Show that the set of real numbers has no nontrivial clopen set without using connectedness of $\mathbb R$.

I tried to show this by showing that an open set in $\mathbb{R}$ is strictly contained in its closure.

My attempt:

Let $U$ is open in $\mathbb{R}$. Then $U$ can be written as union of disjoint open intervals. $U = \bigcup_{n=1}^{\infty}A_n$ , where $A_n = (a_n, b_n)$. Now Closure of $U, Cl(U) = Cl( \bigcup_{n=1}^{\infty}A_n) = \bigcup_{n=1}^{\infty}Cl(A_n)$ [ not true in general ] = $ \bigcup_{n=1}^{\infty}[a_n, b_n] \supsetneq U$. So the problem thus reduces to the problem, whether $Cl(\bigcup_{n=1}^{\infty}A_n) = \bigcup_{n=1}^{\infty}Cl(A_n)$ holds for countably infinite intervals in $\mathbb{R}$. This surely holds for finitely many sets, but doesn't hold for countably infinitely many sets in general. Is this true? or there's another proof of this?

  • $\begingroup$ Why do you assert that $\bigcup_{n=1}^\infty[a_n,b_n]\neq U$? $\endgroup$ Jul 7, 2017 at 13:01
  • $\begingroup$ Because the points a_n, b_n aren't in U $\endgroup$
    – user398623
    Jul 7, 2017 at 13:09
  • 3
    $\begingroup$ It is not true that $\operatorname{cl}\bigcup\limits_{n\in\Bbb N}(a_n,b_n)=\bigcup\limits_{n\in\Bbb N}[a_n,b_n]$. For instance, consider $a_n=\frac1{n+1}$ and $b_n=\frac1n$. $$\bigcup_{n\in\Bbb N}(a_n,b_n)=(0,1)\setminus\{1/n\,:\,n\in\Bbb N,\ n\ge 2\}\\\bigcup_{n\in\Bbb N}[a_n,b_n]=(0,1]\\ \operatorname{cl}\bigcup_{n\in\Bbb N}(a_n,b_n)=[0,1]$$ $\endgroup$
    – user228113
    Jul 7, 2017 at 13:14
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    $\begingroup$ You're right. On the other hand, the answer to your question is negative. Take $A_1=(1/2,1)$, $A_2=(1/4,1/2)$, $A_3=(1/8,1/4)$, and so on. Then $0$ belongs to the closure of the uniont, but not to the union of the closures. $\endgroup$ Jul 7, 2017 at 13:14
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    $\begingroup$ Your proof is actually very close to being correct. You do not need the (false) equation $Cl(U) = Cl( \bigcup_{n=1}^{\infty}A_n) = \bigcup_{n=1}^{\infty}Cl(A_n)$ to prove that $U$ is not closed. You only need the (true) inclusion $Cl(U) = Cl( \bigcup_{n=1}^{\infty}A_n) \supset \bigcup_{n=1}^{\infty}Cl(A_n)$ together with the observation you have already made that the endpoints of each $A_n$ are not elements of $U$. $\endgroup$
    – Lee Mosher
    Jul 7, 2017 at 20:41

1 Answer 1


Let $U, V$ be non-empty open subsets of $\Bbb R$ with $U\cap V=\emptyset$ and $U\cup V=\Bbb R$. Pick $u\in U$, $v\in V$. Wlog. $u<v$. Let $w=\inf\bigl(V\cap [u,v]\bigr)$. If $w\in U$ then $(w-\epsilon,w+\epsilon)\subseteq U$ for some $\epsilon>0$, and so $[w,w+\epsilon)\cap V=\emptyset$, contradicting the definition of infimum. Hence $w\in V$, in particzular $w>u$. But then $(w-\epsilon,w]\subset V$ for some $\epsilon>0$, again contradicting the definition of infimum.


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