This a question from Functional analysis-By Kreyszig
Let $T:X\to X$ be a compact linear operator on a normed space. If dim $X=\infty$, then show that $0 \in \sigma(T)$.
For a operator $T:X \to X$ , he defines the resolvent $\rho(T)$ of T as follows:
if $\lambda \in \rho(T)$, then
(1) $T-\lambda I$ is one-one
(2)$(T-\lambda I)^{-1}$ from the range of $T-\lambda I$ to $X$ is bounded operator.
(3) Range($T-\lambda I$) is dense in X.
And then he defines the spectrum $\sigma$(T) of T as complement of $\rho(T)$ in the complex plane!!
Now the problem is that $X$ is not given to be Banach. I know that this is true in infinite dimensional Banach space because of the fact that identity operator from on $X$ is not compact. Is the question correct if we donot assume X to be Banach.
Thanks in advance!!
Were $0 \not\in \sigma(T)$, then $T$ would be invertible. You don't have to work to hard to figure out that the unit ball of $X$ must be compact. This renders $X$ finite-dimensional.