Connectedness

Definition. A topological space $X$ is disconnected if there exist nonempty disjoint open sets $U, V$ with $X = U \cup V$. Such a pair $\{U, V\}$ is called a separation of $X$.

$X$ is connected if it is not disconnected.

Equivalently, $X$ is connected if and only if the only clopen subsets of $X$ are $\emptyset$ and $X$.

Proof of the equivalence. If $X = U \cup V$ is a separation, then $U = X \setminus V$ is both open and closed, and neither $\emptyset$ nor $X$. Conversely, if $C \subseteq X$ is clopen with $C \notin \{\emptyset, X\}$, then $U = C$ and $V = X \setminus C$ are nonempty disjoint open sets with $U \cup V = X$; a separation. $\blacksquare$

Equivalently again, $X$ is connected if and only if every continuous map $h : X \to \{0, 1\}$ into the two-point space with the discrete topology is constant: preimages of $\{0\}$ and $\{1\}$ are clopen in $X$, so a non-constant $h$ is a nontrivial clopen set, and conversely. This formulation - connectedness as the statement that there are no non-constant maps into the discrete two-point space; is the one that makes the proofs below short.

Example 6.1. $[0, 1]$ is connected. $[0, 1) \cup (2, 3]$ is disconnected (with the subspace topology from $\mathbb{R}$).

Example 6.2. $\mathbb{Q}$ with the subspace topology from $\mathbb{R}$ is totally disconnected: the only connected subsets are singletons.

Proof. Let $A \subseteq \mathbb{Q}$ be connected. As a subset of $\mathbb{R}$ (same subspace topology), $A$ is an interval by Theorem 6.1. If $A$ contained two points $a < b$, it would contain every $c$ with $a < c < b$; including irrational $c$, contradicting $A \subseteq \mathbb{Q}$. So $A$ is a singleton. $\blacksquare$

Theorem 6.1. A subset of $\mathbb{R}$ (with the standard topology) is connected if and only if it is an interval.

(Here an interval is any set $I \subseteq \mathbb{R}$ with the property: if $a, b \in I$ and $a < c < b$, then $c \in I$.)

Proof. ($\Leftarrow$) Intervals are connected. Let $I$ be an interval and suppose $I = U \cup V$ with $U, V$ nonempty, disjoint, and open in $I$. Pick $a \in U$ and $b \in V$; swap the names if necessary so that $a < b$. Since $[a, b] \subseteq I$ (interval property), define

The set is nonempty: $U$ is open in $I$ and contains $a$, so $[a, a + \delta) \subseteq U$ for some $\delta > 0$. Hence $a < c \leq b$, and $[a, c) \subseteq U$ by definition of the supremum. Since $c \in [a, b] \subseteq I$, the point $c$ lies in $U$ or in $V$.

If $c \in U$: openness of $U$ in $I$ gives $\varepsilon > 0$ with $(c - \varepsilon, c + \varepsilon) \cap I \subseteq U$. Then $[a, c + \varepsilon/2] \subseteq U$ (the segment $[a, c)$ from the supremum, the segment $(c - \varepsilon, c + \varepsilon/2]$ from openness, both inside $I$), contradicting $c$ as an upper bound for the set.

So $c \in V$. Openness of $V$ in $I$ gives $\varepsilon > 0$ with $(c - \varepsilon, c + \varepsilon) \cap I \subseteq V$. Choose $x$ with $\max(a, c - \varepsilon) < x < c$; possible since $c > a$. Then $x \in [a, c) \subseteq U$ and $x \in (c - \varepsilon, c) \subseteq V$, contradicting $U \cap V = \emptyset$.

