Show that if $B$ is compact in $X$ then $T(B)$ is closed in $Y$ and

Question

Let $T:X\to Y$ be a closed linear operator where $X,Y$ are normed spaces.Show that if

• $B$ is compact in $X$ then $T(B)$ is closed in $Y$

Solution : Let $y_n=T(x_n)$ be a sequence in $T(B)$ such that $y_n\to y\in \overline {T(B)}$.Then $(x_n)$ is a sequence in $B$ and as $B$ is compact it has a convergent subsequence $(x_{n_k})\to x$.We choose the corresponding terms in the $(y_n)$ sequence and form the subsequence the $(y_{n_k})$ which also converges to $y$.

As $T$ is a closed linear operator from above we conclude that $x\in B$ and $y=Tx\in T(B)$.

Is the solution correct?

Answer 1

Yes, but it would be better to say that we conclude $x\in B$ from $\bar B=B,$ which is by the compactness of $B.$

You could also say that since $C=\{x\}\cup \{x_{n_k}:k\in N\}$ is a closed subset of $B,$ its image $D=\{f(x)\}\cup\{y_{n_k}:k\in N\}$ is a closed subset of $T(B),$ so $y\in \operatorname{Cl}_Y(\{y_{n_k}:k\in N\})\subset \operatorname{Cl}_Y(D)=D\subset T(A).$ This is basically the same thing, without considering whether $y=f(x).$ (Although it is true that $y=f(x).$