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
2007-08-04
08:24:57
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2 answers
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asked by
guyava99
2