Connectedness tells us when a space is made of one piece, but many spaces encountered in [Topology](/page/Topology), [Metric Space](/page/Metric%20Space), and [Continuity](/page/Continuity) are not globally connected. The next natural question is local-to-global in spirit: if the whole space splits apart, what are its largest indivisible pieces? Connected components answer that question. They decompose a [topological space](/page/Topological%20Space) into maximal connected regions, giving a canonical partition that is independent of coordinates, metrics, or chosen paths. The connected component of a point is especially useful because it turns a global property into a pointwise invariant. In analysis, components describe domains of definition for differential equations, isolate pieces of open subsets of Euclidean space, and control where continuous images can move. In algebraic topology, they are the first layer of homotopical information, recorded by the set of path components and compared with connected components through local path connectedness. ## Definition The pointwise problem is to find the whole connected region belonging to a chosen point $x$. Small connected subsets such as $\{x\}$ give too little information, while arbitrary connected subsets through $x$ are not canonical. Taking all connected subsets through $x$ at once produces the largest connected piece forced by the topology. [definition: Connected Component] Let $(X, \tau)$ be a topological space and let $x \in X$. The connected component of $x$ in $X$, denoted $C_x$, is \begin{align*} C_x := \bigcup \{A \subset X : A \text{ is connected and } x \in A\}. \end{align*} [/definition] This definition is primary, but it depends on the ordinary notion of a connected subset. That supporting notion is stated next so the component formula can be read without ambiguity. [definition: Connected Subset] Let $(X, \tau)$ be a topological space. A subset $A \subset X$ is connected if there do not exist nonempty subsets $U,V \subset A$ such that $A = U \cup V$, $U \cap V = \varnothing$, and $U,V$ are open in the [subspace topology](/page/Subspace%20Topology) on $A$. [/definition] The definition uses the subspace topology because a subset should be judged using the open sets it inherits from the ambient space. This makes connectedness intrinsic to $A$ once $A \subset X$ is fixed, and it explains why the union defining $C_x$ searches among connected subsets of $X$. [remark: Empty and Singleton Subsets] The empty subset and every singleton subset of a topological space are connected. For a singleton, there is no way to split its only point between two nonempty disjoint pieces. [/remark] Since $\{x\}$ is connected, the indexing family in the definition of $C_x$ is nonempty and $x \in C_x$. A decomposition is useful only if its pieces can be named independently of the particular point used to find them. Since many different points may produce the same set $C_x$, we also need language for the resulting subsets themselves. [definition: Connected Component of a Space] Let $(X, \tau)$ be a topological space. A subset $C \subset X$ is a connected component of $X$ if there exists $x \in X$ such that $C = C_x$. [/definition] The two definitions are tied together by the pointwise notation: the connected components of $X$ are exactly the distinct sets among the family $(C_x)_{x \in X}$. ## Equivalent Characterisations The union formula is constructive, but in practice one often recognises a component by a maximality property. This shifts attention from the point $x$ to the connected subset itself. [quotetheorem:302] The theorem justifies the common phrase "maximal connected subset." The subtle point is that a union of connected sets need not be connected in general, but it is connected when all the sets in the union share the point $x$. This motivates isolating maximal connectedness as its own recognition criterion, since many arguments identify components by proving that a connected set cannot be enlarged. [definition: Maximal Connected Subset] Let $(X, \tau)$ be a topological space. A subset $C \subset X$ is a maximal connected subset of $X$ if $C$ is connected and whenever $D \subset X$ is connected with $C \subset D$, one has $D = C$. [/definition] The preceding theorem says that connected components and maximal connected subsets are the same objects. The definition is often better for recognition, while the union definition is better for construction. Its limitation is that maximality names the pieces one at a time; it does not yet explain how all points of $X$ are sorted into those pieces. The partition problem asks when two points should belong to the same connected piece. A relation is the right language for this question: it can group points by the existence of a connected subset containing them both, and then its equivalence classes can be compared with the sets $C_x$. This relation is introduced so that the component decomposition can be treated as a genuine partition, not merely as a collection of large connected subsets. [definition: