Proof attempt:

Let there be another clopen set $S$ in which is a proper subset of $\mathbb{R}$. Hence, $ S^c \neq \emptyset $.

We can assert the following statements:

  1. No point of $S$ lies in $S^c$.
  2. No point of $\overline{S}$ lies in $S^c$ [Since the set $S$ is closed, no point of the derived set of $S$ is a member of $S^c$].

  3. No point of $S^c$ lies in $S$.

  4. No point of $\overline{S^c}$ lies in $S$. [$S$ is both open and closed. Hence its complement, i.e. $S^c$ is closed and open. (complement of a closed set is an open set and vice versa).]

Therefore, $S \cap \overline{S^c} =\emptyset$ and $\overline{S} \cap S^c= \emptyset $. Therefore, $S$ and $S^c$ are separated sets. Again, $S\cup S^c =\mathbb{R}$. Therefore, $\mathbb{R}$ is disconnected, a contradiction.

[ $\mathbb{R}$ is connected, since for any $x, y \in \mathbb{R}$ $\implies$ $z \in \mathbb{R}$, where $z$ is any point such that $x<z<y$.]

Is it correct?


I don't know how to react (every answer being downvoted by someone with a better understanding of the subject than mine). I now try to write up a proof ( although very much unoriginal and basically a copy-paste from Rudin).

We can all agree on the fact (regarding $\mathbb{R}$) that for any two numbers $x, y \in \mathbb{R}$ with $x<y$, every number between $x$ and $y$ belongs to $\mathbb{R}$ (can we?).

Suppose, $\mathbb{R}$ can be written as the union of two non-empty separated sets $A$ and $B$ (i.e. by the very definition of separated sets, $A \cap \overline{B}= \emptyset$ and $ \overline{A} \cap B= \emptyset )$

We pick $x \in A$ and $y \in B$ with $x<y$. Define $z= \sup(A \cap [x,y])$. $z$ is going to be a limit point of $A$, $z \in \overline{A}$, therefore $z \notin B$. $\ $ $x\leq z<y$. If $z\notin A$, clearly $z\notin \mathbb{R}$. Again, if $z\in A$, we can find some $t$ between $z$ and $y$ such that $t\notin B$ [since $z$ is not a limit point of $B$]. Consequently, $t \notin \mathbb{R}$.

Being not a union of two separated subsets, $\mathbb{R}$ is connected.

  • $\begingroup$ What is a "derived set" of $S$? And if $S^{c}$ denotes the complement of $S$, why do we need to proof that $S$ is disjoint from its complement (if that's what you mean by "separated")? $\endgroup$ – avs Apr 25 at 21:34
  • $\begingroup$ By "Derived" set, I mean the set of limit points of $S$. $\endgroup$ – Subhasis Biswas Apr 25 at 21:38
  • $\begingroup$ Not only it is disjoint from its complement (a trivial statement), the closure of $S$ is also disjoint from $S^c$. $\endgroup$ – Subhasis Biswas Apr 25 at 21:39
  • $\begingroup$ what is your definition of connected space? $\endgroup$ – dcolazin Apr 25 at 21:40
  • 2
    $\begingroup$ Possible duplicate of Prove $\mathbb{R}$ is connected $\endgroup$ – YuiTo Cheng Apr 26 at 0:54

How about this sleight-of-hand? Assume for a contradiction that $S\subsetneq \mathbb R$ is nonempty and clopen, and consider the function $$ f(x) = \begin{cases} 1 & \text{when }x\in S \\ 3 & \text{when }x\notin S. \end{cases} $$

It is easy to see that (since $S$ and $S^\complement$ are open) $f$ is continuous. And because $S$ is nontrivial, there exist $x_1, x_3\in\mathbb R$ such that $f(x_1)=1$ and $f(x_3)=3$.

Now apply the intermediate value theorem (which is usually proved well before we start worrying about clopen sets) to find an $x_2$ such that $f(x_2)=2$. This flatly contradicts the definition of $f$ above.

  • $\begingroup$ Hmm, someone doesn't like this. Why? (I could understand a downvote because the OP wanted a proof verification and here is an alternative argument instead -- but since the answer by avs which suffers from the same problem didn't get one, I doubt that is it). $\endgroup$ – Henning Makholm Apr 26 at 1:05

This can be done simpler. Suppose there is a nonempty, clopen subset $S$ of $\mathbb{R}$.

Suppose there exists an $x$ in $\mathbb{R}$ that is not in $S$.

Case 1: $(]x, +\infty[ \; \cap \; S)$ is nonempty. (If Case 1 is not the case, then the only other possibility is that $]-\infty, x[ \; \cap \; S$ is nonempty, and this is handled analogously to Case 1, so I won't do it.)

Let $$ x_{S} = \inf \left(\; ]x, +\infty[ \; \cap \; S \;\right) \in \mathbb{R}. $$ Now, since $S$ is clopen, it contains $x_{S}$, but then $x_{S}$ cannot be interior to $S$, so $S$ cannot be open.

Therefore, there is no such $x$, and $S = \mathbb{R}$.

  • 2
    $\begingroup$ Why would such a "right-most" connected component exist? There might be infinitely many connected components, e.g. take $A = \bigcup_{n \in \mathbb{Z}} (n-\frac{1}{4}, n+\frac{1}{4})$. $\endgroup$ – Mark Kamsma Apr 25 at 22:49
  • $\begingroup$ Thank you. Edited. $\endgroup$ – avs Apr 25 at 22:51
  • $\begingroup$ Edited further. $A$ is unbounded and connected. $\endgroup$ – avs Apr 25 at 22:53
  • $\begingroup$ @egreg, better? $\endgroup$ – avs Apr 25 at 23:03
  • $\begingroup$ Not really; there's no reason for $x_S$ not being in the interior of $S$. $\endgroup$ – egreg Apr 25 at 23:15

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