# There are no other clopen sets in $\mathbb{R}$ except for $\mathbb{R}$ and $\emptyset$

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.]

Is it correct?

EDIT:

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 $$x, every number between $$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 $$x. Define $$z= \sup(A \cap [x,y])$$. $$z$$ is going to be a limit point of $$A$$, $$z \in \overline{A}$$, therefore $$z \notin B$$. $$\$$ $$x\leq z. If $$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.

• 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")? – avs Apr 25 at 21:34
• By "Derived" set, I mean the set of limit points of $S$. – Subhasis Biswas Apr 25 at 21:38
• Not only it is disjoint from its complement (a trivial statement), the closure of $S$ is also disjoint from $S^c$. – Subhasis Biswas Apr 25 at 21:39
• what is your definition of connected space? – dcolazin Apr 25 at 21:40
• Possible duplicate of Prove $\mathbb{R}$ is connected – 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.

• 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). – 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 $$S = \mathbb{R}$$.

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