If it exists, the inverse of a compact linear operator in infinite dimensional space cannot be bouded I have been reading some posts on here that I think are related such as this and this. I am still having a tough time coming up with a nice proof for my question. 
Question: If a compact linear operator $T:X \rightarrow X$ on an infinite dimensional normed space $X$ has an inverse which is defined on all of $X$, show that the inverse cannot be bounded.
I found this in $8.3.8$ of Erwin Kreyszig functional analysis. 
 A: Suppose $T^{-1}$ existed and was bounded. Then $T^{-1}(B_X) \subseteq KB_X$ for some $K > 0$. Taking $T$ on both sides yields
$$B_X = T(T^{-1}(B_X)) \subseteq T(KB_X) = KT(B_X).$$
Since $\overline{T(B_X)}$ is compact, it follows that $B_X$ is a closed subset of a compact set, and hence is compact, which implies $X$ is finite-dimensional.
A: It is known that the compact operators $\mathbb{K}(X)$ form a two-sided ideal in the algebra of bounded operators $\mathbb{B}(X)$.
So if $T \in \mathbb{K}(X)$ has an inverse $T^{-1} \in \mathbb{B}(X)$, we have
$$TT^{-1} = I$$
so it would follow that $I \in \mathbb{K}(X)$, and the identity map $I$ is not compact when $X$ is infinite-dimensional.
Therefore $T^{-1}$ cannot be bounded.

Actually, a compact operator $T \in  \mathbb{K}(X)$ when $X$ is a Banach space cannot even be surjective, let alone bijective.
Indeed, a surjective bounded map $T \in \mathbb{B}(X)$ is open by the Open mapping theorem, so $T(B(0,1))$ is an open set. Therefore $T(\overline{B}(0,1))$ contains a closed ball which isn't compact in an infinite-dimensional space $X$. Hence $T(\overline{B}(0,1))$ cannot be precompact so $T$ is not compact.
