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Prove/Disprove that :

$(i)$ Every open Set in $\mathbb R^p$ can be written as the union of countable number of disjoint open Sets.

$(ii)$ Every open subset of $\mathbb R^p$ is the union of a countable collection of closed sets.

I was able to look at some similar posts asking this problem; but one seemed to be using the other and vice versa and seem convoluted.

Unfortunately, I have no idea on how to move forward. Can anyone please help me in preparing a proof for both of these problems?

Thank you for your help.

MathMan
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1 Answers1

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  1. Every open set is union of balls with rational radius and rational center.
  2. Every open ball is a countable union of closed balls.

This gives (ii). For (i), given two points in your open set, say that they are equivalent iff there is a continuous path between them, completely contained in the open set. Argue that this is indeed an equivalence relation, and that its components (equivalence classes) are open. Now use that $\mathbb Q^n$ is dense in $\mathbb R^n$, so there can be no more than countably many equivalence classes.

Andrés E. Caicedo
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  • Thank you but I am afraid that the text I am reading has not even introduced dense sets .. – MathMan Sep 01 '14 at 20:59
  • Surely you can show that every open set contains an element of $\mathbb Q^n$. – Andrés E. Caicedo Sep 01 '14 at 21:00
  • Uhm, so, to prove that every open set in $\mathbb R^p$ contains an element of $\mathbb Q^n$, we need to prove that every open set in $\mathbb R$ contains an element of $\mathbb Q$ ? This should be true since, any open interval contains infinite rational numbers. Am I correct? – MathMan Sep 01 '14 at 21:15
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    Sure, that's one way. And the stronger property you indicate is indeed true. – Andrés E. Caicedo Sep 01 '14 at 21:16
  • Okay .. So, I now say that any two points $x,y$ in the open set are equivalent iff they are connected? And then prove this is an equivalence relation. This seems true since : $~~(a)~ x \sim x ~~(b)~x\sim y \implies y \sim x~~$ $ (c) x \sim y ~, ~y \sim z$ $\implies x \sim z $ . Hence, this is an equivalence relation – MathMan Sep 01 '14 at 21:24
  • having proved that the above represents an equivalence relation, we need to prove that the equivalence classes are open. Can you give me some hints on how to prove it? – MathMan Sep 01 '14 at 21:28
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    If you have a path from $ x $ to $ y $ in your open set, and $ z $ is in a ball centered at $ y $ and contained in the set, then there is a path from $ x $ to $ z $ in the set: First go to $ y $, and then go from $ y $ to $ z $ in a straight line. This shows the component that contains $ x $ is open. – Andrés E. Caicedo Sep 01 '14 at 22:59
  • in the second question , what if we assume these closed sets are pair disjoint? – Idele Oct 12 '16 at 06:19
  • @hctb That's a good question. In that case you can't. This was first shown by Sierpiński. – Andrés E. Caicedo Oct 12 '16 at 12:26
  • @hctb See here for the (key) argument. – Andrés E. Caicedo Oct 12 '16 at 13:28