A metric space that is no countable topological basis

Question

I know that a topology that if a topology is metrizable and separable, then it has a countable basis. Could someone give me an example of topology that is metrizable but has no countable basis ? Probably something like $L^\infty $ but I can't prove that it has no countable basis.

Answer 1

For $L^\infty[0,1]$ you can argue this way: For $t\in (0,1),$ define $f_t=\chi_{[0,t]}.$ Then if $0<s<t<1,$ $\|f_t-f_s\|_\infty=1.$ It follows that the open balls $B(f_t,1/2),$ $t\in (0,1),$ in $L^\infty$ are pairwise disjoint. Thus any basis for $L^\infty$ is uncountable.

Answer 2

A discrete space on uncountable set is metrizable (with a discrete metric). However, any singleton should belong to a basis.

Answer 3

The plane in the jungle river metric is an example, as is $\ell^\infty$ (the sequence space), and the hedgehog metric with uncountably many spines.