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Is a norm on a set a continuous function with respect to the topology induced by the norm?

Is a topology on the set that can make the norm continuous (i.e. the topology that is compatible with the norm) not unique? Is it a superset of the unique topology induced by the norm?

I am asking this question, because I heard (I am also not sure if it is correct) that a topology that can make an inner product continuous is not unique (such a topology is called weak topology on the inner product space?), and is a superset of the topology induced by the inner product.

Thanks and regards! Pointers to some references are appreciated!

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    In general, in a metric space $(X,d)$, the metric $d: X \times X \to \mathbb R$ is (jointly) continuous, i.e., it is continuous w.r.t. the product topology on $X \times X$ (where the topology on $X$ is induced by the metric). Proof uses just the triangle inequality.2012-01-05
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    Yes the norm is a continuous function because the triangle inequality. It's also yes for the second question. Keep in mind that if you add open sets to a given topology (the new topology is said to be finer, and the original is coarser), functions from that space to another topological space that are continuous for the original topology will stay continuous for the new one. For example, if you consider the discrete topology (every subset is open) then the norm is still continuous. However, the topology induced by the norm is the coarsest that make the norm continuous.2012-01-05
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    See also: http://math.stackexchange.com/questions/214939/continuity-in-a-normed-space2015-01-18

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