A space can fail to be connected for a concrete reason: it may split into two regions that do not touch each other topologically. On the real line, the set $(-\infty,0) \cup (0,\infty)$ has two visible pieces, and a continuous path cannot move from one side to the other without leaving the set. The definition of disconnectedness isolates exactly this kind of separation, but it does so using open sets rather than pictures or paths. The point of the definition is not merely to say that a space looks like two pieces. It gives a test that works in arbitrary topological spaces, where there may be no metric, no coordinates, and no path to draw. A disconnection is a way of writing the whole space as two non-empty open regions that are disjoint. Once such a split exists, continuous and topological constructions often reduce to separate questions on the two pieces, because information can live independently on each side of the separation. [example: The Punctured Real Line] Let $X = \mathbb{R} \setminus \{0\}$ with the [subspace topology](/page/Subspace%20Topology) inherited from $\mathbb{R}$. Define \begin{align*} U = (-\infty,0), \qquad V = (0,\infty). \end{align*} The set $U$ is open in $X$ because \begin{align*} U = X \cap (-\infty,0), \end{align*} and $(-\infty,0)$ is open in $\mathbb{R}$. Similarly, \begin{align*} V = X \cap (0,\infty), \end{align*} and $(0,\infty)$ is open in $\mathbb{R}$, so $V$ is open in $X$. Both sets are non-empty, since $-1 \in U$ and $1 \in V$. They are disjoint because no real number satisfies both $x<0$ and $x>0$. Their union is all of $X$: if $x \in X$, then $x \in \mathbb{R}$ and $x \neq 0$, so either $x<0$ or $x>0$; hence $x \in U \cup V$. Conversely, every point of $U \cup V$ is nonzero, so $U \cup V \subset X$. Therefore \begin{align*} U \cap V = \varnothing, \qquad U \cup V = X. \end{align*} Thus $(U,V)$ is a disconnection of $X$, splitting the punctured real line into its negative and positive halves. This example shows why the chosen space matters. In $\mathbb{R}$, the same two rays fail to cover $0$; in $\mathbb{R}\setminus\{0\}$, that point is no longer present, so the two relatively open rays cover the whole space. [/example] This chapter develops disconnected spaces as the negative side of connectedness. We begin with the definition, then translate it into equivalent forms involving clopen sets and continuous maps. After that we examine subspaces, products, components, and examples where the naive visual intuition is reliable, as well as examples where it is not. ## Definition The most robust way to say that a [topological space](/page/Topological%20Space) has fallen into two pieces is to ask for two open pieces that cover the space without overlap. Openness is the right language because it is intrinsic to topology: it does not require distances, straight lines, or a chosen embedding into Euclidean space. The definition below packages the exact failure that occurred in the punctured real line. [definition: Disconnected Space] A topological space $(X,\tau)$ is disconnected if there exist non-empty open subsets $U,V \subsetneq X$ such that \begin{align*} U \cap V = \varnothing, \qquad U \cup V = X. \end{align*} [/definition] When such sets $U$ and $V$ exist, the pair $(U,V)$ is called a disconnection of $X$. The definition is symmetric in $U$ and $V$; the order of the two pieces carries no mathematical content. A disconnection should be thought of as a topological cut made without leaving boundary residue inside the space. Each point of $X$ belongs to exactly one of the two open regions, and neither region accumulates into the other through points of $X$ that still need to be assigned. The natural complementary question is whether such a cut is impossible, since that is the property preserved by many analytic arguments. [definition: Connected Space] A topological space $(X,\tau)$ is connected if it is not disconnected. [/definition] Connectedness does not say that every two points can be joined by a curve. That stronger idea is [path connectedness](/page/Path%20Connectedness), which requires a supply of continuous maps from intervals. Connectedness is weaker and more primitive: it only says that the open-set structure cannot separate the whole space into two non-empty open parts. The two halves of a disconnection are open by definition, but they are also closed, because each is the complement of the other. This suggests searching for a single subset that already contains the whole splitting information. Such a subset is open enough to be one side of a cut and closed enough to make its complement open. [definition: Clopen Subset] Let $(X,\tau)$ be a topological space. A subset $A \subset X$ is clopen in $X$ if $A$ is open in $X$ and closed in $X$. [/definition] The definition asks us to find two open sets at once, but in applications we often