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$A,B$ 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.

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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.

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