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.

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 $\mathbb{R}^{n}$. 
