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

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)%26lt; y %26lt;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%26lt;%26gt;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

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