Re: Uncountable sets in CZF?
From: KRamsay (kramsay_at_aol.com)
Date: 09/04/04
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Date: 04 Sep 2004 05:43:06 GMT
Regarding IZF,
I wrote in <20040830045814.04815.00000438@mb-m17.aol.com>...
[...]
| But most especially, there can't be a surjection from the whole of the
| naturals to the reals.
In article <3c6b9c1e.0408301017.36c9cdec@posting.google.com>,
raf@tiki-lounge.com (Ross A. Finlayson) writes:
|Cantor-Schroeder-Bernstein: it works both ways.
|
|What that means is that one of the reasons that people call the reals
|uncountable is because they've figured out a bijection between the
|reals and the powerset of the naturals, thus they reason that there
|are no bijections between the reals and the naturals, because
|Cantor-Schroeder-Bernstein says the existence of a surjection either
|way between two sets is proof of the existence of a bijection between
|those two sets.
I thought the theorem was about sets A and B where there exist
injections from A to B and from B to A.
I don't know why one would use either result (either Cantor-Bernstein
or the analogous result about surjections) to argue against the
existence of a bijection between the natural numbers and the reals.
It's so much more direct to just use one of the two proofs Cantor
had to show that there is no surjection from the naturals to the
reals. I see no point in complicating the issue.
|That is to say, the existence of a surjection from A to B and from B
|to A implies that A and B are equivalent, and as well from A to B to C
|and C to B to A through composition.
Moreover, in the context of IZF, you can't use Cantor-Schroeder-
Bernstein; it's not a theorem of IZF. The bijection described in
the proof has to be defined by cases. As often is the case, when
a function has been defined by cases, the law of excluded middle
is implicitly invoked to claim that the domain of the function is
the whole original domain. That is, if X is a subset of A and we
say f=g on X and f=h on A-X, then the domain of f is the set of
elements of A that either are members of X or are not members of X.
And of course IZF does not have the law of excluded middle.
|That implies it is not a mathematical fact and to promote the other
|view as gospel, immutable, written in stone, etcetera, would thus be
|deceitful. I'm rather angered that you would suggest the acceptance
|of a mathematical falsehood as mathematical fact. Wouldn't you be?
Not always, no. I already have plenty of experience with seeing people
who are confused like you are offering nonsense as mathematical fact.
If I were angry about it, I would be angry quite a lot of the time.
Of course, it's worse when the person who is confused is so confused,
like you are, that he not only thinks that what he's saying is correct,
but he's really *sure* of its being correct, and that hence everyone
who disagrees with it is wrong. But worse still are the people like
you who not only are confused and sure they're right, but angry
at other people who are sure they're right too.
|Reexamine the claim about there being only one or none proper classes.
Nobody claims there's exactly one proper class. The ordinals are a
proper class, and so is class of all sets. They may believe that there
is no such thing as a proper class, but if they agree there are any,
they agree there are many.
I think you're giving a confused version of the claim that any two
proper classes can be put into one-to-one correspondence.
| Consider why that demands dual representation of the ur-element as
|both zero and Ord, regardless of whether Ord is N. (Steve, infinite
|sets are equivalent.)
|
|Unitize the analog! It's better to have a tool, even a primitive
|tool, to measure sparse points of the continuum, than none, and bar
|further consideration of the matter by fiat.
|
|Why don't you consider that the direct sum of infinitely many copies
|of N is zero?
I can't make any sense out of this.
|From Dave Seaman's giveaway that an "uncountable" set is a countable
|union of countable sets,
You should try to write more precisely (not that I expect you to),
especially when you're allegedly stating what someone else said.
This sounds like you're describing him as saying that each uncountable
set is a countable union of countable sets, which he surely didn't
write. Moreover I doubt he wrote that there exists an uncountable
set that's a countable union of countable sets. No, probably he wrote
something about its being consistent with ZF for there to exist an
uncountable set that's a countable union of countable sets.
|to that the constructivist powerset mapping
|argument is shoddy or fails,
How about getting rid of some of the "shoddy" that you keep
distributing?
|to these metatheoretical hand-wavings of
|yes and no and ignorance fo the excluded middle, we've seen some of
|what I would call progress in the state of discussion of these issues
|here on sci.math, an open public forum where the participants are
|often professional mathematicians and as such loathe to promote
|untruth, particularly in the stark, concrete world of mathematics,
|that is already a paradise with no need for the transfinite:
|inconsistent and thus hellish.
The only appearance of inconsistency is because you're confused!
[...]
|You're intelligent and understand the basic tenets of rational
|discourse, and by now you're probably familiar with the rather few
|arguments of transfinite set theory:
Set theory is a rather huge subject.
Keith Ramsay
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