I am trying to prove this and there is something that I am missing.

Let $E\subseteq\mathbb{R}$ and let us define an equivalence relation on E as follows. If $a\in E$ and $b\in E$ we say that $a\sim b$, if the entire open interval $(a,b)$ is contained in $E$.

Now, if I accept that this equivalence relationship partitions E into a disjoint union of classes, we can easily prove that:

  • an equivalence class is an interval and it's open

  • since $\mathbb{Q}$ is dense in $\mathbb{R}$, and $\mathbb{Q}$ is countable, we can pick a rational to associate with each $(a,b)$

And we are done.

My issue is: since we specify that $(a,b)$ are open disjoint intervals, how can they partition $E$? In particular, what happens to the boundaries $a$ and $b$? We may pick them to be in $E$, sure, but the collection of open intervals bounded by $a_i$ and $b_i$, $i<\infty$, will exclude $a_i$ and $b_i$, thus not covering $E$. So say $E=(1,3)$. The element $2\notin(1,2)\cup (2,3)$. What am I missing?

  • $\begingroup$ In your example, $1.5$ and $2.5$ are in the same equivalence class, for example. $\endgroup$ Sep 21, 2016 at 13:26
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    $\begingroup$ You are missing the fact that $2$ is in the same equivalence class as $1.5$ or $2.5$, since $(1.5,2)$ and $(2,2.5)$ are both inside the open set $(1,3)$. Hence, the decomposition $(1,3) = (1,2) \cup (2,3)$ is not correct. Since $(1,3)$ itself is an interval, the only equivalence class under $\sim$ is $(1,3)$ itself, so the decomposition is trivial. $\endgroup$ Sep 21, 2016 at 13:29
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    $\begingroup$ I think you are really messing up your notations: Is $R$ the real numbers ($\mathbb{R}$)? Shouldn't it be $E \subseteq R$ instead of $E \in R$? Are $Q$ the rational numbers? Could you please specify your topology, in my world $\mathbb{Q}$ is not compact in $\mathbb{R}$. $\endgroup$
    – ctst
    Sep 21, 2016 at 13:40
  • $\begingroup$ @ctst wow I've really messed them up :) I wrote it in a rush. Off course $E\subseteq R$. And I meant dense, not compact. Thanks! $\endgroup$
    – hyperio
    Sep 21, 2016 at 20:14
  • $\begingroup$ Quibble: Your relation isn't symmetric. If $2,3 \in E$ then $(3,2)$ is empty hence $3R2$ but there is no reason to think that $2R3$. $\endgroup$ Sep 21, 2016 at 20:21

1 Answer 1


Let In$[a,b]$ denote the closed interval from $a$ to $b.$ That is In$[a,b]=[a,b]\cup [b,a].$ This removes the need to distinguish the cases $a<b,a=b, a>b.$

For open $ E\subset \mathbb R $ and $ a,b\in E $ let $ a\sim b $ iff In$[a,b]\subset E.$ Obviously $ a\sim b$ iff $b\sim a, $ and also $ a\sim a, $ for any $ a,b\in E.$

Exercise. For any real $a,b,c$ we have $$In[a,b]\cup In[b,c]=In [\min (a,b,c),\max (a,b,c)]=[\min (a,b,c),\max (a,b,c)].$$ Corollary : $\sim$ is transitive.

So $\sim$ is an equivalence relation on $E.$

Exercise.The set of equivalence classes of any equivalence relation on any set comprise a partition of that set.

Note that in your title we implicitly include open half-lines, the whole real line, and the empty set among the open intervals. The members of the partition $E_{/\sim}$ are convex open sets (open because if $x\in E$ then $(x-d,x+d)\subset E$ for some $d>0,$ so $(x-d,x+d)\subset [x]_{\sim}$ ),... but not necessarily bounded open intervals. For example if $E=(1,\infty)$ then $E_{/\sim}=\{E\}.$

If $x,y\in \mathbb R$ with $x<y$ and $(x,y)\in E_{\sim}$ then neither $x$ nor $y$ belongs to $E$.

Another way of describing $E_{/\sim}$ is : For $a\in E, $ let $a^*$ be the set of convex open real set(s) $T$ such that $a\in T\subset E.$ Let $a^{**}=\cup a^*.$ Then $a^{**} \in a^*,$ so $a^{**}$ is the $\subset$-maximum member of $a^*.$ And $a^{**}=[a]_{\sim},$ the $\sim$-equivalence class containing $a.$

  • $\begingroup$ Thank you. Is $In(a,b)\cup In(b,c)=(min(a,b,c),max(a,b,c))$ true? $(1,2)\cup (2,3)\ne(1,3)$ as it does not contain $2$. $\endgroup$
    – hyperio
    Sep 26, 2016 at 9:31
  • $\begingroup$ Right. I was thinking of closed intervals and will edit my A. $\endgroup$ Sep 26, 2016 at 19:28
  • $\begingroup$ Thanks. But in this case, how can a collection of closed intervals partition an open set? $\endgroup$
    – hyperio
    Sep 28, 2016 at 8:26
  • $\begingroup$ Never mind, $a$ and $b$ are chosen to be from $E$. $\endgroup$
    – hyperio
    Sep 29, 2016 at 20:48

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