discover only one candidate piece: a region, an endpoint side, or the preimage of a value. To turn that partial information into a disconnection, we need to know exactly when one subset already forces the complementary piece to be open as well. That is the role of the clopen criterion. [quotetheorem:4997] This theorem is the main working test. Instead of searching for two open pieces at once, it is enough to find one non-empty proper region whose complement is also open. The complement supplies the second half of the disconnection. [example: A Two-Point Discrete Space] Let $X=\{a,b\}$ with the [discrete topology](/page/Discrete%20Topology) $\tau=\{\varnothing,\{a\},\{b\},X\}$. Since $\{a\}\in\tau$, the set $\{a\}$ is open in $X$. Its complement in $X$ is \begin{align*} X\setminus\{a\}=\{b\}. \end{align*} Because $\{b\}\in\tau$, the complement of $\{a\}$ is open, so $\{a\}$ is closed in $X$. Hence $\{a\}$ is clopen. It is non-empty because $a\in\{a\}$, and it is proper because $b\in X$ but $b\notin\{a\}$. Now set \begin{align*} U=\{a\}, \qquad V=\{b\}. \end{align*} Both $U$ and $V$ are open in $X$ since $U,V\in\tau$. They are non-empty because $a\in U$ and $b\in V$. Their intersection is empty, since no element of $X$ is equal to both $a$ and $b$: \begin{align*} U\cap V=\{a\}\cap\{b\}=\varnothing. \end{align*} Their union is the whole space: \begin{align*} U\cup V=\{a\}\cup\{b\}=\{a,b\}=X. \end{align*} Thus $(U,V)$ is a disconnection of $X$. This example shows that disconnectedness is not tied to geometric distance; the topology alone separates the two points. [/example] ## Separations and Clopen Sets ### Relative Openness The definition of disconnectedness uses the entire space. In practice, however, one often studies a subset $A \subset X$ with the subspace topology. The same set may be connected as a space in its own right or disconnected inside the larger environment, and the distinction is controlled by relative openness. To discuss subsets without changing notation every time, we need the standard induced topology. This is the topology that remembers exactly which open sets of the ambient space are visible after restricting attention to the subset. Without it, examples such as $[0,1]\setminus\{1/2\}$ would be judged using the wrong open sets. [definition: Subspace Topology] Let $(X,\tau)$ be a topological space and let $A \subset X$. The subspace topology on $A$ is \begin{align*} \tau_A = \{A \cap O : O \in \tau\}. \end{align*} [/definition] A set open in $A$ need not be open in $X$. This is why $(-\infty,0)$ is open in $\mathbb{R}\setminus\{0\}$ and also open in $\mathbb{R}$, while sets such as $[0,1)$ are open in $[0,1]$ but not in $\mathbb{R}$. What should it mean for a subset to split when the ambient space may not split in the same way? We need a relative version of disconnection that measures openness only inside the subset under study. [definition: Separation of a Subspace] Let $(X,\tau)$ be a topological space and let $A \subset X$. A separation of $A$ is a pair $(U,V)$ of non-empty subsets of $A$ such that $U$ and $V$ are open in the subspace topology on $A$ and \begin{align*} U \cap V = \varnothing, \qquad U \cup V = A. \end{align*} [/definition] A separation of $A$ is not required to use sets open in the ambient space. It uses open sets of $A$, because connectedness is a property of the topological space currently under study. This gives the expected criterion for detecting disconnected subspaces. [quotetheorem:8885] This criterion is mostly a translation, but it prevents a common mistake: checking openness in the wrong topology. The ambient space may refuse to separate a set, while the subspace itself separates after points outside it are ignored. [example: A Closed Interval With a Removed Point] Let $A = [0,1]\setminus\{1/2\}$ with the subspace topology inherited from $\mathbb{R}$, and set \begin{align*} U=[0,1/2), \qquad V=(1/2,1]. \end{align*} We first check that these are open in $A$. Since \begin{align*} A\cap(-1,1/2)=([0,1]\setminus\{1/2\})\cap(-1,1/2)=[0,1/2), \end{align*} we have $U=A\cap(-1,1/2)$. The interval $(-1,1/2)$ is open in $\mathbb{R}$, so $U$ is open in $A$ by the definition of the subspace topology. Similarly, \begin{align*} A\cap(1/2,2)=([0,1]\setminus\{1/2\})\cap(1/2,2)=(1/2,1], \end{align*} so $V=A\cap(1/2,2)$, and $V$ is open in $A$ because $(1/2,2)$ is open in $\mathbb{R}$. Both pieces are non-empty, since $0\in U$ and $1\in V$. They are disjoint: if $x\in U\cap V$, then $x<1/2$ because $x\in U$, and $x>1/2$ because $x\in V$, which is impossible. Hence \begin{align*} U\cap V=\varnothing. \end{align*} They cover $A$ as follows. If $x\in A$, then $0\le x\le 1$ and $x\ne 1/2$. Therefore either $x<1/2$, in which case $x\in U$, or $x>1/2$, in which case $x\in V$. Thus $A\subset U\cup V$. Conversely, every point of $U$ and every point of $V$ lies in $[0,1]$ and is not equal to $1/2$, so $U\cup V\subset A$. Therefore \begin{align*} U\cup V=A. \end{align*} Thus $(U,V)$ is a separation of $A$, and $A$ is disconnected. The point of the example is that the openness is relative: neither $[0,1/2)$ nor $(1/2,1]$ is open in $\mathbb{R}$, but each is open after the missing point $1/2$ is removed and $A$ is given its subspace topology. [/example] ### Relative Clopen Sets The clopen formulation also works inside subspaces. A subset can be clopen in $A$ even when it has boundary points in $X$, because those boundary points may lie outside $A$. This gives a compact way to certify a separation after passing to a subspace. [quotetheorem:8886] The relative clopen criterion turns many examples into short computations. Once a missing point or gap makes one side both open and closed within the subspace, disconnectedness follows. ## Continuous Images and Two-Valued Tests ### Binary Observables Disconnectedness is not only detected by subsets. It is also detected by maps. A space is disconnected precisely when it supports a non-constant continuous map to a two-point discrete space. This viewpoint matters because continuous functions are the language through which topology interacts with analysis. To make the target precise, we isolate the simplest disconnected space: two points with every subset open. A continuous map into this space is the same as a topological yes-or-no test whose yes-set and no-set are both open. The target must be discrete so that both answers can be tested by open preimages. [definition: Two-Point Discrete Space] The two-point discrete space is the topological space $(\{0,1\},\tau)$ where \begin{align*} \tau = \{\varnothing,\{0\},\{1\},\{0,1\}\}. \end{align*} [/definition] The two-point discrete space acts as a detector for clopen subsets. The preimage of $\{1\}$ is open because $\{1\}$ is open, and the preimage of $\{0\}$ is open for the same reason. This suggests a test for disconnectedness that avoids naming a separation directly. Instead of asking for two open pieces, ask whether the space admits a continuous map to the two-point discrete space that really uses both values. Such a map packages the two sides as its fibers, so the next result turns the informal detector idea into an exact characterization. A separation is often awkward to carry around because it names two open sets rather than a single object. The useful replacement is a binary observable: points on one side receive one value and points on the other side receive the other value. The question is whether every genuine split produces such an observable, and whether every nonconstant observable back to the two-point discrete space recovers a split. Continuity is exactly the condition that both fibers are open, while non-constancy is exactly the condition that both sides of the split are present. [quotetheorem:294] This theorem rephrases disconnectedness as the existence of a continuous binary observable. If the space is connected, every continuous yes-or-no measurement is forced to give the same answer everywhere. [example: The Sign Test on the Punctured Real Line] Let $X=\mathbb{R}\setminus\{0\}$, and define $f:X\to\{0,1\}$ by $f(x)=0$ when $x<0$ and $f(x)=1$ when $x>0$. Since every $x\in X$ satisfies exactly one of $x<0$ or $x>0$, this defines $f$ on all of $X$. The two-point discrete space has open sets $\varnothing$, $\{0\}$, $\{1\}$, and $\{0,1\}$. Their preimages are \begin{align*} f^{-1}(\varnothing)=\varnothing. \end{align*} Also, \begin{align*} f^{-1}(\{0\})=\{x\in X:f(x)=0\}=(-\infty,0). \end{align*} Because \begin{align*} (-\infty,0)=X\cap(-\infty,0), \end{align*} and $(-\infty,0)$ is open in $\mathbb{R}$, the set $f^{-1}(\{0\})$ is open in $X$ by the subspace topology. Similarly, \begin{align*} f^{-1}(\{1\})=\{x\in X:f(x)=1\}=(0,\infty). \end{align*} Since \begin{align*} (0,\infty)=X\cap(0,\infty), \end{align*} and $(0,\infty)$ is open in $\mathbb{R}$, the set $f^{-1}(\{1\})$ is open in $X$. Finally, \begin{align*} f^{-1}(\{0,1\})=X. \end{align*} Thus the preimage of every [open set](/page/Open%20Set) in the two-point discrete space is open in $X$, so $f$ is continuous. It is not constant because $-1\in X$ and $1\in X$, while \begin{align*} f(-1)=0. \end{align*} and \begin{align*} f(1)=1. \end{align*} There is no continuous extension $g:\mathbb{R}\to\{0,1\}$ preserving these two side values. If $g(0)=0$, then \begin{align*} g^{-1}(\{0\})=(-\infty,0]. \end{align*} This set is not open in $\mathbb{R}$, because every interval $(-\varepsilon,\varepsilon)$ with $\varepsilon>0$ contains positive points not lying in $(-\infty,0]$. If $g(0)=1$, then \begin{align*} g^{-1}(\{1\})=[0,\infty). \end{align*} This set is not open in $\mathbb{R}$, because every interval $(-\varepsilon,\varepsilon)$ with $\varepsilon>0$ contains negative points not