($\Rightarrow$) Connected sets are intervals. Contrapositive: if $I$ is not an interval, there are $a < c < b$ with $a, b \in I$ and $c \notin I$. Then

are open in $I$, nonempty ($a \in U$, $b \in V$), disjoint, and cover $I$ (as $c \notin I$). So $I$ is disconnected. $\blacksquare$

This theorem is why the intermediate value theorem exists and why $\mathbb{Q}$ is dust: connected subsets of $\mathbb{Q}$, being connected subsets of $\mathbb{R}$, are intervals; but an interval containing two distinct rationals contains the irrationals between them, which $\mathbb{Q}$ does not. So the only connected subsets of $\mathbb{Q}$ are singletons (Example 6.2).

Definition. A space $X$ is path-connected if for any two points $x, y \in X$, there exists a continuous function $\gamma : [0, 1] \to X$ with $\gamma(0) = x$ and $\gamma(1) = y$.

Proposition 6.1. Every path-connected space is connected. The converse is false.

Proof. Let $X$ be path-connected and fix a point $x_0 \in X$. For each $x \in X$ choose a path $\gamma_x : [0, 1] \to X$ from $x_0$ to $x$. Its image is connected (Proposition 6.4 below, applied to the connected interval $[0, 1]$), and all the images contain $x_0$. So

is a union of connected sets with a common point; connected by the pencil lemma (Proposition 6.1.1 below). $\blacksquare$

Proposition 6.1.1 (Pencil lemma). If $\{C_\alpha\}$ is a family of connected subsets of $X$ with $\bigcap_\alpha C_\alpha \neq \emptyset$, then $\bigcup_\alpha C_\alpha$ is connected.

Proof. Let $C = \bigcup_\alpha C_\alpha$ and pick $p \in \bigcap_\alpha C_\alpha$. Let $h : C \to \{0, 1\}$ be continuous into the discrete two-point space. For each $\alpha$, the sets $h^{-1}(0) \cap C_\alpha$ and $h^{-1}(1) \cap C_\alpha$ are disjoint clopen subsets of $C_\alpha$ covering it; connectedness of $C_\alpha$ forces one of them empty, so $h$ is constant on each $C_\alpha$. The constants all agree; each equals $h(p)$. So $h$ is constant on $C$, and $C$ is connected by the two-point characterisation. $\blacksquare$

Example 6.3 (Topologist’s sine curve). Let

$S$ is connected but not path-connected.

Proof that $S$ is connected. Write $A$ for the graph part and $V$ for the vertical segment. The set $A$ is the image of the interval $(0, 1]$; connected by Theorem 6.1; under the continuous map $x \mapsto (x, \sin(1/x))$, so $A$ is connected (Proposition 6.4). We claim $\overline{A} = S$, after which Proposition 6.7 (closures of connected sets are connected) finishes the proof.

$\overline{A} \subseteq S$: if $(x_n, \sin(1/x_n)) \to (x_0, y_0)$ with $x_n > 0$, then either $x_0 > 0$; in which case continuity of $\sin(1/x)$ at $x_0$ forces $y_0 = \sin(1/x_0)$, so the limit is in $A$; or $x_0 = 0$, in which case $|\sin(1/x_n)| \leq 1$ forces $y_0 \in [-1, 1]$, so the limit is in $V$.

$S \subseteq \overline{A}$: take $(0, y) \in V$ and any ball $B$ around it of radius $\varepsilon$ and $x$-reach $\delta$. Since $1/t$ runs over a full period of $\sin$ as $t$ runs over any interval $(0, \delta)$; choose $t$ with $1/t = \arcsin(y) + 2\pi k$ for a sufficiently large integer $k$; there is $x \in (0, \delta)$ with $\sin(1/x)$ within $\varepsilon$ of $y$. So $B \cap A \neq \emptyset$, and $(0, y) \in \overline{A}$.

Proof that $S$ is not path-connected. Suppose $\gamma : [0, 1] \to S$ is a continuous path with $\gamma(0) \in V$ and $\gamma(1) \in A$; write $\pi_x, \pi_y$ for the coordinate projections, which are continuous. The set

is closed (as $V$ is closed and $\gamma$ continuous) and nonempty ($0 \in C$). Let $b = \sup C \in C$; then $\gamma(b) \in V$, say $\gamma(b) = (0, y_b)$, while $\gamma(t) \in A$ for all $t > b$.

