## two nonequivalent complete norms on the same space [closed]

I recently stumbled upon following reasoning:

"Let $X$ be a banachspace with respect to the norms $|\cdot |_1$ and $|\cdot |_2$. Define $|x | := |x |_1 + |x |_2$. Clearly $id: (X,|\cdot|) \to (X,|\cdot|_i)$ is continuous and invertible. Since $(X,|\cdot|)$ is also a banachspace, all three norms must be equivalent by the open mapping theorem."

The problem is, I do not see any reason, why $(X,|\cdot|)$ should be a banachspace but I am unable to construct a counterexample and I think such a counterexample would be rather unintuitive.

I would like to know, wheter or not the above statement is true. If not, is there an easy counterexample?

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It is an exercise to find a counterexample. See Pietro's answer to mathoverflow.net/questions/44716/… Vote to close. – Bill Johnson Feb 28 2011 at 16:35
Bill, you are right. I will delete my answer and vote to close. – Andreas Thom Feb 28 2011 at 16:36
@ANdreas, are you sure one needs the axiom of choice? – Mariano Suárez-Alvarez Feb 28 2011 at 16:42
I agree too, sorry I forgot of that post – Pietro Majer Feb 28 2011 at 16:42
@Mariano: Apparently yes, arsmath points that out in a comment to Pietro Majer's answer Bil Johnson links to above. – Theo Buehler Feb 28 2011 at 16:48
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