Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

up vote 3 down vote accepted
  • A countable unions of separable sets is countable.
  • A subset $K$ with compact closure is separable (consider the covers $(B(x,n^{-1}))_{x\in K}$).
  • $A=\bigcup_{j\geqslant 0}B(0,j)$ and $S(A)=\bigcup_{j\geqslant 0}S(B(0,j))$.
share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.