Connected Sets | Real Analysis

This scene is about connected sets. Move the cut point and check whether the set splits into two open pieces that never touch. A set that cannot be split is connected. An interval never splits: wherever you cut, the cut point itself belongs to neither side. Two far-apart pieces, or the rationals with their dense gaps, split cleanly when the cut lands in a gap. For example, look at one 🧩 jigsaw puzzle. While the pieces interlock, it is a single board. Lying in two completely separate piles, it is already split apart.

Examples & Exercises

Prove completely that the interval $[0,1]$ is connected. That is, show that if $U, V$ are disjoint sets, both open in $[0,1]$ , with $[0,1] = U \cup V$ , then at least one of $U, V$ is empty.

[0,1] = U \cup V,\ U \cap V = \emptyset,\ U, V \text{ open in } [0,1] \implies U = \emptyset \text{ or } V = \emptyset

t = 0.5

: the cut point

0.5

itself belongs to neither side, so the pair does not cover — every cut of

[0,1]

fails this way.

Show proof

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.

Prove completely that if $X$ is path-connected then $X$ is connected. Path-connected means that for any $a, b \in X$ there is a continuous $\gamma : [0,1] \to X$ with $\gamma(0) = a$ and $\gamma(1) = b$ . You may use Problem 1 (connectedness of $[0,1]$ ).

\gamma : [0,1] \to X \text{ continuous},\quad A = \gamma^{-1}(U),\quad B = \gamma^{-1}(V)

[0,1] = A \cup B,\quad A \cap B = \emptyset,\quad A, B \text{ open in } [0,1]

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

t = 1.5

: a space where a separation succeeds — no path can cross between the pieces, for its preimage would separate

[0,1]

Show proof

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.

Prove completely that every subset $A \subseteq \mathbb{Q}$ with at least two points is disconnected. Conclude that $\mathbb{Q}$ is totally disconnected: its only connected subsets are singletons and $\emptyset$ .

\alpha = p + \frac{q - p}{\sqrt{2}}, \qquad \sqrt{2} = \frac{q - p}{\alpha - p} \notin \mathbb{Q}

A = \bigl(A \cap (-\infty, \alpha)\bigr) \cup \bigl(A \cap (\alpha, \infty)\bigr)

The cut

c = \sqrt{2}/2 \approx 0.7071

: being irrational, it hits no rational point, so it covers and splits

\mathbb{Q} \cap [0,1]

Show proof

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.