Take $\delta = 1/3$ and consider the rectangle $U = (-\eta, \eta) \times (y_b - \delta, y_b + \delta) \subseteq \mathbb{R}^2$, a neighbourhood of $(0, y_b)$. By continuity of $\gamma$ there is $\eta' > 0$ with $\gamma((b, b + \eta')) \subseteq U \cap S$. Now $\pi_x \circ \gamma$ is continuous, and the image of the connected interval $(b, b + \eta')$ is a connected subset of $\mathbb{R}$; an interval, by Theorem 6.1; containing positive values (points of $(b, b+\eta')$ map into $A$, whose $x$-coordinates are positive) and with infimum $0$ (as $t \to b^+$, $\pi_x(\gamma(t)) \to \pi_x(\gamma(b)) = 0$). Hence the image contains all of $(0, x_1]$ for some $x_1 > 0$. So for every $x^* \in (0, x_1]$ there is $t^* \in (b, b + \eta')$ with $(\pi_x \circ \gamma)(t^*) = x^*$.

Choose $x^* \in (0, x_1)$ with $\sin(1/x^*) \notin (y_b - \delta, y_b + \delta)$; possible because $1/x$ runs over a full period of $\sin$ on $(1/x_1, \infty)$, so $\sin$ attains values outside any interval of length $2\delta = 2/3 < 2$. For the corresponding $t^*$: $\gamma(t^*) = (x^*, \sin(1/x^*))$ has $\pi_y(\gamma(t^*)) \notin (y_b - \delta, y_b + \delta)$, so $\gamma(t^*) \notin U$; contradicting $\gamma((b, b+\eta')) \subseteq U$. So no such path exists, and $S$ is not path-connected. $\blacksquare$

The moral: connectedness passes to closures (Proposition 6.7), but path-connectedness does not - no path can traverse the infinitely many oscillations and still be continuous at the vertical segment.

Example 6.4. $\mathbb{R}^n$ is path-connected for all $n \geq 1$: any two points are joined by the straight-line path $\gamma(t) = (1-t)p + tq$, continuous as each coordinate is polynomial in $t$. The same paths work in any convex subset of $\mathbb{R}^n$.

Definition. A connected component of $X$ is a maximal connected subset of $X$. The connected components of $X$ form a partition of $X$.

Proposition 6.2. Connected components are closed. They need not be open; in a locally connected space they are always open.

Proof. Let $C$ be a component of $X$; a maximal connected subset. By Proposition 6.7, $\overline{C}$ is connected; it contains $C$, so maximality of $C$ forces $\overline{C} = C$. So components are closed; in every topological space, no separation axioms required.

For openness: if $X$ is locally connected, let $x \in C$ and let $U$ be any open neighbourhood of $x$. Local connectedness gives a connected open neighbourhood $V$ of $x$ with $V \subseteq U$. Now $C \cup V$ is connected (pencil lemma: $C$ and $V$ share the point $x$) and contains $C$, so by maximality $C \cup V \subseteq C$, i.e. $V \subseteq C$. Thus every neighbourhood $U$ of $x$ contains a neighbourhood $V$ of $x$ lying in $C$, which makes $C$ open. $\blacksquare$

Without local connectedness, components can fail to be open: the components of $\mathbb{Q}$ are singletons (Example 6.2), and no singleton is open in $\mathbb{Q}$.

Definition. A path component of $X$ is a maximal path-connected subset. Path components also partition $X$, and each path component is contained in a connected component.

Definition. $X$ is locally connected if for every $x \in X$ and every open neighbourhood $U$ of $x$, there exists a connected open neighbourhood $V$ of $x$ with $V \subseteq U$.

Proposition 6.3. Every open subset of $\mathbb{R}^n$ is locally connected.

