# Show that the lexicographic order topology for $\mathbb{N}\times \mathbb{N}$ is not the discrete

My question is:

Show that the lexicographic order topology for $\mathbb{N}\times \mathbb{N}$ is not the discrete

I have been thinking on the fact that on the discrete topology all singleton sets are open. If can I find one singleton not open the proof is done?

For example $\{(0,0)\}$ is not open because it has not a predecessor

is it right? I don't know how to write it properly

## 3 Answers

For order topologies we also include as (basic) open sets all intervals of the form $$( \leftarrow , a ) = \{ x \in X : x < a \}$$ where $a \in X$. So in the lexicographic order order on $\mathbb{N} \times \mathbb{N}$ we have $\{ \langle 0 , 0 \rangle \} = ( \leftarrow , \langle 0 , 1 \rangle )$. (We similarly include all intervals of the form $( a , \rightarrow ) = \{ x \in X : a < x \}$ as (basic) open sets.)

It is relatively easy to prove that if $<$ is a strict linear order on a set $X$, then $x \in X$ is isolated in the order topology iff

• $x$ has an immediate predeccessor, or is the minimum element; and
• $x$ has an immediate successor, or is the maximum element.

Can you think of points in $\mathbb{N} \times \mathbb{N}$ (points similar to the one you chose, in fact) where one of these two fails?

• for example the point (3,0) does not have an immediate predecessor: (2,x) where $x \in (0, \infty)$ – LFRC Mar 31 '14 at 21:06
• @LFRC: That is correct! – user642796 Mar 31 '14 at 21:07

HINT: In two steps,

1. Show that $\{x\}$ is open if and only if $x$ is a successor of some $y$ and the predecessor of some $z$; or $x$ is the minimal element and it has a successor.

2. Show that for $n>0$, the point $\langle n,0\rangle$ is not a successor in the lexicographic order, and not the minimal element.

$\implies$ Conclude that $\Bbb{N\times N}$ is not dicrete.

Definitely not. First, being closed is not the same as not open. In fact, if every singleton set is open, then every set is both open and closed.

Also, $\{(0,0)\}$ is open, because it's the set of points less than $(1,0)$.

Have you tried looking at $\{(0,1)\}$? Is it open?