Component Equivalence Relation] Let $(X, \tau)$ be a topological space. Define a relation $\sim$ on $X$ by declaring that $x \sim y$ if there exists a connected subset $A \subset X$ such that $x,y \in A$. [/definition] This relation packages the idea that two points lie in the same connected piece whenever some connected set reaches both of them. The important structural question is whether this grouping is consistent enough to form equivalence classes, since that is what would make components a genuine decomposition of the entire space. [quotetheorem:8865] This equivalence-class description is more than a restatement of maximality. It shows why components can be used as the pieces of a decomposition: every point is assigned to exactly one class, and overlap between two components forces equality rather than partial intersection. The hypothesis is built into the relation through connected subsets; without connectedness, arbitrary large subsets through two points would give no meaningful separation information. In examples, this criterion lets us prove that a proposed list of components is complete by showing that no connected subset can cross from one listed piece to another. ## Standard Examples The most familiar components occur in subsets of the real line. They show that components need not be open in the ambient topology, even when they are large intervals. [example: Components of a Disconnected Subset of the Real Line] Let $X = [0,1] \cup [2,3] \subset \mathbb{R}$ with the subspace topology. We show that the connected components are exactly $[0,1]$ and $[2,3]$. The sets $[0,1]$ and $[2,3]$ are connected by the interval connectedness theorem for subsets of $\mathbb{R}$, and their subspace topologies inherited from $X$ agree with their usual subspace topologies inherited from $\mathbb{R}$. Now let $A \subset X$ be connected. If $A$ met both intervals, choose $a \in A \cap [0,1]$ and $b \in A \cap [2,3]$. Since $a \le 1 < 3/2 < 2 \le b$, the two sets \begin{align*} U := A \cap (-\infty,3/2) \end{align*} and \begin{align*} V := A \cap (3/2,\infty) \end{align*} are nonempty, disjoint, and open in the subspace topology on $A$. Also, every point of $X$ lies either in $[0,1] \subset (-\infty,3/2)$ or in $[2,3] \subset (3/2,\infty)$, so every point of $A \subset X$ lies in $U \cup V$. Hence $A = U \cup V$, which is a separation of $A$, contradicting connectedness. Therefore every connected subset of $X$ is contained entirely in $[0,1]$ or entirely in $[2,3]$. If $x \in [0,1]$, then $[0,1]$ is a connected subset of $X$ containing $x$, and no connected subset containing $x$ can include any point of $[2,3]$, so $C_x = [0,1]$. The same argument with the two intervals exchanged gives $C_x = [2,3]$ for every $x \in [2,3]$. Thus the two intervals are precisely the maximal connected pieces of $X$. [/example] This example is a useful warning: a component is maximal among connected subsets, not necessarily among open subsets. In $X$, the two components are both open and closed in the subspace topology, but in $\mathbb{R}$ each is closed and not open. To describe spaces where the component decomposition has the smallest possible pieces, we need a name for the opposite of having large connected regions. This name is useful because small components do not force the topology to be discrete; they only control the connected subsets. [definition: Totally Disconnected Space] Let $(X, \tau)$ be a topological space. The space $X$ is totally disconnected if every connected component of $X$ is a singleton. [/definition] Total disconnectedness does not mean the topology is discrete. It says connected subsets are small, not that every singleton is open. [example: Rational Numbers Are Totally Disconnected] Let $X = \mathbb{Q} \subset \mathbb{R}$ with the subspace topology. We show that for every $q \in \mathbb{Q}$, the connected component of $q$ in $\mathbb{Q}$ is $\{q\}$. Suppose $A \subset \mathbb{Q}$ contains two distinct rational numbers $a<b$. By density of the irrational numbers in $\mathbb{R}$, choose an irrational number $\alpha$ with $a<\alpha<b$. Define \begin{align*} U := A \cap (-\infty,\alpha) \end{align*} and \begin{align*} V := A \cap (\alpha,\infty). \end{align*} Because $a \in A$ and $a<\alpha$, we have $a \in U$, so $U$ is nonempty. Because $b \in A$ and $\alpha<b$, we have $b \in V$, so $V$ is nonempty. The intervals $(-\infty,\alpha)$ and $(\alpha,\infty)$ are open in $\mathbb{R}$, so $U$ and $V$ are open in the subspace topology on $A$. They are disjoint because no real number can be both less than $\alpha$ and greater than $\alpha$. It remains to check that $A=U \cup V$. If $y \in A$, then $y \in \mathbb{Q}$. Since $\alpha$ is irrational, $y \ne \alpha$. Hence either $y<\alpha$, in which case $y \in U$, or $y>\alpha$, in which case $y \in V$. Thus $A \subset U \cup V$, and the reverse inclusion follows from $U,V \subset A$. Therefore $A=U \cup V$, so $A$ is separated by two nonempty disjoint open subsets and is not connected. Thus every connected subset of $\mathbb{Q}$ has