# One question about order topology and subspace topology

everyone! I am learning topology with Munkres's topology book. Some examples of second chapter are very hard for me to understand.

The second question is the example 3 which is on the page 90, the section of subspace topology. The second question says that 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, because it is the intersection of the open set $(3/2, 5/2)$ with $Y$. But in the order topology on $Y$, the set $\{2\}$ is not open. Any basis element for the order topology on $Y$ that contains $2$ is of the form $\{x\;|\;x \in Y \textrm{ and } a<x\leq 2\}$ for some $a \in Y$; such a set necessarily contains points of $Y$ less than $2$.

I do not understand why $Y$ intersecting with an open set $(3/2, 5/2)$ makes $\{2\}$ an open set in the subspace topology on $Y$. I also could not understand why $\{2\}$ is not open in the order topology. Could someone give me a detailed explanation? Thanks~

• Have a look at my answer to your previous similar question. Does it help you to understand this question as well? – wckronholm Mar 15 '14 at 16:18
• Yes, thanks for your help. But I still have some doubts. – congmingniao Mar 15 '14 at 16:51

If $Y$ is a subset of the topological space $X$, then $Y$ has the subspace topology means that a set $U \subset Y$ is open in $Y$ if and only if $U = \mathcal{O} \cap Y$ for some open sent $\mathcal{O} \subset X$.
In your example, $(3/2, 5/2)$ is open in $\mathbb{R}$ and $\{2\} = (3/2, 5/2) \cap Y$, hence $\{2\}$ is open in the subspace topology on $Y$.
• I'm not using a lemma. That first sentence is the definition of the subspace topology and holds for any subset $Y$ of any topological space $X$, not just metric spaces. – wckronholm Mar 15 '14 at 16:51