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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!

Sincerely,

Vien

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2 Answers 2

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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.

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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. –  ramanujan_dirac Feb 3 '13 at 4:35
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the open set (b1,bn),n>1, does not contain b1. this is the answer to ramanujan_dirac –  user167703 Aug 3 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.

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