One of the most basic geometric intuitions is that a space can consist of "one piece" or "several pieces." A circle is one piece; a pair of disjoint circles is two pieces. The challenge for topology is to capture this distinction using only the language of open sets — without reference to distance, coordinates, or any ambient space. The difficulty is more subtle than it first appears. In a [metric space](/page/Metric%20Space), one might try to define "one piece" by saying that any two points can be joined by a curve. This works well for open subsets of $\mathbb{R}^n$, but it fails for general topological spaces. The **topologist's sine curve** — the closure of the graph of $\sin(1/x)$ — is a subset of $\mathbb{R}^2$ that is intuitively "one piece" (it cannot be split into two open halves), yet no continuous path joins the oscillatory part to the vertical segment at the origin. This gap between "cannot be separated" and "can be joined by a path" is the central tension of the theory. [example: The Topologist's Sine Curve] Define the set $S \subset \mathbb{R}^2$ by: \begin{align*} S := \left\{\left(x, \sin\frac{1}{x}\right) : x \in (0, 1]\right\} \cup \left(\{0\} \times [-1, 1]\right). \end{align*} The first component is the graph of the function $g: (0, 1] \to \mathbb{R}$, $g(x) = \sin(1/x)$, which oscillates with increasing frequency as $x \to 0^+$. The second component is the vertical segment $I = \{0\} \times [-1, 1]$. Every point of $I$ is a limit point of the graph: for any $(0, y)$ with $y \in [-1, 1]$, there exists a sequence $x_k \to 0^+$ with $\sin(1/x_k) = y$ (since $\sin$ takes every value in $[-1, 1]$ infinitely often on any interval $(0, \varepsilon)$). Therefore $S = \overline{\{(x, \sin(1/x)) : x \in (0, 1]\}}$, and $S$ is the closure of a connected set (the continuous image of the connected interval $(0, 1]$). However, $S$ is **not** path-connected. Suppose for contradiction that there exists a [continuous](/page/Continuity) path $\gamma: [0, 1] \to S$ with $\gamma(0) = (0, 0)$ and $\gamma(1) = (1, \sin 1)$. Write $\gamma(t) = (\gamma_1(t), \gamma_2(t))$. Let $t_0 = \sup\{t \in [0, 1] : \gamma_1(t) = 0\}$. By continuity of $\gamma_1$ and the fact that $\{t : \gamma_1(t) = 0\}$ is closed, $\gamma_1(t_0) = 0$. For $t > t_0$, we have $\gamma_1(t) > 0$, so $\gamma_2(t) = \sin(1/\gamma_1(t))$. As $t \to t_0^+$, the values $\gamma_1(t) \to 0^+$, so $\sin(1/\gamma_1(t))$ oscillates between $-1$ and $1$ without settling. But $\gamma_2$ is continuous, so $\gamma_2(t) \to \gamma_2(t_0)$ must exist — a contradiction. This space therefore demonstrates that the two natural formalisations of "one piece" — connectedness and path-connectedness — genuinely differ. [/example] The theory of connectedness addresses this by introducing two separate notions. **Connectedness** (the algebraic-topological notion) captures the idea that a space cannot be separated into two open halves. **Path-connectedness** (the geometric notion) captures the idea that any two points can be joined by a continuous curve. Path-connectedness implies connectedness, but not conversely — and understanding when the two notions coincide is one of the main questions in the subject. ## Definition The fundamental definition avoids any mention of paths or distances. Instead, it characterises "one piece" by the absence of a certain kind of partition. [definition: Connected Topological Space] A [topological space](/page/Topology) $(X, \tau)$ is **connected** if there do not exist non-empty [open sets](/page/Open%20Set) $U, V \in \tau$ with: \begin{align*} U \cap V = \varnothing \quad \text{and} \quad U \cup V = X. \end{align*} Such a pair $(U, V)$ is called a **separation** of $X$. Equivalently, $X$ is connected if and only if the only subsets of $X$ that are both open and [closed](/page/Closed%20Set) (i.e., **clopen**) are $\varnothing$ and $X$ itself. A subset $A \subset X$ is **connected** if $A$ is connected in the subspace topology inherited from $X$. [/definition] The equivalence of the two formulations is immediate: if $(U, V)$ is a separation, then $U$ is clopen (it is open by assumption and closed because its complement $V$ is open). Conversely, if $U$ is a non-trivial clopen set ($U \neq \varnothing$, $U \neq X$), then $(U, X \setminus U)$ is a separation. [remark: The Empty Space] The empty space $\varnothing$ satisfies the definition vacuously: there is no separation because there are no non-empty open sets at all. By convention, $\varnothing$ is considered connected in most references, though some authors exclude it. We follow the convention that $\varnothing$ is