As the title says, I have to show that any open set $U\subseteq\mathbb{R}^n$ is a disjoint union of countable many intervals.

Hello, my idea is the following:

Consider any $x=(x_1,...,x_n)\in U\subseteq\mathbb{R}^n, U$ open. Then there exists a $\varepsilon>0$ so that $B(\varepsilon,x)\subseteq U$ because $U$ is open. Since $x$ is in the open interval $$ (a,b)_x:=\left\{(x_1,...,x_n)\in\mathbb{R}^n|a_i<x_i<b_i, i=1,...,n\right\} $$ with $$ a_i:=x_i-\varepsilon,\quad b_i:=x_i+\varepsilon. $$ Because of $\overline{\mathbb{Q}}=\mathbb{R}$, there exist $c_i,d_i\in\mathbb{Q}, i=1,...,n$ with $$ a_i<c_i<x_i,\quad x_i<d_i<b_i. $$ Define $$ [c,d]_x:=\left\{(x_1,...,x_n)\in\mathbb{R}^n|c_i\leq x_i\leq d_i, i=1,...,n\right\}, $$ then $x\in [c,d]_x\subseteq U$.

Now take $x'\neq x$. If $x'$ is not in $[c,d]_x$, one can again construct a closed interval with rational endpoints, i.e., $[e,f]_{x'}$. There exists a $\varepsilon'>0$ with $B(\varepsilon',x')\subseteq U$. When minimizing $\varepsilon'$ to $\varepsilon''$ in such a way that $$ [c,d]_x\cap B(\varepsilon'',x')=\emptyset, $$ $[c,d]_x$ and $[e,f]_{x'}$ are disjoint. If $x'$ is in $[c,d]_X$, choose $[e,f]_{x'}:=[c,d]_x$.

The desired countability is fulfilled because of the countability of $\mathbb{Q}$: There are countable many closed intervals with rational endpoints as constructed above.

I would like to know if my idea to proof is correct or nonsense.

With regards!

  • $\begingroup$ I don't think this result is true for $n>1$. Take a triangle $\{(x,y):\ x>0,y>0,x+y<1\}$. It is clear this is not a finite disjoint union of open intervals. So, it would have to be infinite countable disjoint union instead. If we put two intervals with disjoint closures then the points in their boundary that are inside the triangle cannot be covered. So, we have to keep the intervals away from each other. But then, inside some compact in the interior of the triangle we get a sequence of nested compacts disjoint from the intervals that will have nonempty intersection. So, we can't cover all. $\endgroup$ – OR. Aug 16 '13 at 10:46
  • $\begingroup$ For $n=1$ the connected components are the open intervals (connected opens in $\mathbb{R}$ are open intervals) that form the disjoint union we are looking for. To show they are countable we can just take, for each of them, a rational number inside. So, there are less intervals in the decomposition than there are rational numbers. $\endgroup$ – OR. Aug 16 '13 at 10:50
  • $\begingroup$ In the first comment, third sentence, it should say "If we put two intervals with non-disjoint closures ..." $\endgroup$ – OR. Aug 16 '13 at 10:54
  • $\begingroup$ So to your opinion the task should be: "Show that any open set $U\subseteq\mathbb{R}$ is a disjoint union of intervals? $\endgroup$ – math12 Aug 16 '13 at 11:00
  • $\begingroup$ I think so. Try covering an open triangle with disjoint open 'intervals', where we are understanding intervals as those open rectangles with sides parallel to the axes. $\endgroup$ – OR. Aug 16 '13 at 11:03

It seems the following.

Your proof has a small technical gap (to provide $(a,b)_x\subset B(\varepsilon,x)$ you should take $a_i:=x_i-\varepsilon/\sqrt{n}$,$b_i:=x_i+\varepsilon/\sqrt{n}$ instead of $a_i:=x_i-\varepsilon$, $b_i:=x_i+\varepsilon$) and a principal idea error: you may not cover all points of the uncountable set $U$ by your countable inductive construction.

I see the general situation as follows.

  1. As ABC already wrote in comments, you can cover each non-empty open subset $U$ of $\mathbb R^n$ by (countably many) disjoint open “intervals” only if $n=1$.

  2. By Sierpiński Theorem (see Nuno’s answer), if a continuum $X$ (that is a compact connected Hausdorff space) is a countable union $\bigcup_{i=1}^{\infty} X_i$ of its disjoint closed subsets then at most one of the sets $X_i$ is non-empty. As a corollary, you can cover no non-empty open subset $U$ of $\mathbb R^n$ by countably many disjoint closed “intervals”.

  3. Any non-empty open subset of $\mathbb R^n$ is a disjoint union of countably many “halfclosed” cubes of the form $2^{1-k}(x+[0;1)^n)$, where $k\in\mathbb N$ and $x\in\mathbb Z^n$. I used this construction to prove that each non-empty open subset of the space $\mathbb R^n$ is a disjoint union of $r$ homeomorphic parts if $r\ge 2^{n+1}-1$ or $n=2$ and $r\ge 4$ (see p. 2 of English draft version of my Ukrainian paper “Partitions of subsets of $\mathbb R^n$ onto similar sets” (Nauk. visn. Cherniv. univ., 269. Mathematics – Chernivtsi, Ruta, 2005 – P.88–93)).

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    $\begingroup$ For the half-closed cubes I guess the figure on page 76 conveys the idea. $\endgroup$ – Carsten S Jan 11 '14 at 16:35
  • $\begingroup$ @CarstenS "404 not found". :-( $\endgroup$ – Alex Ravsky Aug 18 '17 at 18:41

Let $x \in U$ and let $U_x$ be the connected component of $U$ that contains $x$ that is - all $ a \in U$ that there is a continuous curve $\gamma:[0,1] \rightarrow U$ so $$\gamma(0)=x,\ \gamma(1)=a$ and $\gamma([0,1]) \subset U$$

Since $U$ is open, $ \ \exists \ r_x > 0 $ so that an open ball $B(x,r_x)$ is contained by $U$.
Open balls are connected sets so $x \in B(x,r_x) \subset U_x$ implies that $x$ is an interior point of $U_x$.
Hence each connected component of $U$ can be represented by an open ball contained by it, and because these components are pairwise disjoints - so are the balls that represents them.

Each ball in this collection of representatives contains a rational point (all coordinates are rational numbers) hence each connected component can be represented by a unique n-length sequence of rational numbers thus at most countable many items in this collection.

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  • $\begingroup$ The question is not to show that $U$ has countably many components, but about the partition of $U$ into disjoint "intervals". $\endgroup$ – Alex Ravsky Aug 18 '17 at 18:37

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