In the space $L^1(0,1)$ take the topology generated by the $L^2$-balls $$B^2_r(f)=\{g\in L^1(0,1):\; \|f-g\|_2<r\}.$$ Is $L^1(0,1)$ separable in this topology?
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No. Let $f_a(x)=|x-a|^{-1/2}$. If we had a countable dense set, then (pigeonhole principle) there would be a $g\in L^1$ with $g-f_a, g-f_b\in L^2$ for some $a\not= b$, so $f_b-f_a\in L^2$, but of course this is false.
answered Dec 2, 2016 at 19:07
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1$\begingroup$ For all it's worth, the L^2 topology breaks L^2 to continuum many disjoint open sets. $\endgroup$ Commented Dec 2, 2016 at 20:22