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I have found the definition of line in metric space.

It is general but has two problems. Considering about $\mathbb R^2$ equipped rectilinear distance, every line by this definition contains a rectangle and is not a string. Besides, it is possible 3 points in a line not collinear.


Here is what I have thought.

Let $<X,d>$ be any metric space.
Definition 0 Point $b$ is said between $a,c$ iff $d(a,b)+d(b,c)=d(a,c)$

Definition 1 A subspace $S$ of $X$ is collinear iff $\forall x\forall y \forall z[\mbox{exist one between the other two}]$

Definition 2 A subspace $L$ is called a line iff it is a maximal collinear subspace in $X$

Definition of collinear set is same as the previous if $|S|\le 3$ but different in otherwise. By this definition a collinear set is a collinear set by the previous definition but the converse is not valid.

According to this definition, it has 2 theorems.

Theorem 1 Every subset of a collinear set is collinear.

Theorem 2 Every collinear set can be extended to a line.

1 holds by definition whereas it does not hold by the previous definition.

2 also holds by the previous definition but requires Zorn's lemma by the current definition.

Hence a Corollary 3 Every set is collinear iff it is included in a line

Finally, by the current definition every line in $\mathbb R^2$ equipped rectilinear distance is actually a curve, no longer contains proper rectangles.

It looks better now. My question:Is it a precise definition?

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1  
Your definition is valid. But there are still some “very unusual” lines in ${\mathbb R}^2$ equipped with the rectilinear distance. Consider the set $\{(0,0), (1,0), (0,1), (1,1)\}$. It is a line according to your definition! –  Yury Nov 21 '12 at 16:40
    
@Yury Well...you are right. It seems the definition of collinear needs to be improved. Thank you. –  Popopo Nov 21 '12 at 16:54
1  
One way to do that is as follows. Now you require that every three points in a collinear set embed isometrically into $\mathbb R$(you state that differently, but your condition is equivalent to that). You can require that every *four* points embed isometrically into $\mathbb R$. Then every collinear set will be isometric to a subset of $\mathbb R$ (“it will be a true line”). –  Yury Nov 21 '12 at 17:02

4 Answers 4

up vote 3 down vote accepted

It is a precise definition, but a global definition based on distance disagrees with the local concept of geodesic:

  • in spaces that are not geodesically complete (like the line or plane with some points removed), the line can have holes

  • the hole can be so large that a line contains only its two endpoints, as in the Euclidean upper half plane $y>0$ with with two $y=0$ points added on the boundary . The "line" between the two boundary points is those points and nothing else.

  • in spaces that have more than one geodesic between two points, such as a cylinder or torus, the collinearity requirement excludes geodesics that wrap many times

  • unions of line segments (with nonempty overlap between any two segments) do not satisfy collinearity for the multiply wrapped geodesics on a cylinder

  • geodesic loops, such as great circles on a sphere or latitudes on a cylinder, present the same problem, where all short enough arcs are metric lines, but the whole loop does not have the collinearity property.

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I'm afraid great circles on a sphere are not straight either. Consider about trisection point $a,b,c$ in a great circle, then $d(a,b)=d(b,c)=d(c,a)$ thus make an equilateral triangle... So the definition needs to be reconsidered in order to include great circles. –  Popopo Nov 22 '12 at 2:23
    
Thanks for the correction. I updated the answer and added another example. –  zyx Nov 22 '12 at 5:46

So you say metric space and I think topology. lines in topology are usually defined by a function from the unit interval to the space. then you can say for any $x_{i}$ the same thing you said for b in defintion 0. I hope that helps.

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Thanks for your notation. –  Popopo Nov 21 '12 at 7:55
3  
Lines are not always straight lines by this definition, @Popopo's definition yields a straight line in metric spaces. Of course that for "straight" we need some sort of measurement, like a metric. –  Asaf Karagila Nov 21 '12 at 7:59

The definition is valid. However, there are still some “very unusual” lines in the space ${\mathbb R}^2$ equipped with the rectilinear distance. For example, the set $\{(0,0),(1,0),(0,1),(1,1)\}$ is a line according to this definition (it is a maximal collinear set).

We can make your definition stronger as follows.

Definition 1. Let us say that a subspace $S$ of a metric space $(X, d)$ satisfies the $k$-collinear condition if for every points $x_1, \dots, x_k$ in $S$ there exist a permutation $\pi$ such that $$\sum_{i=1}^{k-1}d(x_{\pi(i)},x_{\pi(i+1)}) = d(x_{\pi(1)},x_{\pi(k)}).$$

Trivially, every set $S$ satisfies the 2-collinear condition. A set $S$ satisfies the 3-collinear condition precisely when it is collinear according to your definition. Clearly, if a set satisfies the $k$ collinear condition than it also satisfies the $k+1$-collinear condition (we can just let $x_{k+1} = x_k$). However, the set $\{(0,0),(1,0),(0,1),(1,1)\}$ satisfies the 3-collinear condition but not the 4-collinear condition. So in general the 4-collinear condition is strictly stronger than the 3-collinear condition.

Question. Can we get stronger and stronger conditions by increasing $k$? E.g., is 5-collinear condition even stronger than 4-collinear condition?

It turns out that $4$-collinear condition implies $k$-collinear conditions for all $k$. This, in particular, follows from the four-point characterization of tree metrics. This result can be restated as follows:

A set $S$ satisfies the 4-collinear condition if and only if there is an isometric embedding $$\phi:S \hookrightarrow{\mathbb R},$$ i.e. there is a map $\phi:S\to\mathbb R$ s.t. $d(x,y) = |\phi(x) - \phi(y)|$ for every $x$ and $y$ in $S$.

Similarly to your definition, we give the following definition of a line.

Definition 2. A subspace $S$ of a metric space $(X, d)$ is a line if it is a maximal subspace of $(X,d)$ satisfying the 4-collinear condition.

Now every line $S$ in a Banach space (in particular, in ${\mathbb R}^2$ equipped with the rectilinear distance) is a curve. Moreover, there is a natural parametrization $\gamma(t)$ of $S$ ($\gamma:{\mathbb R} \to S$) such that $d(\gamma(s), \gamma(t)) = |s-t|$ for every $s,t\in \mathbb R$.

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Thank you, it sounds good. –  Popopo Nov 26 '12 at 10:43

Yes, it is a precise definition.

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