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I want to show that if we write $\mathbb R$ as a disjoint union of more than one subset of $\mathbb R$, then at least one of the subsets is not open in $\mathbb R$.

I'm sure I can prove this, but not very cleanly, by contradiction: if $\mathbb R$ were a nontrivial disjoint union of open subsets, then we could decompose each subset into its maximally-sized intervals of the form $(-\infty, a)$ or $(b, +\infty)$ or $(a, b)$ with $a < b$; then there exist two neighboring intervals in the overall decomposition, and the neighboring intervals do not contain their common boundary point, which would contradict the disjoint union covering $\mathbb R$.

Is there a better/cleaner/more elegant proof?

feralin
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  • Your proof is not actually correct: there might be infinitely many maximal intervals which are densely ordered, so there are no "neighboring" intervals. – Eric Wofsey Oct 10 '16 at 00:41
  • See Any open subset of R is a at most countable union of disjoint open intervals. – dxiv Oct 10 '16 at 00:42
  • @user251257 I'm not sure your point. I only said that I can decompose open subsets into disjoint unions of open intervals. Did I say something blatantly wrong? :o – feralin Oct 10 '16 at 00:43
  • @EricWofsey that doesn't sound right... any open interval has a positive (or infinite) diameter, so then a set of disjoint intervals must be countable, and thus we can find the "neighboring" interval of any interval. What's wrong with this thought? – feralin Oct 10 '16 at 00:46
  • @feralin ah. Not exactly. My comment is wrong. Sorry. But see Eric's comment. – user251257 Oct 10 '16 at 00:47
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    You might want to take a look at this question: http://math.stackexchange.com/q/1480796/9464 –  Oct 10 '16 at 00:48
  • @Jack thanks for the link! That makes me immediately think of the following proof: since $\mathbb R$ is connected, if $\mathbb R = A \sqcup B$ then $A$ or $B$ is not open. This is much simpler than what I was thinking! – feralin Oct 10 '16 at 00:54
  • @feralin aren't you trying to prove that $\mathbb R$ is connected? – user251257 Oct 10 '16 at 00:57
  • @user251257 no, I'm trying to prove that $\mathbb R$ is not a nontrivial disjoint union of open subsets. This ends up just being a simple consequence of the connectedness of $\mathbb R$, but the fact that $\mathbb R$ is connected is not my end result. – feralin Oct 10 '16 at 00:58
  • @feralin: The rationals are countable, but there are no "neighboring" rationals. – Eric Wofsey Oct 10 '16 at 03:10

3 Answers3

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If this were possible, then it would be possible to write $\mathbb{R}$ as the union of two nonempty disjoint open subsets (since the union of any arbitrary collection of open sets is open). This, however, is impossible. Suppose $\mathbb{R} = V \cup W$, where $V$ and $W$ are nonempty open sets and $V \cap W = \emptyset$.

Now consider an element $x \in \overline{V} \setminus V$. Can we have $x \in W$?

Kaj Hansen
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  • How can I show that $\bar V \setminus V$ is not empty? I think I would need to prove that in the reals, only the empty set and the reals are clopen... – feralin Oct 10 '16 at 00:49
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    More generally, a topologically space $X$ is connected $\iff$ the only clopen sets are $\emptyset$ and $X$. This is fairly straightforward; try proving it yourself first, but there's also discussion here: http://math.stackexchange.com/questions/361387/how-to-prove-that-a-topological-space-is-connected-iff-it-has-exactly-two-clopen – Kaj Hansen Oct 10 '16 at 01:00
  • Hint : Try to prove that $\mathbb{R}$ is connected (using a bounded interval, for instance) first and combine the ideas of the discussion (mainly Kaj Hansen's comment). This will serve you further for all spaces connected by arcs – Duchamp Gérard H. E. Oct 10 '16 at 01:48
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(1) All intervals in $\mathbb{R}$ are connected.

$$\mathbb{R} = \bigcup_{n \in \mathbb{Z}} (-n,n)$$

(2) Consider the diffeomorphism;

$$f: (-a,a) \to \mathbb{R}; \ x \mapsto \frac{ax}{\sqrt{a^2-x^2}}$$

Since all intervals of $\mathbb{R}$ are connected then $(-a,a)$ is connected. Moreover, since $f$ is continuous then $f(-a,a) = \mathbb{R}$ is connected.

Faraad Armwood
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The statement is equivalent to proving that $\mathbb{R}$ is not the union of two disjoint nonempty open sets. Indeed, if $\mathbb{R}$ is the disjoint union of nonempty open sets (more than one), then taking one of them and the union of the others, you obtain $\mathbb{R}$ as the disjoint union of two nonempty open sets.

Suppose $\mathbb{R}=A\cup B$, with $A$ and $B$ disjoint, open and nonempty. Since the complements of open sets are closed, $A$ and $B$ are also closed.

Let $a\in A$ and $b\in B$; without loss of generality, $a<b$.

Consider $S=\{x\in\mathbb{R}:a\ge x, [a,x]\subseteq A\}$. The set $S$ is upper bounded (by $b$) and not empty (because $a\in S$). Therefore $s=\sup S$ exists. Since $A$ is closed and $S\subseteq A$, we have that $s\in A$, because the supremum of a set belongs to its closure.

I claim that $s\in S$. Indeed, if $[a,s]\not\subseteq A$, then there is $t\in [a,s]\cap B$ and $a<t\le s$. Since $B$ is open, there is $\varepsilon>0$ such that $(t-\varepsilon,t+\varepsilon)\subseteq B$. Note that $t-\varepsilon>a$, because $a\in A$. On the other hand, since $s=\sup S$, there is $x\in S$ with $t-\varepsilon<x\le s$, a contradiction because $[a,x]\subseteq A$.

Since $A$ is open, there exists $\delta>0$ such that $(s-\delta,s+\delta)\subseteq A$. But then $[a,s+\delta/2]\subseteq A$, contradicting the fact that $s=\sup S$.

This is the proof that $\mathbb{R}$ is connected. A straightforward modification shows that every interval in $\mathbb{R}$ is connected.

egreg
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