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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 ?

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    $\begingroup$ Have you thought about the case where the $T_n$ are 'approximation' operators? (e.g. $T_nx = (x_1,\dots,x_n,0,0,\dots)$ on $l^p$) $\endgroup$
    – DCM
    Commented Jul 10, 2021 at 17:41
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    $\begingroup$ On Hilbert space every bounded operator is a strong limit of a sequence of compact operators. $\endgroup$
    – Nik Weaver
    Commented Jul 10, 2021 at 20:08
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    $\begingroup$ The in the two previously mentioned examples one has that every operator is the limit of a sequence of operators with finite-dimensional range (called finite-rank operators), which are a fortiori compact operators. In fact the set of finite-rank operators is dense $\mathcal{L}(X)$ in the strong topology, but there cannot be a sequence of finite-rank operators approximating the identity in the strong topology unless the space $E$ has the bounded approximation property, which not all separable Banach spaces have, but is implied by the existence of a Schauder basis. $\endgroup$
    – Robert Furber
    Commented Jul 12, 2021 at 15:44

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