Re: infinity


Tony Orlow wrote:
> Randy Poe said:
> >
> > Tony Orlow wrote:
> > > Randy Poe said:
> > > >
> > > > Tony Orlow wrote:
> > > > > Randy Poe said:
> > > > > > Your mapping is defined in such a way that there is no
> > > > > > x in X such that f(x) = X. This has nothing to do with
> > > > > > "largest finite". Your map, as you defined it, has the
> > > > > > property that no element x is in f(x). Therefore it's
> > > > > > pretty obvious without any references to "largest finite",
> > > > > > that there is no x in X such that f(x) = X.
> > > > > Not within the range of any given X, but in the overall infinite set, there is.
> > > >
> > > > Are you aware that if I claim I have a bijection from a set
> > > > A to a set B, then what I am claiming is that f takes
> > > > elements from set A and maps them to elements of set B?
> > > Sure,
> >
> > Good.
> >
> > > can you enumerate them all?
> >
> > Don't change the subject. If you are aware that this is a
> > requirement for a bijection, why would you claim that you
> > have a bijection from X to P(X) that maps things that aren't
> > in X to things in P(X)?
> I am not changing the subject.

Asking me to enumerate them all is a change of subject, since
the question is REASONING as to whether there is a one-to-
one bijective mapping which takes elements from X and produces
elements of P(X). Whether the map can be ENUMERATED is
irrelevant to whether it exists, and to whether REASONING
can prove that it does or does not exist.

> You want me to tell you the natural that maps to
> the last in the enumeration of the infinite set of subsets in P(*N)?

No I don't, since there is no last.

I've just picked one particular element of P(*N), the
set *N. I don't care where it appears in your mythical
enumeration. I just am asking if it appears at all.

> Fine, I'll
> do that, as soon as you tell me what natural maps to the last in the
> enumeration of infinite set of the evens.

There is no last, in either set. The question is whether
EVERY even is mapped by a natural, without any left out,
and the answer is yes.

Let n be any even natural. Then n/2 is a natural. See? No
discussion of "last natural" or "last even" needed to talk
about "every natural" or "every even".

> My request is no different from yours.

It's quite different. You have a map. I don't care about what
order your map produces elements. To establish whether it is
ONTO P(X), I just need to answer the question as to whether
EVERY element of P(X) is mapped. I really don't care about
identifying a last.

The equivalent request on f:N->E, f(x) = 2x, would be
"is every element of E mapped"? Again, order doesn't matter,
and certainly questions about nonexistent last elements
don't matter.

Do you agree that a bijection from naturals to evens should
map some natural to every even, regardless of the order
in which it does those mappings?

Do you agree that a bijection from any set X to P(X) should
map some element of X to every element of P(X), regardless
of the order in which it does those mappings?
> The subject is the relative ranges of two infinite sets,

No, the subject is the existence of a bijection.

> > there is no x in X such that f(x) = X
> There is no e in E such that f(e) = N.

Eh?

> That is, there is no even such that half of it is more than
> half the naturals.

I don't know what "half the naturals is". Do you disagree
with this statement? For any x in E, x/2 is in N. How about
this one? For every x in N, 2x is in E.

Once more for the slow-witted: If I want to prove that
a map f:A->B is not a bijection, it suffices to find
y in B such that f(x) is not equal to y for any x in A.

If I want to prove that a map f:X->P(X) is not a bijection,
it suffices to find y in P(X) such that f(x) is not equal to
y for any x in X.

I can do that, by choosing y = X.

If I want to prove that a map f:N->E is not a bijection,
it suffices to find y in E such that f(x) is not equal to
y for any x in N.

You can't do that. If you give me any y in E, I will tell
you that f(y/2) = y for the map f = 2x, and that y/2 is
in N.

See how that works? I don't talk about "last", I talk about
EVERY. That's how this REASONING stuff works. It's powerful
stuff. Stop avoiding it, it will save you a lot of time.

- Randy

.

Relevant Pages

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