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Can the topology generated from the lexicographic order on $\mathbb{R}^2$ come from a metric?

I'm assuming (not told otherwise) that the metric on $\mathbb{R}^2$ is the trivial one.

My gut says that it's not the case, because any ball around a point $(x,y)\in\mathbb{R}^2$ will contain the point $(x+\epsilon,y)$ and the distance from this point to $(x,y)$ "should" be infinite (from my understanding of the lexicographic order). But I think I'm confusion order with metric.

Please help me understand this better. And please excuse my English.

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  • $\begingroup$ No - see en.wikipedia.org/wiki/… (not posted as an answer since it's link only). $\endgroup$ – Ethan Bolker Sep 12 '18 at 19:34
  • $\begingroup$ This is the ordered square, $I^2$. This topological space is remarkably not metrizable, and it is an example of how the subspace topology of an ordered space need not be the same as the order tolpology of the subset. $\endgroup$ – Niki Di Giano Sep 12 '18 at 19:42
  • $\begingroup$ For clarification, see the answer to SE-1938870 $\endgroup$ – mlc Sep 12 '18 at 19:59
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Yes, $\mathbb{R}^2$ in the order topology is metrizable. In particular, the bounded metric: $$d(\vec{x} ,\vec{y} )= \min\{|y_2 - x_2|, 1\} \quad \text{if} \quad y_1=x_1, 1 \quad \text{otherwise}$$ Induces the topology.

The ordered square $I^2$ in the order topology, however, is not metrizable, even if it does look like it. The same metric would induce $I^2$ as a subspace of $\mathbb{R}^2$, but the two topologies (subspace and order) are not the same.

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  • $\begingroup$ My quick reading at wikipedia (see my comment) disagrees with your answer. Probably worth thinking about. $\endgroup$ – Ethan Bolker Sep 12 '18 at 19:36
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    $\begingroup$ I knew it would add up to this confusion, I actually wanted to point it out myself. The one you mention is the ordered square, not the same as the ordered plane. $\endgroup$ – Niki Di Giano Sep 12 '18 at 19:38
  • $\begingroup$ Good to have that cleared up for me and for the OP. Thanks. $\endgroup$ – Ethan Bolker Sep 12 '18 at 19:56
  • $\begingroup$ I'm happy to hear that! It is a major source of confusion, mainly because order topologies are among the first spaces that defy some of our intuitions the first time we stuble upon them. $\endgroup$ – Niki Di Giano Sep 12 '18 at 20:07
  • $\begingroup$ You seem to have accidentally omitted part of your definition of the metric $d$. $\endgroup$ – DanielWainfleet Sep 13 '18 at 2:02

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