I've seen many proofs of this theorem. But, unable to think any example showing this.

Suppose, how to write (0,1) as a countable Union of disjoint open intervals.

No idea! I'm stuck

Plz help!

  • 1
    $\begingroup$ "Countable" includes "finite". $\endgroup$
    – saulspatz
    Jan 8 '20 at 21:33
  • 1
    $\begingroup$ and finite includes 1. $\endgroup$ Jan 8 '20 at 22:48
  • $\begingroup$ $\{x\in \Bbb R: \sin(x) >0\}$ is open as the inverse image of an open set under a continuous function: $\sin^{-1}[(0,\infty)]$. Consider that set instead. $\endgroup$ Jan 8 '20 at 22:51

Well, $(0;1)$ is already an open interval. So it is even a union of finitely (namely just one) open interval.

Otherwise, you can also write $$ (0;1) = (0;1) \cup \bigcup_{n\in \mathbb{N}} \emptyset $$ if you desperately want to have a union over infinitely many things :)

  • $\begingroup$ But, all proofs of this theorem targets rational points. $\endgroup$ Jan 8 '20 at 21:36
  • $\begingroup$ What do you mean, by targeting rational points? How would you write the empty set? And the main point is what saulspatz wrote in his comment; Countable includes finite. $\endgroup$ Jan 8 '20 at 21:38
  • $\begingroup$ I've understood what you did but Plz show this for any other open set without using "phi". $\endgroup$ Jan 8 '20 at 21:39
  • $\begingroup$ It is not a phi, but the empty set. Which of my two ways of answering your question don't you understand? $\endgroup$ Jan 8 '20 at 21:44

Your example $(0,1)$ has only one expression as the union of disjoint open sets; i.e. itself. This is obvious because the union any collection of disjoint open intervals is not connected.

There is a more general result. Every open set in $\mathbb R^n$ can be expressed in one and only one way as a countable disjoint union of open connected sets.


$1).\ $ Say $x\sim y$ if there is a connected subspace of $\mathbb R^n$ that contains $x$ and $y$. Check that $\sim$ is an equivalence relation. Then define the components of $\mathbb R^n$ to be the equivalence classes. Take any open set $U\subseteq \mathbb R^n$.The fact that the components of $U$ are maximally connected disjoint sets follows from the definition.

$2).\ $ Components of $U$ are open sets. This follows from the fact that the basis elements in $\mathbb R^n$ are connected. If $x\in C,$ a component of $U,$ then there is a ball $B(x)$ contained in $U.$ And $B(x)\subseteq C$ because $C$ is maximal.

$3).\ $ For uniqueness, note that the $\mathscr A=\{C_{\alpha}\}_{\alpha\in J}$ is an open cover of $\mathbb R^n$. Since $\mathbb R^n$ is Lindelof, $\mathscr A$ has a countable subcover. But the $C_{\alpha}$ are $disjoint$, so the countable subcover must be $\mathscr A$ itself.

  • $\begingroup$ Can you give me an example of any open set from topological space which can be written in that form i.e. countable Union of disjoint open intervals.? $\endgroup$ Jan 9 '20 at 18:57

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