TITLE: Counterexample for the Open Mapping Theorem
QUESTION [28 upvotes]: 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!

REPLY [15 votes]: 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.