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I just read the theorem

"If X is norm linear space and if $X^*$ is separable $X$ is also separable"

from Bachman and Narichi, page 201. To prove it they consider the set $S=\{f \in X^* :\lVert f\rVert=1\}$. And the writer said that $S$ is separable as $S$ is subset of $X^*$ which is separable. How it is possible? Can anyone explain it? I know it is true for open subsets.

Hagen von Eitzen
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Andy
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1 Answers1

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Suppose $X$ is separable. Then there is a countable dense subset $\{x_n\}$. Suppose $S \subset X$.

For each $n$ such that $B(x_n,1) \cap S \neq \emptyset$, create a (possibly finite) sequence $s_{n,k}$ as follows: Choose $k$. If $B(x_n,\frac{1}{k}) \cap S \neq \emptyset$, then select any $s_{n,k} \in B(x_n,\frac{1}{k}) \cap S$.

The set $ \{ s_{n,k} \}$ is countable, and I claim that it is dense in $S$.

Suppose $s \in S$, and $\epsilon>0$. Choose $k$ such that $\frac{1}{k} < \frac{\epsilon}{2}$. Then since $\{x_n\}$ is dense, we can find some $x_n$ such that $\| s-x_n\| < \frac{1}{k}$. Since $s_{n,k} \in B(x_n,\frac{1}{k}) \cap S$ by construction, we have $\| x_n-s_{n,k}\| < \frac{1}{k}$. Hence we have $\|s - s_{n,k} \| \le \| s-x_n\| + \| x_n-s_{n,k}\| < \frac{2}{k} < \epsilon$.

copper.hat
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  • @learnmore: I'm not exactly sure what you are referring to. It is not always true, the construction in the second paragraph may result in a finite sequence. In the last paragraph, since we know that $|s-x_n| < {1 \over k}$ we know that $B(x_n,\frac{1}{k}) \cap S \neq \emptyset$, we know that the sequence element $s_{n,k}$ exists. Hopefully this answers your question? – copper.hat Nov 30 '15 at 05:59
  • I am only asking that how are you sure that such a $k$ exists such that $B(x_n,\frac{1}{k})\cap S\neq\emptyset$ – Learnmore Nov 30 '15 at 16:09
  • @learnmore: There are two points in the answer where this is relevant. In the first, we only start to create a sequence $k \mapsto s_{n,k}$ if $B(x_n,1) $ intersects $S$. The resulting sequence may be finite. In the later part, we choose $s$ and find some $x_n$ such that $s \in B(x_n,{1 \over k})$ (hence $B(x_n,{1 \over k}) $ intersects $S$) and, by construction, we have $s_{n,k} \in B(x_n,{1 \over k})$ . – copper.hat Nov 30 '15 at 16:37