Closed operator bijective 
Definition. Let $X$ and $Y$ normed space. A linear operator $T\colon X\to Y$ is said closed if the graph of operator $T$ is closed in the product topology. That is $T$ is closed if and only if for each sequence $\{x_n\}\subseteq X$ such that $\{x_n\}$ converges in $X$ and $\{Tx_n\}$ converges in $Y$, holds $$\lim_{n\to \infty}T (x_n)=T(\lim_{n\to \infty}x_n).$$

We suppose that $T$ is a bijective closed operator, how I can prove that $T^{-1}$ has closed graph and then it is closed?

Let $y_n \to y$ and $T^{-1}y_n \to x$. We can write $y_n$ as $Tx_n$ for some $x_n$. Hence $Tx_n \to y$ and $x_n =T^{-1}y_n \to x$. Since $T$ is closed this implies $y=Tx$ or $x =T^{-1} y$ as required.
