My book defines the spectrum like this:
Let $H$ be a complex Hilbert space, let $I \in B(H)$ be the identity operator and let $T \in B(H)$. The spectrum of $T$, denoted $\sigma(T)$, is defined to be:
$$\sigma(T)=\{\lambda \in \mathbb{C}: T-\lambda I\text{ is not invertible}\}$$
Later there is a lemma that says that all eigenvalues are in the spectrum:
Let $H$ be a complex Hilbert space and let $T \in B(H)$. If $\lambda$ is an eigenvalue of T then $\lambda \in \sigma(T)$.
But why does the converse not hold? I mean, if $\lambda \in \sigma(T)$, why is not $\lambda$ and eigenvalue? What is wrong with this proof?:
Let $\lambda \in \sigma(T)$, then $T-\lambda I$ is not invertible. Then there is an $x \in H, x \ne0$ such that $T(x)-\lambda x=0$, if not, only 0 will be sent to 0, and then the operator is invertible. But then we have that $T(x)=\lambda x$, and hence $\lambda$ is an eigenvalue.
Do you see where the error is?