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Let $G$ be an open subset of $\Bbb R$. Define the relation $x \sim y $ on $G$ such that $x \sim y$ iff there exists an open interval $I$ such that $x,y \in I$ and $I \subseteq G $

Verify that '$\sim$' is an equivalence relation on $G$.

Deduce that $G$ can be expressed as an union of a countable number of pair wise disjoint open intervals.

Can some one please help me on this especially on how to reduce the result? Is it by using equivalence classes? If so, how do you find equivalence classes?

Seirios
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user83369
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1 Answers1

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The only nontrivial part in proving this is an equivalence relation is transitivity. Recall that if $x\sim y$ and $y\sim z$ (and without loss of generality $x<y<z$) then there are $a,b,c,d$ such that $a<x<y<b$ and $c<y<z<d$.

What can you conclude on the relation between $b$ and $c$? Can you find an interval containing $x$ and $z$ now?

As for the second part, first note that what is an equivalence class? It is an open interval.

But you can't really write down the equivalence classes because you're not given the exact $G$. If $G$ is an interval you will only have one equivalence class, but if $G$ is a much more complicated open set then you will have infinitely many equivalence classes.

However you can still prove that there are only countably many of them, show that you can pick one rational number from each interval. So there is a function from the open intervals into the rational numbers, try to prove it is injective.

Asaf Karagila
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  • Thanks (a,d) would be sufficient to prove transitivity wouldn't it? And for the second part defining a function $ f: \frac{G}{\sim} \to Q $ such that $ f()=y $ where y is a rational in that interval would be enough since it is injective? – user83369 Jul 02 '13 at 16:01
  • Yes on both the questions. But why is this $f$ injective? It's not very difficult, but it still requires proof. – Asaf Karagila Jul 02 '13 at 16:01
  • Since you could find an unique rational number to each interval? I am not sure of how to formally prove thoug – user83369 Jul 02 '13 at 16:06
  • @user83369: Formally speaking you need to show that if two intervals are different then they are sent to different rationals. Use the fact that they are disjoint and that $f$ maps an interval to its element. – Asaf Karagila Jul 02 '13 at 16:54