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As a self-studier, I was reading a proof that any open subset, $U$, of $\mathbb{R}$ is a disjoint union of open intervals. The proof was based on an equivalence relation where $x\sim y$ if $(x,y)$ is contained in $U$.

I have two questions regarding the verification that this is a valid equivalence relation. (Sorry if they are obvious.)

First, for reflexivity, $x\sim x$, is $(x,x)$ the null set?

Second, regarding transitivity, if $x < y < z$ are elements of $U$, with $x\sim y$ and $y\sim z$, then $(x,y)$ and $(y,z)$ are both contained in $U$. My question is how can it be claimed that $x\sim z$, i.e., that $(x,z)$ is in $U$, since $y$ is not in either of the sets (although it was stated that $y$ is an element of $U$).


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The solution is very simple: It's a typo in the definition of the equivalence. The author meant $x\sim y$ if the ${\it closed}\ $ interval $[x,y]$ is contained in $U$. – Christian Blatter Jul 16 '11 at 8:24
@Andrew: For and other approach to show that any open set is the union of... see Royden's Real Analysis 3rd edition p. 42. – leo Jul 16 '11 at 18:21
up vote 1 down vote accepted

$U$ is open then $\mathbb{R}-U$ is close. Suppose that $(x,y)\subseteq U$ and $(y,z)\subseteq U$. Suppose that $y\not\in U$. Since $\mathbb{R}-U$ is closed, there exist a sequence $(y_n)$ in $\mathbb{R}-U$ with $y_n\to y$. Let $r= \min\{(z-y)/2,(y-x)/2\}$. Then there exist $N\in \mathbb{N}$ such that if $n\geq N$, $|y_n-y|\lt r$. Then for $n\geq N$, $y_n\in (x,y)$ or $y_n\in (y,z)$, in any case $y_n\in U$ and $y_n\in\mathbb{R}-U$. This is impossible. Therefore, $y\in U$.

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In general, if $A\subseteq C$ and $B\subseteq C$ then $A\cup B \subseteq C$. Thus $(x,y)\cup (y,z)\cup \{y\} \subseteq U$, i.e. $(x,z)\subseteq U$. – leo Jul 16 '11 at 2:36
Thanks, Leo for your help. I am sorry I was so tardy in expressing my appreciation. Best, Andrew – Andrew Sep 15 '11 at 21:53
You're welcome @Andrew, no problem :) – leo Sep 17 '11 at 1:55
  1. If you check definitions, $(x,y)$ is most likely defined as the set of all $w$ with $x < w < y$. So $(x,x)$ would be the set of all $w$ with $x < w < x$. What set is that?

  2. Try drawing a picture to see what is going on here. More formally, note that to show $(x,z) \subset U$, you have to show that for any $w$ with $x < w < z$, we have $w \in U$. Now necessarily either $w < y$, or $w = y$, or $w > y$. Show that in each case it follows that $w \in U$.

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