# Why is [a, b] not in basis for order topology?

I have started topology from Munkres. Here in section $$14$$, I am stuck up in a definition. It says,

Let $$X$$ be a set with a simple order relation; assume $$X$$ has more than one element. Let B be the collection of all sets of following types:

1. All open intervals $$(a, b)$$ in $$X$$.
2. All intervals of form $$[a, b)$$ of $$X$$
3. All intervals of form $$(a, b]$$ of $$X$$.

The collection B is a basis for order topology on $$X$$.

But here we have omitted $$[a, b]$$. Why?

I have read about standard topology, lower limit topology, upper limit topology. But set of all closed interval don't generate topology. My intuition says that these two are somehow linked.

Let $$X$$ be a set with a simple order relation and $$|X|>1$$. Let $$\mathcal{B}$$ be a collection of the following:

1. Intervals of the form $$(a,b)$$ in $$X$$.

2. Intervals of the form $$[a_0,b)$$, for $$a_0$$ the smallest element of $$X$$, if one exists.

3. Intervals of the form $$(a,b_0]$$ for $$b_0$$ the largest, if one exists.

Then $$\mathcal{B}$$ is a basis for the order topology on $$X$$.

It does not say intervals of the form $$[a,b)$$ and $$(a,b]$$ for any $$a,b$$, but only when the endpoints are the extrema of the order relation.

For example, take $$[0,1]$$ with the subspace topology in $$\mathbb{R}$$ with the standard topology. Then open sets which form a basis are of the form $$(a,b), 0, $$[0,a)$$ and $$(b,1]$$. This is what he means.

The order topology is meant to mimic our understanding of the order topology on $$\mathbb{R}$$. We want sets to be open which are unions of "open intervals", i.e. not containing their endpoints.

Edit: I'll add a bit more.

If you want the question as initially stated, then in fact every subset is open, so this is the discrete topology. To see this, if $$[a,M)$$ and $$(-M,a]$$ are in the basis for each $$a, |a|, then $$\{a\}=[a,M)\cap (-M,a]$$ is an open set for any $$a$$.

An interesting note, though this is a bit more advanced, is that you CAN include one of those half open intervals, either $$[a,b)$$ or $$(a,b]$$, and get a topology which is not trivial or the standard order topology. When we take $$\mathbb{R}$$ and include sets of the form $$[a,b)$$ in the basis we get the so called lower limit topology. This has many interesting properties: it is non-metrizable, it doesn't have a countable basis, and is totally disconnected, all properties we normally associate with $$\mathbb{R}$$ with the order topology.

• I couldn't find notation for a0 and b0 so I wrote it that way in hope that it would be edited. Aug 17, 2015 at 18:00
• Ok. This is nice. Thank you. Aug 17, 2015 at 18:09
• You said "The equivalent statement of $2,3$ in $\mathbb{R}$ are $(-\infty,a)$ and $(b,\infty)$" but I can't interpret this any way that makes it true. $\Bbb R$ has no largest or smallest element, so there is no analog of (2) or (3). In the extended real numbers, there is $\infty$ but then the sets added by (3) look like $(b, \infty]$, not like $(b, \infty)$. In $\Bbb R$, the unbounded interval $(b, \infty)$ simply is not a basic open set in this basis.
– MJD
Nov 20, 2017 at 15:21
• Also you claimed that $[a,b) \cup (a,b] = [a,b]$ and that $[a,b] =\{a\}$ when $a=b$. But when $a=b$, the first claim is wrong because both intervals are empty.
– MJD
Nov 20, 2017 at 15:26
• To your first, that's exactly the point. There is no largest/smallest element, so this is the analog of 2,3 in $\mathbb{R}$ (really, it is the union of all basis elements $(a,b)$, see Munkres for the exact details. $(a,\infty)$ need not be included in a basis for this reason, but it is a subbasis for this basis). For your second comment, yes, it only makes sense when $a<b$. The real reason is $(-\infty,a] \cap [a,\infty) = \{a\}$ is open. I will amend.
– Moya
Nov 20, 2017 at 16:47

I assume your space is connected.You can't include $[a,b]$ ,for arbitrary $a,b$ in your basis.As $[a,b]$ is closed set,being complement of open set $[-m,a)\cup (b,M]$,where $m$ and $M$ are smallest and largest elements respectively(if exist).

As space is connected by assumption,so $[a,b]$ can't be open,otherwise space will be disconnected.As elements of basis elements are open sets.So $[a,b]$ can't belongs to basis,being a closed subset.

• Note that $[a, b]$ is open if it is the entire set. Aug 18, 2015 at 15:17