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I've wondered about the following question :

Is there an (explicit?) example of a vector space $X$, two complete norms $\|\cdot\|_1$ and $\|\cdot\|_2$ on $X$, and a sequence $(x_n) \subseteq X$ such that $x_n$ converges to $x$ with respect to $\|\cdot\|_1$, $x_n$ converges to $y$ with respect to $\|\cdot\|_2$, but $x \neq y$?

Obviously, this would imply that $\|\cdot\|_1$ and $\|\cdot\|_2$ are not equivalent. In fact, these two statements are equivalent, which is a consequence of the Open Mapping Theorem.

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    No, they are arbitrary norms. Would it be better to use another notation?2012-09-25

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Bill Johnson's example in MathOverflow seems answer the question.

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    I recently stumbled upon this question and very interesting example. However, it does not seem to me that in the example given, the second norm is complete. Does anybody have verification of this?2018-10-25
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Not sure but I would guess no. Maybe see if $(X, ||.||_1 + ||.||_2)$ is also a Banach space then you'll probably get a contradiction if you assume x is not y.

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    Yes, perhaps choice is unavoidable here. I'm curious to see if it's the case!2012-09-26