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While going through this question in math stackexchange I have come across the fact that "if $(X,d)$ is a separable metric space, then for any subset $Y\subseteq X$, the subspace $(Y,d)$ is separable" which is mentioned in the answer by 1015 to the same question.

After reading that question, I have stuck in a question. My question is : is it true that

For a normed linear space $X$, $X$ is separable iff $S_X=\{x\in X \mid ||x||=1\}$ is separable.

I managed to get the same result for the unit ball $B_X=\{x\in X \mid ||x||<1\}$ But not quite sure about $S_X$. I think it is true for $S_X$ also.

Please help me to solve this problem. Thnx in advance.

One side of the problem follows directly from the fact I mentioned at the very beginnng i.e. if $X$ is separable then $S_X$ is separable since it is a subset of $X$.

For the other part I am completely stuck. I have taken a countable dense subset $D=\{x_1,...,x_n\}$ of $S_X$ but can not find a way to construct a countable dense subset of $X$.

Please help me. Thnx again.

Community
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usermath
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1 Answers1

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You can take the set $\mathbb{Q}\cdot D = \{r \, x_i \mid r \in \mathbb{Q}, i \in \mathbb{N}\}$: for an arbitrary $x \in X$, you have $x = \|x\| \cdot x/\|x\|$, with $\|x\| \in \mathbb{R}$ and $x/\|x\| \in S_X$. The former can be approximated with an element from $\mathbb{Q}$ and the latter with an element from $D$.

gerw
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  • Thank you for your answer. I still have a confusion about what do you mean by "approximated" in both the cases. Will you please explain? It will be of great help to understand the solution. Thanx again – usermath Sep 11 '15 at 03:49
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    For every $\varepsilon > 0$, you find $q \in \mathbb{Q}$ with $| |x|-q | \le \varepsilon$ and $n \in \mathbb{N}$ with $| x/|x| - x_n| \le \varepsilon$. Now, you can estimate $|q , x_n - x|$ by using the triangle inequality. – gerw Sep 11 '15 at 09:23
  • Thank you for your answer. I think I have understood how you want to approach. Please see if it is correct or not. We choose $x_n \in D$ and $q\in \mathbb Q$ such that $||x-x_n||x||||<\frac{\epsilon}{||x||+1}||x||,|||x||-q|<\frac{\epsilon}{||x||+1}$ And then $||qx_n-x||\le||x-||x||x_n||+||x_n|||q-||x|||<\frac{\epsilon}{||x||+1}(||x||+1)$ (Since $||x_n||=1$) And hence $||qx_n-x||<\epsilon$ which is we required. Am I correct in this?Thnx again – usermath Sep 11 '15 at 16:25
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    Yes, this is correct. – gerw Sep 11 '15 at 17:36
  • Thank you so much for your answer. – usermath Sep 12 '15 at 04:08