I'm a little confused about compact operators and whether or not they are invertible. Just hoping someone here can clear up my confusion:
Let $T$ be a compact operator on a Banach space $X$. Since $T$ is compact, we know that $0$ is in the spectrum of $T$. This implies that either $T$ is not invertible, or if it is then the inverse operator isn't bounded. Now, if $T$ did have an inverse $T^{-1}$, then $T$ would be a bijective bounded operator. Hence, by the Banach Open mapping theorem, the map $T^{-1}$ would be bounded. This seems to imply that no compact operator is invertible. But it seems that if $\{x_{n}\}$ is a basis for $X$, then define our operator $T$ so that $Tx_{n}=\lambda_{n}x_{n}$, where, say $\lambda_{n}=2^{-n}$. This operator $T$ is approximated by finite-rank operators, and hence is compact. Also $T^{-1}$ exists, but it obviously isn't bounded.
What is going on here? I feel like I'm going in circles.