In Munkres Book, Pg 90, I came across this question in Example 2:
Let $Y$ be the subset $[0,1)\cup\{2\}$ of $\mathbb{R}$. In the subspace topology on $Y$, the one-point set $\{2\}$ is open as it is the intersection of the open set $(\frac{3}{2},\frac{5}{2})$ with $Y$. But in the order topology on $Y$, the set $\{2\}$ is not open. Any basis element for the order topology om $Y$ that contains $\{2\}$ is of the form $\{x\mid x\in Y \text{ and } a<x\leq 2\}$
for some $a\in Y$; such a set necessarily contains points of $Y$ less than $2$.
I am a beginner with topology and lately the notion of open sets and closed sets have been confusing me a lot. It would be great help if someone can clarify the following doubts I have.
1) What about set $[0,1)$? Is it half-open or open for subspace topology of $Y$ as it could be the intersection of open interval of $(-\frac{1}{2}, 1)$ of $\mathbb{R}$ with $Y$? And what about order topology? I think it should be half-open for it.
2) I understand one-point set $\{2\}$ is open for subspace topology on $Y$. In the order topology, will it be half-open again based on the basis element used to describe it? Or will it be closed?
3) The variable $a$ described in the problem, what values can it take? I think it should be 1. Is that right? Or is $a=0$?