# Characterization of closed linear maps

I found this in a book : "A linear operator $$T:X\rightarrow Y$$ between normed linear spaces X and Y is said to be closed provided whenever $$(x_n)$$ is a sequence in X, if $$x_n\rightarrow x$$ and $$T(x_n)\rightarrow y$$, then $$y=T(x)$$."

I assumed $$T$$ to be a closed map (as in maps closed sets in $$X$$ to closed sets in $$Y$$) and proved the above without using the linearity of $$T$$. However, I am having trouble proving the converse. I assumed $$A$$ is closed in $$X$$. I have to show $$T(A)$$ is closed in $$Y$$. For that I chose a sequence $$(T(x_n))$$ in $$T(A)$$ with $$T(x_n)\rightarrow y$$. But I don't know if $$(x_n)$$ converges or not to some $$x$$ in order to apply the given condition.

Any help is appreciated.

Edit: The forward implication that I claimed to have proved is incorrect

• Wait a minute. There is something wrong with your argument. You cannot deduce that $y=Tx$, even if you know that $T$ maps closes sets to closed sets. If in a closed set $A$ you have $x_n\to x$, and you know that the sequence $Tx_n$ converges to some $y$, then you know that $y$ in $T(A)$ but you don't know whether it is $Tx$ or not. – uniquesolution Jun 1 at 8:29
• That's right. I mistakenly assumed $T$ to be injective in my argument. – Hrit Roy Jun 1 at 8:41