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In $1D$ case, $\mathbb{R}^1$ as a typical case of $1D$ Euclidean space is totally ordered. I was wondering if any $1D$ Euclidean space $E^1$ is ordered as well? What is it

For higher dimensional case, are both $\mathbb{R}^n$ and $E^n$ ordered? What are the orders?

Thanks and regards!


I meant for an order for an Euclidean space:

that can induce or be compatible with the topology of the Euclidean space;

or that is defined in their definition, for example, the order of $\mathbb{R}^1$ is the one defined in one of $\mathbb{R}^1$'s definition as an ordered field with supreme property;

or that is used by default or most commonly.

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What do you mean by 'ordered'? Every set can be ordered (in many ways), but surely this is not what you ar after... – Mariano Suárez-Alvarez Aug 19 '10 at 14:56
I meant for an order that is defined in their definition, or defined by default, or used most commonly. For example, the order of R1 is the one defined in one of R1's definition as an ordered field with supreme property. – Tim Aug 19 '10 at 15:01 perhaps? – Aryabhata Aug 19 '10 at 15:04
@Moron: the lex order puts different weights on different dimensions, which seems not true for Rn or En. – Tim Aug 19 '10 at 15:20
@Tim: Your question is not clear, really. Are you just looking for an ordering (dictionary order will do in that case) or some kind of ordering which induces a topology and somehow that is derived based on the single dimensional case? – Aryabhata Aug 19 '10 at 16:14
up vote 8 down vote accepted

I think the OP is asking whether, for $n > 1$, the topology on $\mathbb{R}^n$ is an order topology.

The answer is no. Here is one simple way to see this: recall that a point $x$ in a topological space $X$ is a cut-point if $X \setminus x$ is disconnected. For $n > 1$, $\mathbb{R}^n$ has no cut-points. However, for any order topology, all but at most two points are cut-points, namely the points corresponding to the largest and smallest elements of the order, if any. [To be clear, these points may still be cut points, as for instance with $\mathbb{Q} \cap [0,1]$.] Otherwise $(-\infty,x)$, $(x,\infty)$ gives a separation of $X$. Since $\mathbb{R}^n$ has more than two points (!), for $n > 1$ the Euclidean topology is not an order topology.

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Thanks! Nice to know about the cut point property of order topology. If not required to induce the Euclidean topology, is there some order that have been seen to used commonly on Euclidean space? – Tim Aug 19 '10 at 23:30
@Tim: I can only say what others have said above: there are lots of ways of putting an ordering on any infinite set. Some of them are "products", in various senses, of the standard ordering on $\mathbb{R}$, in particular the lexicographic ordering. But I don't know how to put an ordering on $\mathbb{R}^n$ that has anything to do with it being Euclidean $n$-space and not just a set. – Pete L. Clark Aug 20 '10 at 6:13

If you are meaning a locally Euclidean whose topology is the one given by a linear order then $S^1$ is a counter-example.

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