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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. – Gregor Bruns Nov 17 '12 at 1:26
@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? – Popopo Nov 17 '12 at 1:41
Should every line be "infinite" in both directions? That's basically just a judgment call. – Benjamin Dickman Nov 17 '12 at 1:57
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. – Gregor Bruns Nov 17 '12 at 1:58
One may want to consider a great circle to be a "straight line" on a sphere, which your definition would exclude. – Rahul Nov 17 '12 at 2:15

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