Proof. Let $U \subseteq \mathbb{R}^n$ be open, $x \in U$, and let $N$ be any neighbourhood of $x$ in $U$. Openness gives $\varepsilon > 0$ with $B(x, \varepsilon) \subseteq U$; shrinking along the neighbourhood $N$, there is $\delta > 0$ with $B(x, \delta) \subseteq N \cap U$. The open ball $B(x, \delta)$ is convex; the straight-line segment between two of its points stays inside it, by the triangle inequality $|(1-t)p + tq - x| \leq (1-t)|p - x| + t|q - x| < \delta$; hence path-connected (linear paths), hence connected. So $B(x, \delta)$ is the required connected open neighbourhood of $x$ inside $N$. $\blacksquare$

Example 6.5. The topologist’s sine curve is connected but not locally connected.

Proof that $S$ is not locally connected. Let $p = (0, y) \in V$ and consider the basic neighbourhood $N = B(p, \varepsilon) \cap S$ with $\varepsilon < 1$. The set of $x$-coordinates of graph points in the ball,

meets the condition periodically: as $x \to 0^+$, the value $1/x$ sweeps through infinitely many full periods of $\sin$, so $E$ has infinitely many connected components accumulating at $0$. Choose a boundary point $x_1 \in (0, 1)$ between two components of $E$, with $E$ on both sides: then no point of $N$ has $x$-coordinate $x_1$ (graph points at $x_1$ sit on the boundary of the ball, and vertical-segment points have $x = 0$). Hence

splits $N$ into two nonempty, disjoint, open subsets; the first contains $(0, y) \in V$, the second contains the component of $E$ above $x_1$. So $N$ is disconnected. Since every neighbourhood of $p$ in $S$ contains such an $N$, no neighbourhood of $p$ is connected: $S$ is not locally connected at any point of $V$. $\blacksquare$

Proposition 6.4. The continuous image of a connected space is connected. Therefore connectedness is a topological invariant.

Proof. If $f : X \to Y$ is continuous and $X$ is connected, suppose $f(X) = U \cup V$ with $U, V$ open and disjoint in $f(X)$. Then $f^{-1}(U)$ and $f^{-1}(V)$ are open, disjoint, and cover $X$, contradicting connectedness. $\blacksquare$

Proposition 6.5 (Intermediate Value Theorem). If $f : X \to \mathbb{R}$ is continuous and $X$ is connected, then $f$ attains every value between any two of its values. This generalises the classical IVT from $\mathbb{R}$ to any connected space.

Proof. $f(X)$ is connected (Proposition 6.4) and is a subset of $\mathbb{R}$, so by Theorem 6.1 it is an interval. An interval containing two values contains every value between them. The classical IVT is the case $X = [a, b]$, connected by Theorem 6.1. $\blacksquare$

Proposition 6.6 (Products). The product of connected spaces is connected (with the product topology), for finite and even arbitrary index sets.

Proof. Two factors. Fix $(x_0, y_0) \in X \times Y$. For each $y \in Y$ consider

The two pieces are homeomorphic to $Y$ and $X$ respectively, hence connected, and they share the point $(x_0, y)$. By the pencil lemma, $T_y$ is connected. Every $T_y$ contains the common set $\{x_0\} \times Y \ni (x_0, y_0)$, and $X \times Y = \bigcup_y T_y$ since $(x, y) \in X \times \{y\} \subseteq T_y$. The pencil lemma once more: $X \times Y$ is connected.

Arbitrary products. Let $P = \prod_{i \in I} X_i$ with each $X_i$ connected, and fix $a = (a_i) \in P$. Let $D$ be the set of points agreeing with $a$ in all but finitely many coordinates, and for a finite $F \subseteq I$ let

which is connected by repeated application of the two-factor case. Each $D_F$ contains $a$, so $D = \bigcup_F D_F$ is connected by the pencil lemma. And $D$ is dense in $P$: a basic open set $\prod_i U_i$ (with $U_i = X_i$ for all $i$ outside some finite $F$) meets $D_F$; pick any point of $\prod_{i \in F} U_i$ in the restricted coordinates and $a$ elsewhere. So $D \subseteq P \subseteq \overline{D}$, and $P$ is connected by Proposition 6.7. $\blacksquare$

Note where the two factors are used: the pencil lemma welds the slices to a common spine, and the closure argument promotes finite support to arbitrary support. Neither step needs any separation axioms.

