A question about quotient space of $R(T^{n})$ I am reading a paper about spectral theory. The author says it is easy to see the following proposition: 
For $T\in L(X)$, if dim$(R(T^{d})/R(T^{d+1}))<\infty$, then
$R(T^{d})$ is closed if and only if $R(T^{d+1})$ is closed . Here, $L(X)$ denotes the algebra of all bounded linear operators on Banach space $X$.
But, actually, I do not understand why it is correct.
 A: Assume first that $R(T^{d+1})$ is closed in $X$. Since $R(T^d)/R(T^{d+1})$ has finite dimension, one can find a finite-dimensional subspace $E\subset X$ such that $R(T^d)=E\oplus R(T^{d+1})$. So, to show that $R(T^d)$ is closed in $X$, it is enough to prove the following "well known" fact.
$\bf Fact\; 1.$ If $Z$ is a normed space and $E, F$ are linear subspaces of $Z$ such that $F$ is closed in $Z$ and $\dim E<\infty$, then $E+F$ is closed in $Z$.
To check this, one may assume that $E\cap F=\{ 0\}$. First, note that since $F$ is closed in $E\oplus F$ and $\dim E<\infty$, the canonical projection $\pi_E :E\oplus F\to E$ is continuous. This is easy to check by induction on $\dim E$, using the fact that a linear functional is continuous if and only if its kernel is closed. Now, let $(z_n)$ be a sequence in $E\oplus Z$ converging to some $z\in Z$. Then the sequence $(\pi_E(z_n))$ is Cauchy in $E$ by the continuity of $\pi_E$, hence convergent to some $e\in E$ because $\dim E<\infty$. But then $z_n-\pi_E(z_n)\to z-e$, and since $F$ is closed it follows that $z-e\in F$. Hence, we get $z=e+(z-e)\in E+F$. This shows that $E+F$ is indeed closed in $Z$.
Conversely, assume that $R(T^d)$ is closed in $X$ and that $X$ is a Banach space. Then $Z=R(T^d)$ is a Banach space and $R(T^{d+1})=R(T_{| Z})$. So, to show that $R(T^{d+1})$ is closed in $X$, it is enough to prove the following second "well known" fact.
$\bf Fact \;2.$ If $Z$ is a Banach and $S\in\mathcal L(Z)$ is such that $R(S)$ has finite codimension in $Z$, then $R(S)$ is closed in $Z$.
To check this, choose a finite-dimensional subspace $E\subset Z$ such that $Z=E\oplus R(S)$. Then $E$ is closed in $Z$, so the quotient space $Z/E$ is a perfectly defined Banach space. Denote by $\pi :Z\to Z/E$ the canonical quotient map and by $\widetilde \pi$ the restriction of $\pi$ to $R(S)$. Then $\widetilde \pi$ is of course continuous, and it is a linear isomorphism from $R(S)$ onto $Z/E$, by the very definition of $E$. (At this point, we don't know that $\widetilde\pi^{-1}$ is continuous). Moreover, the operator $\pi\circ R : Z\to Z/E$ is onto, again by the definition of $E$. By the open mapping theorem (recall that $Z$ and $Z/E$ are Banach spaces), one can find some constant $C$ such that the following holds: for any $u\in Z/E$, one can find $z\in Z$ with $\Vert z\Vert\leq C\Vert u\Vert$ such that $\pi\circ S(z)=u$. But for any such $z$ we have $S(z)=\widetilde\pi^{-1} (u)$ since $\pi\circ S=\widetilde\pi\circ S$, and $\Vert S(z)\Vert\leq \Vert S\Vert \,\Vert z\Vert\leq C\Vert S\Vert\, \Vert u\Vert$. So we have shown that $\widetilde\pi^{-1} :Z/E\to R(S)$ is continuous. It follows that $\widetilde\pi$ is a topological linear isomorphism from $R(S)$ onto $Z/E$. Since $Z/E$ is a Banach space, we conclude that $R(S)$ is also a Banach space under the norm induced by $Z$; in other words, $R(S)$ is closed in $Z$.
