finite dimensional range implies compact operator Let $X,Y$ be normed spaces over $\mathbb C$. A linear map $T\colon X\to Y$ is compact if $T$ carries bounded sets into relatively compact sets (i.e sets with compact closure). Equivalently if $x_n\in X$ is a bounded sequence, there exist a subsequence $x_{n_k}$ such that $Tx_{n_k}$ is convergent.
I want to prove that if $T\colon X\to Y$ has finite dimensional range, then it's compact.
 A: Hint: Every finite dimensional normed space is isomorphic to $\mathbb R^n$, so any bounded subset is pre-compact (sets with compact closure). 
A: if $T$ is bounded and dim rangeT is finite ,the operator $T$ is compact.(kreyszig p.407)
A: Here is a direct proof.
Proof. Since $T$ has finite rank, Im$T$ is a finite-dimensional normed space. Furthermore, for any bounded sequence $\{ x_n \}$ in $X$, the sequence $\{ T x_n \}$ is bounded in Im$T$, so by Bolzano-Weierstrass theorem this sequence must contain a convergent subsequence. Hence $T$ is compact. $\square$
A: You require the boundedness of $T$. Fact is that boundedness + finite rank of $T$ gives compactness of $T$. This you can find in any good book in Functional Analysis. I would recommend J.B Conway
A: Assuming $T$ is continuous, you know that if $||x||=1$, then $T(x) \leq M$ for some $M \in \mathbb{R}$. So $T(B^{X}_{1}(0)$) (unit ball in $X$) is contained in $\overline{B^{\text{Im}(T)}_{M}(0)}$, which is compact since the image is finite dimensional and hence isomorphic to $\mathbb{R}^{n}$. 
