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I have a problem stated,

"Let $X$ be a normed space. Prove that the topology generated by norm is exactly the coarsest topology on $X$ s.t. the norm and all translations are continuous."

Here's what I've done so far, it seems easy to prove that the norm and all translations are continuous in the topology generated by norm since it would be the same if we consider $X$ as a metric space. The coarsest topology part, however, seems not immediate. Actually, I'm a little confused here.

I think the coarsest topology for the norm map and the translations to be continuous must be the Topology generated by the norm map (since it's really vague about the topology in which the translations are continuous). But I read in Is the norm topology the same as the initial topology generated by the norm function? that Topology generated by norm and Topology generated by the norm map are not the same. Can anyone show me how to solve this problem? THank you

T C
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1 Answers1

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In the other post you are referring to there isn't the requirement that translations are continuous. That is the difference. See also @Omnomnomnom's answer. If you have continuous translations you can easily define a continuous distance function. Here is a proof:

Take a point $x$ then by assumption the map $$ N\colon X \to \Bbb R , \ y \mapsto \Vert x-y \Vert $$ is continous since it is the composition of a translation and the norm. So for any $\varepsilon > 0$ the set $N^{-1}(\varepsilon) = \{y\in X \mid \Vert x-y \Vert < \varepsilon \} = B_{\varepsilon}(x)$ is open.

Now how do you define a topology on a normed space? Usually you say it is a metric space and in a metric space you define a set to be open if all points of it contain an open ball. So every open set is the union of open ball(the form the base of the topology).

user60589
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