Let $X$ be a topological vector space (TVS), and $Y \subset X$ be a closed linear subspace. Let $T: X \to Y$ be a bounded linear operator such that $T^2 = T$ and $T(X) = Y$. Define $S := I - T$, then we could define $Z := S(X)$ such that $X = Y\oplus Z$. I am wondering if $Z$ is necessarily closed? If yes, how could I show it? Could anyone give me some hint on this?
Let $\pi: X \to Y$ be the projection to $Y$, then $Z = \pi^{-1}(\{0\})$. But I think in general TVS, singleton might not be closed, am I correct? This in the class was listed as a corollary of Closed graph theorem, but I don't see it.
I think we indeed need the Closed graph theorem, as $X$ may not be $T_1$ (If $X$ is $T_1$, then the singleton is closed).
Note that $S$ is the projection from $X$ to $Z$. Denote the graph of $S$ by $G_S := \{(x,Sx): x\in X \}$. Then by the Closed Graph theorem, $G_S$ is closed. For any $y \notin Z$, $y \neq 0$ and $y \neq Sx$ for any $x \in X$. Then $(y,y) \notin G_S$. By the closeness of $G_S$, there exists some neighborhood $V$ of $y$ such that $V \times V \subset (G_S)^c$. For any $z \in V \cap Z$, $(z,z) = (z,Sz) \in G_S$, a contradiction. So such $z$ does not exist. Therefore, $V \subset Z^c$, showing that $Z$ is closed.