It is well-known that the space of compact operators over Banach spaces is closed within the norm topology. My question:

Let $X$ be a Banach space.

Considering the strong topology (defined by seminorms) on $\mathcal{L}(X)$ i.e., $\{p_x(T):=Tx \text{ for all } T\in \mathcal{L}(X),\, x\in X \}$. Do we still have the same result over the strong topology.

$\textbf{Question}$:

$(T_n)_{n\in\mathbb{N}}$ a sequence of compact operators which converges strongly to $T\in \mathcal{L}(X)$, is $T$ compact ?

everybounded operator is a strong limit of a sequence of compact operators. $\endgroup$finite-rank operators), which area fortioricompact operators. In fact the set of finite-rank operators is dense $\mathcal{L}(X)$ in the strong topology, but there cannot be asequenceof finite-rank operators approximating the identity in the strong topology unless the space $E$ has thebounded approximation property, which not all separable Banach spaces have, but is implied by the existence of a Schauder basis. $\endgroup$