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Let $A$ and $B$ be normed vector spaces and let $S\in \mathscr{K}(A,B)$ be a compact operator.

Question: How does it follow that the image of $S$ is separable?

Thanks for the help.

Davide Giraudo
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LSt
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2 Answers2

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  • A countable union of separable sets is separable (if $S_j$ is separable, let $\left(x^{j}_n\right)_{n\geqslant 1}$ be a dense sequence in $S_j$; then $\left\{x_n^{j},n,j\geqslant 1\right\}$ is countable and dense in $\bigcup_{j\geqslant 1}S_j$).
  • A subset $K$ with compact closure is separable (consider the covers $\left(B\left(x,n^{-1}\right)\right)_{x\in K}$).
  • $A=\bigcup_{j\geqslant 0}B\left(0,j\right)$ and $S(A)=\bigcup_{j\geqslant 0}S\left(B\left(0,j\right)\right)$.
Davide Giraudo
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  • Giraudo and how see countable unions of separable sets is separable – user89940 Feb 19 '19 at 01:56
  • @user89940 I have edited. – Davide Giraudo Feb 19 '19 at 08:02
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    thanks, but think a litlte more, i'm confusing a little.... if i said $S$ is separable, so exist $F$ countable that $\overline{F}=S$, so $S$ must be closed? because $\overline{F}$ is a close subset. what means say $R$ subset not close is separable, means that $\overline{R}$ is separable? . because in this case if $T$ compact operator, say $Im(T)$ is separable, not always $Im(T)$ is closed. thanks – user89940 Feb 19 '19 at 18:30
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The space is a countable union of balls centered in zero: $$A = \bigcup_{n\in N}B(0,n).$$

The image of $B(0,n)$ is precompact, therefore, separable. Countable union of separable sets is separable.

TZakrevskiy
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