# Point spectrum of $T$ implies residual spectrum of $T'$?

In Reed and Simon, Methods of Modern Mathematical Physics, v.1, Functional Analysis, we find the following proposition in Chapter VI.3:

Let $X$ be a Banach space and $T \in \mathscr{L}(X)$. If $\lambda$ is in the point spectrum of $T$, then $\lambda$ is in either the point or the residual spectrum of $T'$.

I'm confused by the either/or. It seems to me $\lambda$ ought to be in the residual spectrum of $T'$ if it's in the point spectrum of $T$, as follows.

Suppose $x \in \text{ker} \, (\lambda - T) \setminus \{0\}$. Suppose $\varphi = (\lambda - T')(\psi)$, where $\psi \in X^{*}$ is arbitrary. Then $\varphi(x) = 0$ since $$\varphi(x) = \psi((\lambda - T)x) = \psi(0) = 0.$$ Consequently, $\varphi \in \text{span}\{x\}^{\perp}$ and, thus, $R(\lambda - T') \subseteq \text{span}\{x\}^{\perp}$. By the Hahn-Banach Theorem, $\text{span} \{x\}^{\perp}$ is a proper closed subspace of $X^{*}$ and, thus, $R(\lambda - T')$ is not dense. By definition, $\lambda$ is in the residual spectrum of $T'$.

Am I making a mistake in the argument somewhere?

Part of the definition of residual spectrum is that $\lambda$ is not an eigenvalue, i.e. not in the point spectrum. Your argument is perfectly good for showing that $\lambda$ is either in the point spectrum or the residual spectrum of $T'$, but you did nothing to rule out the possibility that it is in the point spectrum.