at most one point. Since $\{q\}$ is connected and contains $q$, the largest connected subset of $\mathbb{Q}$ containing $q$ is exactly $\{q\}$. Therefore every connected component of $\mathbb{Q}$ is a singleton, so $\mathbb{Q}$ is totally disconnected. [/example] This boundary case is important in analysis because it separates connectedness from density. The set $\mathbb{Q}$ is dense in $\mathbb{R}$, but its components are singletons. Components defined by connected subsets can be much larger than the regions reachable by continuous curves. To compare these two notions, we introduce the path-based analogue, which asks for an actual continuous path from one point to another. [definition: Path Component] Let $(X, \tau)$ be a topological space and let $x \in X$. The path component of $x$ is the set of all points $y \in X$ for which there exists a continuous map \begin{align*} \gamma: [0,1] &\to X \end{align*} with $\gamma(0)=x$ and $\gamma(1)=y$. [/definition] Every path component lies inside a connected component, because the image of a connected interval under a continuous map is connected. The reverse inclusion can fail without additional local hypotheses. [example: Topologist's Sine Curve] Let \begin{align*} S := \{(x,\sin(1/x)) \in \mathbb{R}^2 : 0 < x \le 1\} \end{align*} and let $X = \overline{S} \subset \mathbb{R}^2$. First identify the closure. Since $\sin(1/x)\in[-1,1]$ for every $0<x\le 1$, every [limit point](/page/Limit%20Point) of $S$ with first coordinate $0$ must lie in $\{0\}\times[-1,1]$. Conversely, if $y\in[-1,1]$, choose $\theta\in[-\pi/2,\pi/2]$ with $\sin\theta=y$, and define \begin{align*} x_n := \frac{1}{\theta+2\pi n} \end{align*} for all sufficiently large $n$ so that $\theta+2\pi n>0$. Then $x_n\to 0$ and \begin{align*} \sin(1/x_n)=\sin(\theta+2\pi n)=\sin\theta=y. \end{align*} Hence $(x_n,\sin(1/x_n))\to(0,y)$, so \begin{align*} X=S\cup(\{0\}\times[-1,1]). \end{align*} The set $S$ is connected because it is the image of the connected interval $(0,1]$ under the continuous map $x\mapsto (x,\sin(1/x))$, and $X=\overline{S}$ is connected because the closure of a connected subset is connected. Therefore $X$ has exactly one connected component. Now suppose, for contradiction, that a path $\gamma:[0,1]\to X$ joins a point of the vertical segment to a point of $S$. Write \begin{align*} \gamma(t)=(u(t),v(t)). \end{align*} Then $u$ and $v$ are continuous, $u(t)\ge 0$ for all $t$, $u(0)=0$, and $u(1)>0$. Let \begin{align*} a:=\sup\{t\in[0,1]:u(t)=0\}. \end{align*} Continuity gives $u(a)=0$, while $u(t)>0$ for every $t\in(a,1]$. Choose $b>a$ with $u(b)>0$. For all sufficiently large $n$, the two positive numbers \begin{align*} r_n:=\frac{1}{\pi/2+2\pi n} \end{align*} and \begin{align*} s_n:=\frac{1}{3\pi/2+2\pi n} \end{align*} satisfy $0<r_n,s_n<u(b)$. By the intermediate value property applied to $u$ on $[a,b]$, there are $t_n,w_n\in(a,b]$ such that $u(t_n)=r_n$ and $u(w_n)=s_n$. Since $\gamma(t_n),\gamma(w_n)\in S$, their second coordinates are forced to be \begin{align*} v(t_n)=\sin(1/r_n)=\sin(\pi/2+2\pi n)=1 \end{align*} and \begin{align*} v(w_n)=\sin(1/s_n)=\sin(3\pi/2+2\pi n)=-1. \end{align*} Also $r_n\to0$ and $s_n\to0$. If a subsequence of $t_n$ converged to some $c>a$, then continuity would give $u(c)=0$, contradicting $u(t)>0$ on $(a,1]$; hence a subsequence satisfies $t_n\to a$, and the same argument gives a subsequence with $w_n\to a$. Continuity of $v$ at $a$ would then give both $v(a)=1$ and $v(a)=-1$, a contradiction. Thus no path in $X$ joins a point of $S$ to a point of $\{0\}\times[-1,1]$. The space has one connected component but more than one path component, so connected components can be strictly larger than path components. [/example] The example explains why [Path Connectedness](/page/Path%20Connectedness) is treated separately from connectedness. Components measure separation by open sets, while path components measure reachability by continuous curves. ## Properties A central reason components matter is that they behave well under continuous maps. A continuous image of a connected set remains connected, so components constrain where maps can send connected pieces. [quotetheorem:4998] This result is often used in contrapositive form: if two points land in different components of $Y$, then they cannot lie in the same connected subset of $X$ mapped continuously into $Y$. For components to be useful as pieces of a space, we also need to know how they sit topologically inside the ambient space. The first general fact is closedness: a component already contains all of its limit points that can be attached without producing a separation. [quotetheorem:302] Closedness should not be confused with openness. Components can fail to be open, and this failure is common in spaces that are not locally connected. [example: Components Need Not Be Open] Let $X=\mathbb{Q}\subset\mathbb{R}$ with the subspace topology. From the computation in the preceding example, every connected subset of $\mathbb{Q}$ has at most one point, so the connected component of