I think any finite dimensional normed space is separable so the answer to the above would be yes. But I am not 100% sure since since this is $L^\infty$ which is not separable. Fundamentally I am not grasping the reason why $L^\infty$ is not separable.
2026-04-03 16:47:02.1775234822
Is a finite dimensional subspace of $L^\infty$ separable?
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Any finite-dimensional subspace of a TVS $X$ over $\Bbb R$ is homeomorphic to some $\Bbb R^n$ and thus (locally compact and) separable. This is a standard fact proved in most text books (e.g. Rudin's functional analysis or Dunford and Schwarz).