# Understanding Order topology on a particular set

Problem

Consider the set $$X= \{1,2\}×\mathbb{Z}^+$$ with the dictionary order as an ordered set with a smallest element.

Doubt

1. I am not able to understand what kind of intervals form the basis of this topology.

2. Also I am not able to understand how this topology is not a discrete topology

• It is defined the way you have defined it. It says that (2,1) is an exception. – blue boy Nov 4 '18 at 7:10

Aha, the OP means $$\mathbb{Z}^+ = \{1,2,3,4,5,\ldots\}$$ as the second component...

Consider what the order looks like: $$(1,1) < (1,2) < (1,3) < \ldots <$$ is just a copy of $$\mathbb{Z}^+$$ order-wise (with minimum $$(1,1)$$) and we have the same ordering for the points starting with $$2$$: $$(2,1) < (2,2) < (2,3) < \ldots <$$ and all points of the form $$(1,n)$$ are smaller than any point of the form $$(2,m)$$, so we have two copies of $$\mathbb{Z}^+$$, one all to the left of the other copy.

The order topology on it is generated by open intervals $$((i,n),(j,m))$$ with $$(i,n) < (j,m)$$, plus (for the minimum $$(1,1)$$) all sets of the form $$[(1,1), (i,n))$$ as well.

But if $$(i,n)$$ is in $$\{1,2\} \times \mathbb{Z}^+$$, and $$n \ge 2$$ then $$\{(i,n)\} = ((i,n-1),(i,n+1))$$ and so those singleton sets are open and the order topology is locally at those $$(1,n)$$ the discrete topology (lots of isolated points).

Also, $$\{(1,1)\} = [(1,1), (1,2))$$ so the minimum is also an isolated point.

Only $$(2,1)$$ is interesting: it's not the minimal element of the whole set, so a basic open neighbourhood of it looks like $$((1,n), (2,n))$$, an open interval with left endpoint smaller than $$(2,1)$$ so this point can only be of the form $$(1,n)$$ by the definition of the lexicographic order, and the right hand side we can just assume to be the smallest point above it, $$(2,2)$$. But this open interval never only contains just $$(2,1)$$, but always has a whole tail $$(1,n+1), (1, n+2), \ldots$$ as well.

So indeed at this one point $$(2,1)$$ the space is shown to be non-discrete, it's the only non-isolated point and the whole set $$\{1\} \times \mathbb{Z}^+$$ is in fact a sequence converging to it. A space is discrete iff all points are isolated points and in such a space there are no sequences that converge (except those that are eventually constant).