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.