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Let $E$ be an open set in $\mathbb{R}$. Fix $x\in E$.

I have proved that statement is true when $\{y\in \mathbb{R}|(x,y)\subset E\}$ is bounded above and $\{z\in \mathbb{R}|(z,x)\subset E\}$ is bounded below.

If at least one of those above are not bounded, $E$ must be equal to one of;

1.$\mathbb{R}$

2.$\{r\in \mathbb{R}|r<k\}$for some $k\in \mathbb{R}$

3.$\{r\in \mathbb{R}|k<r\}$for some $k\in \mathbb{R}$

Are these sets can be the union of an at most countable collection of disjoint 'segments'?

Cameron Buie
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Katlus
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  • Yes: each of those sets is a segment. In this context segment simply means order-convex set: $A\subseteq\Bbb R$ is a segment iff $[a,b]\subseteq A$ whenever $a,b\in A$ and $a\le b$. – Brian M. Scott Aug 12 '12 at 12:29
  • @Brian Is 'segment' generally referring to 'order-convex set'? – Katlus Aug 12 '12 at 12:34
  • If a definition of segment is '$(a,b)$ for $a,b\in \mathbb{R}$', can't those be the union of at most countablr 'segments'? – Katlus Aug 12 '12 at 12:36
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    Hint: there is a rational number in every segment. – Karolis Juodelė Aug 12 '12 at 12:38
  • @Karolis How actually that could be a hint..? I know $\mathbb{Q}$ is 'dense' in $\mathbb{R}$, hence contains a countable base, but is that related to this? Would you be more specific? – Katlus Aug 12 '12 at 12:41
  • @Katlus: Segment can have various meanings depending on context. If you take it to mean a set of the form $(a,b)$ with $a,b\in\Bbb R$, then $\Bbb R$ can be written as a countable union of segments, but not as a countable union of pairwise disjoint segments. – Brian M. Scott Aug 12 '12 at 12:42
  • You can't write {\bf R} as a union of disjoint finite segments. If one of the segments is, say, $I=(17,42)$, then you can't have an open interval containing 42 and disjoint from $I$. – Gerry Myerson Aug 12 '12 at 12:47
  • @Katlus, note that boundedness is not a topological property. $(0, 1)$ is homeomorphic to $\mathbb{R}$. If you can prove a topological property from $(0, 1)$ but not for $\mathbb{R}$, you must have made a mistake. – Karolis Juodelė Aug 12 '12 at 13:19
  • Related: http://math.stackexchange.com/questions/98923/open-sets-of-mathbbr1-and-axiom-of-choice – Asaf Karagila Aug 12 '12 at 13:25
  • @Karolis I used that argument when '$\forall x\in \mathbb{R}$,${y\in \mathbb{R}|(x,y)\subset E}$ and ${z\in \mathbb{R}|(z,x)\subset E}$ are bounded above and below respectively. Once the existence of 'collection of disjoint segments' is guaranteed I can check its cardinality. I was asking 'existence of such collection'. – Katlus Aug 12 '12 at 14:26

3 Answers3

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I don’t know what argument you used, but here’s the easiest one that I know.

Let $U$ be a non-empty open subset of $\Bbb R$. Define a relation $\sim$ on $U$ as follows: for $x,y\in U$, $x\sim y$ iff either $x\le y$ and $[x,y]\subseteq U$, or $y\le x$ and $[y,x]\subseteq U$. It’s not hard to check that $\sim$ is an equivalence relation. For $x\in U$ denote by $U(x)$ the $\sim$-equivalence class of $x$. Then $U(x)$ is order-convex and open in $\Bbb R$.

  1. $U(x)$ is order-convex. Suppose that $y,z\in U(x)$ with $y<z$. If $x\le y$, then $[y,z]\subseteq[x,z]\subseteq U(x)$. If $z\le x$, then $[y,z]\subseteq[y,x]\subseteq U(x)$. And if $y<x<z$, then $[y,z]=[y,x]\cup[x,z]\subseteq U(x)$.

  2. $U(x)$ is open. Suppose that $y\in U(x)$. Suppose that $x<y$. $U$ is open, so there are $u,v\in\Bbb R$ such that $y\in(u,v)\subseteq U$. Let $w\in (y,v)$ be arbitrary. Then $[x,w]=[x,y]\cup[y,w]\subseteq U$, so $y\in(x,w)\subseteq U(x)$. The case $y<x$ is entirely similar, and the case $y=x$ is trivial.

Thus, $\{U(x):x\in U\}$ is a partition of $U$ into order-convex open sets. Since each must contain a rational numbers, there are only countably many of these sets.

Note that the result fails if you insist on having bounded open intervals. If $(a,b)$ is one of the intervals in the decomposition of $U$, no other interval can contain either $a$ or $b$. Thus, if $U$ contains an open ray, $U$ cannot be decomposed into pairwise disjoint bounded open intervals.

