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I am reading lower limit topology on Wikipedia, which states that the lower limit topology

[...] is the topology generated by the basis of all half-open intervals $[a,b)$, where a and b are real numbers. [...] The lower limit topology is finer (has more open sets) than the standard topology on the real numbers (which is generated by the open intervals). The reason is that every open interval can be written as a (countably infinite) union of half-open intervals.

I cannot see how to write $(a,b)$ as a countably infinite union of half-open intervals.

Rodrigo de Azevedo
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Geoffrey Critzer
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5 Answers5

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If $M = (a,b)\cap \mathbb{Q}$ then $$(a,b) = \bigcup _{c\in M} [c,b)$$

nonuser
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$$(a,b) = \bigcup_{n=1}^\infty \, \left[\, \left(1-\frac{1}{n}\right)\, a + \frac{1}{n} b ,\, b\right) $$

Mutantoe
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Lee Mosher
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    You might want to start with $n=2$. If it means anything at all, $[b,b) = \emptyset$ is not an open interval. – aschepler Jun 10 '18 at 22:53
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    @aschepler Isn't $\emptyset$ the trivial/degenerate open (and closed) interval? Or, at least, you can define it to be... (It's certainly open, by defintion.) – étale-cohomology Jun 11 '18 at 01:51
  • @étale-cohomology If the empty set is an open interval and if the answer to OP's question is yes, then what are the half-open intervals which, as a union, form this empty set? – Alfe Jun 11 '18 at 09:18
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    @Alfe Maybe the union of the empty set (considered as the trivial/degenerate half-open interval) with itself? (I don't really know!) – étale-cohomology Jun 11 '18 at 10:01
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    @Alfe: The same problem would arise if you use open intervals as a basis for the topology on $\mathbb{R}$. In either case, you have to allow for the "trivial union" of basis elements (i.e., the union of 0 sets—0 is, of course, a countable number) to obtain $\emptyset$ as a union of basis elements. – Michael Seifert Jun 11 '18 at 15:24
  • @étale-cohomology The question was about a countably infinite amount of half-open intervals (for whatever reason). Of course, this is a corner case and could be handled differently, if it needs to. It would be nice if everything fell into place but the empty set is probably out of it. – Alfe Jun 12 '18 at 10:57
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$$(a,b) = \bigcup \{[x,b): a < x < b \}$$

Every $[x,b) \subseteq (a,b)$ whenever $a < x < b$ for the right to left inclusion, and on the other hand, if $a < x < b$, $x \in [x,b)$, which shows the left to right inclusion. If you want a countable union at all cost (topologies are closed under all unions, but maybe you're doing measure theory?) then take $x$ to be all rationals in $(a,b)$ so

$$(a,b) = \bigcup \{[q,b): q \in \mathbb{Q} \text{ and } a < q < b\}$$

but then the proof of equality is a little more involved: the right to left inclusion stays the same, but if $p \in (a,b)$ we first pick $q\ in \mathbb{Q}$ with $a < q < p$, and note that $p \in [q,b)$ which is a subset of the right hand side.

Henno Brandsma
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    Yes, thank you. I do insist on writing (a,b) as a COUNTABLE union of half open intervals. What I am actually trying to determine is if the smallest sigma-field containing the half open intervals is the collection of all open sets in R. – Geoffrey Critzer Jun 10 '18 at 15:44
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    @GeoffreyCritzer yes, in $\sigma$-algebras it's common to use the rationals as a way to make arbitary unions countable. – Henno Brandsma Jun 10 '18 at 15:48
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One can write

$$ (a,b) = \bigcup_{n\geq 1}\left[a+\frac{b-a}{2n},b\right) $$

One inclusion is straighforward. For the non trivial one, pick $t \in (a,b)$, and $n$ large enough such that $t > a +\frac{b-a}{2n}$, so that $t \in \left[a+ \frac{b-a}{2n},b\right)$.

Mutantoe
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qualcuno
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Here's an example of a union of disjoint intervals:

$$ (a,b) = \bigcup_{n=1}^\infty\left[a+\frac{b-a}{n+1}, a+\frac{b-a}n\right) $$

I wonder, why nobody presented it yet?

CiaPan
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  • Because there was no need for it. It is rare that all open sets can be written as disjoint unions of the elements of a particular basis. For example, most open sets in $\Bbb R^2$ cannot be expressed as the disjoint union of open disks. That all open intervals can be expressed as a countable union of disjoint half-open intervals is simply a minor curiousity, not something particularly useful to this topology. – Paul Sinclair Jun 11 '18 at 17:05