Clopen Sets

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 $\varepsilon > 0$.

Now shrink to $\delta = \min(\varepsilon/2, (b-c)/2) > 0$ (positive since $c < b$): then $c + \delta \in [a,b]$, exceeds $c$, and lies in $U$, so it belongs 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.

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.

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.