connected. [/remark] The definition is purely negative — it says what a connected space is *not* (separable into two open halves), rather than what it *is*. This makes connectedness difficult to verify directly. The following equivalent characterisations provide more workable criteria. [quotetheorem:294] Characterisation (3) is often the most useful in practice. To show that a space is connected, it suffices to show that every continuous integer-valued function on it is constant. To show that a space is *disconnected*, it suffices to exhibit a continuous surjection onto the two-point discrete space $\{0, 1\}$. Characterisation (2) is the topological generalisation of the [Intermediate Value Theorem](/page/Continuity): a continuous real-valued function on a connected space cannot "jump" from one value to another without passing through every intermediate value. In fact, for subsets of $\mathbb{R}$, the Intermediate Value Theorem is precisely the statement that connected subsets are intervals — a result we establish below. ## Connected Subsets of the Real Line The most concrete setting for connectedness is the real line $\mathbb{R}$ with its standard [topology](/page/Topology). Here, the topological notion of "one piece" coincides exactly with the order-theoretic notion of convexity: a subset of $\mathbb{R}$ is connected if and only if it is an interval. This result is fundamental because it provides the bridge between the abstract topological definition and the classical Intermediate Value Theorem from real analysis. It also serves as the base case for many inductive arguments about connectedness in $\mathbb{R}^n$. [quotetheorem:295] The "if" direction (intervals are connected) is the essence of the Intermediate Value Theorem. The "only if" direction (connected subsets are intervals) uses the completeness of $\mathbb{R}$: if $X$ contains points $a < b$ but misses some $c \in (a, b)$, then $X \cap (-\infty, c)$ and $X \cap (c, \infty)$ form a separation of $X$. This characterisation fails in the rationals $\mathbb{Q}$. The set $\mathbb{Q}$ is totally disconnected: the only connected subsets are singletons and the empty set. Indeed, for any two rationals $a < b$, the irrational number $\sqrt{2}$ (or any irrational between $a$ and $b$) provides a point $c \notin \mathbb{Q}$ that splits $\{q \in \mathbb{Q} : a \le q \le b\}$ into two relatively open halves $\{q : a \le q < c\}$ and $\{q : c < q \le b\}$. The completeness of $\mathbb{R}$ — the fact that every bounded set has a supremum — is what prevents this splitting. [example: A Connected Subset of $\mathbb{R}$ That Is Not Path-Connected Does Not Exist] In $\mathbb{R}$, connectedness and path-connectedness coincide for all subsets. If $X \subset \mathbb{R}$ is connected, then $X$ is an interval, and any two points $a, b \in X$ are joined by the path $\gamma: [0, 1] \to X$ defined by $\gamma(t) = (1-t)a + tb$. The image of $\gamma$ is the interval $[\min(a,b), \max(a,b)] \subset X$ (since $X$ is convex). This is a special feature of $\mathbb{R}$. In $\mathbb{R}^2$, the topologist's sine curve (discussed above) shows that the two notions genuinely differ. [/example] ## Preservation Under Continuous Maps A central theme in topology is that continuous maps preserve topological structure. For connectedness, this takes a particularly clean form: the continuous image of a connected space is connected. This is one of the most frequently used results in all of topology — it allows us to establish connectedness of complicated spaces by exhibiting them as continuous images of simpler ones. [quotetheorem:296] The hypothesis that $f$ is continuous cannot be weakened to measurable or Borel. The indicator function $\mathbb{1}_{\mathbb{Q}}: \mathbb{R} \to \{0, 1\}$ maps the connected space $\mathbb{R}$ onto the discrete (disconnected) space $\{0, 1\}$, and it is Borel measurable but not continuous. Note also that the theorem says nothing about the *preimage* of a connected set. The preimage of a connected set under a continuous map need not be connected: the map $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x^2$ sends $\mathbb{R}$ to $[0, \infty)$, and the preimage of the connected set $\{1\}$ is $\{-1, 1\}$, which is disconnected. This theorem immediately yields a powerful method for proving connectedness: to show that a space $Y$ is connected, find a connected space $X$ and a continuous surjection $f: X \to Y$. [example: The $n$-Sphere Is Connected] For $n \ge 1$, the unit sphere $S^n = \{x \in \mathbb{R}^{n+1} : |x| = 1\}$ is connected. Consider the map $f: \mathbb{R}^{n+1} \setminus \{0\} \to S^n$ defined by $f(x) = x / |x|$. This map is continuous (as a composition of continuous functions) and surjective. Since $n \ge 1$, the domain $\mathbb{R}^{n+1} \setminus \{0\}$ is path-connected (hence connected): any two points $a, b \in \mathbb{R}^{n+1} \setminus \{0\}$ can be joined by a polygonal path avoiding the origin (if the straight-line segment from $a$ to $b$ passes through the origin, detour via a third point not on the line through $a$ and $b$; such a point exists since $n + 1 \ge 2$). Therefore $S^n = f(\mathbb{R}^{n+1} \setminus \{0\})$ is the continuous image of a connected space, hence connected. For $n = 0$, $S^0 = \{-1, 1\} \subset \mathbb{R}$ is a two-point discrete space, which is disconnected. This matches the fact that $\mathbb{R} \setminus \{0\}$ is disconnected (it has two connected components, $(-\infty, 0)$ and $(0, \infty)$). [/example] ## Closure, Union, and Algebraic Operations Connectedness is preserved under several natural set-theoretic operations: taking closures, forming unions with common points, and (as we shall see in a later section) taking products. These results provide a toolkit for building connected sets from simpler connected pieces. ### Closure Preserves Connectedness A natural question arises when working with connected subsets: if we add limit points to a connected set, can we accidentally introduce a separation? The answer is no. [quotetheorem:297] The "more general" statement is important: we need not take the full closure. Any set squeezed between a connected set and its closure inherits connectedness. This is because the key property of $Y$ being dense in $Z$ forces any continuous integer-valued function on $Z$ to be constant: it must be constant on the dense subset $Y$ (by connectedness of $Y$), and continuity then forces it to be constant on all of $Z$. This result was already used implicitly in the opening example: the topologist's sine curve $S$ is the closure of the graph $\{(x, \sin(1/x)) : x \in (0, 1]\}$. The graph is connected (as the continuous image of the connected interval $(0, 1]$), so its closure $S$ is connected by this theorem. ### Unions of Overlapping Connected Sets The union of two connected sets need not be connected: $(-\infty, 0)$ and $(0, \infty)$ are both connected, but their union $\mathbb{R} \setminus \{0\}$ is not. The failure is that the two sets are disjoint — there is no "bridge" between them. When connected sets do overlap, their union remains connected. [quotetheorem:298] The hypothesis requires *pairwise* non-empty intersection. A weaker condition — that the union has no separation — is not sufficient without further structure. However, a useful variant weakens pairwise intersection to a "chain" condition: if $\mathcal{A} = \{A_\alpha\}_{\alpha \in \Lambda}$ is a family of connected sets with the property that there exists a single $A_0 \in \mathcal{A}$ meeting every other member ($A_0 \cap A_\alpha \neq \varnothing$ for all $\alpha$), then $\bigcup_\alpha A_\alpha$ is connected. This is because the pairwise intersection condition can be replaced with a common "hub." [example: $\mathbb{R}^n$ Is Connected for $n \ge 1$] We verify that $\mathbb{R}^n$ is connected using the union theorem. For each $v \in \mathbb{R}^n$, the line $L_v := \{tv : t \in \mathbb{R}\}$ through the origin is homeomorphic to $\mathbb{R}$ (via the map $t \mapsto tv$), hence connected. Every two such lines share the origin: $L_v \cap L_w \ni 0$ for all $v, w$. Since $\mathbb{R}^n = \bigcup_{v \in \mathbb{R}^n} L_v$ and the family satisfies pairwise intersection (each line contains $0$), $\mathbb{R}^n$ is connected. Alternatively, the same argument works with the family of line segments from the origin to each point: the segment from $0$ to $x$ is the continuous image of $[0,1]$ under $t \mapsto tx$, hence connected, and all these segments share the origin. [/example] ## Connected Components Given an arbitrary topological space, different "pieces" of the space can be made precise through the notion of connected components. The fundamental question is: can every topological space be canonically decomposed into maximal connected pieces? The answer is yes, and the decomposition is unique, but the resulting pieces may fail to be open — a subtlety that motivates the notion of local connectedness discussed later. [definition: Connected Component] Let $X$ be a topological space and $x \in X$. The **connected component** of $x$, denoted $C_x$, is the union of all connected subsets of $X$ containing $x$: \begin{align*} C_x := \bigcup \{A \subset X : A \text{ is connected and } x \in A\}. \end{align*} [/definition] This