Which of the following is true in normed linear spaces? Let $X$ and $Y$ be normed linear spaces, and let $T : X \to Y$ be any bijective linear map with closed graph. Then which one of the following statements is TRUE?


*

*The graph of $T$ is equal to $X \times Y $

*$T^{-1}$  is continuous

*The graph of  $T^{-1}$  is closed

*$T$ is continuous
I have no idea. I think 1 should be true as $f$ is bijective which means graph should be $X \times Y$. But the given option is 3. Why?
 A: Proof of (3):
$$\text{Graph of $T^{-1}$}=\{(y,T^{-1}y)\in Y \times X: y \in Y\}$$ Let $(y_n,T^{-1}y_n) \in \text{Graph of $T^{-1}$}$ such that $(y_n,T^{-1}y_n) \longrightarrow (y,x)$. Then $y_n \to y$ and $T^{-1}y_n \to x$
Take $x_n= T^{-1}y_n$. Bijectivity of $T$ implies $Tx_n \to y$. Since graph of $T$ is closed, $Tx=y$. That is $x=T^{-1}y$ and hence the result follows 

Counterexample for (1): 
Already given by Arthur in the comment. The graph in this example is a straight line which is obviously not equal to $\Bbb R \times \Bbb R$ 
A: $(1)$ fails to be true since for each $x \in X$, there is exactly one pair $(x,y) \in G(T)$ (namely, $y= Tx$). In particular, if $Y$ has more than one point then for $y' \neq Tx$, $(x,y') \not \in G(T)$ and hence $G(T) \neq X \times Y$.
$(2)$ and $(4)$ fail in general, but are true if you assume $X$ and $Y$ to be complete (this is a combination of the closed graph theorem and the inverse mapping theorem).
However, $(3)$ is true. Define $S: X \times Y \to Y \times X$ by $S(x,y) = (y,x)$. Then it is obvious that $S$ is a homeomorphism. We additionally have that $G(T^{-1}) = S(G(T))$ (it is a good exercise to check this). In particular, if $G(T)$ is closed then so is $G(T^{-1})$ since homeomorphisms map closed sets to closed sets.
