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No, it is not sequential (hence non-metrisable) unless $X$ is finite-dimensional. Otherwise, let $(z_n)_{n=1}^\infty$ be a linearly independent sequence that is dense in $X$. For each $k$ we may consider the subspace $Z_k$ of $B(X)$ comprising operators mapping ${\rm span}\{z_1, \ldots, z_k\}$ to itself that are 0 on some complement of ${\rm span}\{z_1, \ldots, z_k\}$ (so that $\dim Z_k = k^2$). We may then choose a finite $\tfrac{1}{k}$-net $T_{k,j}$ of the sphere $\{T\in Z_k\colon \|T\|=k\}$. Let $S$ be the union of all the nets picked above. We claim that 0 is in the SOT-closure of $S$.

Indeed, suppose that $U$ is a SOT-open neighbourhood of 0. Let $x_1, \ldots, x_n\in X$ and let $\varepsilon > 0$ be such that $$\{T\in B(X)\colon \max_{i\leqslant n} \|Tx_i\| < 2\varepsilon \}\subseteq U.$$ Take $k$ with $1/k <\varepsilon$. When $k$ is large enough, by the density of $(z_n)_{n=1}^\infty$ there must be $T_k\in Z_k$ with $\|T_k\|=k$ such that $\|T_kx_i\|<\varepsilon$ for all $i$. Pick $j$ so that $\|T_{k,j} - T_k\|\leqslant\tfrac{1}{k}$. Thus, $$\|T_{k,j} x_i\| = \|T_{k,j} x_i - T_kx_i + T_k x_i\|\leqslant \tfrac{1}{k}+\varepsilon<2\varepsilon, $$ that is, $T_{k,j}\in U$.

Consequently, the set $S$ is not SOT-closed however it is SOT-sequentially closed.

No, it is not sequential (hence non-metrisable) unless $X$ is finite-dimensional. Otherwise, let $(z_n)_{n=1}^\infty$ be a linearly independent sequence that is dense in $X$. For each $k$ we may consider the subspace $Z_k$ of $B(X)$ comprising operators mapping ${\rm span}\{z_1, \ldots, z_k\}$ to itself that are 0 on some fixed complement of ${\rm span}\{z_1, \ldots, z_k\}$ (so that $\dim Z_k = k^2$). We may then choose a finite $\tfrac{1}{k}$-net $T_{k,j}$ of the sphere $\{T\in Z_k\colon \|T\|=k\}$. Let $S$ be the union of all the nets picked above. We claim that 0 is in the SOT-closure of $S$.

Indeed, suppose that $U$ is a SOT-open neighbourhood of 0. Let $x_1, \ldots, x_n\in X$ and let $\varepsilon > 0$ be such that $$\{T\in B(X)\colon \max_{i\leqslant n} \|Tx_i\| < 2\varepsilon \}\subseteq U.$$ Take $k$ with $1/k <\varepsilon$. When $k$ is large enough, by the density of $(z_n)_{n=1}^\infty$ there must be $T_k\in Z_k$ with $\|T_k\|=k$ such that $\|T_kx_i\|<\varepsilon$ for all $i$. Pick $j$ so that $\|T_{k,j} - T_k\|\leqslant\tfrac{1}{k}$. Thus, $$\|T_{k,j} x_i\| = \|T_{k,j} x_i - T_kx_i + T_k x_i\|\leqslant \tfrac{1}{k}+\varepsilon<2\varepsilon, $$ that is, $T_{k,j}\in U$.

Consequently, the set $S$ is not SOT-closed however it is SOT-sequentially closed.

No, it is not sequential (hence non-metrisable) unless $X$ is finite-dimensional. Otherwise, let $(z_n)_{n=1}^\infty$ be a linearly independent sequence that is dense in $X$. For each $k$ we may consider the subspace $Z_k$ of $B(X)$ comprising operators mapping ${\rm span}\{z_1, \ldots, z_k\}$ to itself. Moreover, we may choose a finite $\tfrac{1}{k}$-net $T_{k,j}$ of the sphere $\{T\in Z_k\colon \|x\|=k\}$. Let $S$ be the union of all the nets picked above. We claim that 0 is in the SOT-closure of $S$.

Indeed, suppose that $U$ is a SOT-open neighbourhood of 0. Let $x_1, \ldots, x_n\in X$ and let $\varepsilon > 0$ be such that $$\{T\in B(X)\colon \max_{i\leqslant n} \|Tx_i\| < 2\varepsilon \}\subseteq U.$$ Take $k$ with $1/k <\varepsilon$. When $k$ is large enough, by the density of $(z_n)_{n=1}^\infty$ there must be $T_k\in Z_k$ with $\|T_k\|=k$ such that $\|T_kx_i\|<\varepsilon$ for all $i$. Pick $j$ so that $\|T_{k,j} - T_k\|\leqslant\tfrac{1}{k}$. Thus, $$\|T_{k,j} x_i\| = \|T_{k,j} x_i - T_kx_i + T_k x_i\|\leqslant \tfrac{1}{k}+\varepsilon<2\varepsilon, $$ that is, $T_{k,j}\in U$.

