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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.

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2 Answers 2

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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.

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  • $\begingroup$ I couldn't find notation for a0 and b0 so I wrote it that way in hope that it would be edited. $\endgroup$
    – divya
    Aug 17, 2015 at 18:00
  • $\begingroup$ Ok. This is nice. Thank you. $\endgroup$
    – divya
    Aug 17, 2015 at 18:09
  • $\begingroup$ 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. $\endgroup$
    – MJD
    Nov 20, 2017 at 15:21
  • $\begingroup$ 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. $\endgroup$
    – MJD
    Nov 20, 2017 at 15:26
  • $\begingroup$ 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. $\endgroup$
    – Moya
    Nov 20, 2017 at 16:47
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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.

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

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