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How can a norm define a set in a vector space. I don't understand for example how 2 different norms can define a same open set.

It's not intuitive to me.

An open set doesn't need a norm to be open (or to even exist)

aribaldi
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    How do you define an open set? – Bernard Jun 08 '16 at 13:48
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    It's true that an open set doesn't need a norm. However, if you have a normed space $(X, |\cdot |)$, then the norm will in a natural way distinguish certain subsets $U\subseteq X$ that have the properties we expect open sets to have. Therefore we call those distinguished sets "the open sets of $X$ induced by the norm". – Arthur Jun 08 '16 at 13:54

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Are you talking about open balls or open sets? If $|| \cdot ||$ is a norm on a vector space $V$, then an open ball in $V$ (with respect to the norm $|| \cdot ||$) is a set of the form $$ \{ v \in V : ||v - v_0|| < r \}$$ for a fixed $v_0 \in V, 0 < r \in \mathbb{R}$. An open set (with respect to $|| \cdot ||$) is a set which is equal to a union of open balls in the sense of $|| \cdot ||$.

If $V$ is a finite dimensional real vector space, then $V$ can be identified with a finite product of $\mathbb{R}$s, which gives $V$ a topology of open sets (product topology). As you say, these open sets are defined independently of any norm.

But it is a standard result any two norms $|| \cdot ||_1, ||\cdot ||_2$ on such a space $V$ are equivalent, in the sense that any open set in the sense of $|| \cdot ||_1$ is also an open set in the sense of $|| \cdot ||_2$, and vice versa. In fact, the open sets from either of these norms are the exactly the same as the open sets of $V$ from the product topology.

This is not saying that open balls in the sense of one norm are also opens ball in the sense of the other. Rather, any set $S \subseteq V$ which is equal to a union of various open balls in the sense of $|| \cdot ||_1$, can also be expressed as a union of other various open balls in the sense of $|| \cdot ||_2$, and vice versa.

D_S
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  • could you provide a reference to the proof of the fact "any open set in the sense of $||\cdot||{1}$" is also an open set in the sense of $||\cdot||{2}$ and vice versa? – johnny09 Mar 17 '19 at 23:29
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    https://math.mit.edu/~stevenj/18.335/norm-equivalence.pdf – D_S Mar 17 '19 at 23:49
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If $\|\cdot\|$ is a norm on $X$, there is the induced metric $(x,y)\mapsto \|x-y\|$, which defines a topology with the collection of all sets of the form $$ \{y\in X : \|x-y\| < r\},$$ (with $r>0$) as a basis. A subset of $X$ is open with respect to this norm off it can be written as a union of basis elements.

If two norms generate the same topology, then the open sets will be the same regardless of which norm we are using.

Math1000
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With a norm you have a metric and thus a notion of distance. From this you can create balls. B(x,r) is the set of all vectors distance less than r from x. B(x,r)=B(x,s) with (WLOG) r

Jacob Wakem
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