# Proof Verification: any open set can be written as countable union of open intervals

I came up with the following proof that open sets can be written as a countable union of (not necessarily disjoint) open intervals, but I’m uncertain about one step I took. I’ve looked at Any open subset of $\Bbb R$ is a at most countable union of disjoint open intervals. [Collecting Proofs], but I not sure if my proof is among them.

Recall that the definition of open subset is that for every point $$x$$ in an open subset $$U$$, there exists some $$\delta_x$$ such that the neighborhood around the point, $$(x-\delta_x, x+\delta_x)$$ lies fully within $$U$$. That means $$U$$ can be written as $$U=\bigcup_{x\in U}(x-\delta_x, x+\delta_x)$$ which you can verify yourself by proving both $$\subseteq$$ and $$\supseteq$$. This does not guarantee a countable union, so we are not done. Let's make the following tweak: find a number, $$\delta_x'\leq \delta_x$$ such that $$x-\delta_x'$$ is rational. If $$x-\delta_x' = x_1-\delta_{x_1}'$$ for some $$x_1\neq x$$, then replace both intervals with the interval $$(x-\delta_x', \max\{x+\delta_x', x_1+\delta_{x_1}'\})$$. If we union over all these intervals, we get countably many open intervals (because the rationals are countable) that union to $$U$$: $$U= \bigcup_{x\in U}(x-\delta_x',\max\{\ldots\})$$ and we are done.

My concern is about the step where I combine potentially uncountably many intervals into one using $$\max$$. Is this allowed? Is the overall proof valid? If not, is there an easy fix, or should I scrap the proof?

• as the answer below I think that this "proof" is wrong or incomplete at better. There are better proofs about this, by example that there is a countable basis for the standard topology of $\Bbb R$ Sep 21, 2019 at 20:08
• Expanded my answer with a quite direct argument. Sep 21, 2019 at 22:04

IMO, the replacement step is vague and there are better proofs (see your link): if you care about the disjointness it's best to use the connected components of $$U$$. These are open (generalised) intervals and automatically disjoint.

If you don't care about disjointness use that $$\Bbb R$$ is second countable and thus hereditarily Lindelöf, so that every union of open sets can be thinned out to one of a countable subfamily.

Or: If $$x \in U$$ there is a $$\delta_x$$ with $$(x-\delta_x, x+\delta_x) \subseteq U$$ and as there is a rational between every two reals we find $$q_x, r_x \in \Bbb Q$$ such that $$x - \delta_x < q_x < x < r_x < x+\delta_x$$. Doing this for every $$x \in U$$ we get $$U=\bigcup \{ (q_x,r_x) : x \in X\}$$ and this union is actually at most countable because $$\Bbb Q^2$$ is a countable set.

• @D.R. The answer by Brian (107 votes) seems to be most popular, mine using connected components is also among them (less popular). Your current proof cannot be "saved" I think. Sep 21, 2019 at 21:09
• @D.R. You indeed don't know how many instances you have where you're replacing. Does the process ever stop? Is it some transfinite process? etc. Sep 21, 2019 at 21:15
• @D.R. Why not use that we choose both endpoints of the interval as rationals, then the union is countable rightaway.No further processing needed. Just add a small argument why this can be done. The intervals are not disjoint but you don't care about that. Sep 21, 2019 at 21:18
• Could you write up that up a little more detailed? I was thinking about that but I just didn't want to deal with two deltas (one below and one above). If it fixes the problem, that would be wonderful! If you add a bit to your answer I would appreciate that very much. :)
– D.R.
Sep 21, 2019 at 21:24
• I have a concern: what if the rationals repeat? Would the 'replacement process' be just use one of them?
– D.R.
Sep 22, 2019 at 4:22