---- > [!definition] Definition. ([[order topology]]) > Let $X$ be a set having [[strict order relation|simple order relation]] lt;$ with more than one element. Let $\mathscr{B}$ be the collection of all sets of the following types: > 1. All [[open interval|open intervals]] $(a,b)$ in $X$; > 2. All [[half-open interval|intervals of the form]] $[a_{0},b)$, where $a_{0}$ is the smallest element (if any) of $X$. > 3. All intervals of the form $(a, b_{0}]$, where $b_{0}$ is the largest element (if any) of $X$. > The collection $\mathscr{B}$ is a [[basis for a topology|basis]] for a [[topological space|topology]] on $X$, called the **order topology**. > \ > **Remark.** If $X$ has no smallest resp. largest element, there are no sets of type (2) resp. (3). > [!justification] > We must show $\mathscr{B}$ is indeed a [[basis for a topology|basis]]— we'll showing covering and [[nestles in|nestling]]. > \ > Covering: Let $x \in X$. If $x$ is the smallest resp. largest element of $X$ (if any) it lies in any set of type (2) resp. (3). Else, there exist $a<x$ and $b>x$, and so $x \in (a,b) \in \mathscr{B}$. > \ > Nestling: Let $B_{1}, B_{2}$ in $\mathscr{B}$ with $x \in B_{1} \cap B_{2}$. It suffices to show that $B_{1} \cap B_{2} \in \mathscr{B}$. Proceed by cases. If at least one of $B_{1}$, $B_{2}$ are of type (1) then so is $B_{1} \cap B_{2}$. Else if $B_{1}$ and $B_{2}$ have the same type of (2) or (3), $B_{1} \cap B_{2}$ again has that type. Else (WLOG) $B_{1}$ has type (2) and $B_{2}$ has type (3), in which case $B_{1} \cap B_{2}$ has type (1). > [!example] Examples. > 1. In some sense, the **order topology** is a generalization of the [[standard topology on the real line|standard topology]] on $\mathbb{R}$. It is immediate to verify that the definition specializes. > 2. Consider the set $\mathbb{R} \times \mathbb{R}$ in the [[dictionary order relation|dictionary order]]. To avoid ambiguous notation, denote its general element $\begin{bmatrix} x \\ y{}\end{bmatrix}$. This set has no smallest/largest element, so the **order topology** on it has as [[basis for a topology|basis]] the collection of all [[open interval]]s of the form $(\begin{bmatrix} a \\ b{}\end{bmatrix}, \begin{bmatrix} c \\ d{}\end{bmatrix})$ for $a < c$, and for $a=c$ and for $a=c$ and $b < d$. > 3. The [[order topology]] on $\mathbb{N}$ is actually the [[discrete topology]], for every singleton is [[open set|open]]: if $n > 1$ then we have $\{ n \} = (n-1,n+1)$, while if $n=1$ we have $\{ n \}=[n,n+1)$. > [!basicexample] > > **Equip $\mathbb{R}^{2}$ with the [[dictionary order relation|dictionary ordering]]. Then the [[order topology]] $\tau$ on $\mathbb{R}^{2}$ coincides with the [[product topology]] $\tau'$ on $\mathbb{R}^{2}=\mathbb{R} \times \mathbb{R}$, where the first factor is equipped with the [[discrete topology]] and the second factor is equipped with the [[standard topology on the real line|standard topology]].** > **Lemma.** The [[dictionary order relation|dictionary order topology]] on $\mathbb{R}^{2}$ has as [[basis for a topology|basis]] $\mathscr{B}_{2}^{(\tau)}$ the collection of dictionary-order-[[open interval|intervals]] that look like $\{ x \} \times I, \text{ where } I \text{ is a nonempty open interval of } \mathbb{R}.$ **Proof of lemma.** We will employ the [[basis-nestling characterization of comparing topologies]] (problem 3) to show that $\mathscr{B}_{2}^{(\tau)}$ [[nestles in]] the basis $\mathscr{B}_{1}^{(\tau)}$ from the provided definition, and vice versa. So, suppose $(x_{1}, x_{2}) \in X$ is an element of $B_{1}^{(\tau)} \in \mathscr{B}_{1}^{(\tau)}$: there exist $a,b,c,d \in \mathbb{R}$ such that $\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} \in (\begin{bmatrix} a \\ b \end{bmatrix}, \begin{bmatrix} c \\ d \end{bmatrix}),$ that is, ($a < x_{1} < c$) or ($x_{1}=a$ and $x_{2} > b$) or ($x_{1}=c$ and $x_{2} < d$). We proceed by cases to