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.
Suppose $Y$ is a subset of $X$, show the order topology on $Y$ is never stronger than, but may be weaker than, the relative topology on $Y$ induced by the order topology on $X$.
Source: Folland, Real Analysis, exercise 4.9.
Can anybody help please? I have no idea how to do this problem.
HINT: Let $\tau$ be the order topology on $Y$, and let $\tau_Y$ be the subspace topology on $Y$ that it inherits from $X$. You’re asked to show that $\tau\subseteq\tau_Y$, and that there are examples in which $\tau\subsetneqq\tau_Y$.
Showing that $\tau\subseteq\tau_Y$ is straightforward: you need only show that for each $y\in Y$, $\{z\in Y:z<y\}$ and $\{z\in Y:y<z\}$ are in $\tau_Y$. (Why?) To show that $\{z\in Y:z<y\}$, say, is in $\tau_Y$, you must find an open set $U$ in $X$ such that $U\cap Y=\{z\in Y:z<y\}$. There’s a very natural candidate for that $U$; what is it?
Finding an example in which $\tau\subsetneqq\tau_Y$ is perhaps a little less straightforward. You can use $\Bbb R$ with its usual order as your starting point. Try to find a set $Y\subseteq\Bbb R$ with a maximum element $y_0$ with the property that $\{y_0\}$ is in $\tau_Y$ but not in $\tau$. I’ve left an example in the spoiler-protected block below if you really can’t think of one.