The Order Topology for $X$ is the smallest Topology in which order is continuous I'm reading General Topology by John L. Kelley and am new to Topology. I came across this problem and have been struggling. It is chapter 1 problem I. a)
He first begins by defining the order topology as follows:
Let $X$ be a set linearly ordered by $<$ which is anti-symetmric so $x<x$ is false. The order topology $\tau$ on $X$ has a subbase of all sets of the form $\{x|x<a\}$ or $\{x|a<x\}$. For some $a\in X$.
From this, I was able to gather that a base for $\tau$ is the finite intersections of sets of the form $\{x|x<a\}$ or $\{x|a<x\}$.
The Question:
The order topology on $X$ is the smallest topology in which order is continuous in the following sense: if $a,b\in X$ and $a<b$ then there are neighborhoods $U$ of $a$ and $V$ of $b$ such that for $x\in U$ and $y\in V$, then $x<y$.
My attempt:
First I figured that we should show that $\tau$ is actually order continuous. Suppose $a,b\in X$, then for each neighborhood $U_1$ of $a$ and $U_2$ of $b$, there are members $V_1$ and $V_2$ of the base such that $a\in V_1\subset U_1$ and $b\in V_2\subset U_2$. This is by the definition of a base. Then if $a<b$, ...
Basically, I am stuck and would appreciate any pointers. Also, how do we show that it is the smallest such topology?
 A: First, let's prove that in the order topology on $X$, the order is continuous - in the meaning of "continuous" the book gave.
We have to prove that if $a,b \in X$ and $a<b$, then there are neighborhoods $U$ of $a$ and $V$ of $b$ such that $\forall x \in U$ and $\forall y \in V$, $x<y$.
Two cases:

*

*If $\exists c \in X, a < c < b$, we define
$U = \{z \mid z < c\}$ and $V = \{z \mid c < z\}$.
$a < c$ so $a \in U$.
$U$ is open by definition of the order topology, so $U$ is a neighborhood of $a$.
Similarly, $V$ is a neighborhood of $b$.
If $x \in U$ and $y \in V$, $x < c < y$, so $x < y$.

*Otherwise, define
$U = \{z \mid z < b\}$ and $V = \{z \mid a < z\}$.
$a < b$ so $a \in U$.
$U$ is open by definition of the order topology, so $U$ is a neighborhood of $a$.
Similarly, $V$ is a neighborhood of $b$.
If $x \in U$ and $y \in V$, $X$ is linearly ordered and antisymmetric, so either $x < a$, $x=a$, or $a < x$.
The last case is not possible, because there is no $x$ such that $a<x<b$, and $x=b$ or $b<x$ are impossible by definition of $U$.
So we have $x \le a$, and similarly $b \le y$. As $a<b$, this proves $x<y$.

Now let's prove the second part, i.e this topology is the smallest topology where order is continuous (with the specific meaning they gave to that).
We have to prove that in any topology where order is continuous, $\forall a \in X$ the sets $\{x \mid x < a\}$ and $\{x \mid a < x\}$ are open.
$\forall a \in X, \forall c < a, \exists U_c$ a neighborhood of $c$ and $V$ a neighborhood of $a$ such that $\forall x \in U_c, \forall y \in V, x<y$.
So $\exists O_c$ an open neighborhood of $c$, $O_c \subset U_c$ which implies $\forall x \in O_c, \forall y \in V, x<y$.
In particular, $\forall x \in O_c, x < a$.
Define $W = \bigcup_{c<a} O_c$.
$W$ contains any $c<a$, and $\forall x \in W, \exists c, x \in O_c$, so $x<a$.
So $W$ is exactly $\{x \mid x<a\}$.
And $W$ is open, being an union of open sets.
The proof for $\{x \mid a<x\}$ is open is similar.
