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From: cjh@petsd.UUCP (Chris Henrich)
Newsgroups: net.math
Subject: Re: How Many Continuous Functions Are There
Message-ID: <656@petsd.UUCP>
Date: Tue, 8-Oct-85 13:12:37 EDT
Article-I.D.: petsd.656
Posted: Tue Oct  8 13:12:37 1985
Date-Received: Sat, 12-Oct-85 12:13:48 EDT
References: <310@ihnet.UUCP> <10556@ucbvax.ARPA> <3368@pur-ee.UUCP>
Reply-To: cjh@petsd.UUCP (PUT YOUR NAME HERE)
Organization: Perkin-Elmer DSG, Tinton Falls, N.J.
Lines: 42
Summary: 

[]
In article <3368@pur-ee.UUCP> yena@pur-ee.UUCP (Hao-Nhien Qui Vu)
writes:
[ quoting another article in which it is asserted that the set
of sequences of polynomials has cardinality C ]
>Nope.  Consider: Each real number is the limit of a sequence of
>rational numbers, but reals are uncountable and rationals are.
>
>The reason is that each limit is (sort of) defined by a countable
>subset of (in your case) polynomials, so there are as many limits
>as there are countable subsets (actually, there are less, but that
>can be taken care of easily [details are left for homework :-) ]).
>There are C polynomials ---> there are aleph-0 * C countable subsets.

In fact the number of countable subsets is C ** aleph-0.
>
>Incidentally, C = aleph-1.
...
>and so aleph-0 * C = aleph-2.

Actually aleph-0 * aleph-1 = aleph-1.  But also,
C ** aleph-0 = aleph-1.  Proof:

C ** aleph-0 = aleph-1 ** aleph-0
             = ( 2 ** aleph-0) ** aleph-0
             = 2 ** (aleph-0 ** aleph-0)
             = 2 ** aleph-0
             = aleph-1.

References: College-level books on set theory.  One by Halmos,
_Naive_Set_Theory_, is quite good.
By "Naive" he means that he does not go into the most
perplexing issues.  If you want something a good deal more
difficult, try the appendix in Kelley, _General_Topology_.
Regards,
Chris

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