compact operator in banach spaces

Question

If $T$ be a linear operator between from one Banach space $X$ to another $Y$, then $T$ is compact iff $T(B)$ is relatively compact in $Y$ for every bounded subset $B$ of $X$.

Then, the book that I am reading (Applied Analysis by Hunter) says that:

We do not require the range of $T$ to be closed, so $T(B)$ need not be compact even if $B$ is a closed bounded set.

But, I am not quite understanding this. How does the closedness of $T(X)$ relate to the closedness of $T(B)$? Also, how is the closedness of $B$ relevant at all? I mean, regardless of whether $T(X)$ is closed or not, we do not expect $T(B)$ to be closed when $B$ is closed. Do we?

Answer 1

Continuous linear operators are generally not closed mappings. Even from $\mathbb R^2$ to $\mathbb R,$ the function $f((x,y))=x$ maps the closed subset $\{(x,1/x):x\ne 0\}$ of $\mathbb R^2$ onto a non-closed subset of $\mathbb R.$ The author is emphasizing this. A careless reading of the def'n might give the impression that a compact operator is a closed operator, which is generally false. My example $f$ is a compact operator...... A compact operator maps a bounded set onto a set whose closure is compact.

Some books use "completely continuous operator" to mean "compact operator".

Answer 2

If $T(X)$ were closed, then relatively closed subsets are also closed in $Y$, since the intersection of closed sets is closed. Between Banach spaces an (injective) linear operator has closed range if and only if it is bounded below. Since closed subspaces of Banach spaces are Banach this would tell you that T(B) is closed.