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1.What is the precise definition of affine normed space?Do there exist normed spaces which are not affine?

2.Suppose $V$ is normed space,according to the absract definition of affine space,there must exist a set $A$ together with a vector space V with a transitive free action of $V$ on $A$.What is the set $A$,is $A=V$ as a set?

3.Is every Euclidean space a aafine normed space?

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    $\begingroup$ I guess it is just an affine space, associated to a vector space that comes with a norm. The true notion is that of a "normed vector space", its affine counterpart is not deep. $\endgroup$
    – Giuseppe Negro
    Commented Oct 21, 2019 at 9:01
  • $\begingroup$ I think $A$ could be any set, you might take $A=V$, but the fact that $V$ acts on itself turns $V$ into an affine space, the notion of the action of the additive group of the normed vector space on the "point-space" is crucial $\endgroup$
    – Peter Melech
    Commented Oct 21, 2019 at 9:31
  • $\begingroup$ @Peter Melech,does $A$ has a relationship with $V$? $\endgroup$
    – math112358
    Commented Oct 22, 2019 at 6:52
  • $\begingroup$ @math112358 Yes, the free action of $(V,+)$ on $A$ is the relationship. $\endgroup$
    – Peter Melech
    Commented Oct 22, 2019 at 9:05

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A subset $A$ of a vector space $V$ is called affine if it satisfies any of the following equivalent conditions:

  • There is a $p \in A$ such that the set $A - p := \{v -p\mid v \in A\}$ is a vector subspace of $V$.
  • For every $p \in A$, the set $A - p$ is a vector subspace of $V$.
  • For every pair of points $p, q \in A$ and $t$ in the field of $V$, $tp + (1-t)q \in A$. That is, if $A$ contains two points, it also contains the line running through them.

You could develop the concept of "affine" without reference to a vector space. Euclid did that for real affine spaces of 1, 2, and 3 dimensions over two thousand years ago. He called them "lines", "planes" and "space", respectively. But generally, it is nicer to have a ready-made linear structure built-in instead of having to create one from a bunch of geometric axioms.

Note that any vector space is automatically affine. More generally if $U$ is subspace of $V$ and $p \in V$, then $U + p$ is an affine space.

There is no such thing as a "normed affine space", as "norm" refers to the distance from a point to the origin, and affine spaces do not contain an origin or any other distinguished point. There is no property of the affine space itself that will differentiate between two points within it. The only way to identify a specific point in an affine space is by reference to the containing vector space. However, if the vector space $V$ has a norm, then there is a metric on any affine subspace, measuring the distance between any two points in the affine space.

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  • $\begingroup$ I wonder whether the definition you mentioned above is equivalent to the following definition.See math.stackexchange.com/questions/379724/… $\endgroup$
    – math112358
    Commented Oct 23, 2019 at 6:39
  • $\begingroup$ It is equivalent. My definition is a specific instantiation of that definition. The vector space $V$ in my definition and that one differ. Let me call mine $\hat V$, which is just some larger extension of $V$. They allow $A$ to be an arbitrary set, and then must define a compatible addition from $A\times V \to A$. I require $A \subset \hat V$ such that $A + V = A$. The addition they need is thus satisfied. My method may seem more restrictive, but in fact, you can always define such a $\hat V$ from their definition, so it really isn't. $\endgroup$
    – Paul Sinclair
    Commented Oct 23, 2019 at 14:51
  • $\begingroup$ If $A$ is an arbitrary set,when we talk about the Euclidean space $\Bbb R^n$,what does $A$ stand for? $\endgroup$
    – math112358
    Commented Oct 23, 2019 at 15:08
  • $\begingroup$ $A$ doesn't stand for anything until you define it. Can you please explain a little more about what you are asking? $\endgroup$
    – Paul Sinclair
    Commented Oct 23, 2019 at 23:38
  • $\begingroup$ I mean that how to check that $\Bbb R^n$ is an affine space by using the the abstract definition(the form of a triple $(A,V,\phi)$.) $\endgroup$
    – math112358
    Commented Oct 24, 2019 at 2:01

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