Let X be a Normed linear space & S(x) ={x;norm x=1}.Show X is separable iff S(x) is separable?

Answer 1

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Answer 2

I'll do one direction for you. Suppose X is separable, and we have a countable dense subset T. Now take the set T' of normalized vectors in T-{0}. We claim that T' is a dense subset of S(X) also.

Indeed, let e>0 and x in S(X). Then the open ball U(x, e) in X contains an element y of T: i.e. || y - x || < e. Let y' be the normalization of y. This gives:

|| y' - x || <= || y' - y || + || y - x || < || y' - y || + e.

Now, || y' - y || = || y' - ||y|| * y' || = | 1 - ||y|| | = | ||x|| - ||y|| | < e.

So the above inequality < 2e, and U'(x, 2e) must contain an element of T'.

Answer 3

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Answer 4

Just at a guess:

Take a countably dense subset of S. Look at the set of all rational number scalar multiples of members of S. That would be a countably dense subset of the whole space, no?

Conversely, take a countably dense subset of the X. Normalize each of its elements. Wouldn't the result be a countably dense subset of S?