If a nonempty set of real numbers is open and closed, is it $\mathbb{R}$? Why/Why not?

Question

In other words, are $\emptyset$ and $\mathbb{R}$ the only open and closed sets in $\mathbb{R}$? Why/Why not?

I tried by assuming a set is equal to its interior points and contains its limit points.

A bounded set will not do since stuff like $[1,4]$ and $\{5\}$ will not work, though that is not really proof. Help please?

Anyway, it must then be unbounded.

If $a$ is a real number then $(a,\infty)$, $(-\infty,a)$, $[a,\infty)$ and $(-\infty,a]$ don't seem to cut it so it must be $\mathbb{R}$.

Answer 1

A space $X$ is connected if the only subsets of $X$ with empty boundary are $X$ and the empty set. Alternatively, $X$ is connected if the only subsets of $X$ which are both open and closed are $X$ and the empty set. Therefore, your first question asks if there exists any subset of $\mathbb R$ that is connected. Indeed, yes, there is. An interval is connected. Your second question seems to ask if $\mathbb R$ is connected. Yes, it is. For a proof, see here.

Answer 2

You're trying to prove that $\mathbb{R}$ is connected. You can do as follow : if a set $\mathbf{A}$ is both open and closed, then you can check that $\mathbb{1}_\mathbf{A}$ (the characteristic function of $\mathbf{A}$) is continuous, because $\mathbb{1}_\mathbf{A}^{-1}(O)$ where O is open is either $A$, its complement, the empty set or $R$ depending on $1$ and/or $0$ being in $O$, and all these set are open.

But $\mathbb{1}_\mathbf{A}$ only takes values $1$ or $0$, so it's easy to see that if it's continuous, then it's constant (if it is not, by the Intermediate Value Theorem, then it should also take all values between $0$ and $1$). Hence it's either always $1$ or $0$. And so $\mathbf{A}$ is either $\mathbb{R}$ or the empty set.

Answer 3

Because $\mathbb R$ is connected. Asume $U$ is a nonempty proper open and closed subset of $\mathbb R$, that is $V:=\mathbb R\setminus U$ is also nonempty and open. Let $u\in U$, $v\in V$. Wlog. $u<v$. Let $a=\sup([u,v]\cap U)$. As $[u,v]\cap U$ is a nonempty bounded closed set, $a\in([u,v]\cap U)$. Hence $u\le a<v$ (as $v\notin U$) and $U$ contains some $\epsilon$ neighbourhood of $a$. But then $\min\{a+\frac12\epsilon, v\}\in U\cap[u,v]$, contradiction to $a$ being the supremum.

Answer 4

Let $S$ be a nonempty subset of $\mathbb{R}$ that is both open and closed. Let's say that $S$ contains the element $s$. We want to show that no element of $\mathbb{R} - S$ can be bigger than or less than $s$, immediately implying that it is empty.

Suppose that there are elements of $\mathbb{R} - S$ bigger than $s$. We can construct the set $X$ of all such elements, and consider its infimum. Use the fact that $S$ is open and closed to derive a contradiction. A similar argument works to show there are no elements of $\mathbb{R} - S$ smaller than $s$.

Note: This argument crucially uses the structure of $\mathbb{R}$ in asserting the existence of infimums and supremums.

Answer 5

Suppose $X \subset \mathbb{R}$ is nonempty, open and closed. Let $x_0 \in X$. Finally suppose that $X \neq \mathbb{R}$; then there is some $y \not\in X$; WLOG we can assume that $y > x_0$.

Then the set $Z = \{ x \in \mathbb{R} : x > x_0, x \not\in X \}$ is bounded below (by $x_0$) and nonempty ($y \in Z$). Therefore $\inf Z = z$ exists.

• Suppose $z \in X$. Then since $X$ is open, it contains an open neighborhood $(z - \epsilon, z + \epsilon)$. This contradicts the definition of $z = \inf Z$, because there would be a sequence $z_n > z$, $|z - z_n| < \frac{1}{n}$, $z_n \in Z \Rightarrow z_n \not \in X$. This is not possible, because $[z, z + \epsilon) \subset X$.

• Suppose $z \not \in X$. Then since $X$ is closed, its complement is open, therefore there is an open neighborhood $(z - \epsilon, z + \epsilon)$ contained in $\mathbb{R} \setminus X$. Then $z - \frac{\epsilon}{2}$ contradicts the $\inf$ definition of $z$.

It follows that $X = \mathbb{R}$.

Answer 6

Here's yet another proof, which works by constructing a border point if $A$ is clopen nonempty proper subset of $\mathbb{R}$:

Be $A\subset\mathbb R$ both open and closed, but neither empty nor $\mathbb R$. Then there exist points $a\in A$ and $b\in\mathbb R\setminus A$.

Now construct two sequences $(a_n)$ and $(b_n)$ as follows:

$a_0=a$, $b_0=b$. For any $n$, be $c_n=(a_n+b_n)/2$. If $c_n\in A$, then $a_{n+1}=c_n$, $b_{n+1}=b_n$, else $a_{n+1}=a_n$, $b_{n+1}=c_n$.

Quite obviously for all $n$, $a_n\in A$ and $b_n\notin A$. Also $\lim_{n\to\infty}\left|a_n-b_n\right| = \lim_{n\to\infty}2^{-n}\left|a-b\right| = 0$. Therefore there exists exactly one point $x$ so that $\min(a_n,b_n)\le x\le\max(a_n,b_n)$ for all $n$ (nested intervals).

Since $\lim_{n\to\infty}a_n=x$ and $\lim_{n\to\infty} b_n=x$, we have in every open neighbourhood of $x$ both points in $A$ (namely $a_n$ for sufficiently large $n$) and in the complement of $A$ (namely $b_n$ for sufficiently large $n$). Thus $x$ is a border point of $A$, in contradiction that $A$ is both open and closed, and thus cannot have any border points.

Answer 7

I think I have a simpler proof using the border of $A$:

Let $A \subseteq R$ nonempty, open and closed, and assume $A \neq \mathbb{R}$.

Then $\partial A = \overline A \cap \overline{(\mathbb{R} \setminus A)} = $ ($A=\overline A$ since $A$ closed, same for $\overline{\mathbb{R} \setminus A}$) $ = A \cap (\mathbb{R} \setminus A) = \emptyset$ (the border is empty)

However $x = \sup A(=\overline A) \in \partial A$, a contradiction. $(\forall r>0) \langle x-r, x+r \rangle$ contains points both in $A$ and in $A \setminus \mathbb{R}$.