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From: bhuber@sjuvax.UUCP (B. Huber)
Newsgroups: net.math
Subject: Re: How Many Continuous Functions Are There
Message-ID: <2369@sjuvax.UUCP>
Date: Fri, 11-Oct-85 10:57:53 EDT
Article-I.D.: sjuvax.2369
Posted: Fri Oct 11 10:57:53 1985
Date-Received: Mon, 14-Oct-85 03:37:08 EDT
References: <310@ihnet.UUCP>, <10556@ucbvax.ARPA> <3368@pur-ee.UUCP>
Organization: St. Joseph's University, Phila. PA.
Lines: 48

It looks like nothing except an explicit one - to - one correspondence will
satisfy you guys.

Since it is pretty clear that the set C(R,R) of continuous functions with
real values defined on all the real numbers has at least aleph-one elements,
I will give you an injection from C(R,R) into the set R of real numbers.  
Because the cardinality of R is aleph-one, the Schroder-Bernstein theorem (which
is intuitively obvious) asserts that the cardinality of C(R,R) is at most
aleph-one, thereby finishing the demonstration that its cardinality is
exactly aleph-one.

First, any continuous function is uniquely determined by its values on the
subset Q of rational numbers, because every other real number is a limit of
rationals.

Q is enumerable.  So do it: count the rationals, letting q1 be the first,...
and qn the nth.

Express all real numbers in base two.       

Now here's the injection:
Let f: R----->R be a continuous function.  f(q1) is expressed as one string of
zeros and ones in base two; corresponding to each digit is its 'place', which
ranges among the integers. (The place is the power of two represented by the
digit.)  let r1 be the real number which has a zero in every even place, and
in place 2k-1, has the digit appearing in the kth place of f(q1).  That is, we
scatter the digits of f(q1) by inserting zeros between any two successive ones.
   Next, scatter the digits of f(q2) among the places in r1 which are odd      
multiples of two, in a systematic manner (as we did to f(q1) above).  Call the
result r2.  From r2 you can recover both f(q1) and f(q2), by looking in the
odd places for the digits of f(q1), and in the odd multiple of two places for
the digits of f(q2).
   Continue in this manner, scattering the digits of f(qn) in sequence among
the places of r(n-1) which are odd multiples of 2**(n-1), to create rn.  
f(q1), ..., f(qn) can all be recovered from rn.

We have thus managed to encode complete information about all of the 
values f(q1), ..., f(qn), ... in one 'limit' real, r(infinity), in such a
way that examination of r(inifinity) reveals the value of any f(qn), for
all positive integers n.  Since this sequence of values uniquely determines
f, we see that there are no more continuous functions than there are real
numbers.

I apologize for the length of this rather simple explanation; but in light
of the preceding discussion, an unassailable and elementary proof seemed
important.

--Bill Huber, St Joseph's University, Philadelphia.