lying in $[0,\infty)$. In both cases, continuity fails, so removing $0$ is exactly what makes the sign test continuous. [/example] ### Images of Connected Spaces A connected domain has no internal open split, so a continuous map should not be able to manufacture one after mapping into another space. If the image could be separated, the inverse images of the two separating pieces would pull that separation back to the original domain. This obstruction is the basic reason connectedness is preserved by continuous images. [quotetheorem:296] The contrapositive is not a general test for disconnectedness of the domain, but the theorem is a powerful obstruction: a connected domain cannot be continuously mapped onto a disconnected image. Why does this obstruction recover the familiar [intermediate value theorem](/theorems/180)? Because connected subsets of $\mathbb{R}$ are intervals, so a continuous real-valued function on a connected interval cannot skip a value between two attained values. [quotetheorem:629] The [intermediate value theorem](/theorems/629) says that a continuous real-valued function on an interval cannot jump across a value. In topological language, the image of a connected interval is connected, and connected subsets of $\mathbb{R}$ are intervals. ## Intervals and the Real Line The real line is where the idea is most visible. A subset of $\mathbb{R}$ is connected exactly when it has no gaps, which is to say it is an interval in the order-theoretic sense. This section connects the open-set definition with the familiar geometry of intervals. A gap in a subset of the real line is witnessed by two points of the set with a missing point between them. The interval property rules this out and is the right order-theoretic companion to connectedness. We isolate it because it translates topology into order. [definition: Interval Subset of $\mathbb{R}$] A subset $I \subset \mathbb{R}$ is an interval if for every $x,y \in I$ with $x < y$ and every $z \in \mathbb{R}$ satisfying $x < z < y$, we have $z \in I$. [/definition] This definition includes open intervals, closed intervals, half-open intervals, rays, singletons, and the empty set. For connectedness questions, non-empty intervals are the essential examples. The classification theorem below says that, in $\mathbb{R}$, this order property is exactly the same as connectedness. [quotetheorem:295] This theorem explains why the punctured real line is disconnected: it fails the interval property. It also explains why $[0,1]$, $(0,1)$, $[0,1)$, and $\mathbb{R}$ are connected despite their different endpoint behaviour. [example: The Rational Numbers Are Disconnected] Let $A=\mathbb{Q}$ with the subspace topology inherited from $\mathbb{R}$. Choose the irrational number $\alpha=\sqrt{2}$, and define \begin{align*} U=\mathbb{Q}\cap(-\infty,\alpha), \qquad V=\mathbb{Q}\cap(\alpha,\infty). \end{align*} The set $(-\infty,\alpha)$ is open in $\mathbb{R}$, so $U=A\cap(-\infty,\alpha)$ is open in $A$ by the definition of the subspace topology. Similarly, $(\alpha,\infty)$ is open in $\mathbb{R}$, so $V=A\cap(\alpha,\infty)$ is open in $A$. Both sets are non-empty: $0\in\mathbb{Q}$ and $0<\sqrt{2}$, so $0\in U$; also $2\in\mathbb{Q}$ and $\sqrt{2}<2$, so $2\in V$. They are disjoint, because if $x\in U\cap V$, then $x<\alpha$ from $x\in U$ and $x>\alpha$ from $x\in V$, which is impossible. Hence \begin{align*} U\cap V=\varnothing. \end{align*} They cover $\mathbb{Q}$ because every rational number $q$ satisfies exactly one of $q<\alpha$ or $q>\alpha$: equality $q=\alpha$ cannot occur since $\alpha=\sqrt{2}\notin\mathbb{Q}$. Thus $q\in U$ in the first case and $q\in V$ in the second case, so $\mathbb{Q}\subset U\cup V$. Conversely, every point of $U$ and every point of $V$ is rational by definition, so $U\cup V\subset\mathbb{Q}$. Therefore \begin{align*} U\cup V=\mathbb{Q}. \end{align*} Thus $(U,V)$ is a separation of $\mathbb{Q}$, and $\mathbb{Q}$ is disconnected. This example shows that density in $\mathbb{R}$ does not imply connectedness: although rational numbers lie arbitrarily close to every real number, the irrational cut at $\sqrt{2}$ separates $\mathbb{Q}$ into two relatively open pieces. [/example] The rational numbers are more than disconnected; every pair of distinct rational points can be separated by an irrational cut. Ordinary disconnectedness only asks for one global split, so it misses how completely the rationals fall apart at smaller scales. To measure that stronger failure of connectedness, we ask whether any connected subset can contain two different points. If every attempted two-point connected piece can be cut apart, then the only connected subsets left are singletons and the empty set. That is the condition called total disconnectedness. [definition: Totally Disconnected Space] A topological space $(X,\tau)$ is totally