# $F:A\rightarrow B$ be compact operator. Show closure subset of $B$ contains $0$.

$$A,B$$ are Banach spaces, $$A$$ infinite dimensional and $$F:A\rightarrow B$$ a compact linear operator. Show that the closure of $$\{Fx:||x||=1\}\subset B$$ contains $$0$$.

I've managed in a previous assignment, to show there is a sequence $$(a_n)_n$$ in $$A$$, such that $$||a_n||=1$$ and $$||a_n-a_m||\geq 1$$ for all $$n,m\geq 1$$, however I'm having trouble continuing with the above exercise.

Hint: Assume for the contrary that there exists $$\varepsilon>0$$ such that $$\|Fx\|\geq\varepsilon$$ whenever $$\|x\|=1$$. Then show that $$\|Fa_n-Fa_m\|\geq\varepsilon$$ for all $$n,m\geq1$$. Use this to obtain a contradiction.