Clopen Sets | Real Analysis

Sets that are both open and closed — clopen. In

\mathbb{R}

only

\varnothing

and

\mathbb{R}

qualify (proved in full), but change the ambient space and the same set's fate changes:

[0,1]

is merely closed in

\mathbb{R}

, yet clopen inside

[0,1]\cup[2,3]

Examples & Exercises

Let $A \subseteq \mathbb{R}$ be both open and closed (clopen), with $\emptyset \ne A \ne \mathbb{R}$ . Derive a contradiction, proving completely that the only clopen subsets of $\mathbb{R}$ are $\emptyset$ and $\mathbb{R}$ itself.

\text{clopen in } \mathbb{R} \implies A = \emptyset \text{ or } A = \mathbb{R}

X = \mathbb{R}

A = (0,1)

: open but not closed — no nontrivial clopen set exists in

\mathbb{R}

Show proof

If $A$ is clopen then so is its complement $A^c$ : $A$ open makes $A^c$ closed, and $A$ closed makes $A^c$ open. Since $A \ne \mathbb{R}$ we have $A^c \ne \emptyset$ , and since $A \ne \emptyset$ , $A$ itself is nonempty.
Hence $\mathbb{R} = A \cup A^c$ is a decomposition into two disjoint nonempty open sets — a separation of $\mathbb{R}$ . Now we show $\mathbb{R}$ is connected. Suppose $\mathbb{R} = U \cup V$ with $U, V$ disjoint nonempty open; pick $a \in U$ , $b \in V$ (renaming so $a < b$ ).
Put $c = \sup(U \cap [a,b])$ (nonempty, containing $a$ ; bounded above by $b$ ). $U$ is not only open but also closed, since $U = \mathbb{R} \setminus V$ . Now $c$ lies in the closure of $U \cap [a,b]$ , so $c \in \overline{U} = U$ . We have $c \le b$ , and $c \ne b$ (as $b \in V$ and the sets are disjoint), so $c < b$ . Since $U$ is open, $(c-\varepsilon, c+\varepsilon) \subseteq U$ for some $\varepsilon > 0$ ; but then $c + \varepsilon/2 \in [a,b]$ exceeds $c$ while belonging to $U \cap [a,b]$ — contradicting $c = \sup$ . Hence no separation of $\mathbb{R}$ exists.
This contradicts the separation from the second step, so the assumed nontrivial clopen set $A$ cannot exist: the only clopen subsets of $\mathbb{R}$ are $\emptyset$ and $\mathbb{R}$ . The scene's $A = (0,1)$ is open but not closed, dodging this theorem — closing it would require adding the missing limit points $0, 1$ , which would break openness.

For the subspace $X = [0,1] \cup [2,3] \subseteq \mathbb{R}$ and $A = [0,1]$ , prove completely that $A$ is clopen in $X$ but not open in $\mathbb{R}$ .

[0,1] = X \cap (-1/2,\, 3/2) = X \cap [-1/2,\, 3/2]

x = 1

\varepsilon = 0.5

: even at the boundary point

1

(1-0.5, 1+0.5) \cap X = (0.5, 1] \subseteq A

, so the ball stays in

A

relative to

X

— openness in

X

made visible.

Show proof

Openness in $X$ : check $[0,1] = X \cap (-1/2,\, 3/2)$ . The inclusion $\subseteq$ is clear; conversely a point of $X \cap (-1/2,\, 3/2)$ lies in $[0,1]$ or in $[2,3] \cap (-1/2,\, 3/2)$ , and the latter is empty, so it lies in $[0,1]$ . Since $(-1/2,\, 3/2)$ is open in $\mathbb{R}$ , $[0,1]$ is open in $X$ .
Closedness in $X$ : the same argument gives $[0,1] = X \cap [-1/2,\, 3/2]$ (since $[2,3] \cap [-1/2,\, 3/2] = \emptyset$ ). As $[-1/2,\, 3/2]$ is closed in $\mathbb{R}$ , $[0,1]$ is closed in $X$ . Hence $[0,1]$ is clopen in $X$ .
Not open in $\mathbb{R}$ : $1 \in [0,1]$ , yet every ball $(1-\varepsilon, 1+\varepsilon)$ contains $1 + \varepsilon/2 \notin [0,1]$ . So the same set $[0,1]$ is clopen in $X$ while in $\mathbb{R}$ it is merely closed, not open — clopenness is relative to the ambient space.

Let $X$ be a topological space and $A \subseteq X$ clopen. Prove completely that if $\emptyset \ne A \ne X$ then $X$ is disconnected.

X = A \cup A^c, \qquad A \cap A^c = \emptyset, \qquad A,\, A^c \ne \emptyset \text{ open in } X

X = (0,1) \cup (1,2)

A = (0,1)

X = A \cup A^c

is a separation into disjoint nonempty opens — showing

X

is disconnected.

Show proof

If $A$ is clopen then so is its complement $A^c = X \setminus A$ : complementation swaps openness and closedness ( $A$ open makes $A^c$ closed; $A$ closed makes $A^c$ open).
Since $\emptyset \ne A \ne X$ , both $A$ and $A^c$ are nonempty. Moreover $X = A \cup A^c$ , $A \cap A^c = \emptyset$ , and both are open in $X$ — a separation of $X$ . Hence $X$ is disconnected.
The converse holds too: if $X = U \cup V$ is a separation then $U = X \setminus V$ is also closed, hence clopen. So "a nontrivial clopen set exists" is equivalent to "disconnected". Problem 1 above is the special case $X = \mathbb{R}$ — since $\mathbb{R}$ is connected, no nontrivial clopen set can exist there.