Re: Cardinal arithmetic question

From: Herman Rubin (hrubin_at_odds.stat.purdue.edu)
Date: 06/07/04 

Date: 7 Jun 2004 14:14:35 -0500

In article <0406061501230.16548-100000@gandalf.math.ukans.edu>,
Fred Galvin <galvin@math.ukans.edu> wrote:
>On Sun, 6 Jun 2004, Stephen J. Herschkorn wrote:

>> Fred Galvin wrote:

>> >Defining cardinals as a kind of ordinals is very inappropriate and
>> >outlandish when working without the axiom of choice.

>> Kunen, who defines cardinals as such, appears to disagree.

>In that case you should believe Kunen; he is an expert on set theory,
>I am not.

If you read any work on independence of the axiom of
choice, you will disagree with Kunen. There is no
good "definition" of cardinal number, but it is not
necessary to have one. In fact, before von Neumann
came up with a canonical family of ordinal numbers,
there was a similar problem with getting canonical
representations for them. This was after Zermelo
had proved that the Axiom of Choice implied the
Well-Ordering Theorem, and after Frankel showed
that Choice could not be proved in ZFU.

But it does not matter. One can define the operations
with sets or order types to get the results, except
in SOME cases where the axiom of choice is needed.

>> In his development of cardinals, he is careful to point out which
>> results depend on AC and which do not. He does note, however, that
>> his "book is concerned mainly with set theory with AC." Who knows
>> in what context the OP (not quoted here) was working?

>Probably ZFC. But I wasn't replying to the OP, I was replying to A N
>Niel's opinion as to which of the OP's boring homework problems needed
>AC and which didn't.

>> I understand Fred's point, though (and Fred is surely the expert here).

>No, he isn't.

>> Without AC, not every set has a cardinality with this definition. One
>> can still discuss the existence of bijections and injections between
>> non-well-ordered sets, but reference to cardinalities is desirable since
>> it simplifies the discussion.

It simplifies the discussion. See H. Rubin and J. Rubin,
_Equivalents of the Axiom of Choice, II_, or P. Howard
and J. Rubin, _Consequences of the Axiom of Choice_.

>> Or does it? One can always use the curly comparison symbols (which
>> denote the existence of injections and bijections).

>If you're really interested in seeing how much of cardinal arithmetic
>can be developed without AC, it would be a good idea to dafine
>cardinal numbers of arbitrary sets, not just well-orderable sets. Let
>me undelete the OP's homework problems:

>> (a) Prove k+u=max(k,u)=ku for all infinite cardinals k and u
>> (b) Prove (k^(u))^(v)=(k)^(uv) for all cardinals k,u and v
>> (c) Prove 2^(k)=(k)^(k) for all infinite cardinals k (Hint:
>> 2^k=2^(kk) and k< 2^(k) )

>If only well-orderable sets are allowed to have cardinals, then (a)
>does not require AC. On the other hand, (b) and (c) don't exactly make
>sense without AC, since 2^k is not necessarily a cardinal. Of course
>you could restate them with "curly comparison symbols".

On the other hand, (b) was well known before there were any
canonical forms, as was the meaning of (c). I do not know
if max(k,u) is always defined for infinite cardinals k and u
neither of which is well-ordered, and it is weaker than Choice
in any case. It is certainly bounded by k+u, and if k=2k holds
for all k, it is max (k,u).
theorems shown to imply Choice. 

-- 
This address is for information only.  I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu         Phone: (765)494-6054   FAX: (765)494-0558

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