# Compact operator with closed range has finite dimensional range

I have to prove that:

If $$X,Y$$ be Banach Spaces, and $$T\in B(X,Y)$$ is a compact operator, then $$T(X)$$ is closed in $$Y$$ if and only if $$\dim T(X)<\infty$$.

Can anybody help me with this proof, please? There is surely some property I haven't thought about, but I'm getting really weird right now... Thank you!

Let $Z=T(X)$. Then $Z$ is also a Banach space, as a closed subspace of a Banach space, and $T:X\to Z$ is onto, and hence open, due to Open Mapping Theorem. If $Z$ were infinite dimensional, then $T$ would not be compact, as open sets in infinite dimensional spaces are not pre-compact.

• How can I see that open sets cannot be precompact in infinite dimensional spaces?
– Marc
Jan 16, 2014 at 22:32

Hint: Try the open mapping theorem.

• Ok, I just proved that T must be open as it is surjective if you consider it as an operator from $X$ to $T(X)$, and $T(X)$ is a Banach Space as it is closed in Y. But how can I go on with it? Because $T(B_X)$ should then be a open neighbourhood of $0$, but if $dim(T(X))<\infty$ this cannot be, I think. So I would have proved that the dimension is not finite.
– Marc
Jan 15, 2014 at 15:41
• It will be a neighbourhood, but not an open one. This is not a problem. But it is also compact by assumption, so it follows that the unit ball of $T(X)$ is compact. Does that ring a bell? Jan 15, 2014 at 16:16
• It only can be compact if the dimension of $T(X)$ is finite. But how do we know that the unit ball in $T(X)$ is compact? I'm not seeing something that's obvious, sure...
– Marc
Jan 15, 2014 at 16:57

Hint: an infinite dimensional Banach space is never $\sigma$-compact (by Baire category theorem).