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

Question

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.

Answer 1

Read Munkres carefully. He writes:

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 $(a,b), 0<a<b<1$, $[0,a)$ 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 $a, |a|<M$, 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.

Answer 2

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.