# The closure of $(0,1)$ in the lower-limit topology on $\mathbb{R}$

I've been given the topology $\tau_l$ on $\mathbb{R}$ generated by the subbasis consisting of all half open intervals $[a,b)$. I've concluded therefore that one can define the topology as:

$$\tau_l := \{ U \subset \mathbb{R} : U = {\textstyle \bigcup_i} [a_i, b_i) \}$$

I've then been asked to find the closure of $(0,1)$ in this topology, which is defined as the smallest closed set containing that interval; to that end I'm trying to characterise the closed sets and I thought for finding then the following would help me: $$\left(\bigcup_i [a_i,b_i) \right)^c = \bigcap_i ([a_i,b_i)^c) = \bigcap_i ((-\infty,a_i) \cup [b_i,\infty)).$$ But from here I don't know where to go; you can't just swap the union and intersections, can you?

Or am I going about this the wrong way? My friend said the answer is actually $[0,1)$ so I expect to find that all open sets are closed too.

• If you feel like you have a good grasp of the opens of this topology, it might help you to realize that the complement of the closure is the interior or the complement. Thus, to solve this question, it suffices to find the interior of $(0,1)^c$. – Mees de Vries Jun 1 '15 at 12:51
• Ahh I see! Thank you! – NecroJ Jun 1 '15 at 13:03

Well, it's relatively easy to see that $[0,1)$ is closed. This is because $[1,\infty)$ and $(-\infty, 0]$ are open sets, and ther complement, $[0,1)$, is therefore closed.

That said, there are two things you may be tempted to conclude from this which are simply NOT true:

• Just because every set of the sub-basis are closed sets, that does not mean that all open sets of the topology are both open and closed. This is because a union of open sets is always open, but a union of closed sets may not be closed.
• Just because $[0,1)$ is a closed set, that does not mean that it is the closure of $(0,1)$. The closure is the smallest closed set that contains $(0,1)$. This means that the closure of $(0,1)$ will be a subset of $[0,1)$, but a superset of $(0,1)$, meaning that the closure of $(0,1)$ may be either $(0,1)$ (if the set is closed) or $[0,1)$ if it is not.
• Hmm I see.. does this mean that you just have to 'see' [0,1) as the closed set you're looking for? I mean isn't there something more systematic? Otherwise I have to go through potentially any set that is greater than or equal to (0,1) in size until it is closed right? – NecroJ Jun 1 '15 at 13:04
• @JohnnyBreen A great deal of maths is about 'seeing' things that help you. It's called intuition, and it develops through a lot of practice. – 5xum Jun 1 '15 at 13:06
• I guess you meant $(-\infty,0)$, not $(-\infty,0]$. – celtschk Jul 16 '16 at 17:51
• I think he did mean that, celtschk. But why isnt the notation $[-\infty,0)$ in a lower limit topology? That after all is the shape of the topology. If this is not possible/allowed then can we extend a set that far left in the first place? Perhaps then only finite length intervals are allowed in the topology? – CogitoErgoCogitoSum Apr 10 '18 at 1:26

The collection of half-open intervals is closed under finite intersection and $\mathbb R$ can be written as a union of half-open intervals. This allows the conclusion that the collection is not only a subbase but also a base for the topology.

For any $x\notin[0,1)$ we can find easily a half-open interval $[a,b)$ s.t. $[0,1)\cap[a,b)=\varnothing$.

For any $x\in[0,1)$ we cannot find such an interval so the answer of your friend is correct: $[0,1)$ is here the closure of $(0,1)$.

Indeed any interval $[a,b)$ is closed and open in this topology, but your conclusion that all open sets are closed is incorrect. E.g. $(0,1)=\bigcup_{n=2}^{\infty}[\frac1{n},1)$ is open (as a union of open sets), but it is not closed (since it does not coincide with its closure, as shown above).

• Thank you for this. I do make lots of oversights when I do maths. I'm not proud of it. – NecroJ Jun 1 '15 at 13:17