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I've to prove that the set of all subjective bounded linear operator is a closed subset of B(X, Y).

I tried with a sequence $T_n$ of surjective bounded maps, suppose it converges to some T $\in$ B(X, Y).

I've to show now T is surjective and bounded.

Let y $\in$ Y, for each $T_n$ there exists $x_n \in$ X such that $T_n(x_n)=y$.

Can someone help me how to proceed from here?

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1 Answer 1

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This is false. Take $Y$ to be the scalar field. Let $f$ be any non-zero continuous linear functional on $X$ And $f_n=\frac f n$. Then $f_n \to 0$ in operator norm and each $f_n$ is surjective.

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  • $\begingroup$ Oh thanks, What about openness of this subset? $\endgroup$
    – Sushant
    Dec 6, 2020 at 5:18

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