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? 



Separability of $L^1$ in $L^2$ topology
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.


