Show that graphs of a contraction map and its inverse are both closed.

Question

The following is from Conway's book on Functional Analysis that states that being Y codomain of T Banach is necessary. Green underlined explanation can be find in here. However I am struggling the red-underlined section :

$\newcommand{\un}[2]{\color{#1}{\underline{\color{black}{\text{#2}}}}}$ The preceeding example shows that the domain of the operator in the Closed Graph Theorem must be assumed to be complete. The next example (due to Alp Eden) shows that the range must also be assumed to be complete. Let $\mathscr{X}$ be a separable infinite-dimensional Banach space and let $\left\{e_{i}: i \in I\right\}$ be a Hamel basis for $\mathscr{X}$ with $\left\|e_{i}\right\|=1$ for all $i$. Note that a Baire Category argument shows that $\un{green}{$I$ is uncountable. }$ If $x \in \mathscr{X}$, then $x=\sum_{i} \alpha_{i} e_{i}, \alpha_{i} \in \mathbb{F}$, and $\alpha_{i}=0$ for all but a finite number of $i$ in $I$. Define $\|x\|_{1} \equiv \sum_{i}\left|\alpha_{i}\right|$. It is left as an exercise for the reader to show that $\|\cdot\|_{1}$ is a norm on $\mathscr{X}$. Since $\left\|e_{i}\right\|=1$ for all $i,\|x\| \leqslant \sum_{i}\left|\alpha_{i}\right|=\|x\|_{1} .$ Let $\mathscr{Y}=\mathscr{X}$ with the norm $\|\cdot\|_{1}$ and let $T: \mathscr{Y} \rightarrow \mathscr{X}$ be defined by $T(x)=x$. Note that it was just shown that $T: \mathscr{Y} \rightarrow \mathscr{X}$ is a contraction. $\un{red}{Therefore gra $T$ is closed and hence so is gra $T^{-1}$}$. But $T^{-1}$ is not continuous because if it were, then $T$ would be a homeomorphism. Since $\mathscr{X}$ is separable, it would follow that $\mathscr{Y}$ is separable. But $\mathscr{Y}$ is not separable. To see this, note that $\left\|e_{i}-e_{j}\right\|_{1}=2$ for $i \neq j$ and since $I$ is uncountable, $\mathscr{Y}$ cannot be separable.

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My questions:

1- How $\operatorname{gra} Τ$ is closed?

2- How $\operatorname{gra} Τ^{-1}$ is closed?

Answer 1

Those are general results.

Let $X, Y$ be normed vector spaces. If $T: X \rightarrow Y$ is continuous then $\operatorname{gra} T$ is closed.

Proof: Let $\{(x_n, Tx_n)\}_n$ be a sequence in $\operatorname{gra} T$ converging to $(x,y) \in X \times Y$. Then $x_n \to x$ and, since $T$ is continuous, $Tx_n \to Tx$. But $Tx_n \to y$, so we have that $y=Tx$, which means $(x,y) \in \operatorname{gra} T$. $\square$

Note that all contractions are continuous.

The second result we need is also general.

Let $X, Y$ be metric spaces. If $S \subset X \times Y$ is closed, then $S^t=\{(y,x) : (x,y) \in S \}$ is also closed.

Proof: There are several way to prove this result. One of them is : Note that $Tr: X \times Y \rightarrow Y \times X$ defined as $Tr((x,y)) = (y,x)$ is a homeomorphism, and so, since $S$ is closed, $S^t = Tr(S)$ is also closed.

Another way to prove this result is to consider a sequence $\{(y_n,x_n)\}_n$ in $S^t$ converging to $(y,x) \in Y \times X$. Then $\{(x_n,y_n)\}_n$ is in $S$ converging to $(x,y)$. Since $S$ is closed, we have that $(x,y) \in S$. So $(y,x) \in S^t$. So $S^t$ is closed. $\square$.

An immediate consequence of the previous result is

Let $X, Y$ be normed vector spaces and $T: X \rightarrow Y$ any one-to-one linear operator. If $\operatorname{gra} T$ is closed then $\operatorname{gra} T^{-1}$ is also closed.

Proof: Just note that $\operatorname{gra} T^{-1}=\{(y,x): (x,y) \in \operatorname{gra} T\}$ and apply the previous result. $\square$