N, and
> > likewise there are not enough members in *N to enumerate all the
> > subsets in *N.
> But, as the keepers of that standard say, what holds for the finite
> case does not necessarily hold for the infinite case. Once you have
> infinite sets, neither one ends, and there is always a natural for
> any subset, and always a subset for any natural. Isn't this the way
> the standard treatment of bijection goes for infinite sets?
> >
> >
> > > Please try to frame your objection in a more operative manner.
> > > How do you know there aren't a large enough infinity of bits for
> > > the power set vs the values in the set?
> >
> > Because I can prove it (and it's a very old proof). A powerset of
> > a nonempty set contains more elements that the set. Can you prove
> > otherwise?

> No, I fully agree with that conclusion. However, there is nothing
> that precludes a bijection between any ordered infinite set and its
> power set, despite the different sizes.

Since 'more' in the context above means that there cannot be any
bijection, To is now claiming that where there *are* reasons why no
bijections can exist, there also are no reasons why bijections cannot
exist.

This is life in TOmatics.
.

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