Does a compact linear operator on an infinite dimensional Banach space have a bounded inverse?

Question

Suppose $X$ is an infinite dimensional Banach space, and $A$ is a compact linear operator from $X$ to $X$. If $A$ is invertible and $A^{-1}$ is bounded, then $A A^{-1}=I$ ($I$ is the identity operator on $X$) is compact, which is impossible according to Riesz's Lemma. In this way we prove that $A$ doesn't have a bounded inverse. Whereas on the other hand, if $A$ is invertible, then $A^{-1}$ is indeed bounded according to the Inverse Operator Theorem. What's wrong?

Answer 1

Nothing is wrong. It just happens that there are no invertible compact operators from an infinite dimensional Banach space $X$ into itself.

Answer 2

A compact operator $A$ defined in an infinite dimensional space does not have an inverse, since the image of a the closed ball $B$ is contained in a compact space $C$, and if $A^{-1}$ is bounded (continuous), $A^{-1}(C)$ is compact, this is implies that $B\subset A^{-1}(C)$ is compact, contradiction.