Consequently, the set $S$ is not SOT-closed however it is SOT-sequentially closed.

No, it is not sequential (hence non-metrisable) unless $X$ is finite-dimensional. Otherwise, let $(z_n)_{n=1}^\infty$ be a linearly independent sequence that is dense in $X$. For each $k$ we may consider the subspace $Z_k$ of $B(X)$ comprising operators mapping ${\rm span}\{z_1, \ldots, z_k\}$ to itself that are 0 on some complement of ${\rm span}\{z_1, \ldots, z_k\}$ (so that $\dim Z_k = k^2$). We may then choose a finite $\tfrac{1}{k}$-net $T_{k,j}$ of the sphere $\{T\in Z_k\colon \|T\|=k\}$. Let $S$ be the union of all the nets picked above. We claim that 0 is in the SOT-closure of $S$.

Indeed, suppose that $U$ is a SOT-open neighbourhood of 0. Let $x_1, \ldots, x_n\in X$ and let $\varepsilon > 0$ be such that $$\{T\in B(X)\colon \max_{i\leqslant n} \|Tx_i\| < 2\varepsilon \}\subseteq U.$$ Take $k$ with $1/k <\varepsilon$. When $k$ is large enough, by the density of $(z_n)_{n=1}^\infty$ there must be $T_k\in Z_k$ with $\|T_k\|=k$ such that $\|T_kx_i\|<\varepsilon$ for all $i$. Pick $j$ so that $\|T_{k,j} - T_k\|\leqslant\tfrac{1}{k}$. Thus, $$\|T_{k,j} x_i\| = \|T_{k,j} x_i - T_kx_i + T_k x_i\|\leqslant \tfrac{1}{k}+\varepsilon<2\varepsilon, $$ that is, $T_{k,j}\in U$.

Consequently, the set $S$ is not SOT-closed however it is SOT-sequentially closed.

No, it is not sequential (hence non-metrisable) unless $X$ is finite-dimensional. Otherwise, let $(z_n)_{n=1}^\infty$ be a countable, linearly independent dense subset of $X$. For each $k$ we may pick a subspace $Z_k$ of $B(X)$ that comprises operators mapping ${\rm span}\{z_1, \ldots, z_k\}$ to itself. Moreover, we may choose a finite $\tfrac{1}{k}$-net $T_{k,j}$ of the sphere $\{T\in Z_k\colon \|x\|=k\}$. Let $S$ be the union of all the nets picked above. We claim that 0 is in the SOT closure of $S$.

Indeed, suppose that $U$ is a SOT open neighbourhood of 0. Let $x_1, \ldots, x_n\in X$ and let $\varepsilon > 0$ be such that $$\{T\in B(X)\colon \max_{i\leqslant n} \|Tx_i\| < 2\varepsilon \}\subseteq U.$$ Take $k$ with $1/k <\varepsilon$. When $k$ is large enough, by density of $(z_n)$ there must be $T_k\in Z_k$ with $\|T_k\|=k$ such that $\|T_kx_i\|<\varepsilon$ for all $i$. Pick $j$ so that $\|T_{k,j} - T_k\|\leqslant\tfrac{1}{k}$. Thus, $$\|T_{k,j} x_i\| = \|T_{k,j} x_i - T_kx_i + T_k x_i\|\leqslant \tfrac{1}{k}+\varepsilon<2\varepsilon, $$ that is, $T_{k,j}\in U$.

Consequently, the set $S$ is not SOT closed however it is SOT sequentially closed.

No, it is not sequential (hence non-metrisable) unless $X$ is finite-dimensional. Otherwise, let $(z_n)_{n=1}^\infty$ be a linearly independent sequence that is dense in $X$. For each $k$ we may consider the subspace $Z_k$ of $B(X)$ comprising operators mapping ${\rm span}\{z_1, \ldots, z_k\}$ to itself. Moreover, we may choose a finite $\tfrac{1}{k}$-net $T_{k,j}$ of the sphere $\{T\in Z_k\colon \|x\|=k\}$. Let $S$ be the union of all the nets picked above. We claim that 0 is in the SOT-closure of $S$.

Indeed, suppose that $U$ is a SOT-open neighbourhood of 0. Let $x_1, \ldots, x_n\in X$ and let $\varepsilon > 0$ be such that $$\{T\in B(X)\colon \max_{i\leqslant n} \|Tx_i\| < 2\varepsilon \}\subseteq U.$$ Take $k$ with $1/k <\varepsilon$. When $k$ is large enough, by the density of $(z_n)_{n=1}^\infty$ there must be $T_k\in Z_k$ with $\|T_k\|=k$ such that $\|T_kx_i\|<\varepsilon$ for all $i$. Pick $j$ so that $\|T_{k,j} - T_k\|\leqslant\tfrac{1}{k}$. Thus, $$\|T_{k,j} x_i\| = \|T_{k,j} x_i - T_kx_i + T_k x_i\|\leqslant \tfrac{1}{k}+\varepsilon<2\varepsilon, $$ that is, $T_{k,j}\in U$.

Consequently, the set $S$ is not SOT-closed however it is SOT-sequentially closed.