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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 $a=0$?

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First of all, let me say that there are either open sets or closed sets in topological spaces, there are no 'half-open sets', you can have however half-open 'intervals' if the set is ordered and in the order topology these half-open intervals are open sets.

Having said that, here are the answers:

1) [0,1) is an open set of $Y$ with the subspace topology. It is also open in $Y$ with the order topology on $Y$.

2) {2} is an open set of $Y$ with the subspace topology. It is a closed set of $Y$ with the order topology being the complement of $[0,1)$ in $Y$.

3) $a$ can take any value in [0,1).

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  • $\begingroup$ Thanks a lot. I didn't know that half-open 'intervals' are considered as open sets. $\endgroup$ – Khushboo May 21 '15 at 17:38
  • $\begingroup$ But I have a doubt. For 2) you said based on the complement of open set $[0,1)$, $\{2\}$ is closed. But can't we use the same argument to describe $\{2\}$ as closed set in subspace topology, even though it is open? $\endgroup$ – Khushboo May 21 '15 at 17:52
  • $\begingroup$ Yes, in fact [0,1) and {2} both are closed and also open sets (also called 'clopen' sets) of $Y$ in the subspace topology. This also means that $Y$ with the subspace topology is not connected. But $Y$ with the order topology is connected. $\endgroup$ – user135988 May 21 '15 at 20:15
  • $\begingroup$ @Khushboo Also, I would like to point out that, when I said 'half-open intervals' are open it was in the sense of the section on the order topology in Munkres. I didn't mean that the sets like $[a,b)$ or $(b,a]$ are open in $\mathbb R$, they are not open. You should read this section 'the order topology' more carefully to understand what it is meant by `half-open intervals'. $\endgroup$ – user135988 May 21 '15 at 20:38
  • $\begingroup$ Okay I will read it again. Thanks for the clarification. $\endgroup$ – Khushboo May 21 '15 at 20:39

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