Order Topology

Last updated Nov 1, 2022

Let $(T, \leq)$ be a Total Ordering. Then the Order Topology on $T$ is the topology generated by the Open Rays on $T$.

Let $(T, \leq)$ be a Total Ordering. Then the collection $$\mathcal{B} := T \cup {( \leftarrow, a) : a \in T } \cup {(a, \rightarrow) : a \in T} \cup {(a,b) : a \leq b, a,b \in T}$$ is a Topological Basis for $T$.

We will show first that the $\mathcal{B}$ is a Topological Basis for its generated topology $\tau$ on $T$. We know for all $x \in T$, there is a $B \in \mathcal{B}$ so that $x \in B$ since $T \in \mathcal{B}$. Suppose $B_{1}, B_{2} \in \mathcal{B}$ and $B_{1} \cap B_{2} \neq \emptyset$. Let $x \in B_{1} \cap B_{2}$. We work through the various possibilities for $B_{1}$ and $B_{2}$:

1. If $B_{1} = T$, then $x \in B_{2} \subset B_{1} \cap B_{2} = B_{2}$. A simlar argument holds if $B_{2} = T$.
2. Suppose $B_{1} = ( \leftarrow, a)$ for some $a \in T$.
   1. If $B_{2} = T$, see (1).
   2. If $B_{2} = ( \leftarrow, b)$ for some $b \in T$, then $x \in ( \leftarrow, \min(a,b)) \subset B_{1} \cap B_{2}$.
   3. If $B_{2} = (b, \rightarrow)$ for some $b \in T$, then $b < a$ since $B_{1} \cap B_{2} = \emptyset$. Then $x \in (b, a) \subset B_{1} \cap B_{2}$.
   4. If $B_{2} = (b,c)$ for $b < c \in T$, then $x \in (b, \min(a,c)) \subset B_{1} \cap B_{2}$.
3. A similar argument to (2) holds if $B_{1} = (a, \rightarrow)$ for some $a \in T$:
   1. If $B_{2} = T$, see (1).
   2. If $B_{2} = (b, \rightarrow)$ for some $b \in T$, then $x \in (\max(a,b), \rightarrow) \subset B_{1} \cap B_{2}$.
   3. If $B_{2} = ( \leftarrow, b)$ for some $b \in T$, see (2.2).
   4. If $B_{2} = (b,c)$ for $b < c \in T$, then $x \in (\max\limits(a,b), c) \subset B_{1} \cap B_{2}$.
4. Suppose $B_{1} = (a, b)$ for some $a < b \in T$.
   1. If $B_{2} = T$, see (1).
   2. If $B_{2} = ( \leftarrow, c)$ for some $c \in T$, see (2.3).
   3. If $B_{2} = (c, \rightarrow)$ for some $c \in T$, see (3.4).
   4. If $B_{2} = (c,d)$ for some $c < d \in T$, then $x \in (\max\limits(a,c), \min(b,d)) \subset B_{1} \cap B_{2}$.

Therefore, $\tau$ has Topological Basis $\mathcal{B}$.

Since $\mathcal{B}$ contains all Open Rays in $T$, $\tau$ contains the Order Topology on $T$. On the other hand, since $(a, b) = (\leftarrow, b) \cap (a, \rightarrow)$ for $a \leq b \in T$, we know the Order Topology contains $\mathcal{B}$, so it also contains $\tau$. Thus $\tau$ is the Order Topology and $\mathcal{B}$ is a Topological Basis for it. $\blacksquare$