Re: Aleph One Sets

From: Apollo Hogan (apollo_at_math.berkeley.edu)
Date: 10/20/04 

Date: Wed, 20 Oct 2004 21:28:10 +0000 (UTC)

In article <2tnjt2F225n3jU1@uni-berlin.de>,
Mike Oliver <mike_lists@verizon.net> wrote:
>Apollo Hogan wrote:
>> In article <cl39iv0r3s@drn.newsguy.com>,
>> Daryl McCullough <daryl@atc-nycorp.com> wrote:
>
>>>Here's a cardinality question for you (the answer *might* be aleph_1)
>>>
>>>Define an equivalence class on sets of reals as follows: R1 ~~ R2 if
>>>there exists an order-preserving bijection between R1 and R2. What is
>>>the cardinality of the set of equivalence classes of ~~?
>>
>>
>> Well, if 2^aleph_0 = aleph_beta, then there will be at least beta different
>> equivalence classes (one of each cardinality), so, for example, in a model
>> where 2^aleph_0 = aleph_(omega_2), there are (at least) aleph_2 different
>> equivalence classes.
>
>Nice!
>
>But one can do better--it's not hard to show directly that there are
>at least 2^{aleph_0} distinct classes. Embed length-omega strings of
>zeroes and ones thusly: For each 0, put a copy of omega in
>an interval; for each 1, a copy of omega* (that's omega reversed).

Nice. I was trying to think of something like this but I was using
Q (rationals) and N, but adjacent copies of Q collapse into one copy :-/

>Here's what I'm wondering: Can you actually embed 2^{2^aleph_0}?
>I don't immediately see why or why not.

I don't understand this question. Do you mean are there 2^{2^{aleph_0}}
many equivalence-classes? I think this should be independent of ZFC. But
I really don't know. (It is consistent with MA+not CH that all omega_1-dense
subsets of R are order-isomorphic.)

--Apollo

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