Proposition 6.7 (Closure). If $A \subseteq X$ is connected, then any set $B$ with $A \subseteq B \subseteq \overline{A}$ is connected. In particular, the closure of a connected set is connected.

Proof. Let $h : B \to \{0, 1\}$ be continuous into the discrete two-point space; by the two-point characterisation it suffices to show $h$ is constant. The restriction $h|_A$ is constant (say $0$), since $A$ is connected. The set $h^{-1}(0)$ is closed in $B$ (preimage of the closed - indeed clopen; subset $\{0\}$ of the discrete space) and contains $A$, so it contains $\overline{A} \cap B = B$: every point of $B$ is a limit point of $A$, and $h$ is $0$ on all of $A$, so by continuity $h$ is $0$ on all of $B$. Hence $h$ is constant, and $B$ is connected. $\blacksquare$

Definition. A space is totally disconnected if its only connected subsets are singletons. Examples include $\mathbb{Q}$ (with the subspace topology from $\mathbb{R}$) and the Cantor set $C$.

Proposition 6.8. The Cantor set $C$ is totally disconnected, compact, uncountable, and perfect (every point is a limit point). It is homeomorphic to $\{0, 1\}^{\mathbb{N}}$ with the product topology.

Proof. Write $C = \bigcap_{n=0}^{\infty} C_n$, where $C_0 = [0, 1]$ and $C_{n+1}$ deletes the open middle third of each interval of $C_n$. Equivalently, $C$ is the set of reals in $[0, 1]$ admitting a base-3 (ternary) expansion using only the digits $0$ and $2$: surviving stage $n$ means the first $n$ ternary digits can be chosen in $\{0, 2\}$, and the nested intersection is exactly those $x$ with such an expansion (the ambiguous case, a trailing digit $1$, corresponds to an endpoint that also has an all-$\{0,2\}$ expansion, e.g. $1/3 = 0.1_3 = 0.0222\ldots_3$).

Compact. Each $C_n$ is a finite union of closed intervals, hence closed, and $C$ is an intersection of closed sets, hence closed and bounded in $\mathbb{R}$; compact by Heine—Borel.

Perfect. The endpoints of the deleted middle thirds belong to every $C_n$; an endpoint of a gap at stage $n$ is an endpoint of a retained interval at every later stage; so they lie in $C$. Now let $x \in C$ and $\varepsilon > 0$. Choose $n$ with $3^{-n} < \varepsilon$; then $x$ lies in some stage-$n$ interval $J$ of length $3^{-n}$. The two endpoints of $J$ are in $C$, and at least one of them differs from $x$, giving a point of $C$ within $\varepsilon$ of $x$. So every point of $C$ is a limit point of $C$.

Uncountable. Define $\varphi : \{0, 1\}^{\mathbb{N}} \to C$ by $\varphi((a_n)) = \sum_{n=1}^{\infty} 2 a_n 3^{-n}$. This is the ternary digit map, and it is a bijection: injective, because if the sequences first differ at index $k$, then

so no two sequences map to the same point (the usual expansion ambiguity never arises for digits restricted to $\{0, 2\}$); surjective, because every point of $C$ has a ternary expansion with digits only $0$ and $2$, and that expansion is the required preimage. So $|C| = |\{0, 1\}^{\mathbb{N}}| = |\mathcal{P}(\mathbb{N})|$, uncountable by Cantor’s theorem.

Homeomorphism. The map $\varphi$ is continuous: if two sequences agree on the first $N$ digits, their images differ by at most $\sum_{n > N} 2 \cdot 3^{-n} = 3^{-N}$, which is exactly convergence in the product topology (a basic open set restricts finitely many coordinates). A continuous bijection from the compact space $\{0,1\}^{\mathbb{N}}$ (a product of finite discrete spaces) to the Hausdorff space $C \subseteq \mathbb{R}$ is a homeomorphism; the criterion proved in the compactness page. So $C \cong \{0, 1\}^{\mathbb{N}}$.

