# Multiplication not continuous in $B(H)$ in the strong operator topology

I'm reading this answer by t.b., and I'm only interested in the case when $X$ is an infinite-dimensional Hilbert space. Regarding Question 2, in the first bullet point he claims the following:

"Given such a set $U(S,x_1,\ldots,x_n, \varepsilon)$, pick $y_i$ in such a way that $\|Sx_i - y_i \| \lt \varepsilon$ and that $x_1,\ldots,x_n,y_1,\ldots,y_n$ are linearly independent."

I don't quite get how you can get such $y_i$'s. Could someone clarify a bit why this is the case?

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Let $v_1,\ldots,v_n$ be linearly independent unit vectors in $X$ such that $\mathrm{span}\{v_1,\ldots,v_n\}\cap\mathrm{span}\{x_1,\ldots,x_n,Sx_1,\ldots,Sx_n\}=\{0\}$. These exist because $X$ is infinite dimensional. Let $y_k=Sx_k+\dfrac{\varepsilon}{2}v_k$.