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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}|rfor some $k\in \mathbb{R}$

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

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

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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 $a\le b$.2012-08-12
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    @Brian Is 'segment' generally referring to 'order-convex set'?2012-08-12
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    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'?2012-08-12
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    Hint: there is a rational number in every segment.2012-08-12
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    @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?2012-08-12
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    @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.2012-08-12
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    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 $I$.2012-08-12
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    @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.2012-08-12
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    Related: http://math.stackexchange.com/questions/98923/open-sets-of-mathbbr1-and-axiom-of-choice2012-08-12
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    @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'.2012-08-12

3 Answers 3

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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. 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, 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. $U$ 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 $[x,v]=[x,y]\cup[y,w]\subseteq U(x)$, so $y\in(x,w)\subseteq U(x)$. The case $y 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.

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    Should the equivalence relation be $\leq$?2012-08-12
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    @M.B.: $\leq$ is not an equivalence relation.2012-08-12
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    No, but as it stands it is either $x and $[x,y] \subseteq U$ or $y and $[y,x] \subseteq U$. What about $x\sim x$?2012-08-12
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    @M.B.: Yes, that was a typo: I thought $\le$ and typed $<$. Fixed.2012-08-12
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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.

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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.