Relations among the five spectral parts

If $T_{\lambda }$ is not injective then it is not bounded below: a bounded-below $T_{\lambda }$ satisfies $c\mathrm{\parallel }x\mathrm{\parallel }\le \mathrm{\parallel }{T}_{\lambda }x\mathrm{\parallel }$ for $c>0$, so $T_{\lambda }x=0$ forces $x=0$. Hence ${\sigma }_p(T)\subseteq {\sigma }_{ap}(T)$, which is the second inclusion of claim 2.

If $\lambda \in {\sigma }_r(T)$ then, by definition, $T_{\lambda }$ is injective with non-dense range. Hence $\lambda \in {\sigma }_{cp}(T)$ and $\lambda \mathrm{\notin }{\sigma }_p(T)$, so ${\sigma }_r(T)\subseteq {\sigma }_{cp}(T)\setminus {\sigma }_p(T)$. Conversely, if $\lambda \in {\sigma }_{cp}(T)\setminus {\sigma }_p(T)$, then $T_{\lambda }$ has non-dense range and is injective, which is exactly $\lambda \in {\sigma }_r(T)$. Thus ${\sigma }_r(T)={\sigma }_{cp}(T)\setminus {\sigma }_p(T)$.

If $\lambda \mathrm{\notin }\sigma (T)$ then $T_{\lambda }$ is invertible, hence bounded below and of dense range; so $\lambda \mathrm{\notin }{\sigma }_{ap}(T)\cup {\sigma }_{cp}(T)$.

Claim 3, other inclusion: let $\lambda \in \sigma (T)$. If $T_{\lambda }$ is not bounded below then $\lambda \in {\sigma }_{ap}(T)$. If it is bounded below, then by [L3] it is injective with closed range; were the range also dense, closedness would give range $=X$, so $T_{\lambda }$ would be bijective with bounded inverse, hence invertible, contradicting [L4]; therefore the range is not dense and $\lambda \in {\sigma }_{cp}(T)$.

Claim 3, one inclusion: ${\sigma }_{ap}(T)\cup {\sigma }_{cp}(T)\subseteq \sigma (T)$ is the contrapositive of [step 1.3].

Claim 1: for $\lambda \in \sigma (T)$ the operator $T_{\lambda }$ is not invertible by [L4]; by [L1] it falls into exactly one of the three classical cases, so the three sets are disjoint and their union is $\sigma (T)$. The covering half of claim 3 is [step 1.4] and the reverse half is [step 2.1]; claim 2 consists of [step 1.1] and [step 1.2].