Open sets in the order topology

Question

I'm having trouble understanding the order topology. If we consider $R_{\ge 0} \times R_{\ge 0}$ given the order topology coming from the lexicographic order, my understanding is that the sets $[2,3] \times (2,3)$, $(2,3) \times [2,3]$, $(2,3) \times (2,3)$ are all open. For the set $[2,3] \times (2,3)$, for any $\langle x,y \rangle \in [2,3] \times (2,3)$, the nbhd $((x,\dfrac{2+y}{2}),(y,\dfrac{3+y}{2})) \subset [2,3]\times(2,3)$ and contains points $p,q$ such that $p < \langle x,y\rangle < q$. Thus, the set $[2,3] \times (2,3)$ is open. $(2,3) \times [2,3]$ and $(2,3)\times(2,3)$ follow by similar logic. However, the set $[0,2) \times [0,\infty)$ is not open because there does not exist a open nbhd around $\langle 0,0\rangle$. Is my understanding correct?

Answer 1

This is actually a great question for increasing your understanding of the lexicographic order. To recall, in this order, $(a,b) < (c,d)$ if and only if either $a<c$ or $a=c$ and $b<d$. (This is exactly how we sort words when alphabetizing, for example in a dictionary—hence the name "lexicographic".)

So for your first remark: it's not actually true (when $x\ne y$) that $$ \bigg(\bigg(x,\dfrac{2+y}{2}\bigg),\bigg(y,\dfrac{3+y}{2}\bigg)\bigg) \subset [2,3]\times(2,3). $$ Take for example $x=2.1$ and $y=2.4$. Note that every single element of the form $(2.3,z)$ for $z\in\Bbb R_{\ge0}$ is in the interval $\big( (2.1,2.2), (2.4,2.7) \big)$. In particular, $(2.3,4)$ is in that interval, but is not in $[2,3]\times(2,3)$.

A modification of your idea works, however: any $(x,y) \in [2,3]\times(2,3)$ is contained in the neighborhood $$ \bigg(\bigg(x,\dfrac{2+y}{2}\bigg),\bigg(x,\dfrac{3+y}{2}\bigg)\bigg) \subset [2,3]\times(2,3) $$ (note the single difference). Another way of phrasing this idea: an arbitrary union of open sets is always open, and $$ [2,3]\times(2,3) = \bigcup_{x\in[2,3]} \big( \{x\}\times(2,3) \big) $$ where each $\{x\}\times(2,3)$ is open by the above argument.

The same argument (almost exactly) shows that $(2,3)\times(2,3)$ is also an open set in the lexicographic topology.

On the other hand, $(2,3) \times [2,3]$ is not an open set, because it does not contain a neighborhood of $(2.9,2)$ (for example). The elements "just less than" $(2.9,2)$ are of the form $(2.9,y)$ where $y$ is just less than $2$—and no such elements are in $(2,3) \times [2,3]$.

It turns out that the set $[0,2) \times [0,\infty)$ is in fact open. Can you prove this using these ideas?