Let C[0,1] denote the set of continuous functions over [0,1]. Is C[0,1] seperable?

Question

A set M is seperable if there is a countable dense subset P of M.

I'd like to run my argument by you guys.

Here is my solution: I believe that C[0,1] is not seperable. To prove this I would like to have an uncountable discrete subset. Now, any a in [0,1] has a unique infinite expression mod 2. For a = 0.(a_1)(a_2)(a_3)... there is a continuous function f_a on [0,1] s.t. f_a(y) = 0 whenever y = 1/2 +...+1/(2^n) and a_n = 0, f_a(y) = 1 whenever y = 1/2 +...+1/(2^n) and a_n = 1 For any other y, 1/2 +...+1/(2^n)< y <1/2 +...+1/(2^n) + 1/2^[n+1]) for some positive integer n. Hence we define f_a(y) = [a_(n+1) - a_n]/2^(n+1)(x - [1/2 +...+1/(2^n)])

It is easy to see that the distance between f_a and f_b is 1 for a<>b. Furthermore, there are as many such functions as sequences over {0,1}.

Hopefully the idea behind my solution is expressed clearly enough.

I'd appreciate any other solutions

I'm sorry for not being clear. the set C[0,1] is a set of continuous functions from [0,1] into R. Hence it has a metric d(f,g) defined by max 0<=t<=1 abs(f(t) - g(t)). I hope that now the question is clear.

I was a bit sloppy in expounding my attempted proof ;( Before going anywhere it is very important to emphasize that seperable means dense + countable subset F of C[0,1]

First, I'd like to state a hypothetical: If there exists an **uncoutable** subset P of C[0,1] with the property that for any f,g in P d(f, g) =1, then C[0,1] could not be seperable: Every dense subset D would have to have some f* s.t. d(f*,f) <1/4 for every f in C[0,1] and hence in P. For any other function g in P s.t. g<>f, d(f*, g) >1/4. Hence to be 'near' each function/point in P D would have to have at least as many members as P. Hence D is uncountable.

In the attempt to produce such P, I decided to built a unique function out of every a in [0,1]. First express a as an infinite expension mod 2. This infinite expension is uniques mod 2. For example a = 0.1011... then f_a(0) = 0, f_a(1/2) = 1 *because a_1=1* f_a(1/2 +1/(2)^2) = 0 and f_a(1/2 +...1/(2)^n) = 0 if a_n =0 and 1 otherwise.

To make f_a continuous simply connect the 0's and 1's with lines. One should imagine the function to be like a mountain ridge with ever-increasing steapness and occilation.

Answer 1

You have guessed wrong; C[0,1] is separable. Here's a hint that may lead you to a proof: Any function in C can be uniformly approximated by polynomials.

Answer 2

My dear fellow, in general an infinite dimensional vector space does not have a natural topology on it. What topology are you considering on C[0,1]? Alternatively: what is the distance between functions you are talking about? :-(