Counterexample for the Open Mapping Theorem I would like to ask a counterexample for the classical theorem in functional analysis: the open mapping theorem in the case that $Y$ is Banach, but $X$ is not Banach to show that the completeness of X is crucial. 
In details, find a continuous linear mapping $T:X \to Y$ such that $T(X)=Y$ and $Y$ is Banach but $T$ is not open.
If we can construct this, we could get an interesting example: there exists a bijective linear (contiuous) mapping between two normed space $X$ and $Y$, and only one of them is Banach.
The counterexamples for the case when $Y$ is not Banach is simple, but I didn't come up if I need $X$ is not Banach and $Y$ is Banach.
Thanks!
 A: Choose an infinite dimensional Banach space $\left( Y,\|\cdot\| \right)$ with a non-continuous linear form $\phi$.
Take $X$ to be the same vector space $Y$, with the norm $\|x\|_ \phi:=\|x\|+|\langle \phi,x\rangle|.$ Take $T$ the identity. Then $T$ is continuous, as $ \|x\| \leq \|x\|_ \phi $. However,  $ \|\cdot\|_ \phi $ can't be complete, or it would be equivalent to $ \|\cdot\|$ by the open mapping theorem, and $\phi$ would be continuous w.r.to $\|\cdot\|$.
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Rmk: existence of a non-continuous linear form on any infinite dimensional normed space. On any infinite dimensional normed space $Y$ you can define a non-continuous linear form as the projection map on the quotient of $Y$ on a non-closed linear subspace $N$ of codimension 1 (hence dense), identifying the quotient $Y/N$ with $\mathbb{R}$. Such a linear subspace can be defined starting from a Hamel basis. Indeed, let $\{u_\lambda\}_{\lambda\in\Lambda}$ be a Hamel basis for $Y,$ and pick $\lambda_0\in\Lambda$. We may assume $\inf_{\lambda\in\Lambda}\|u_\lambda\|=0$ because $\Lambda$ is infinite and one can re-normalize the basis. Define $N:=\mathrm{span}\left( \{u_\lambda-u_{\lambda_0}\, : \,\lambda\in\Lambda\}\right).$ Then $N$ is of codimension 1 and $\bar N=Y.$ Of course, the existence of a Hamel basis requires the Zorn lemma.
A: Problem 5.31 of Folland's Real Analysis provides such an example.
Let $Y$ be a Banach space, $Z$ any normed space, and let $S : Y \to Z$ be an unbounded, everywhere defined linear operator (you need the axiom of choice to construct such a thing).  Let $X \subset Y \times Z$ be the graph of $S$: $X = \{(y,Sy) : y \in Y\}$; by the closed graph theorem $X$ is not complete.  Let $T : X \to Y$ be the map $T(y,Sy)=y$.  $T$ is bijective and bounded but its inverse cannot be bounded, so it is not open.
A: Minh, this is not a new answer since you have already a satisfying number of them, but it's a riddle for you, since you are interested in this topic (too long to be posted as a comment).
"Theorem". All Banach norms on a real vector space $X$ are equivalent.
"Proof". (sketch). Let $\|\cdot\|_1$ and  $\|\cdot\|_2$ two Banach norms on $X$. Consider  $\|\cdot\|_3:=\|\cdot\|_1+ \|\cdot\|_2. $ Prove that it is actually a norm. Prove that a sequence converge to $x\in X$ w.r.to  $\|\cdot\|_3$ if and only if it converges to $x$ both w.r.to  $\|\cdot\|_1$ and w.r.to  $\|\cdot\|_2$. Prove that a sequence is Cauchy wrto  $\|\cdot\|_3$ if and only if it is Cauchy both w.r.to  $\|\cdot\|_1$ and w.r.to  $\|\cdot\|_2$. Deduce that $\|\cdot\|_3$ is complete. Apply the OMT tho the identity from $(X, \|\cdot\|_3)$ to  $(X, \|\cdot\|_1)$ and from $(X, \|\cdot\|_3)$ to  $(X, \|\cdot\|_2)$, and deduce that the three norms are equivalent.
However, if all Banach norms on $X$ are equivalent, all linear forms are continuous, and in infinite dimension there are non-continuous linear forms.
A: Take a normalized Hamel basis $y_a$, $a\in A$,  for $Y$ a separable infinite dimensional Banach space, so that $A$ has cardinality the continuum.  Write $A$ as a disjoint union of $B$ and a sequence $a_n$.  Define $T$ from the linear span $X$  of the unit vector basis $e_a$, $a\in A$, of $\ell_1(A)$ to $Y$  by setting $T e_{a_n} = n^{-1} y_{a_n}$ and $T e_a = x_a$ for $a\in B$. Then $T$ has norm one and is surjective and injective, but is not open.
