# Verifying an equivalence relation

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~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~x, is (x,x) the null set?

Second, regarding transivity, if x < y < z are elements of U, with x~y and y~z, then (x,y) and (y,z) are both contained in U. My question is how can it be claimed that x~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).

Thanks.

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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

## 2 Answers

$U$ 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 close, 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 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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