Spectrum of compact operators on an infinite dimensional normed space

Question

The question is as follows:

Let $T:X \to X$ be a compact linear operator on a normed space. If the $dim X= \infty$ then show that $0 \in \sigma(T)$.

My attempt:

Suppose on the contrary that $0 \notin \sigma(T)$. Then it follows that $0 \in \rho(T)$, the resolvent set. This would imply that $(T-0I)^{-1}$ exists, that is $T$ is invertible.

How can i use the compactness of $T$ to get a contradiction ?

Answer 1

If there exists a continuous linear operator $T^{-1} : X\to X$ then $T$ will be a homeomorphism hence $B=T^{-1} (\overline{T(B)} )$ should be compact but this is impossible. $B$ - denotes the unit ball.

Answer 2

If $0 \notin \sigma_p(T)\cup\sigma_r(T)$ , then $T: X \to Im(T)$ is a bijective operator and $Im(T)$ is a dense subspace of $X$. The operator $T^{-1}: Im(T) \to X$ is not bounded, since $T^{-1}T = I_X$ is not compact. Therefore $0 \in \sigma_c(T)$.