definition makes sense because the collection being unioned is non-empty (it contains $\{x\}$, which is connected). By the union theorem for overlapping connected sets (every set in the collection contains $x$, so any two of them intersect), $C_x$ is itself connected. It is therefore the largest connected subset of $X$ containing $x$. [quotetheorem:302] Property (5) — that connected components are closed — follows from the fact that the closure of a connected set is connected: $\overline{C_x}$ is connected and contains $x$, so by maximality $\overline{C_x} \subset C_x$, forcing $C_x = \overline{C_x}$. A crucial point: connected components need **not** be open. In a space where components are open, the components form a clopen partition, and the space decomposes cleanly. But in general, the components can be quite pathological. [example: Connected Components of the Cantor Set] The **Cantor set** $\mathcal{C} \subset [0, 1]$ is constructed by iteratively removing open middle thirds: \begin{align*} \mathcal{C} := [0,1] \setminus \bigcup_{n=0}^{\infty} \bigcup_{k=0}^{3^n - 1} \left(\frac{3k+1}{3^{n+1}}, \frac{3k+2}{3^{n+1}}\right). \end{align*} Equivalently, $\mathcal{C}$ consists of all $x \in [0, 1]$ whose base-3 expansion uses only the digits $0$ and $2$. We claim that the connected component of every point $x \in \mathcal{C}$ is $\{x\}$ itself — that is, $\mathcal{C}$ is **totally disconnected**. To see this, let $x, y \in \mathcal{C}$ with $x \neq y$. We must show that $\{x, y\}$ is not contained in any connected subset of $\mathcal{C}$. Since $x \neq y$, the open interval $(x, y)$ has positive length. In particular, $(x, y)$ contains one of the removed open intervals (at some stage of the Cantor construction, the interval between $x$ and $y$ gets split). That is, there exists $c \in (x, y) \setminus \mathcal{C}$. Then: \begin{align*} \mathcal{C} \cap (-\infty, c) \quad \text{and} \quad \mathcal{C} \cap (c, \infty) \end{align*} are two disjoint non-empty relatively open subsets of $\mathcal{C}$ (they are open in the subspace topology because $(-\infty, c)$ and $(c, \infty)$ are open in $\mathbb{R}$), and $x$ lies in the first while $y$ lies in the second. Any connected subset of $\mathcal{C}$ containing both $x$ and $y$ would inherit this separation — contradicting connectedness. Therefore every connected component is a singleton, and $\mathcal{C}$ is totally disconnected. Despite this, $\mathcal{C}$ is uncountable (it bijects with $[0,1]$ via the map sending a base-3 expansion in digits $\{0, 2\}$ to a base-2 expansion), compact (closed and bounded in $\mathbb{R}$), and has no isolated points (every point is a limit point). It is homeomorphic to the product $\{0, 1\}^{\mathbb{N}}$. [/example] The Cantor set example shows that a space can be uncountable and have no isolated points while still being totally disconnected. This is in sharp contrast with connected spaces, where the absence of isolated points is a minimal requirement (a connected Hausdorff space with more than one point has no isolated points). ## Path-Connectedness The topological definition of connectedness captures the algebraic idea that a space cannot be split into open halves. But there is a second, more geometric notion: can any two points be joined by a continuous curve? This idea — **path-connectedness** — is stronger than connectedness and more directly useful in many applications, particularly in algebraic topology and complex analysis. [definition: Path-Connected Space] A topological space $X$ is **path-connected** if for every pair of points $x, y \in X$, there exists a continuous map $\gamma: [0, 1] \to X$ with $\gamma(0) = x$ and $\gamma(1) = y$. Such a map $\gamma$ is called a **path** from $x$ to $y$. [/definition] The relationship between the two notions is asymmetric: path-connectedness always implies connectedness, but the converse fails in general. [quotetheorem:300] The implication is a direct consequence of the preservation theorem: if $X$ were disconnected, there would exist a continuous surjection $f: X \to \{0, 1\}$. For any path $\gamma: [0, 1] \to X$ from a point in $f^{-1}(0)$ to a point in $f^{-1}(1)$, the composition $f \circ \gamma: [0, 1] \to \{0, 1\}$ would be a continuous surjection from $[0, 1]$ onto $\{0, 1\}$ — but $[0, 1]$ is connected, so continuous maps from $[0, 1]$ to a discrete space must be constant. This is a contradiction. The topologist's sine curve (from the opening example) is the standard counterexample showing that the converse fails: it is connected but not path-connected. ### When Do the Two Notions Coincide? The