construct $B_{2}^{(\tau)} \in \mathscr{B}_{2}^{\tau}$ satisfying $\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \in B_{2}^{(\tau)} \subset B_{1}^{(\tau)}$: >- ($a < x_{1} < c$): Take any [[open interval]] $I$ in $\mathbb{R}$ (e.g., $I=\mathbb{R}$) containing $x_{2}$; then $\{ x_{1} \} \times I$ belongs to $(\begin{bmatrix}a \\ b\end{bmatrix}, \begin{bmatrix}c \\ d\end{bmatrix})$, for we have not changed the fact that $a<x_{1}<c$. >- ($x_{1}=a$ and $x_{2} > b$): Pick some $\varepsilon>0$ and take the [[open interval]] $I:=(b, x_2 + \varepsilon)$ in $\mathbb{R}$; then $\{ x_{1}\} \times I$ satisfies $\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \in \{ x_{1} \} \times I \subset B_{1}^{(\tau)}$. >- ($x_{1}=c$ and $x_{2} < d$): Pick some $\varepsilon > 0$ and take the [[open interval]] $I:=(x_{2} - \varepsilon, d)$ in $\mathbb{R}$; then $\{ x_{1}\} \times I$ satisfies $\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \in \{ x_{1} \} \times I \subset B_{1}^{(\tau)}$. The casework has exhibited that the [[topology generated by a basis|topology generated by]] $\mathscr{B}_{2}^{(\tau)}$ is [[comparable topologies|finer]] than that generated by $\mathscr{B}_{1}^{(\tau)}$. Of course, every element of $\mathscr{B}_{2}^{(\tau)}$ is itself an element of $\mathscr{B}_{1}^{(\tau)}$, and thus, given $\begin{bmatrix} x_1 \\ x_2\end{bmatrix} \in B_{2}^{(\tau)}$ we immediately witness $\begin{bmatrix} x_1 \\ x_2\end{bmatrix} \in B_{1}^{(\tau)} \subset B_{2}^{(\tau)}$ for some $B_{1}^{(\tau)} \in \mathscr{B}_{1}^{(\tau)}$ (namely, $B_{1}^{(\tau)}=B_{2}^{(\tau)}$). From this the lemma follows. > **Proof of statement.** We want to show that each of $\tau,\tau'$ is [[comparable topologies|finer]] than the other. From [[basis-nestling characterization of comparing topologies]] (problem 3), it suffices to show that their [[basis for a topology|bases]] $\mathscr{B}^{(\tau)}$ and $\mathscr{B}^{(\tau')}$ mutually [[nestles in|nestle]], where $\mathscr{B}^{(\tau)}=\mathscr{B}_{2}^{(\tau)}$ from the **lemma**. > In light of [[product of factor bases is basis of product topology]], we will consider the [[basis for a topology|basis]] elements of $(\mathbb{R}^{2}, \tau')$ to be products of the form $D \times I \in \mathscr{B}^{(\tau')}$, where $D$ is any subset of $\mathbb{R}$ and $I$ is an [[open interval]] in $\mathbb{R}$. > So, let $\begin{bmatrix}x_{1} \\ x_{2}\end{bmatrix} \in \mathbb{R}^{2}$ be contained by a basis element $B^{(\tau)} \in \mathscr{B}^{(\tau)}$, i.e., $\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} \in \{ a \} \times I = B^{(\tau)}$ for some $a \in \mathbb{R}$ and $I$ an [[open interval]] of $\mathbb{R}$. But notice that $B^{(\tau)} \in \mathscr{B}^{(\tau')}$, since it is the product of a subset of $\mathbb{R}$ (the singleton $\{ a \}$) and an [[open interval]] of $\mathbb{R}$. We thus have $\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} \in B^{(\tau')} \subset B^{(\tau)}$ for some $B^{(\tau')} \in \mathscr{B}^{(\tau')}$ and therefore may conclude that $\textcolor{Skyblue}{\tau' \supset \tau}$. > Next, let $\begin{bmatrix}x_{1} \\ x_{2}\end{bmatrix} \in \mathbb{R}^{2}$ be contained by a basis element $B^{(\tau')} \in \mathscr{B}^{(\tau')}$, i.e., $\begin{bmatrix}x_{1} \\ x_{2}\end{bmatrix} \in A \times I=B^{(\tau')}$for $A$ some subset of $\mathbb{R}$ and $I$ an [[open interval]] in $\mathbb{R}$. Since $x_{1} \in A$, it is clear that $\begin{bmatrix}x_{1} \\ x_{2}\end{bmatrix} \in \{ x_{1} \} \times I \subset A \times I=B^{(\tau')}$ But $\{ x_{1} \} \times I$ belongs to $\mathscr{B}^{(\tau)}$. And so $\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} \in B^{(\tau)} \subset B^{(\tau')}$ for some $B^{(\tau)} \in \mathscr{B^{(\tau)}}$ and therefore may conclude that $\textcolor{Skyblue}{\tau' \subset \tau}$. This completes the two-way inclusion. ^705189 ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```