Re: Questions on sets with infinitely many elements
- From: "george" <greeneg@xxxxxxxxxx>
- Date: 10 Mar 2007 12:21:43 -0800
On Mar 9, 11:04 pm, Phil <toob-head...@xxxxxxxxxxxxx> wrote:
No one would be impressed if we said, "If x is the last odd number, and
we take the number preceding x, namely w, then (w+2) is also an odd
number, so x is not the last odd number."
OF COURSE they would.
Compare to "If x is the last natural number, adn we take the
number preceding x, namely w, then w+2 is also a natural
numeber, so x is not the last natural number". THAT IS a
proof. EVERYbody (who knows anything) is impressed.
My proof that x is not the last odd number is PURE
CRAP!!!
Hardly. But that also hardly matters; what is abundantly clear
is that you simply don't know what a proof is. Would you like
to learn?
But I see "proofs" like that in mathematics, especially where
infinity is involved, all the time ...
That doesn't matter; since you are not understanding them, it
doesn't matter that you are seeing them.
....
It works in the sense that I now have a set with a member which both has
an immediate predecessor, and is preceded by infinitely many members.
Idiot, PLEASE: you had THAT in THE FIRST reply you ever
got to your ORIGINAL message, on Mar.7. I ought to know,
since I wrote it. Just in case you didn't read it the first time,
here it is again:
Yes, it is possible for an infinite well-ordered set to have a last
element.
The canonical examples are infinite "successor ordinals".
If you define x as the set of natural numbers, then x U {x}
is a set whose last element is infinite (x) and whose earlier elements
are all finite naturals. But this ordering is not INHERENT to this
set;
this is the ordering IMPOSED by the membership relation. YOU COULD
permute this set into any one of uncountably many OTHER DIFFERENT
orders, if you felt like it.
(4) Element x is followed, in set A, by elements (x+1), (x+2), (x+3), ...
Yes; under one definition of addition (ordinal addition), x U {x}
MEANS x+1.
x+2 would just be (x+1)+1, and the last element of x+2 would be x+1.
So if you had this set, which we have here called x+2, its members
would
be {0,1,2,...,x, x+1}.
The last element of this set, x+1, is a member which has an immediate
predecessor (x), and which is preceded by infinitely many members.
My big question is this: can I link or align each member of the set:
{a_1, a_2, a_3, ...., b, c}
with the natural numbers, as in:
{a_1, a_2, a_3, ...., b, c}
| | | | |
{ 1, 2, 3, ...., x, y}
Not if you map a_n to n for every natural n, no.
However, you don't have to map them that way.
You could map a_69 to 70 and a_n to n+1 for all n>69,
and then you could let x be 0 and y be 69.
So b would be mapped to x=0 and c would be mapped to y=69,
and all the infinitely many a's would be mapped to all the other
infinitely many natural numbers.
So that x is a natural number,
one which is preceded by infinitely many
natural numbers?
It's YOUR mapping. You can have x preceded
by WHATEVER NATNUMS YOU LIKE,
in WHATEVER ORDER YOU LIKE.
But since you are re-ordering infinitely many different things,
some of your orders might not be communicatable finitarily.
.
- References:
- Questions on sets with infinitely many elements
- From: Phil
- Re: Questions on sets with infinitely many elements
- From: Ken Pledger
- Re: Questions on sets with infinitely many elements
- From: Phil
- Re: Questions on sets with infinitely many elements
- From: David Marcus
- Re: Questions on sets with infinitely many elements
- From: G . Frege
- Re: Questions on sets with infinitely many elements
- From: Phil
- Re: Questions on sets with infinitely many elements
- From: Ken Pledger
- Re: Questions on sets with infinitely many elements
- From: Phil
- Questions on sets with infinitely many elements
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