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

Question

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.

Answer 1

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.

Answer 2

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).