Re: why does professor david c ullrich have to put people down to feel good about himself?

On Dec 19, 5:04 am, David C. Ullrich <dullr...@xxxxxxxxxxx> wrote:
On Thu, 18 Dec 2008 22:48:32 -0800 (PST), lwal...@xxxxxxxxx wrote:
"In the course of teaching a math for liberal arts course, I noticed
that there are really only two infinite cardinals that one can
encounter in real life, aleph-null and c.  The alternative set theory
of Vopenka also has only these two cardinals."
He "noticed" that, did he? I guess my life is not "real",
then.

The reason that I've mentioned Holmes is because
Ullrich often points out that tommy1729's
comments are too incoherent to be a theory. But
with Holmes, this is one objection he can't make.

It's a fact that there exist Lebesgue measurable subsets
of R which are not Borel sets. How does one prove that?
The simplest proof, and in fact the only proof I know,
goes like this: You show that the class of Borel sets
has cardinality c. You note that every subset of the
Cantor set is Lebesgue measurable, hence the class
of Lebesgue measurable sets has cardinality 2^c.
Since 2^c > c the two classes are not the same.
That's curious, isn't it? The cardinal 2^c actually appearing
in the proof of an important theorem of analysis. I
must be hallucinating, after all Holmes "noticed" that
that proof I just sketched doesn't exist.

Let's see what Holmes has to say about this:

"It is clear that the reals can be realized as sets using cuts in the
rationals. Continuous functions from R->R can be realized by
restricting them to the rationals. Intervals in the reals can be
coded, and so can Borel sets at each stage in their hierarchy."

So Holmes seems to imply that only Borel sets are
important, and so it's not necessary to consider
sets that aren't Borel at all. Obviously, Ullrich
disagrees with this sentiment.

According to Holmes, PST contains the bare minimum
needed for analysis. PST proves the existence of
finite powersets, as well as the proper class P(N),
and demonstrates a way to construct R, also as a
proper class.
Ah. So those large sets do exist, they're just proper
classes. Why in the world would somone object
to "large" sets, yet be happy with proper classes,
which are larger than any set is allowed to be?

Actually, no class of cardinality 2^c exists in PST.

In PST, we have:
-- Finite sets
-- Countably infinite sets
-- Proper classes of the continuum cardinality

Thus, there is no class of all Lebesgue measurable
subclasses of R in PST. This is proved via his
Limitation of Size axiom, which states that all
proper classes have the same cardinality (as R).

Why distinguish between sets and proper classes? By
doing so, we have an answer to Ullrich's question:

Powerset would be restricted so that only the sets
that need to have powersets would have them.
Exactly how do you propose to do this restriction?
Definition: The set S "needs to have a power set"
if ______.

The answer, at least in PST, is, the class S has a
powerclass if and only if S is a set.

Proof of <-: If S is a set, then Class Comprehension
gives us the powerclass P(S). (Of course, P(S) may
either be a proper class or a set.) QED <-.

Proof of ->: If S has a powerclass, then P(S) is a
class whose elements are the subclasses of S, one of
which is S itself. But Holmes defines a set to be an
element of a class. Therefore S is a set. QED ->.

Exactly the same as those who have only three
cardinals, 1, 2, and "many", except at a higher
level. I've never seen any reason why a number
system with only those three numbers is
better than the standard one.

An obvious reference to that famous tribe of Bushmen
with only one, two, and many.

What we gain is that those who are philosophically
opposed to ZFC and the large sets whose existence
it proves will have a more acceptable theory.
What do we gain _mathematically_? Or, _why_
are such people "philosophically opposed" to
such things?

By "philosophically opposed," I mean people such as
intuitionists, constructivists, and finitists. It is
unreasonable to expect them to accept classical
analysis, or require them to reject any theory that
doesn't provide the minimum for _classical_ analysis.

Requiring a finitist to accept infinite sets is like
forcing an adherent of ZFC to accept various large
cardinals that he/she might not accept, such as
inaccessible, hyperinaccessible, Mahlo, etc.

I mean really. Suppose I said I was philosophically
opposed to numbers greater than 2, and decided
to investigate that system where the only numbers
were 1, 2, and "many". You ask what's wrong with
including the number 3, and all I can say is I'm
"philosophically opposed" to it. That would seem
somewhat arbitrary, wouldn't it?

Why don't all ZFC adherents accept the existence of
Mahlo or huge cardinals?

All theories, including ZFC, have cardinals whose
existence can't be proved. For finitists, aleph_0
is such a cardinal. For Holmes and PST, beth_2 is
such a cardinal. For ZFC, the first inaccessible
is such a cardinal.

At this point you ask me why I don't just go
ahead and allow 3 and 4 as numbers -
wouldn't that be simpler? I say I'm philosophically
opposed to 3 and 4, although I can't give
any reason for this. You ask me if I see any
mathematical problsms caused by inserting
3 and 4. I don't, I'm just "opposed" to them.

And there are no mathematical problems caused by
allowing inaccessibles, but they are still not
part of ZFC.

I don't know why intuitionists are philosophically
opposed to classical analysis. They just are, and
some people may be philosophically opposed to 2^c
or 2^(2^c) the for the same reason.

If a Bushman proposed a set theory in which two
was the largest cardinal, how would I respond? I
most likely ignore the thread, since I ignore
threads that contradict small finite theory.

I have more to say on this soon....
.

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