gap between connectedness and path-connectedness is closed in many natural settings. The most important result is that for open subsets of Euclidean space, the two notions are equivalent. [quotetheorem:301] The key to this result is the local structure of $\mathbb{R}^n$: every point has an open ball neighbourhood, and open balls are convex (hence path-connected). This allows one to build paths locally and piece them together. More precisely, fix a point $a \in U$ and let $W = \{x \in U : \text{there exists a path in } U \text{ from } a \text{ to } x\}$. One verifies that $W$ is both open and closed in $U$. It is open because if $x \in W$ and $B(x, r) \subset U$, then every point of $B(x, r)$ can be reached by concatenating the path from $a$ to $x$ with a straight-line segment in $B(x, r)$. It is closed because if $x \in U \setminus W$ and $B(x, r) \subset U$, then no point of $B(x, r)$ can be in $W$ (otherwise we could reach $x$ by a path through $B(x, r)$). Since $U$ is connected and $W$ is non-empty (it contains $a$), we conclude $W = U$. This argument fails for non-open sets: the topologist's sine curve is a closed (hence not open) connected subset of $\mathbb{R}^2$ that is not path-connected. The crucial property used is that $U$ has a basis of path-connected open sets — a property that $\mathbb{R}^n$ enjoys but that general topological spaces need not. [example: The Deleted Comb Space] Consider the "comb space" in $\mathbb{R}^2$: the horizontal base $[0, 1] \times \{0\}$ together with vertical teeth at $x = 1/n$: \begin{align*} Y := ([0, 1] \times \{0\}) \cup \bigcup_{n=1}^{\infty} \left\{\frac{1}{n}\right\} \times [0, 1]. \end{align*} This set is path-connected: every point $(1/n, t)$ on a tooth can be joined to the origin by travelling down the tooth to $(1/n, 0)$ and then along the base. Now form the **deleted comb space** by adjoining the single limit point $(0, 1)$: \begin{align*} X := Y \cup \{(0, 1)\}. \end{align*} The point $(0, 1)$ is a limit point of $Y$: the tooth tips $(1/n, 1) \in Y$ satisfy $(1/n, 1) \to (0, 1)$ as $n \to \infty$. Therefore $X \subset \overline{Y}$, and since $Y \subset X \subset \overline{Y}$ with $Y$ connected, the closure theorem guarantees that $X$ is connected. However, $X$ is **not** path-connected. We show that no continuous path in $X$ joins $(0, 1)$ to any point of $Y$. Suppose for contradiction that $\gamma: [0, 1] \to X$ is a path with $\gamma(0) = (0, 1)$ and $\gamma(1) \in Y$. Write $\gamma(s) = (\gamma_1(s), \gamma_2(s))$. The points of $X$ with first coordinate $0$ are $(0, 0)$ (which lies on the base $[0, 1] \times \{0\} \subset Y$) and $(0, 1)$ (the adjoined limit point). The segment $\{0\} \times (0, 1)$ is not in $X$, so the only point with $\gamma_1 = 0$ and $\gamma_2 > 0$ is $(0, 1)$ itself. For the path to leave $(0, 1)$, it must eventually reach a tooth $\{1/n\} \times [0, 1]$ or the base. Let $s_0 = \inf\{s \in [0, 1] : \gamma(s) \neq (0, 1)\}$. For $s$ slightly larger than $s_0$, $\gamma(s) \neq (0, 1)$, so $\gamma(s)$ lies in $Y$. Any point of $Y$ with $\gamma_2(s) > 0$ must lie on some tooth $\{1/n\} \times [0, 1]$. As $s \to s_0^+$, continuity of $\gamma$ requires $\gamma(s) \to (0, 1)$, so $\gamma_2(s) \to 1$ and $\gamma_1(s) \to 0$. Since $\gamma_1(s) > 0$ for $s$ near $s_0$ with $\gamma_2(s)$ near $1$ (the path is near the tops of the teeth), the path must visit teeth $\{1/n\} \times [0, 1]$ for arbitrarily large $n$ (because $\gamma_1(s) \to 0$). But each tooth is a closed segment, and moving from one tooth to another requires passing through the base $[0, 1] \times \{0\}$ (where $\gamma_2 = 0$). Thus $\gamma_2$ would need to oscillate between values near $1$ (on the upper parts of the teeth) and $0$ (on the base) infinitely often as $s \to s_0^+$. This contradicts the continuity of $\gamma_2$ at $s_0$, since $\gamma_2(s_0) = 1$. The deleted comb space therefore provides another example — distinct from the topologist's sine curve — of a connected space that is not path-connected, reinforcing that the equivalence of the two notions genuinely requires the set to be open. [/example] ## Products of Connected Spaces A fundamental question in the study of product spaces is whether connectedness is a productive property: if each factor is connected, is the product connected? The answer is affirmative, and it holds for arbitrary (not just finite) products — a result that is considerably more difficult than the finite case. ### Finite Products The finite case admits a direct proof using the union theorem. [quotetheorem:299] The idea is to express $X \times Y$ as a union of