(Fake proof) Bounded linear operators $\mathcal{L}(X,Y)$ between separable Banach spaces $X$, $Y$ is itself a separable Banach space?

Question

Fake proof: Let $X$ and $Y$ be separable Banach spaces. We know that the space $\mathcal{L}(X,Y)$ of bounded operators $X \to Y$ endowed with the operator norm is itself a Banach space. (Cf. [1][2][3].) So we need to show that $\mathcal{L}(X,Y)$ is separable.

It is known (cf. p.14 of Alexander S. Kechris, Classical Descriptive Set Theory) that the closed unit ball of $\mathcal{L}(X,Y)$ with the subspace topology inherited from the strong operator topology of $\mathcal{L}(X,Y)$ is separable. (Cf. [4][5].)

However, if the unit ball of a normed vector space is separable, then the entire vector space must be separable. (Cf. [6][7].) Therefore $\mathcal{L}(X,Y)$ is separable (in the strong operator topology), so "$\mathcal{L}(X,Y)$ is a separable Banach space".

Why proof is fake (?): The main issue is that the operator norm topology on $\mathcal{L}(X,Y)$ is being conflated with the strong operator topology on $\mathcal{L}(X,Y)$. Is that correct?

The strong operator topology on $\mathcal{L}(X,Y)$ is separable but need not be (complete) metrizable. In contrast, the operator norm topology on $\mathcal{L}(X,Y)$ is (complete) metrizable but need not be separable. (Cf. [8][9].) Is that correct?

E.g. in general the strong operator topology on $\mathcal{L}(X,Y)$ is generated by semi-norms [10] in some sense related to the operator norm, but in general is not even metrizable [11] and thus clearly not induced by the operator norm and not a Banach space. Cf. also this question [12].

Answer 1

Yes, you are correct. Separability for different topologies bears no relation.

You can see this clearly when $X=Y=\ell^2(\mathbb N)$. You can embed $A=\ell^\infty(\mathbb N)$ into $L(X)$ as multiplication operators. That is $$M_xy=(x_ny_n).$$ This embedding is isometric when you consider the operator norm. As $A$ is not separable, neither is $L(X)$. On the other hand, $L(X)$ is separable in the strong/weak operator topology since $X$ is separable and the finite-rank operators are dense in $L(X)$ in the weak/strong operator topology.