3
$\begingroup$

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!


Update:

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.

  • 0
    @Moron: some order that can induce or be compatible with the topology, just as in 1D case.2010-08-19

2 Answers 2

9

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.

  • 2
    @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.2010-08-20
0

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