connected "crosses." Fix a point $(a, b) \in X \times Y$. For each $x \in X$, the "cross" through $(x, b)$ is: \begin{align*} T_x := (X \times \{b\}) \cup (\{x\} \times Y). \end{align*} Each $T_x$ is connected: it is the union of two connected sets ($X \times \{b\} \cong X$ and $\{x\} \times Y \cong Y$) that share the point $(x, b)$. Moreover, every pair $T_x, T_{x'}$ intersects at $(x, b)$ and $(x', b)$ (both lie on $X \times \{b\}$, which is contained in every $T_x$). Since $X \times Y = \bigcup_{x \in X} T_x$ and the family has pairwise non-empty intersection, the union theorem gives connectedness. By induction, any finite product $X_1 \times X_2 \times \cdots \times X_k$ of connected spaces is connected. ### Arbitrary Products The generalisation to arbitrary products requires a more delicate argument. The difficulty is that in an infinite product $\prod_{\alpha \in A} X_\alpha$ with the [product topology](/page/Product%20Topology), the basic open sets constrain only finitely many coordinates. One cannot simply induct. [quotetheorem:963] The standard approach proceeds as follows. Fix a point $a = (a_\alpha)_{\alpha \in A} \in \prod X_\alpha$. For each finite subset $F \subset A$ and each choice of coordinates $(x_\alpha)_{\alpha \in F}$, define the "slice" $S_F$ that agrees with $a$ outside $F$ and is free to vary within $F$: \begin{align*} S_F := \left\{(y_\alpha) \in \prod_{\alpha \in A} X_\alpha : y_\alpha = a_\alpha \text{ for all } \alpha \notin F\right\}. \end{align*} Each $S_F$ is homeomorphic to $\prod_{\alpha \in F} X_\alpha$, which is connected by the finite case. All slices contain the point $a$, so the union $\bigcup_F S_F$ (over all finite $F \subset A$) is connected by the union theorem. This union is dense in $\prod X_\alpha$ because the product topology has a basis of sets constraining only finitely many coordinates. Since the closure of a connected set is connected, and the closure of a dense set is the whole space, $\prod X_\alpha$ is connected. This theorem would fail if the product were equipped with the box topology instead of the product topology. In the box topology, the product $\mathbb{R}^{\mathbb{N}} = \prod_{n=1}^{\infty} \mathbb{R}$ is disconnected: the set of bounded sequences and its complement are both open in the box topology (each is a union of boxes), providing a separation. This illustrates why the product topology — which only constrains finitely many coordinates at a time — is the "correct" topology for infinite products. ## Local Connectedness Connected components are always closed, but they need not be open. When a space has the property that connected components of open sets are again open, the local structure is much better behaved. This leads to the notion of local connectedness. The motivation comes from pathological examples. The topologist's sine curve is connected, but it is not "connected in a neighbourhood" of the points on the vertical segment $\{0\} \times [-1, 1]$: every sufficiently small open neighbourhood of $(0, 0)$ in $S$ is disconnected (it splits into pieces along the wild oscillation). This suggests that connectedness should have a *local* counterpart. [definition: Locally Connected Space] A topological space $X$ is **locally connected** if for every point $x \in X$ and every open set $U$ containing $x$, there exists a connected open set $V$ with $x \in V \subset U$. Equivalently, $X$ is locally connected if and only if the connected components of every open subset of $X$ are open. [/definition] Neither local connectedness nor connectedness implies the other: - $\mathbb{R}^n$ is both connected and locally connected. - The topologist's sine curve is connected but **not** locally connected. - The discrete space $\{0, 1\}$ (with the discrete topology) is locally connected but **not** connected. - The rationals $\mathbb{Q}$ are neither connected nor locally connected. [example: The Topologist's Sine Curve Is Not Locally Connected] Returning to the topologist's sine curve $S$ from the opening, consider the point $p = (0, 0) \in S$. We show that no connected open neighbourhood of $p$ exists within $S$. Take any open set $U$ in $\mathbb{R}^2$ containing $p$, and consider $V = U \cap S$. Since $U$ is open, there exists $\varepsilon > 0$ such that $B(p, \varepsilon) \subset U$. The portion of $S$ inside $B(p, \varepsilon)$ consists of: - The segment $\{0\} \times (-\varepsilon, \varepsilon)$ (part of the vertical segment), and - Infinitely many arcs of the graph $\{(x, \sin(1/x))\}$ for small $x > 0$. These arcs enter and leave $B(p, \varepsilon)$, creating infinitely many "disconnected