Proving that a certain definition of neighborhoods forms a neighborhood topology

general-topology

This is Exercise 2 from Section 2.1 on page 21 of Topology and Groupoids, by Brown.

Exercise:

Let $\leq$ be an order relation on the set $X$. Let $x \in X$ and $N \subseteq X$. We say that $N$ is a neighborhood of $x$ if there is an
open interval $I$ of $X$ such that $x \in I \subseteq N$.

Prove that these neighbourhoods of points of $X$ form a neighbourhood
topology on $X$. This topology is called the order topology on $X$.
What is the order topology on $\mathbb{R}$?

More information:

The axioms for a neighborhood topology are given as follows (page 20 same book):

  1. If $N$ is a neighborhood of $x$, then $x \in N$.
  2. If $N$ is a subset of $X$ containing a neighborhood of $x$, then $N$ is a neighborhood of $x$.
  3. The intersection of two neighborhoods of $x$ is again a neighborhood of $x$.
  4. Any neighborhood $N$ of $x$ contains a neighborhood $M$ of $x$ such that $N$ is a neighborhood of each point of $M$.

My attempt:

To prove that neighborhoods as defined here form a neighborhood topology, we need to verify that the neighborhoods in this exercise satisfy the four axioms given earlier.

  1. Assume that $N$ is a neighborhood of $x$. Since there is an open interval $I$ of $X$ such that $x \in I \subseteq N$, we see that $x \in N$, which means the first axiom is satisfied.
  2. Assume that $M$ is a neighborhood of $x$, and $M \subseteq N$. There is an open interval $I$ of $X$ such that $x \in I \subseteq M \subseteq N$. This implies that $x \in I \subseteq N$, which shows that $N$ is a neighborhood of $x$, so the second axiom is satisfied.
  3. Assume that $M$ and $N$ are neighborhoods of $x$. Then there are open intervals $I_N$ and $I_M$ such that $x \in I_M \subseteq M$ and $x \in I_N \subseteq N$. Let $I = I_M \cap I_N$. Clearly $x \in I \subseteq M \cap N$. If we can show that $I$ is an open interval, then it follows that $M \cap N$ is a neighborhood of $x$, so the third axiom is satisfied.
  4. Assume that $N$ is a neighborhood of $x$. There is an open interval $I$ of $X$ such that $x \in I \subseteq N$. Let $M = I$. Clearly $M$ is a neighborhood of $x$, because trivially, $x \in I \subseteq M$. We want to show that $N$ is a neighborhood of each point of $M$. Let $y \in I$, and suppose $I$ is written in the form $I = (a, b)$. Let $\delta = \min \{y – a, b – y \}$. Then $(y – \delta, y + \delta) \subseteq I \subseteq N$, which shows that $N$ is a neighborhood of $y$. Since $y$ was an abritrary selection from $I = M$, we conclude that $N$ is a neighborhood of each point of $M$.

As for the question "What is the order topology on $\mathbb{R}$?", I think this is called the "usual topology" on $\mathbb{R}$.

Questions and comments:

I have searched for "order topology" on StackExchange and Google but found nothing that addresses this exercise. I know almost nothing about topology beyond the neighborhood axioms mentioned above.

I think 1. and 2. are probably okay. I think 3. is fine as long as the intersection of two open intervals is an open interval, but I don't know how to prove this. I haven't even been given a definition of "open interval" in this abstract setting, so I'm not quite sure what I'm working with.

For 4., my concern is that I'm using more properties than $X$ and $\leq$ necessarily have. I understand that in topology in general, I cannot assume properties of the real numbers that I'm used to taking for granted, but I'm not sure if I'm doing that here, and if so, how to fix it.

Thanks for any help.

Best Answer

Since there is no generally accepted standard topology on arbitrary partially ordered sets, I suspect that by order relation he means a linear (or total) order on $X$. In that case an open interval in $X$ is simply a set of the form

$$(a,b)=\{x\in X:a<x<b\}\;,$$

where $a,b\in X$ and $a<b$. These sets form a base for a topology on any linearly ordered set, and the topology is indeed called the order topology; a space endowed with a linear order and topologized with the associated order topology is a linearly ordered topological space, or LOTS for short. The order topology on $\Bbb R$ is indeed the usual Euclidean topology, which is also induced by the metric $d(x,y)=|x-y|$.

Let $(a,b)$ and $(c,d)$ be open intervals in $X$. Then

$$(a,b)\cap(c,d)=\begin{cases} \varnothing,&\text{if }b\le c\text{ or }d\le a\\ (c,b),&\text{if }a\le c<b\le d\\ (c,d),&\text{if }a\le c<d<b\\ (a,b),&\text{if }c<a<b\le d\\ (a,d),&\text{if }c<a<d<b\;, \end{cases}$$

and you can verify that these are the only possibilities. Thus, the intersection of two open intervals is an open interval, provided that it is non-empty. (Technically, it’s an open interval anyway, since $(a,b)=\varnothing$ if $a\ge b$.)

Added: I have now seen Brown’s definitions, and it turns out that he is indeed talking about what I would call a linear order. However, his definition of open interval includes the open rays $(\leftarrow,a)$ and $(a,\to)$ for $a\in X$ as well as $\varnothing$ and $X$ itself. That adds a few possibilities, like $(a,b)\cap(c,\to)$, but they are easily analyzed along the same lines. For instance, $(a,b)\cap(c,\to)$ is $\varnothing$ if $b\le c$, $(c,b)$ if $a\le c<b$, and $(a,b)$ if $c<a$. It does not change the generated topology, since $X$ and the open rays are unions of open intervals in the stricter sense.

Your answer to (4) doesn’t quite work, because there is not reason to think that arithmetic operations are meaningful in an arbitrary linearly ordered space $X$. You’re also working too hard: if $y\in M(= I)$, then $y\in I\subseteq N$, so $N$ is a nbhd of $y$.

Otherwise your answers are fine, apart from the acknowledged gap in (3).

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