José Carlos Santos
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Brian M. Scott
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  • Should the equivalence relation be $\leq$? – M.B. Aug 12 '12 at 13:25
  • @M.B.: $\leq$ is not an equivalence relation. – tomasz Aug 12 '12 at 13:26
  • No, but as it stands it is either $x<y$ and $[x,y] \subseteq U$ or $y<x$ and $[y,x] \subseteq U$. What about $x\sim x$? – M.B. Aug 12 '12 at 13:28
  • @M.B.: Yes, that was a typo: I thought $\le$ and typed $<$. Fixed. – Brian M. Scott Aug 12 '12 at 13:30
  • @BrianM.Scott:I have one doubt here. Each $U(x)$ contains rational nos. and hence there are countable many of these sets. Exception: For irrational no. $x\in \mathbb R$, we note that $\mathbb R= \cup (x-1,x+1)=\cup I_x$. Since each $I_x$ contains rational nos. and so should their collection $\implies $ we found a bijection from set of irrational nos. to rational nos. How to explain this paradox? – Koro Feb 04 '21 at 01:53
  • @Koro: No paradox: your intervals are not pairwise disjoint. You can choose a rational from each one of them, but since there are uncountably many intervals and only countably many rationals, you’ll end up choosing the same rational number for infinitely many different intervals. It’s only when the intervals are pairwise disjoint that choosing one point from each is guaranteed to give you distinct points. – Brian M. Scott Feb 04 '21 at 01:58
  • @BrianM.Scott: Thanks for your response. Suppose I define $Q_x=$ set of all rational nos. in $U(x)$. I agree that $Q_x$ is countable. Now I define a function: $f:U\to \cup_{x\in U} Q_x $ by $f(t)=Q_t$ then even if it is bijection, how can we claim $\cup Q_x$ is countable despite noting that the union in $Q_x$ is over $U$ which may be uncountable as well. Uncountable collection of countable sets is not in general countable. I'm looking forward to finding a bijection from $\cup U(x)$ to a countable set to prove the result. – Koro Feb 04 '21 at 02:04
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    @Koro: I really don’t understand what you’re trying to do, but it appears to me that you don’t understand the argument in the answer. $\mathscr{U}={U(x):x\in U}$ is a pairwise disjoint family of non-empty open sets. Every non-empty open set contains a rational number. Since the sets are pairwise disjoint, no two of them can contain the same rational number: no matter how I pick a rational number from each of them, I must pick a different rational for each of them. Thus, no matter how I pick these rationals, I’m constructing an injection from $\mathscr{U}$ into $\Bbb Q$. And $\Bbb Q$ is ... – Brian M. Scott Feb 04 '21 at 02:15
  • ... countable, so $\mathscr{U}$ must also be countable. That’s all there is to it. – Brian M. Scott Feb 04 '21 at 02:15
  • @BrianM.Scott: I see it now. From $\mathscr{U} $ to $\mathbb Q$!. Thanks a lot sir. – Koro Feb 04 '21 at 04:03
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    @Koro: You’re very welcome; I’m glad that you were able to get it sorted. – Brian M. Scott Feb 04 '21 at 04:04
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Let $E$ be an open subset of $\mathbb{R}$. We can assume without loss of generality that E is nonempty. Consider a real number $x \in E$. In the context of this problem, a segment $(a_1,b_1)$ means the set of all $p \in \mathbb{R}$ such that $a_1 \lt p \lt b_1$. $ $Let $y = inf \lbrace a \in E \mid (a,x) \subseteq E \rbrace$ and $w = sup \lbrace a \in E \mid (x,a) \subseteq E \rbrace$. Then $E$ is the union of all such segments $(y,w)$ for all points $x \in E$. Since each of these segments is open, it remains to show that they are at most countable. First, note that the rationals are known to be dense in the reals. Also, the rationals are a countable set. Thus, let $\lbrace r_n \rbrace$ be an ordering of the rational numbers. Now order the segments $(y,w)$ such that $(y_1,w_1)$ precedes $(y_2,w_2)$ if and only if the least $i$ such that $r_i \in (y_1,w_1)$ is less than the least $j$ such that $r_j \in (y_2,w_2)$. $r_i$ and $r_j$ exist because the rationals are dense. This is an ordering of the segements whose union is $E$, showing that the segments are at most countable.

Andrew
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Choose any open set $U \subset \mathbb{R} = \bigcup \limits_{s \in S} s$ where $S$ is a set of disjoint segments. As $\mathbb{Q}$ is dense in $\mathbb{R}$, every segment of $\mathbb{R}$ contains some $q \in \mathbb{Q}$. This leads to an injection from $S$ to $\mathbb{Q}$ from which follows that $S$ is countable.

Karolis Juodelė
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