Connected Sets

Suppose $[0,1] = U \cup V$ is a separation, with $U, V$ both nonempty.

Renaming if needed, assume $0 \in U$ ($0$ lies in one of $U, V$, and the symmetry $x \mapsto 1 - x$ makes the two cases identical).

Put $S = \{x \in [0,1] : [0,x] \subseteq U\}$ and $c = \sup S$.

Since $0 \in S$, $S \ne \emptyset$; it is bounded above by $1$, so $c \in [0,1]$ exists by completeness.

First, $c \in U$.

Since $V$ is open in $[0,1]$, its complement $U = [0,1] \setminus V$ is closed in $[0,1]$ — that is, $U$ equals its own closure in the subspace: $\overline{U} \cap [0,1] = U$.

If $c = 0$ then $c = 0 \in U$. If $c > 0$, then for every $\varepsilon > 0$ the definition of supremum gives $x \in S$ with $c - \varepsilon < x \le c$.

That $x$ satisfies $x \in [0,x] \subseteq U$, so the neighborhood $(c-\varepsilon, c+\varepsilon)$ meets $U$, and hence $c \in \overline{U}$.

Therefore $c \in \overline{U} \cap [0,1] = U$.

Next, $[0,c] \subseteq U$: if $y < c$ then the definition of supremum gives $x \in S$ with $y < x \le c$, so $y \in [0,x] \subseteq U$; and the endpoint $c$ itself lies in $U$ by the previous step.

Now $c = 1$. Suppose instead $c < 1$.

Since $c \in U$ and $U$ is open in $[0,1]$, there is $\delta > 0$ with $(c-\delta, c+\delta) \cap [0,1] \subseteq U$.

Put $\delta' = \min(\delta/2, (1-c)/2) > 0$ (positive since $c < 1$). Then $[0, c+\delta'] = [0,c] \cup (c, c+\delta'] \subseteq U$.

So $c + \delta' \in S$, yet $c + \delta' > c = \sup S$ — a contradiction. Hence $c = 1$.

Hence $[0,1] = [0,c] \subseteq U$.

But a separation requires $V \ne \emptyset$, so some $v \in V$ exists.

Then $v \in [0,1] \subseteq U$ and $v \in V$, giving $U \cap V \ne \emptyset$ — contradicting disjointness.

Therefore no separation of $[0,1]$ exists, and $[0,1]$ is connected.

This is why every cut in the scene fails: a cut point $t \in [0,1]$ can belong to neither open side, and this proof shows the same phenomenon defeats every open pair, not just interval cuts.

Let $X$ be path-connected, and suppose for a contradiction that $X = U \cup V$ is a separation: $U, V$ open in $X$, disjoint, and both nonempty.

Pick $a \in U$ and $b \in V$.

By path-connectedness there is a continuous $\gamma : [0,1] \to X$ with $\gamma(0) = a$ and $\gamma(1) = b$.

Put $A = \gamma^{-1}(U)$ and $B = \gamma^{-1}(V)$.

Since $\gamma$ is continuous and $U, V$ are open in $X$, both $A$ and $B$ are open in $[0,1]$: if $t_0 \in A$ then $\gamma(t_0) \in U$; openness of $U$ gives $\varepsilon > 0$ with $U$ containing the $\varepsilon$-neighborhood of $\gamma(t_0)$; continuity then sends a whole $\delta$-neighborhood of $t_0$ into that $\varepsilon$-neighborhood, so the $\delta$-neighborhood lies in $A$. The same argument works for $B$.

Check the four separation conditions in $[0,1]$.

Cover: for every $t \in [0,1]$, $\gamma(t) \in X = U \cup V$, so $t \in A \cup B$; that is, $[0,1] = A \cup B$.

Disjoint: if $t \in A \cap B$ then $\gamma(t) \in U \cap V = \emptyset$, impossible; so $A \cap B = \emptyset$.

Nonempty: $0 \in A$ (since $\gamma(0) = a \in U$) and $1 \in B$ (since $\gamma(1) = b \in V$).

Thus $[0,1] = A \cup B$ is a separation of $[0,1]$.

This contradicts Problem 1 — $[0,1]$ is connected, so it admits no separation.

Hence no separation $X = U \cup V$ of $X$ can exist, and $X$ is connected.

The converse is false: the closure of the topologist's sine curve is connected but not path-connected.

Let $A \subseteq \mathbb{Q}$ have at least two points, and pick $p, q \in A$ with $p < q$.

Put $\alpha = p + \frac{q - p}{\sqrt{2}}$.

Since $0 < \frac{1}{\sqrt{2}} < 1$ (that is, $1 < \sqrt{2} < 2$), $\alpha$ is a strict convex combination of $p$ and $q$:

$p < \alpha < q$.

The number $\alpha$ is irrational. Suppose $\alpha \in \mathbb{Q}$.

Then $\alpha - p \in \mathbb{Q}$, and $\alpha - p = \frac{q - p}{\sqrt{2}} > 0$, so it is nonzero; rearranging gives $\sqrt{2} = \frac{q - p}{\alpha - p}$.

Since $q - p \in \mathbb{Q}$ is nonzero, this ratio is rational — contradicting the irrationality of $\sqrt{2}$.

Hence $\alpha \notin \mathbb{Q}$; in particular, $\alpha$ equals no element of $A$.

Put $U = A \cap (-\infty, \alpha)$ and $V = A \cap (\alpha, \infty)$.

Open: $(-\infty, \alpha)$ and $(\alpha, \infty)$ are open rays in $\mathbb{R}$, so $U$ and $V$ are open in the subspace $A$.

Disjoint: $x < \alpha$ and $x > \alpha$ are incompatible, so $U \cap V = \emptyset$.

Cover: if $a \in A$ then $a \in \mathbb{Q}$, so $a \ne \alpha$, and therefore $a < \alpha$ (so $a \in U$) or $a > \alpha$ (so $a \in V$). That is, $A = U \cup V$.

Nonempty: $p \in U$ (as $p < \alpha$) and $q \in V$ (as $\alpha < q$).

Thus $A = U \cup V$ is a separation of $A$, so $A$ is disconnected.

Since $A$ was an arbitrary subset of $\mathbb{Q}$ with at least two points, every connected subset of $\mathbb{Q}$ has at most one point — $\mathbb{Q}$ is totally disconnected.

The scene's cut $\sqrt{2}/2$ is exactly this $\alpha$ for $p = 0$, $q = 1$: $\alpha = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$. Between any two rationals an irrational slips in and splits the set.