disconnected if every connected subset of $X$ has at most one point. [/definition] Total disconnectedness is a much stronger condition than disconnectedness. A disconnected space may still contain large connected pieces; for example, $[0,1] \cup [2,3]$ is disconnected, but each interval inside it is connected. [example: The Rational Numbers Are Totally Disconnected] Let $C \subset \mathbb{Q}$ be a connected subset. We show that $C$ cannot contain two distinct points. Suppose, for contradiction, that $x,y \in C$ with $x<y$. Choose an irrational number $\alpha$ such that $x<\alpha<y$, and define \begin{align*} U=C\cap(-\infty,\alpha) \end{align*} and \begin{align*} V=C\cap(\alpha,\infty). \end{align*} The set $U$ is open in $C$ because $(-\infty,\alpha)$ is open in $\mathbb{R}$ and $U=C\cap(-\infty,\alpha)$. Similarly, $V$ is open in $C$ because $(\alpha,\infty)$ is open in $\mathbb{R}$ and $V=C\cap(\alpha,\infty)$. Both sets are non-empty: since $x<\alpha$, we have $x\in U$, and since $\alpha<y$, we have $y\in V$. They are disjoint. If $z\in U\cap V$, then $z<\alpha$ because $z\in U$, and $z>\alpha$ because $z\in V$, which is impossible. Hence \begin{align*} U\cap V=\varnothing. \end{align*} They cover $C$. If $z\in C$, then $z\in\mathbb{Q}$. Since $\alpha$ is irrational, $z\ne\alpha$, so either $z<\alpha$ or $z>\alpha$. In the first case $z\in U$, and in the second case $z\in V$. Thus $C\subset U\cup V$. Conversely, since $U\subset C$ and $V\subset C$, we have $U\cup V\subset C$. Therefore \begin{align*} U\cup V=C. \end{align*} So $(U,V)$ is a separation of $C$, contradicting the connectedness of $C$. Thus no connected subset of $\mathbb{Q}$ contains two distinct points. The empty set and every one-point subset are connected because they cannot be written as the union of two non-empty disjoint open subsets. Therefore the connected subsets of $\mathbb{Q}$ are exactly the empty set and the one-point subsets. [/example] ## Components A disconnected space may split once, but the first split is rarely the whole story. After separating a space into two pieces, each piece might split again. Components organize the maximal connected regions that remain after all possible separations have been accounted for. The definition uses maximality rather than a chosen decomposition, because a space can have many different disconnections. A component is the largest connected subset containing a given point. This gives a canonical object attached to each point. [definition: Connected Component] Let $(X,\tau)$ be a topological space and let $x \in X$. The [connected component](/page/Connected%20Component) of $x$ in $X$ is the union of all connected subsets of $X$ that contain $x$. [/definition] The union in the definition is taken over a family of connected sets sharing the point $x$. That common point is what prevents the union from falling apart. This is the local mechanism behind components: each point belongs to one largest connected region, and these regions form the intrinsic connected pieces of the space. Components give a canonical replacement for a disconnection. A disconnection depends on a chosen cut; the component decomposition is intrinsic. [example: Components of Two Disjoint Intervals] Let \begin{align*} X=[0,1]\cup[2,3]\subset\mathbb{R} \end{align*} with the subspace topology. We show that the connected components of $X$ are exactly $[0,1]$ and $[2,3]$. First define \begin{align*} A=X\cap(-1,3/2) \end{align*} and \begin{align*} B=X\cap(3/2,4). \end{align*} Since $(-1,3/2)$ and $(3/2,4)$ are open in $\mathbb{R}$, the sets $A$ and $B$ are open in $X$ by the definition of the subspace topology. We compute \begin{align*} A=([0,1]\cup[2,3])\cap(-1,3/2)=[0,1] \end{align*} because every point of $[0,1]$ lies between $-1$ and $3/2$, while every point of $[2,3]$ is at least $2>3/2$. Similarly, \begin{align*} B=([0,1]\cup[2,3])\cap(3/2,4)=[2,3] \end{align*} because every point of $[2,3]$ lies between $3/2$ and $4$, while every point of $[0,1]$ is at most $1<3/2$. The sets $A$ and $B$ are non-empty, since $0\in A$ and $2\in B$. They are disjoint, because no real number can lie in both $[0,1]$ and $[2,3]$. Thus \begin{align*} A\cap B=\varnothing. \end{align*} They cover $X$, since \begin{align*} A\cup B=[0,1]\cup[2,3]=X. \end{align*} Therefore $(A,B)$ is a separation of $X$. Now let $C\subset X$ be connected. If $C$ met both $[0,1]$ and $[2,3]$, then $C\cap A$ and $C\cap B$ would be non-empty. They are open in $C$ because $A$ and $B$ are open in $X$, and \begin{align*} (C\cap A)\cap(C\cap B)=C\cap(A\cap B)=C\cap\varnothing=\varnothing. \end{align*} Also, \begin{align*} (C\cap A)\cup(C\cap B)=C\cap(A\cup B)=C\cap X=C. \end{align*} So $(C\cap A,C\cap B)$ would be a separation of $C$, contradicting connectedness. Hence every connected subset of $X$ lies entirely in $[0,1]$ or entirely in $[2,3]$. The sets $[0,1]$ and $[2,3]$ are connected because they are intervals in $\mathbb{R}$, by *Connected Subsets of $\mathbb{R}$*. Since no connected subset of $X$ can properly enlarge either interval by adding a point from the other interval, each is a maximal connected subset of $X$. Thus the connected components of $X$ are $[0,1]$ and $[2,3]$. This example shows why components are not the same as points: a disconnected space may have components that are whole connected intervals. [/example] Components often feel like pieces of a space, so one expects them to have some permanence under limits. They need not be open in arbitrary spaces, but they are always closed: if a point lies in the closure of a component, adding that point cannot create a separation from the component. This makes components more rigid than an arbitrary decomposition into connected sets. [quotetheorem:302] This closedness result is often the first structural fact that prevents components from behaving like arbitrary pieces. Extra hypotheses, such as local connectedness, are needed to make components open. ## Products and Sums ### Disjoint Unions Disconnectedness interacts with constructions. If a space is built as a disjoint union, disconnectedness is usually built in. If a space is built as a product, connectedness is more stable than disconnectedness: products of connected spaces remain connected, while a disconnected factor makes the product disconnected when the other factor is non-empty. The construction that deliberately places spaces side by side without contact is the topological disjoint union. It is the categorical version of saying that two systems coexist without sharing points or neighbourhoods. We define it because it is the model example of a space whose disconnectedness comes from its construction. [definition: Topological Disjoint Union] Let $\{(X_i,\tau_i) : i \in I\}$ be a family of topological spaces. The topological disjoint union is the set $\bigsqcup_{i \in I} X_i$ equipped with the topology in which a subset $O \subset \bigsqcup_{i \in I} X_i$ is open if and only if $O \cap X_i$ is open in $X_i$ for every $i \in I$. [/definition] This construction makes each summand open. If at least two summands are non-empty, any one summand and the union of the rest form a disconnection. The theorem below records this built-in split. [quotetheorem:8887] Disjoint unions are the cleanest source of disconnected spaces because the separation is part of the construction. Products are subtler because points in different factors vary simultaneously. ### Products Products require a topology that makes coordinate projections continuous without forcing infinitely many independent open restrictions at once. This choice is what makes products compatible with central compactness and connectedness theorems. Before comparing connected and disconnected factors, we need this topology explicitly. [definition: Product Topology] Let $\{(X_i,\tau_i) : i \in I\}$ be a family of topological spaces. The [product topology](/page/Product%20Topology) on $\prod_{i \in I} X_i$ is the topology generated by sets of the form $\prod_{i \in I} U_i$, where $U_i \in \tau_i$ for every $i \in I$ and $U_i = X_i$ for all but finitely many indices $i$. [/definition] The product topology is not just a technical choice; it is what makes coordinate-wise topology compatible with global connectedness. The next theorem is needed to rule out a tempting but false intuition: adding more coordinates does not automatically introduce more ways to disconnect a space. If every factor resists a binary clopen split, then the whole product still resists one. [quotetheorem:299] This theorem gives a strong warning: a product can be high-dimensional and complicated without being disconnected. Rectangles, cubes, tori, and infinite products of intervals remain connected because no coordinate supplies a clopen split. If one factor already has a separation, the product has a natural candidate for a separation: keep the disconnected coordinate on one side or the other and let all other coordinates vary freely. The only possible obstruction is emptiness in the remaining factors, since an empty companion factor would collapse the whole product. With non-empty factors, the coordinate split becomes a genuine global split. [quotetheorem:8888] The theorem is useful in examples because disconnectedness propagates through non-empty products. A single binary split in one coordinate separates the whole product into two slabs. [example: A Disconnected Cylinder] Let \begin{align*} X=(\mathbb{R}\setminus\{0\})\times S^1, \end{align*} where $S^1=\{(a,b)\in\mathbb{R}^2:a^2+b^2=1\}$ has the subspace topology. Define \begin{align*} U=(-\infty,0)\times S^1 \end{align*} and \begin{align*} V=(0,\infty)\times S^1. \end{align*} The set $(-\infty,0)$ is open in $\mathbb{R}\setminus\{0\}$ because \begin{align*} (-\infty,0)=(\mathbb{R}\setminus\{0\})\cap(-\infty,0), \end{align*} and $(-\infty,0)$ is open in $\mathbb{R}$. Also $S^1$ is open in itself. Hence $U$ is open in $X$ by the product topology. The same argument gives \begin{align*} (0,\infty)=(\mathbb{R}\setminus\{0\})\cap(0,\infty), \end{align*} so $(0,\infty)$ is open in $\mathbb{R}\setminus\{0\}$, and therefore $V$ is open in $X$. Both pieces are non-empty: $(-1,(1,0))\in U$ because $-1<0$ and $(1,0)\in S^1$, while $(1,(1,0))\in V$ because $1>0$ and $(1,0)\in S^1$. They are disjoint. If $(t,p)\in U\cap V$, then $t<0$ from $(t,p)\in U$ and $t>0$ from $(t,p)\in V$, which is impossible. Hence \begin{align*} U\cap V=\varnothing. \end{align*} They cover $X$. If $(t,p)\in X$, then $t\in\mathbb{R}\setminus\{0\}$ and $p\in S^1$. Since $t\ne 0$, either $t<0$ or $t>0$. In the first case $(t,p)\in U$, and in the second case $(t,p)\in V$, so $X\subset U\cup V$. Conversely, every point of $U$ and every point of $V$ has first coordinate in $\mathbb{R}\setminus\{0\}$ and second coordinate in $S^1$, so $U\cup V\subset X$. Therefore \begin{align*} U\cup V=X. \end{align*} Thus $(U,V)$ is a disconnection of $X$. Geometrically, removing the central real-coordinate value $0$ leaves one cylindrical region with negative first coordinate and another with positive first coordinate. [/example] ## Path Connectedness and Failure Modes A common first guess is that disconnected means no path connects certain points. That is often true in familiar Euclidean examples, but it is not the definition. Path connectedness is stronger than connectedness, and its failure does not always imply disconnectedness. The path-based notion is important because many analytic and geometric arguments produce actual curves. It deserves its own definition before we compare it with disconnectedness. The definition asks for a continuous parametrized route between any two points. [definition: Path Connected Space] A topological space $(X,\tau)$ is path connected if for every $x_0,x_1 \in X$ there exists a continuous map $\gamma: [0,1] \to X$ such that $\gamma(0)=x_0$ and $\gamma(1)=x_1$. [/definition] The definition raises the comparison question that motivates this part of the chapter: does the ability to travel by paths force the open-set notion of being in one piece? The next theorem says yes. Any proposed separation would divide the image of a path into two separated pieces, contradicting the connectedness of the interval that parametrizes the path. [quotetheorem:300] This result explains why intervals, convex subsets of $\mathbb{R}^n$, and many manifolds are connected: they are path connected. Yet it does not classify all connected spaces. [example: The Topologist's Sine Curve] Let \begin{align*} G=\{(x,\sin(1/x)):0<x\leq 1\}\subset \mathbb{R}^2. \end{align*} Then \begin{align*} X=G\cup(\{0\}\times[-1,1]). \end{align*} We first show that $X$ is the closure of $G$. Since $-1\leq \sin(1/x)\leq 1$ for every $0<x\leq 1$, every point of $G$ lies in $[0,1]\times[-1,1]$, so every [limit point](/page/Limit%20Point) of $G$ has second coordinate in $[-1,1]$. If $(x,y)$ is a limit point of $G$ with $x>0$, then the map $t\mapsto \sin(1/t)$ is continuous at $x$, so $y=\sin(1/x)$ and $(x,y)\in G$. Thus the only possible limit points of $G$ outside $G$ have the form $(0,y)$ with $y\in[-1,1]$. Conversely, fix $y\in[-1,1]$. Choose $\alpha\in[-\pi/2,\pi/2]$ with $\sin\alpha=y$. For all sufficiently large integers $n$, the number \begin{align*} \theta_n=\alpha+2\pi n \end{align*} is positive, so \begin{align*} x_n=1/\theta_n \end{align*} satisfies $0<x_n\leq 1$ and $x_n\to 0$. Also \begin{align*} \sin(1/x_n)=\sin(\theta_n)=\sin(\alpha+2\pi n)=\sin\alpha=y. \end{align*} Hence $(x_n,\sin(1/x_n))=(x_n,y)\to(0,y)$, so every point of $\{0\}\times[-1,1]$ lies in the closure of $G$. Therefore $X=\overline{G}$. The set $(0,1]$ is connected because it is an interval in $\mathbb{R}$, and the map \begin{align*} h:(0,1]\to\mathbb{R}^2,\qquad h(x)=(x,\sin(1/x)) \end{align*} is continuous. Hence $G=h((0,1])$ is connected by *Continuous Image of a [Connected Space](/page/Connected%20Space)*. Since the closure of a connected subset is connected, $X=\overline{G}$ is connected. We now show that $X$ is not path connected. Suppose there were a continuous path $\gamma:[0,1]\to X$ from $(0,0)$ to a point of $G$, say \begin{align*} \gamma(0)=(0,0),\qquad \gamma(1)=(a,\sin(1/a)) \end{align*} with $0<a\leq 1$. Write \begin{align*} \gamma(t)=(u(t),v(t)). \end{align*} Then $u$ and $v$ are continuous, $u(0)=0$, and $u(1)=a>0$. Let \begin{align*} t_0=\sup\{t\in[0,1]:u(t)=0\}. \end{align*} The set $\{t:u(t)=0\}$ is closed because $u$ is continuous, so $u(t_0)=0$. Also $t_0<1$, and by the definition of $t_0$ we have $u(t)>0$ for every $t\in(t_0,1]$. Therefore, for $t>t_0$, \begin{align*} v(t)=\sin(1/u(t)). \end{align*} Choose $t_1\in(t_0,1]$. Since $u(t_0)=0$ and $u(t_1)>0$, the intermediate value theorem gives points $s_n,r_n\in(t_0,t_1]$ such that \begin{align*} u(s_n)=\frac{1}{\pi/2+2\pi n} \end{align*} and \begin{align*} u(r_n)=\frac{1}{3\pi/2+2\pi n} \end{align*} for all sufficiently large $n$. These target values tend to $0$. If $\varepsilon>0$, then $u$ has a positive minimum on $[t_0+\varepsilon,t_1]$, so for all sufficiently large $n$ the points $s_n$ and $r_n$ must lie in $(t_0,t_0+\varepsilon)$. Hence $s_n\to t_0$ and $r_n\to t_0$. For those same indices, \begin{align*} v(s_n)=\sin(\pi/2+2\pi n)=1 \end{align*} and \begin{align*} v(r_n)=\sin(3\pi/2+2\pi n)=-1. \end{align*} Continuity of $v$ at $t_0$ would force both sequences $v(s_n)$ and $v(r_n)$ to converge to $v(t_0)$, but one sequence is constantly $1$ and the other is constantly $-1$. This contradiction shows that no path in $X$ joins $(0,0)$ to a point of the oscillating graph $G$. Thus $X$ is connected but not path connected. The example separates the open-set notion of being in one piece from the stronger requirement that points can be joined by continuous paths. [/example] There is also a local version of path connectedness that repairs the converse in many familiar spaces. In Euclidean open sets and manifolds, connectedness and path connectedness agree under this local hypothesis. The definition asks for path connected neighbourhoods at arbitrarily small scales. [definition: Locally Path Connected Space] A topological space $(X,\tau)$ is locally path connected if for every $x \in X$ and every open neighbourhood $O \subset X$ of $x$, there exists an open set $P \subset X$ such that $x \in P \subset O$ and $P$ is path connected. [/definition] The definition is designed to answer the failure exposed by the topologist's sine curve: connectedness alone may not produce paths, but perhaps connectedness plus enough local paths does. The next theorem is the precise repair. It says that local path connectedness makes the open-set and path-based pictures agree component by component, which is why connected open subsets of Euclidean spaces behave pathwise. [quotetheorem:1058] For analysis, this theorem explains why open connected subsets of $\mathbb{R}^n$ can usually be treated with polygonal paths and curve-based arguments. The open-set definition remains the foundation, but local geometry supplies paths. ## Beyond and Connected Topics Disconnected spaces sit at the entrance to several larger theories. The first direction is connectedness itself: many arguments in [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology) use connected intervals and continuous images to prove intermediate value results and classify [connected subsets of the real line](/theorems/295). A second direction is functional analysis. In [Cambridge III Functional Analysis](/page/Cambridge%20III%20Functional%20Analysis), disconnectedness appears less as a picture of separated sets and more through clopen subsets, spectra, and idempotent decompositions. A [compact space](/page/Compact%20Space) with non-trivial clopen subsets often gives function algebras that split into direct product pieces. A third direction is complex analysis. Domains in [Cambridge IB Complex Analysis](/page/Cambridge%20IB%20Complex%20Analysis) are usually required to be open and connected, because [analytic continuation](/page/Analytic%20Continuation) and contour deformation behave component by component. If an open set is disconnected, a [holomorphic function](/page/Holomorphic%20Function) on it is simply a compatible collection of holomorphic functions on its components, with no analytic relation between different components. A fourth direction is elementary real analysis. The interval structure developed in [Cambridge IA Analysis Notes](/page/Cambridge%20IA%20Analysis%20Notes) is the one-dimensional model for connectedness: the absence of gaps is exactly what allows limit and continuity arguments to pass through every intermediate value. The final direction is total disconnectedness and zero-dimensional topology. Spaces such as $\mathbb{Q}$, the [Cantor set](/page/Cantor%20Set), and many profinite spaces have enough clopen sets to separate points. They are disconnected in a much stronger sense, and their topology is governed by clopen partitions rather than intervals or paths. ## References Androma, [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology). Androma, [Cambridge III Functional Analysis](/page/Cambridge%20III%20Functional%20Analysis). Androma, [Cambridge IA Analysis Notes](/page/Cambridge%20IA%20Analysis%20Notes). Androma, [Cambridge IB Complex Analysis](/page/Cambridge%20IB%20Complex%20Analysis). James R. Munkres, *Topology* (2000). Stephen Willard, *General Topology* (1970). Walter Rudin, *Principles of Mathematical Analysis* (1976).