Total disconnectedness. Let $A \subseteq C$ be connected. As a subset of $\mathbb{R}$, $A$ is an interval (Theorem 6.1). But $C$ contains no interval of positive length: the total length of $C_n$ is $(2/3)^n$, which tends to $0$, while $A \subseteq C \subseteq C_n$ for every $n$. An interval of positive length cannot fit inside sets of arbitrarily small total length, so $A$ is a single point. $\blacksquare$

The digit map is the structural insight: the Cantor set is exactly the countable product of two-point spaces, with the middle-thirds construction supplying the embedding into $[0,1]$.

Problem. Show that $\operatorname{GL}(2, \mathbb{R})$ (the set of $2 \times 2$ invertible real matrices with the subspace topology from $\mathbb{R}^4$) is disconnected.

Problem. For which $n$ is $\mathbb{R}^n \setminus \{0\}$ path-connected?

Connectedness asks whether a space comes in separate pieces. A connected space cannot be split into two non-empty open sets that are disjoint, which means there is no “gap” that separates it. Intuitively, you can travel between any two points without leaving the space. The intermediate value theorem is the most familiar consequence: a continuous function on a connected domain must take every value between its endpoints.

Path-connectedness is a stronger and more intuitive version: a path-connected space has a continuous path between every pair of points. Every path-connected space is connected, but the converse fails for exotic spaces like the topologist’s sine curve, which is connected but has no continuous path crossing the oscillating part. In $\mathbb{R}^n$, connected subsets are exactly the intervals and their higher-dimensional analogues. Connectedness is the topological tool for proving that certain configurations are impossible: if you can show a space is disconnected, you have proven that it splits into distinct parts that cannot be continuously deformed into each other.

Mistake 1: Assuming that connected implies path-connected. Every path-connected space is connected, but the converse is false. The topologist’s sine curve is a classic example of a connected space that is not path-connected. Do not assume that a space is path-connected just because it is connected.

Mistake 2: Confusing connectedness with convexity. Convexity is a geometric property that implies path-connectedness (and hence connectedness) in $\mathbb{R}^n$, but connectedness does not imply convexity. For example, the unit circle $S^1$ is connected but not convex. A set can be connected without being convex.

Mistake 3: Forgetting that the continuous image of a connected set is connected. If $f: X \to Y$ is continuous and $X$ is connected, then $f(X)$ is connected. This property is often used to prove that certain sets are connected, but it is not an “if and only if” statement. The preimage of a connected set under a continuous map need not be connected.

Mistake 4: Assuming that a disconnected set can be written as a union of disjoint open sets in the ambient space. A set $A$ is disconnected if it can be written as $A = (U \cap A) \cup (V \cap A)$ where $U$ and $V$ are disjoint open sets in the ambient space. The sets $U$ and $V$ are open in the ambient space, not necessarily in $A$. Do not confuse the subspace topology with the ambient topology.

Mistake 5: Confusing total disconnectedness with disconnectedness. A space is totally disconnected if its only connected subsets are singletons. A space can be disconnected without being totally disconnected. For example, $[0, 1] \cup [2, 3]$ is disconnected but not totally disconnected, since $[0, 1]$ and $[2, 3]$ are connected subsets.

• Compactness: Compactness and connectedness are both fundamental topological properties that characterize space structure.

• Topological Spaces: Topological spaces provide the framework for defining connectedness through separation properties.

• Continuity and Homeomorphisms: Continuous functions preserve connectedness, and homeomorphisms preserve all topological invariants.

• Quantum Mechanics

• Graph Theory

• Classical Mechanics

• Electromagnetism

• Closed Sets, Closure, Interior, and Boundary
• Common Pitfalls
• Compactness
• Continuity and Homeomorphisms