# Why is the image of a compact operator separable?

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.

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• 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 $S(A)=\bigcup_{j\geqslant 0}S(B(0,j))$.
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.