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Yesterday one my classmate asked me for the the definition of line. I have found it in coordinate geometry, in vector space, and in metric space.

The definition in metric space is quite general which has only one axiom(every 3 points collinear) but has a problem: it indeed satisfied by real line, but line segment and ray are also satisfies this condition in themselves when be considered as a metric subspace.

So should an extra axiom be added in? It is $\forall x\forall r \in \mathbb R \exists y,z(d(y,x)+d(x,z)=d(y,z)\land d(y,x)>r \land d(x,z)>r)$ and can make sure every line is infinite long in both two directions.

Besides, is it enough to add this axiom only? e.g. line should be linear continuum.

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    A line segment/ray does not fit the definition of a line (an induced line) in your link. It does not consist of *all* points that satisfy the condition.2012-11-17
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    @GregorBruns I understood, line segment/ray in $\mathbb R$ does not satisfies this. But if consider $[0,1]$ as an metric subspace of $\mathbb R$, then it is a line in itself whereas it is no longer a line after the extra condition be added in. So is it necessary to add the extra condition?2012-11-17
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    *Should* every line be "infinite" in both directions? That's basically just a judgment call.2012-11-17
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    It depends on what you want. If you consider, e.g., $[0,1]$ as a subspace in its own right, then the line segment $[0,1]$ (in $\mathbb{R}$) is really a line in the subspace $[0,1]$: It goes 'from one end of the space to the other'. So if you want to consider the situation *relative* to the ambient space, you could add the axiom. If you care about definitions without reference to the embedding, then you should probably not.2012-11-17
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    One may want to consider a great circle to be a "straight line" on a sphere, which your definition would exclude.2012-11-17
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    @RahulNarain Okay, you are right. But if we do not add something, it is possible a straight line has endpoint when the whole metric space has. Doesn't it a matter?2012-11-17
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    How does it matter?2012-11-17
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    @RahulNarain According to our common sense, a straight line should be straight indeed, but in addition should not has endpoint which is also the different aspect from line segment and ray.2012-11-17
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    You shouldn't expect generalizations of mathematical concepts to arbitrary spaces to continue to follow "common sense", which is really just a collection of prejudices absorbed from the simple cases you have looked at so far. Some of the properties you are used to will continue to hold, but certainly others will fail. For example, in $\mathbb R^2$ under the [taxicab metric](https://en.wikipedia.org/wiki/Taxicab_geometry), the "line" through two points contains a whole rectangle.2012-11-17
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    Common sense also says [multiplication should be commutative](https://en.wikipedia.org/wiki/Matrix_multiplication) and [$1+1$ is not $0$](https://en.wikipedia.org/wiki/Modular_arithmetic), but you know how that goes...2012-11-17
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    @RahulNarain Yes, not all semigroup are abel; $[1]_2+[1]_2= [0]_2$. And for taxicab metric on $\mathbb R^2$, a straight line indeed is not even a 'line'. (I have made a mistake that considered the whole space as a straight line)2012-11-17
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    @RahulNarain And in a taxicab metric space there can be 3 points which are in a straight line but not collinear, is it also plausible?2012-11-17
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    I don't know what "collinear" means if not "in a straight line".2012-11-17
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    @RahulNarain The site I cited says:"A set S of points is collinear if some pair of points in S induce a line that contains all of S"2012-11-17
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    OK, so the points $(0,0)$, $(1,1)$, $(2,0)$ are contained in the line between $(0,0)$ and $(2,1)$ but are not themselves collinear. What do you want me to tell you? If you don't like the consequences of their definition, don't work in their system. *shrug*2012-11-17
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    @RahulNarain Er...is '*maximal collinear set*' a good definition of *line*?2012-11-17

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