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From: yena@pur-ee.UUCP (Anthony T Yen)
Newsgroups: net.math
Subject: Re: How Many Continuous Functions Are There
Message-ID: <3368@pur-ee.UUCP>
Date: Mon, 7-Oct-85 14:17:09 EDT
Article-I.D.: pur-ee.3368
Posted: Mon Oct  7 14:17:09 1985
Date-Received: Tue, 8-Oct-85 04:09:14 EDT
References: <310@ihnet.UUCP>, <10556@ucbvax.ARPA>
Organization: Electrical Engineering Department , Purdue University
Lines: 24


  >> 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