Order topology

Total order

1. Total order

   A relation $<$ on a set $X$ is called a total (or linear) order if the following is satisfied:

   • (O1): For every $x,y\in X$ with $x\ne y$, we have that $x<y$ or $y<x$ (comparability).
   • (O2): For every $x\in X$, we have that $x\not<x$ (nonreflexivity).
   • (O3): if $x<y$ and $y<z$, then $x<z$ (transitivity).

2. Totally ordered set

   In this case, the pair $(X,<)$ is called a totally ordered set. If $x<y$ or $x=y$, we write $x\leq y$. An example of a total order is given by $X=\mathbb{R}$ with usual order.

Intervals

1. Intervals

   If $X$ is a totally ordered set, and $a,b\in X$, we will use the following notation for intervals:

   $$\begin{align*} [a,b] &=\{x\in X\mid a\leq x\leq b\}\\ (a,b) &=\{x\in X\mid a< x< b\}\\ (a,b] &=\{x\in X\mid a< x\leq b\}\\ [a,b) &=\{x\in X\mid a\leq x< b\}\\ (a,\infty) &=\{x\in X\mid x>a\}\\ (-\infty,b) &=\{x\in X\mid x<b\} \end{align*}$$

The order topology

1. Order topology

   Let $X$ be a totally ordered set with more than one element. Let $\mathcal{S}$ be the collection:

   $$\mathcal{S}=\{(-\infty,a)\mid a\in X\}\cup\{(b,\infty)\mid b\in X\}.$$

   Then $\mathcal{S}$ is subbase for a topology on $X$, called order topology.

Base for the order topology

1. Base for the order topology

   Let $X$ be a totally ordered set with more than one element. Let $\mathcal{B}$ be the collection of all intervals of the form $(a,b)$, together with those of the form $[m,b)$, in case $X$ has a minimum $m$, and those of the form $(a,M]$, in case $X$ has a maximum $M$. Then $\mathcal{B}$ is a base for the order topology on $X$.

2. Proof

   The set of finite intersections of the intervals in the subbasis can be obtained as union of elements of the intervals described. $\Box$

Dictionary order

1. Dictionary order

   Let $X$ and $Y$ be totally ordered sets. We define an order relation on the cartesian product $X\times Y$ by:

   $$(x,y)<(x',y'),$$

   if either:

   • $x< x'$, or
   • $x=x'$ and $y< y'$.

2. The dictionary order is a total order

   The order on $X\times Y$ just described is a total order, called the dictionary order.

Proof that the dictionary order is a total order

• Let $(x,y)\ne (x',y')$ with $(x,y), (x',y')\in X\times Y$. If $x\ne x'$, then one of the pairs is less than the other. Otherwise $x=x'$, and since the pairs are distinct, we must have $y\ne y'$, and so (O1) follows.
• If we had $(x,y)<(x,y)$, then it would follow that either $x<x$ or $x=x,y<y$. Since none of these is possible, we conclude that (O2) is true.
• The proof of (O3) is left as an exercise.

Examples

• The order topology on $\mathbb{R}$ is the same as the standard topology.
• The order topology on $\mathbb{R}\times \mathbb{R}$ with the dictionary order is different from the usual metric topology on $\mathbb{R}^2$.
• The order topology on $\{0,1\}\times \mathbb{N}$ has almost all singletons as open sets.

Exercises

1. Let $(X,<)$ be a totally ordered set, and let $Y\subseteq X$. If, for $a,b\in Y$ we define $a<'b$ whenever $a<b$ in $X$, show that $(Y,<')$ is a totally ordered set.
2. If $X$ is a totally ordered set, and $Y\subseteq X$, show that $Y$ can have at most one smallest element.
3. If $X$ is a totally ordered set, and we have that $x<y$ and there are no $z\in X$ such that $x<z<y$, we say that $y$ is an immediate succesor of $x$. Show that any $x\in X$ has at most one immediate succesor.
4. Let $(X,<)$ be a totally ordered set with no minimum element, and let $\mathcal{C}=\{(a,\infty)\mid a\in X\}$. Show that $\mathcal{C}$ is a basis for a topology on $X$. Is $\tau_{\mathcal{C}}$ in general, the same as the order topology?

Links

• Total order - Wikipedia, the free encyclopedia
• Order topology - Wikipedia, the free encyclopedia