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


You should not confuse the terms "closed linear operator" in the sense of the definition you mentioned, and a "closed map" in a topological sense, which is the second sense you assumed to be equivalent to the first, i.e. mapping closed sets to closed sets. They are actually not related, and it is not true that every closed linear operator is also a closed map, i.e., maps closed sets to closed sets.

For further reference, see here.

  • $\begingroup$ I see! That's a relief. But then was I wrong in stating that one implies the other? $\endgroup$ – Hrit Roy Jun 1 at 8:26
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    $\begingroup$ 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. $\endgroup$ – uniquesolution Jun 1 at 8:29
  • $\begingroup$ That's right. I mistakenly assumed $T$ to be injective in my argument. $\endgroup$ – Hrit Roy Jun 1 at 8:41

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