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

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.

• No - see en.wikipedia.org/wiki/… (not posted as an answer since it's link only). – Ethan Bolker Sep 12 '18 at 19:34
• 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. – Niki Di Giano Sep 12 '18 at 19:42
• For clarification, see the answer to SE-1938870 – mlc Sep 12 '18 at 19:59

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.
• You seem to have accidentally omitted part of your definition of the metric $d$. – DanielWainfleet Sep 13 '18 at 2:02