Strong open neighborhood contains an infinite dimensional subspace

Question

Suppose $H$ is $l^{2}$ space, let $B(H)$ denote the space of all continuous linear maps. How to prove that every open neighborhood of $0\in B(H)$, as in strong operator topology, contain an infinite dimensional subspace of $B(H)$?

Answer 1

Let $U$ be an open neighbourhood of $0$ in the S.O.T.

Since sets of the form $$U(T_0, x, \varepsilon) = \{T: \| Tx - T_0x\| \leq \varepsilon\}$$ form a subbase for the S.O.T. we can find $x_1, \dots, x_n$ and $\varepsilon > 0$ such that $$\bigcap_{i=1}^n \{T : \|Tx_i\| \leq \varepsilon\} \subseteq U$$ This means that it is enough to find an infinite dimensional subspace $X \subseteq B(H)$ such that $T \in X$ implies that $Tx_i = 0$ for each $i=1, \dots, n$.

Let $Y = \operatorname{span}\{x_i : i = 1, \dots, n\}$. Then $H = Y \oplus Y^\perp$ and we have a continuous orthogonal projection $P: H \to Y^\perp$. Since $Y$ is finite dimensional, the space $B(Y^\perp, H)$ is infinite-dimensional. Let $X = \{S \circ P : S \in B(Y^\perp, H)\} \subseteq B(H)$. It is straightforward to check that $X$ is the desired subspace of $B(H)$.