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.
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.