If $X$ is a linearly ordered set, the topology $\mathcal{T}$ generated by the sets $\{x:x<a\}$ and $\{x:x>a\}$ ($a \in X$) is called the order topology.
If $a, b \in X$ and $a<b$, there exist $U,V \in \mathcal{T}$ with $a \in U$, $b \in V$, and $x<y$ for all $x \in U$ and $y \in V$. Show the order topology is the weakest topology with this property.
Source: Folland, Real Analysis, exercise 4.9.
Can anybody help please? I have no idea how to do this problem.
Let $\tau$ be any topology on $X$ with the stated property; what you’re being asked to prove is that $\tau\supseteq\mathcal{T}$, which amounts to proving that the sets $\{x\in X:x<a\}$ and $\{x\in X:a<x\}$ belong to $\tau$ for each $a\in X$. For convenience let $(\leftarrow,a)=\{x\in X:x<a\}$ and $(a,\to)=\{x\in X:a<x\}$.
HINT: Fix $a\in X$. For each $x<a$ there are $U_x,V_x\in\tau$ such that $x\in U_x$, $a\in V_x$, and $y<z$ whenever $y\in U_x$ and $z\in V_x$. In particular, $y<a$ for every $y\in U_x$. Now use the fact that any union of open sets is open to show that $(\leftarrow,a)\in\tau$. Then find a similar argument to show that $(a,\to)\in\tau$.