Is there a descrete uncountable subset of C[0,1]-->R with its usual metric?

OK, so C[0,1] is seperable. But this seems to clash with the following observation. Please tell me where my reasoning went wrong.





For every a∈[0,1] a can be expressed uniquely as an infinite expension mod 2. That is a = (a_1)/2 + (a_2)/2^2 + ... +(a_n)/2^n... where each a_i is either 0 or 1. Define f_a(x) = 0 whenever x = 0 or x = 1/2 + 1/2^2 + ... +1/2^n, for some n and a_n = 0. Similarly, define f_a(x) = 1 whenever x = 1/2 + 1/2^2 + ... +1/2^n, for some n and a_n = 1. For all other x, 1/2 + ... +1/2^n %26lt; x %26lt; 1/2 + ... +1/2^n + 1/2^(n+1) for some n≥0. Define f_a(x) to be the point on the line, connecting f_a(1/2 + ... +(1/2^n ) and f_a(1/2 + ... +(1/2^n +1/2^(n+1)).





For example, if a = 0.01... then f_a(0) = 0, f(1/2) = 0, and f(1/2 + 1/4) = 1. , for , 0%26lt;x%26lt;1/2 f_a(x) = 0 and for 1/2%26lt;x%26lt;1/2 +1/4 = 3/4 f_a(x) = (4)(x -1/2).





It seems that f_a is continuous with domain [0,1] and that d(f_a, f_b) = 1 with a not equal to b. Hence the set of all f_a is a descrete uncountable set.

Is there a descrete uncountable subset of C[0,1]--%26gt;R with its usual metric?
What about f_a when a=2/3? You see at once that f_(2/3) cannot be continuous at 1.





I also would like to point out to the other answerers that a separable metric space does not admit an uncountable discrete subset. Since C[0,1] with the uniform norm is a metric space, the existence of a discrete uncountable set would indeed be quite remarkable, and when I say remarkable I intend PLAINLY FALSE.





Finally, may I correct the spelling of you, dear asker. Englishmen write separable, discrete, expansion, NOT seperable, descrete, expension.
Reply:I'm no Englishman either (Englishman singular, Englishmen plural). Report It

Reply:Note that f_a is continuous if and only if "a" has finite expansion, so you have only countably many of such functions. On the other hand, if you apply your construction to C[0,1) (i.e. 1 excluded) you actually find an uncountable discrete subset: C[0,1) is not separable. Report It

Reply:The best answer I can come up with is that it isn't true in general that discrete, uncountable sets aren't separable.





There are obscure reference papers to this on the internet. See, for example, reference 8 at:





http://www.u-aizu.ac.jp/official/researc...
Reply:If I may expand a bit on GI's remarks, the definition of a separable space is that it contains a countable dense subset. If a space is not separable, this means it does not contain a countable dense subset. Equivalently, every dense subset is uncountable. This is *not* logically equivalent to saying there is an uncountable nowhere dense subset.





Now, it may actually be true that a space with an uncountable nowhere dense subset is not separable-- I don't really know one way or the other, because no immediate counterexample springs to mind. I'm just saying that it is not an immediate logical consequence of the definition, and so, if true, it would require proof.

daisy