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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?

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  • $\begingroup$ 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 $\Bbb R$ $\endgroup$
    – Masacroso
    Commented Sep 21, 2019 at 20:08
  • $\begingroup$ Expanded my answer with a quite direct argument. $\endgroup$ Commented Sep 21, 2019 at 22:04

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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.

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  • $\begingroup$ @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. $\endgroup$ Commented Sep 21, 2019 at 21:09
  • $\begingroup$ @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. $\endgroup$ Commented Sep 21, 2019 at 21:15
  • $\begingroup$ @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. $\endgroup$ Commented Sep 21, 2019 at 21:18
  • $\begingroup$ 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. :) $\endgroup$
    – D.R.
    Commented Sep 21, 2019 at 21:24
  • $\begingroup$ I have a concern: what if the rationals repeat? Would the 'replacement process' be just use one of them? $\endgroup$
    – D.R.
    Commented Sep 22, 2019 at 4:22

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