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From: usenet@ucbvax.ARPA (USENET News Administration)
Newsgroups: net.math
Subject: Re: How Many Continuous Functions Are There
Message-ID: <10585@ucbvax.ARPA>
Date: Tue, 8-Oct-85 05:30:38 EDT
Article-I.D.: ucbvax.10585
Posted: Tue Oct  8 05:30:38 1985
Date-Received: Thu, 10-Oct-85 06:39:40 EDT
References: <310@ihnet.UUCP> <10556@ucbvax.ARPA> <3368@pur-ee.UUCP>
Reply-To: tedrick@ucbernie.UUCP (Tom Tedrick)
Organization: University of California, Berkeley
Lines: 28

In article <3368@pur-ee.UUCP> yena@pur-ee.UUCP (Anthony T Yen) writes:
>
>  >> Since the cardinality of the set of polynomials
>  >> mapping R -> R is C, and any continuous function
>  >> mapping R -> R is the limit of a sequence of
>**>> polynomials (the card of the set of such sequences
>**>> again being C), therefore the card of the set of
>  >> continuous functions is also 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.
>
>Incidentally, C = aleph-1.
>( C <= the number of finite subsets of reals = aleph-1
>and C >= the number of reals = aleph-1 --> C = aleph-1)
>and so aleph-0 * C = aleph-2.
>
>================================================
>Hao-Nhien Qui Vu ( pur-ee!vu ) Purdue University

I don't understand what is wrong with the original argument.
I'm too tired to deal with this now, can someone check this out?