Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

in section 14, Munkres introduces the order topology, and gives this example:

The set $X$ = {1,2} $\times \mathbb{Z}_{+}$ in the dictionary order is an example of an ordered set with a smallest element. Denoting 1 $\times$ $n$ by $a_{n}$ and 2 $\times$ $n$ by $b_{n}$, we can represent $X$ by $a_{1},a_{2}, \cdots ; b_{1}, b_{2}, \cdots$. The order topology on $X$ is $not$ the discrete topology. Most one-point sets are open, but there is an exception - the one-point set $\{ b_{1}\}$. Any open set containing $b_{1}$ must contain a basis element about $b_{1}$ (by definition), and any basis element containing $b_{1}$ contains points of the $a_{i}$ sequence.

I understand that given a set $X$, the discrete topology on $X$ is the collection of $all$ subsets of $X$. And it looks like they're assuming that the topology is generated by a certain basis, and that the "definition" that Munkres is using to justify the first part of the last sentence is the one on page 78, i.e. the a subset $U$ of $X$ is open in $X$(i.e. in the topology) if for each $x \in U$, there is a basis element $B \in$ B such that $x \in B$ and $B \subset U$, where B is the basis. The part that's confusing me is the bolded part in the previous paragraph. Why would this imply that $b_{1}$ is not open?

Thank you for any help/clarification!



share|cite|improve this question
up vote 4 down vote accepted

As you mentioned, let $\mathcal{B}$ be a basis. $U$ is open if and only if for all $x$ in $U$ there exists a $B \in \mathcal{B}$ such that $x \in B \subset U$.

Now for your example, let $U = \{b_1\}$, $x = b_1$. If $\{b_1\}$ was actually open, then there would exists a basis open set $B$ such that $b_1 \in B \subset \{b_1\}$. However all basis open sets containing $b_1$ contains some $a_i$. So it not possible that $B \subset \{b_i\}$. The definition of being open according to the basis of the topology is not fulfilled so $\{b_1\}$ can not be open.

By the way, I interpreted the question as : how does the bold statement imply $\{b_1\}$ is not open? I assumed you know how to prove the bold statement.

share|cite|improve this answer
Thanks for the answer! – Vien Nguyen Aug 19 '12 at 0:03
@William: How can you say "However all basis open sets containing $b_1$ contains some $a_i$?", for e.g. the open set $(b_1, b_n)$ where $n \gt 1$, is a basis element containing $b_1$, but does not contain $a_i$. I am extremely confused. Thanks in advance. – user23238 Feb 3 '13 at 4:35
the open set (b1,bn),n>1, does not contain b1. this is the answer to ramanujan_dirac – user167703 Aug 3 '14 at 8:09

If $\{b_1\}$ was to be open, then as you say, some basis element $B$ would have to exist so that $b_1 \in B \subset \{b_1\}$ so that $B = \{b_1\}$. But as he argues, this is not a basis element for the order topology.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.