strands" near the vertical segment. More precisely, for small $\delta > 0$, the set $S \cap (\{x : 0 < x < \delta\} \times \mathbb{R}) \cap B(p, \varepsilon)$ consists of infinitely many disjoint arcs. The connected component of $p$ in $V$ is contained in $\{0\} \times (-\varepsilon, \varepsilon)$, which does not contain any point with positive first coordinate. Therefore $V$ is not connected (it contains points of positive first coordinate that are separated from $p$ within $V$). Since every neighbourhood of $p$ in $S$ is disconnected, $S$ is not locally connected at $p$. [/example] Local connectedness has an important consequence for the structure of connected components. [quotetheorem:1003] This means that in a locally connected space, the partition into connected components is a partition into clopen sets. Each component is an "open-closed island," cleanly separated from the others. This is the best possible decomposition: the components form a discrete partition of $X$. ## Standard Arguments in Connectedness Working with connectedness requires a small set of recurring techniques. This section collects the standard methods and explains when each is appropriate. ### Showing a Space Is Connected: The Clopen Argument The most common method for proving that a space $X$ is connected proceeds by contradiction. Suppose $U \subset X$ is a non-empty clopen set; the goal is to show $U = X$. **Template.** Let $U \subset X$ be clopen and non-empty. Using the specific structure of $X$ (e.g., a metric, a group operation, a continuous surjection from a known connected space), show that $U$ must contain all points of $X$. Conclude $U = X$, so the only clopen sets are $\varnothing$ and $X$. This was the argument used to show that connected open subsets of $\mathbb{R}^n$ are path-connected: the set of points reachable by a path from a fixed base point is shown to be both open and closed, hence equal to the whole space. ### Showing a Space Is Connected: The Continuous Surjection Method To show $Y$ is connected, find a connected space $X$ and a continuous surjection $f: X \to Y$. This was used to show that $S^n$ is connected (as the image of $\mathbb{R}^{n+1} \setminus \{0\}$) and that graphs of continuous functions are connected. ### Showing a Space Is Disconnected: Constructing a Separation To show $X$ is disconnected, exhibit either: 1. A partition of $X$ into two disjoint non-empty open sets, or equivalently, 2. A continuous surjection $f: X \to \{0, 1\}$ (where $\{0, 1\}$ has the discrete topology). Method (2) is often more practical: define a function that is "locally constant" on each piece. [example: $\mathbb{R} \setminus \{0\}$ Is Disconnected] Define $f: \mathbb{R} \setminus \{0\} \to \{0, 1\}$ by: \begin{align*} f(x) := \begin{cases} 0 & \text{if } x < 0, \\ 1 & \text{if } x > 0. \end{cases} \end{align*} This function is continuous because the preimages $f^{-1}(0) = (-\infty, 0)$ and $f^{-1}(1) = (0, \infty)$ are both open in $\mathbb{R} \setminus \{0\}$. Since $f$ is a continuous surjection onto a disconnected space, $\mathbb{R} \setminus \{0\}$ is disconnected. [/example] ### Building Connected Sets: The Chain Method To show that a large set is connected, decompose it into connected pieces that form a "chain" — each piece overlaps the next. This is a direct application of the union theorem. The method is especially useful for showing that manifolds and CW complexes are connected: decompose into connected open cells or chart domains, each overlapping its neighbours. ### Preserving Connectedness Under Limits The intersection of connected sets need not be connected (two overlapping discs can have a disconnected intersection once their overlap shrinks). However, a decreasing sequence of *compact* connected subsets has connected intersection: if $K_1 \supset K_2 \supset \cdots$ are compact and connected, then $\bigcap_{n=1}^{\infty} K_n$ is connected. This is a consequence of [compactness](/page/Compact%20Space) and the finite intersection property. The compactness hypothesis cannot be dropped. Consider the following subsets of $\mathbb{R}^2$: for each $n \ge 1$, define \begin{align*} A_n := \{(x, y) \in \mathbb{R}^2 : x \le -1\} \cup \{(x, y) \in \mathbb{R}^2 : x \ge 1\} \cup \{(x, y) \in \mathbb{R}^2 : -1 \le x \le 1 \text{ and } y \ge n\}. \end{align*} Each $A_n$ is connected: the left half-plane $\{x \le -1\}$ and the right half-plane $\{x \ge 1\}$ are each connected, and the horizontal bridge $\{-1 \le x \le 1, y \ge n\}$ is connected (it is convex). These three pieces share common points — for instance, $(-1, n)$ lies in the left half-plane and