Open mapping $\iff$ bounded inverse $\iff$ closed graph I've studied closed graph theorem. In particular I first saw the open mapping theorem, then the bounded inverse theorem and finally the closed graph. I saw the proof that if a linear operator is closed then is bounded. Then my notes state ($X,Y$ Banach spaces):
$$ T:X\to Y  \text{ linear,  then } \\ T \text{ closed} \iff \textit{G}(T)\text{ closed }$$
Now I've some doubts. This statement looks very trivial to me but when I try to prove this I have necessarily to define a norm in $X \times Y$, e.g. $ ||\cdot||_{X\times Y}=||\cdot||_X+||\cdot||_Y$ and prove it, but this is not a general proof. I give definitions:
\begin{align}
& T \text{ is closed if } x_n\to x \text{ and } Tx_n \to y \implies y=Tx \\&
G(T)=\{ (x,y)\in X \times Y: Tx=y \}
\end{align}
So the point is this, my notes proved: $$ T  \text{ closed} \iff T \text{ bounded}$$
Then they say $$ T \text{ closed} \iff G(T) \text{ closed}$$
Hence the first theorem can be restated as $$ T \text{ bounded} \iff G(T) \text{ closed}$$
I want to clarify the second $\iff$.
Moreover it concludes by saying that:
$$ \text{Open mapping } \iff \text{Bounded inverse }\iff \text{Closed graph} $$
Im not sure about this. I tryed to prove that$$ \text{Closed graph} \implies \text{Open mapping }$$
But I only managed to prove that if $C$ is closed then $T(C)$ is closed. But$$ T(C^c) \ne T(C)^c$$ unless T is injective, so that I cannot conclude my proof.
 A: This is an answer for the part which you say in the comments you are yet to prove. It sounds like your main source of confusion is the topology/choice of norm on the product space so let me spend some time clarifying that first.
Given topological spaces $X$ and $Y$, there is a natural topology $X \times Y$ on the product space called the product topology. In the case of normed spaces, we would like to have that this product space $X \times Y$ with the product topology is again a normed space for some norm (i.e. there is a norm which induces the product topology). There are lots of norms that do this. As I mentioned in the comments $\|(x,y)\|_p = (\|x\|^p + \|y\|^p)^{\frac{1}{p}}$ for $p \in [1, \infty)$ and $\|(x,y)\|_\infty = \max\{\|x\|,\|y\|\}$ are all examples of norms on the product space that induce the product topology (in particular, they are equivalent norms). In fact, when $X$ and $Y$ are Banach spaces then these norms also make $X \times Y$ a Banach space.
As a result of this, it is normal that when you are talking about Banach spaces and you want to define the product space $X \times Y$, then you just assume that it comes with any one of the norms I've mentioned unless the choice for some reason matters (a lot of the time it doesn't). In this case, since closedness of $G(T)$ only depends on the topology of $X \times Y$, which norm we pick of the ones mentioned definitely doesn't matter.
So what this means is that you are being asked to prove that $T$ is closed if and only if $G(T)$ is closed in the product space for any one of the norms $\|\cdot \|_p$, $p \in [1, \infty]$ on $X \times Y$. For definiteness, you can just assume that we have decided to work with $p = \infty$.
The trick is to prove the following lemma, whose proof I'll leave as an exercise here.

Lemma: $(x_n,y_n) \to (x,y)$ in $X \times Y$ if and only if $x_n \to x$ in $X$ and $y_n \to y$ in $Y$.

Now assume $T$ is closed. We want to show that $G(T)$ is closed which means we want to show that if $(x_n, Tx_n) \to (x,y)$ then $(x,y) \in G(T)$. By the Lemma, $x_n \to x$ in $X$ and $Tx_n \to y$ in $Y$ and so since $T$ is closed $y = Tx$ which means $(x,y) \in G(T)$ as desired.
Conversely, assume $G(T)$ is closed. To check that $T$ is closed we need to take a sequence $x_n$ such that $x_n \to x$ and $Tx_n \to y$ and show that $Tx = y$. By the Lemma, we know that $(x_n, Tx_n) \to (x,y)$ in $X \times Y$. But $(x_n, Tx_n) \in G(T)$ and $G(T)$ is closed so that $(x,y) \in G(T)$ also. This means exactly that $Tx = y$ as desired.