any $q\in\mathbb{Q}$ is exactly $\{q\}$. We show that $\{q\}$ is not open in $\mathbb{Q}$. If $\{q\}$ were open in the subspace topology, then there would be an open set $O\subset\mathbb{R}$ such that \begin{align*} \{q\}=O\cap\mathbb{Q}. \end{align*} Since $q\in O$ and $O$ is open in $\mathbb{R}$, there exists $\varepsilon>0$ such that $(q-\varepsilon,q+\varepsilon)\subset O$. Choose a positive integer $n$ with $n>1/\varepsilon$. Then \begin{align*} 0<\frac{1}{n}<\varepsilon. \end{align*} The number $q+1/n$ is rational because $q\in\mathbb{Q}$ and $1/n\in\mathbb{Q}$, and it satisfies \begin{align*} q<q+\frac{1}{n}<q+\varepsilon. \end{align*} Thus $q+1/n\in O\cap\mathbb{Q}=\{q\}$, but $q+1/n\ne q$, a contradiction. Therefore no singleton component $\{q\}$ is open in $\mathbb{Q}$. Connected components in $\mathbb{Q}$ are singletons, so this space has components that are closed by the general closedness theorem for connected components, but not open. [/example] The example shows that openness of components requires more than the definition of connectedness. The missing local condition is that small connected neighbourhoods are available around every point, so that a component contains an open neighbourhood of each of its points. [definition: Locally Connected Space] Let $(X, \tau)$ be a topological space. The space $X$ is locally connected if for every $x \in X$ and every open set $U \subset X$ with $x \in U$, there exists an open connected set $V \subset X$ such that $x \in V \subset U$. [/definition] Local connectedness rules out the kind of fine fragmentation seen in $\mathbb{Q}$ and in oscillatory examples. It makes components visible as open regions, which is exactly what many analytic arguments need when they work component by component on open domains. [quotetheorem:1003] The hypothesis of local connectedness is doing real work here: it supplies connected open neighbourhoods small enough to stay inside a given open set. The conclusion can fail in spaces such as $\mathbb{Q}$, where every component is closed but no singleton component is open. When the hypothesis holds, components are not merely abstract maximal subsets; they are open regions on which local arguments can be run independently. For open subsets of Euclidean space, this becomes even stronger. Open balls in $\mathbb{R}^n$ are path connected, so open subsets of $\mathbb{R}^n$ are locally connected and their components are open. In this setting, one can also ask whether connectedness agrees with the more geometric idea of joining points by continuous paths. [quotetheorem:301] This Euclidean theorem is special to open subsets of $\mathbb{R}^n$. It explains why analysts often move freely between connected components and path-connected domains when working with open regions in Euclidean space. The openness assumption should not be suppressed: outside locally connected or Euclidean-open settings, connected components and path components can differ sharply. At the page level, the final test is conceptual rather than another formal card. A [connected space](/page/Connected%20Space) has a single component, while a disconnected space has more than one maximal connected piece. Thus components refine connectedness rather than replace it: they record exactly how the space breaks apart once global connectedness fails. ## Relationship to Other Concepts Connected components sit between raw topology and geometric reachability. They use only [Open Set](/page/Open%20Set), [Closed Set](/page/Closed%20Set), and subspace topology, so they make sense in every topological space. Path components require maps from intervals, so they carry extra geometric information. In metric spaces, components interact with balls and distances but are not determined by distance alone. A metric space can have many small components despite being dense in a connected completion, as $\mathbb{Q} \subset \mathbb{R}$ shows. This makes components a useful diagnostic for the difference between metric closeness and topological connectedness. In analysis on open sets $U \subset \mathbb{R}^n$, components often become the natural domains on which local arguments become global. A locally constant [continuous function](/page/Continuous%20Function) is constant on each connected component. A differential equation with local uniqueness cannot transport information across different components unless boundary or coupling data connects them externally. In algebraic topology, $\pi_0(X)$ usually denotes the set of path components rather than the set of connected components. For locally path connected spaces, path components and connected components coincide, so $\pi_0(X)$ records the same decomposition. Outside that setting, the topologist's sine curve shows that one connected component can contain several path components. ## References James R. Munkres, *Topology* (2000). John L. Kelley, *General Topology* (1955). Stephen Willard, *General Topology* (1970). [Topology](/page/Topology). [Connectedness](/page/Connectedness). [Path Connectedness](/page/Path%20Connectedness).