Counterexample for the Open Mapping Theorem I would like to ask a counterexample for the open mapping theorem:
Find a discontinuous linear mapping $T:X \to Y$ such that $T(X)=Y$ and $X,\;Y$ are Banach but $T$ is not open.
Could you help me with this problem?
Thanks!
 A: The idea is quite simple:
Construct the linear operator with your desired properties on a Hamel Basis!
(Note that for getting discontinuous one must not use an ONB!)
For convenience let $X,Y=\mathcal{l}^1(\mathbb{N})$.
Choose your Basis as follows:
$B=\{b_1,\ldots\}\cup C$ (with $C$ extends to a Basis)
Define your operator an that:
$T(e_n):=c_n e_n$ and $T(c):=1 c$
Then your operator is surjective for $c_n\neq 0$!
a) $T$ continuous (and open): $c_n:=1$;
b) $T$ not continuous (but open): $c_n:=n$;
c) $T$ not open (not! continuous*): $c_n:=\frac{1}{n}$;
d) $T$ neither continuous nor open: $c_n:=n\quad n\in 2\mathbb{N}_0$ and $c_n:=\frac{1}{n}\quad n\in 2\mathbb{N}_0+1$.
a) That is just the identity operator, so continuous and open.
b) Take $x_n:=\frac{1}{n}e_n$. Then $x_n\to 0$ but $T x_n\nrightarrow T 0$.
c) Take $x_n:=1 e_n$. Then $\{x_n\}\subseteq B_{1+\delta}$ but never $T\{x_n\}\subseteq B_r$.
d) Works analogeously.
*Note, since the open mapping theorem holds you cannot construct: $T$ not open (but continuous).
