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enter image description here enter image description here In this proof line -3: "${^\perp}(F^\perp)$ is closed linear span of $F$ and this subspace of $X$ is separable." I know ${^\perp}(F^\perp)=\overline{\operatorname{span}(F)}$, but we only know $F$ is countable. How to show $span(F)$ is countable to satisfy ${^\perp}(F^\perp)$ is separable?

Here is some definitions: If $A\subseteq X$, then we denote $$A^{\perp}=\{x^*\in X^*:\langle a,x^*\rangle=0 \textrm{ for all }a\in A\}.$$ Similarly, if $B\subseteq X^*$, then we denote $$^{\perp}B=\{x\in X:\langle x,b^*\rangle=0 \textrm{ for all }b^*\in B\}.$$

Thank you!

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The linear span of a countable set $F$ is separable : pick a countable dense set $D \subseteq \mathbb{F}$, say $\mathbb{Q}$ or $\mathbb{Q} + i\mathbb{Q}$ depending on whether the field of reals or complex numbers is used, and then the set

$$\{ \sum_{i=1}^n q_i \cdot x_i: n \in \mathbb{N} ,\forall_{1\le i \le n}: q_i \in D, x_i \in F\}$$

is countable (a union of images of the countable sets $\mathbb{Q}^n \times F^n$ over all finite $n$) and dense in $\operatorname{span}(F)$. And if $\operatorname{span}(F)$ is separable, so is its closure, obviously.

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  • $\begingroup$ Thanks for your reply! Can you be more specific why the set $$\{ \sum_{i=1}^n q_i \cdot x_i: n \in \mathbb{N} ,\forall_{1\le i \le n}: q_i \in D, x_i \in F\}$$ is dense and countable? $\endgroup$ – Answer Lee Sep 14 '17 at 21:26
  • $\begingroup$ I had a small argument about why its countable (as a countable union of countable sets) and we can approximate a term $r\cdot x$ ($r \in \mathbb{F})$ by a sequence $q_i \cdot x$ (from our countable set above) where the $q_i \to r$ are from $D$, which is dense, by continuity of scalar multiplication. We can do this for each of the finitely many terms in an element of the span of $F$. $\endgroup$ – Henno Brandsma Sep 14 '17 at 21:44
  • $\begingroup$ I am sorry. I just begin to learn analysis by myself I don't quite understand (a union of images of the countable sets $\mathbb{Q}^n \times F^n$ over all finite $n$) to get the set you give is countable. Can you give me more details? Appreciate!! $\endgroup$ – Answer Lee Sep 14 '17 at 22:51
  • $\begingroup$ @AnswerLee What do you know about countable sets? $\endgroup$ – Henno Brandsma Sep 15 '17 at 3:28
  • $\begingroup$ @ HennoBrandsma A finite set or an infinite set with the same cardinality with $\mathbb{N}$. $\endgroup$ – Answer Lee Sep 15 '17 at 3:52

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