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I am currently reading up about norm topology, I have a background in functional analysis but I do not know anything about topology, aside from that topology is a collection of open sets with some properties on a set.

For example:

The norm topology or uniform topology or uniform operator topology is defined by the usual norm $\|x\|$ on $B(H)$. It is stronger than all the other topologies below.

What does it mean for a norm to "define" a topology? A norm is a function which measures distance, a topology is a collection of open sets, what is the intersection between these two concepts?

Can someone who is conversant in both fields please elaborate on what it means for an algebra of bounded linear operators to have a topology, specifically what a norm topology is about?

Fraïssé
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1 Answers1

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There is a metric space structure on a normed linear space $(X, \|\cdot \|)$: define the distance between $x, y\in X$ by

$$d(x, y) = \|x-y\|.$$

On the other hand, whenever you have a metric on $X$, you can use that to define a topology on $X$: A set $U \subset X$ is called open if for all $x\in U$, there is $\epsilon >0$ so that

$$ d(x, y) < \epsilon \Rightarrow y\in U.$$

That is, the "open" ball $\{ y: d(x, y) < \epsilon\}$ is also contained in $U$. One can check that the collection of all such sets form a topology on $X$. This is usually called a metric topology. But now your metric comes from a norm, it's also called the norm topology.

  • So can I think of the norm topology as a set of $\epsilon$ open balls contained in $X$ as defined by the metric? I can see how this is an interesting fact but how exactly does it help us in proving convergence of operators and such? – Fraïssé Sep 29 '15 at 01:09
  • @Fraïssé: More precisely, those open balls form a base for the norm topology. That means it contains all unions of those open balls. – joriki Jan 16 '24 at 12:44