# open singleton set in order topology $\{1,2\}\times \mathbb{Z}_+$

Consider the following example from Munkres's Topology:

Example 4: The set $X = \{ 1,2 \} \times \mathbb{Z}_{+}$ in the dictionary order is another example of an ordered set with a smallest element. Denoting $1 \times n$ by $a_n$ and $2 \times n$ by $b_n$, we can represent $X$ by $$a_1, a_2, \ldots; b_1, b_2, \ldots.$$ The order topology on $X$ is not the discrete topology. Most one-point sets are open, but there is an exception --- the one-point set $\{ b_1 \}$. Any open set containing $b_1$ must contain a basis element about $b_1$ (by definition), and any basis element containing $b_1$ contains points of the $a_i$ sequence.

I also read this post, but I still have some confusing about the concept open in the last statement.

What I understand:

To prove $b_1$ is not open, we need to show $b_1\notin \tau$, where $\tau$ is the order topology in $X$, so $\tau$ includes all set in the form of $(a,b),[a_0,b),(a,b_0]$. If we can prove $\{b_1\}$ is not in all these three forms, then it is not open.

Since $b_1$ is neither the minimum of $X$ nor maximum, let's assume $\{b_1\}=(x,y)$, then $x$ must be included in $a_i$ sequence, and $y$ must be included in $b_i$ sequence, so $x$ must be $a_i$ for some $i$, then since $a_i$ sequence is infinite, the interval $(x,y)$ can never be singleton, but left side $\{b_1\}$ is singleton, contradiction!

Does my idea work? any mistakes? Thanks in advance ^_^

• Yes, your argument is fine. You could make it a bit more explicit by saying that as $x$ must be $a_i$ for some $i$, then $(x, y)$ includes the elements $a_{i+1}, a_{i+2}, \ldots$ that are not in the singleton set $\{b_1\}$. – Rob Arthan Oct 9 '16 at 22:23

Unfortunately, your argument doesn't quite work.

First of all, a notational correction: What you need to prove is that $\{b_1\}$ (the singleton set containing only $b_1$) is not open, not that the point $b_1$ is not open.

With that correction, your argument goes

To prove $\{b_1\}$ is not open, we need to show $\{b_1\}\notin \tau$, where $\tau$ is the order topology in $X$, so $\tau$ includes all set in the form of $(a,b),[a_0,b),(a,b_0]$. If we can prove $\{b_1\}$ is not in all these three forms, then it is not open.

Your first sentence is correct -- $\tau$ includes all sets of those forms (assuming that $a_0$ and $b_0$ represent the smallest and largest elements of the set, if any). However, those are not the only sets in $\tau$, so just showing that $\{b_1\}$ is not of one of those forms is not sufficient to show that it's not open.

Here's a more familiar example to think about: the order topology on $\mathbb R$. This contains all sets of the form $(a,b)$ (there's no largest or smallest element, so there are no open sets of the form $[a_0,b)$ or $(a,b_0]$. However, the set $(0,1) \cup (2, 3)$ is not of this form, and yet it is open.

The point is that the intervals of the forms you wrote down form a basis for the order topology; but they do not form all of the open sets. To show that $\{b_1\}$ is not open, you have to show that there is NO basis set of one of those forms that contains the point $b_1$ and is contained in the set $\{b_1\}$. In your argument, if you replace the phrase "let's assume $\{b_1\}=(x,y)$" with "let's assume $b_1 \in (x,y)\subseteq \{b_1\}$," you should be on your way.