I am trying to show counter examples to the Open Mapping Theorem. In this particular case, I am trying to show that both spaces need to be Banach.
First the OMT:
Let $X, Y$ be Banach spaces. Let $T : X \rightarrow Y$ be a surjective bounded linear operator. $T$ is an open mapping.
Now, for the counterexample, let $X = l^1$ equipped with $||\cdot||_1$. Let $Y = l^1$ equipped with $||\cdot||_\infty$. Claim: Y is not complete. Let $T: X \rightarrow Y$ be the identity operator defined by $Tx = x$. Clearly T is bijection and is bounded (thus continuous). Claim: T is not an open mapping.
Basically, the main question is that $X$ is Banach, $Y$ is not and I want to show that the operator is not an open mapping.
I am also having a hard time proving that $Y$ is not complete (I know its not, I just cant prove it).