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.

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
