I'm reading through the book Linear Analysis by Bollobás. I'm having a bit of trouble understanding part of the proof of theorem 7 in chapter 13, which states:
Let $T$ be a compact operator and assume $\lambda \neq 0$ is not an eigenvalue of $T$. Then $\lambda \notin \sigma(T)$.
The proof relies on the following lemma:
Lemma 5: Let $T : X \rightarrow X$ be compact, and set $S = Id-T$. Then $SX$ is a closed subspace of $X$.
The part of the proof I don't understand goes as follows.
Let $T:X \rightarrow X$ be compact, and set $S = Id-T$. Let $Y_n = S^n(X)$. By lemma 5, the subspaces $Y_n$ are closed.
I understand why $Y_1 = S(X)$ is closed, but I don't understand why the rest of the $Y_n$ are closed. It doesn't seem to follow directly from lemma 5.
I suppose you could say $S^n = (Id-T)^n = Id - \sum_{k=1}^n \binom{n}{k} (-1)^{k-1} T^k$, and since the collection of compact operators is a subspace, lemma 5 applies here. But, this seems too complicated to just be glossed over. Is this the way I should think about it, or is there a simpler justification?
Since $T$ is compact, the restriction of $T$ to $S(X)$ is compact. So, applying the lemma 5 you get that $S\bigl(S(X)\bigr)$ is closed; in other words, $S^2(X)$ is closed. And now you can start all over again: since $T$ is compact, the restriction of $T$ to $S^2(X)$ is compact and so…