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From: janw@inmet.UUCP
Newsgroups: net.math
Subject: Re: How Many Continuous Functions Are Th
Message-ID: <5700015@inmet.UUCP>
Date: Wed, 9-Oct-85 13:58:00 EDT
Article-I.D.: inmet.5700015
Posted: Wed Oct  9 13:58:00 1985
Date-Received: Sat, 12-Oct-85 06:26:59 EDT
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Nf-From: inmet!janw    Oct  9 13:58:00 1985


>  >> 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. 
>[etc.]

The marked proposition in parentheses is perfectly correct.
To prove this, it is sufficient to assign a unique real number
to every sequence of real numbers. (And then apply this twice:
to the sequence of coefficients of a polynomial, then
to the sequence of polynomials). Proof follows:

We can assume all the numbers to be in [0,1) (the extra step of 
going from this to unlimited numbers is quite easy).
Then a sequence S of them can be represented by an infinite matrix
of decimal digits such that s[i,j] is the digit in the i-th position
in the j-th member of the sequence. Traversing this matrix in the
usual diagonal fashion : concatenating the finite diagonals
in the order of increasing length (s[1,1]s[2,1]s[1,2]s[3,1] ...),
we obtain a digit sequence uniquely defined for the matrix;  pre-
ceded by "0.", this yields the real number corresponding to S.

		Jan Wasilewsky