the bridge, while $(1, n)$ lies in the right half-plane and the bridge — so their union $A_n$ is connected by the union theorem. The sequence is nested: $A_1 \supset A_2 \supset \cdots$, since the bridge $\{-1 \le x \le 1, y \ge n+1\}$ is contained in $\{-1 \le x \le 1, y \ge n\}$. Each $A_n$ is closed (as a union of closed sets), but not compact (each is unbounded). The intersection is: \begin{align*} \bigcap_{n=1}^{\infty} A_n = \{(x, y) : x \le -1\} \cup \{(x, y) : x \ge 1\}, \end{align*} because the bridge $\{-1 \le x \le 1, y \ge n\}$ contributes no points to the intersection (for any fixed $(x, y)$ with $-1 \le x \le 1$, choosing $n > y$ gives $(x, y) \notin A_n$). This intersection is disconnected: the open strip $\{-1 < x < 1\}$ separates the two half-planes. The function $f(x, y) = \operatorname{sgn}(x)$ restricts to a continuous surjection from $\bigcap A_n$ onto $\{-1, +1\}$. ## Connectedness in Analysis and Geometry The abstract theory of connectedness becomes a concrete tool when applied to specific mathematical settings. In analysis, the most important applications arise from the interaction between connectedness and continuous functions — particularly the fact that continuous real-valued functions on connected spaces satisfy the Intermediate Value Theorem. In geometry and algebra, connectedness constrains the structure of groups and fibres. ### The Intermediate Value Theorem The classical Intermediate Value Theorem from real analysis is a direct consequence of the connectedness of intervals and the preservation of connectedness under continuous maps. [quotetheorem:629] This is an immediate consequence of the theorems already established: $[a, b]$ is a connected subset of $\mathbb{R}$, so $f([a, b])$ is a connected subset of $\mathbb{R}$, hence an interval. Since $f(a)$ and $f(b)$ lie in this interval, every value between them does as well. The importance of this perspective is that it generalises: the Intermediate Value Theorem holds for continuous real-valued functions on *any* connected topological space, not just on intervals. This is precisely the content of the equivalent characterisation of connectedness stated earlier. ### Constancy on Connected Domains A continuous function from a connected space into a discrete space must be constant. This principle — which is the defining property of connectedness from characterisation (3) — has numerous applications: - If $f: U \to \mathbb{R}$ is continuous on a connected open set $U \subset \mathbb{R}^n$ and $f(x) = 0$ for all $x$ in some non-empty open subset $V \subset U$, then $f = 0$ on all of $U$ provided the zero set $f^{-1}(0)$ is also open (which holds if, for example, $f$ is locally constant on $f^{-1}(0)$, or if $f$ is analytic). - A locally constant function on a connected space is globally constant. This is because the level sets of a locally constant function are clopen, and on a connected space the only clopen sets are $\varnothing$ and the whole space. - The rank of a continuous family of matrices is constant on connected components. If $A: X \to \mathbb{R}^{m \times n}$ is continuous and $X$ is connected, then $\operatorname{rank}(A(x))$ is constant on each connected component of the set where it is locally constant. [example: Sign of the Determinant on $\mathrm{GL}_n(\mathbb{R})$] The general linear group $\mathrm{GL}_n(\mathbb{R}) = \{A \in \mathbb{R}^{n \times n} : \det A \neq 0\}$ is an open subset of $\mathbb{R}^{n^2}$ (the preimage of $\mathbb{R} \setminus \{0\}$ under the continuous map $\det$). It is disconnected: the continuous function $\operatorname{sgn} \circ \det: \mathrm{GL}_n(\mathbb{R}) \to \{-1, +1\}$ is a surjection onto a two-point discrete space. The two connected components are: \begin{align*} \mathrm{GL}_n^+(\mathbb{R}) &:= \{A \in \mathbb{R}^{n \times n} : \det A > 0\}, \\ \mathrm{GL}_n^-(\mathbb{R}) &:= \{A \in \mathbb{R}^{n \times n} : \det A < 0\}. \end{align*} These are path-connected (hence connected, and since $\mathrm{GL}_n(\mathbb{R})$ is open in $\mathbb{R}^{n^2}$, connected components are the same as path components). The identity matrix $I$ lies in $\mathrm{GL}_n^+(\mathbb{R})$, and this component is precisely the set of matrices that can be continuously deformed to the identity — a fact that underpins the theory of orientations on manifolds. [/example] ## References - Munkres, J. R., *Topology* (2nd edition, 2000). - Willard, S., *General Topology* (1970). - Engelking, R., *General Topology* (revised edition, 1989). - Kelley, J. L., *General Topology* (1955). - Steen, L. A. and Seebach, J. A., *Counterexamples in Topology* (1978).