Why do we distinguish the continuous spectrum and the residual spectrum? As we know, continuous spectrum and residual spectrum are two cases in the spectrum of an operator, which only appear in infinite dimension. 
If $T$ is a operator from Banach space $X$ to $X$, $aI-T$ is injective, and $R(aI-T)$ is not $X$. If $R(aI-T)$ is dense in $X$, then $a$ belongs to the continuous specturm, it not, $a$ belongs to the residual spectrum. 
I want to know why do we care about whether $R(aI-T)$ is dense, thanks. 
 A: As pointed out by M. Reed, B. Simon in the section VI.3 of their "Methods of Modern Mathematical Physics: Functional Analysis":

The reason that we single out the residual spectrum is that it does not occur for a large class of operators, for example, for self-adjoint operators.

A: The point $\lambda\in\mathbb{C}$ belongs to the spectrum of operator $T$ if the operator $T_\lambda:=T-\lambda I$ is not invertible. 
What can prevent $T_\lambda$ from being invertible? Recall that we are working in Banach space $X$ so invertibility is equivalent to bijectivity. Thus we need to study reasons why operator $T_\lambda$ can't be bijective. We can distinguish two cases:


*

*the operator $T_\lambda$ is not injective

*operator $T_\lambda$ is injective but not surjective
Now we discuss these cases.
1) The first one is the most common. In this case $\mathrm{Ker}(T_\lambda)$ is non-trivial so $T_\lambda$ is not invertible, and we say that $\lambda$ is in the point spectrum. If $X$ is finite dimensional this is the only possible case for operator not to be bijective. The reason is that an injective operator on a finite dimensional space is automatically surjective. But in case $X$ is infinite dimensional there are examples of injective but not surjective operators!
2) In the second case we have injective but not surjective operators. This means that the image of the operator $\mathrm{Im}(T)$ (which is a linear subspace) is not the whole space $X$. If $X$ is finite dimensional it is impossible for the operator $T_\lambda$ to be injective but not surjective, so this is not the case. If $X$ is infinite dimensional there two possibilities for the subspace $\mathrm{Im}(T_\lambda)$ not to be the whole $X$. Here we have two cases:
2.1) $\overline{\mathrm{Im}(T_\lambda)}=X$, speaking informally $T_\lambda$ is "almost surjective". In this case we say that $\lambda$ is in continuous spectrum.
2.2) $\overline{\mathrm{Im}(T_\lambda)}\neq X$, speaking informally $T_\lambda$ is "essentially non-surjective", even the closure of its image is a proper subspace of $X$! In this case we say that $\lambda$ is in the residual spectrum.
There are other classifications of points of the spectrum, but this one is the most common.
