Let $X$ be a Banach space and $B(X)$ be the space of bounded operators on $X$. 


Suppose that the strong operator topology on $B(X)$ is separable and that the cardinal number of $B(X)$ is continuum. 



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> **Question**. Can we conclude the norm topology on $X$ is separable?
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**Remark.** It was proved by [Tomek Kania](https://mathoverflow.net/questions/302286/if-the-cardinality-of-bx-the-space-of-operators-on-x-is-continuum-must) that the assumption $|B(X)|=\mathfrak{c}$ is not solely enough to get the separability assertion. 



Does separability of the strong operator topology imply separability of the underlying space?
Let $y$ be a norm-one vector in $X$. Consider the evaluation map $T\mapsto Ty$, which is a (linear) surjection from $B(X)$ onto $X$. This map is SOT-norm continuous. Indeed, suppose that $T\_n\to T$ in SOT. In particular, $\|T\_ny - Ty\|\to 0$ as $n\to \infty$.


A continuous image of a separable space is separable, hence $X$ is separable.


