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A connected component of a topological space $X$ is a nonempty connected subset that is maximal under subset inclusion. If $C$ is a connected component and $C\subseteq D\subseteq X$ with $D$ a connected subset, then $D=C$. Maximality means that no larger connected subset contains $C$; it does not mean largest cardinality among all connected subsets.

For example, the connected components of $[-2,2]\setminus \{0\}$ in the usual subspace topology of the real line are $[-2,0)$ and $(0,2]$. Each is a real interval, and no connected subset of the real line can meet both without containing the missing point $0$. For the finite set $\{-2,0,2\}$ in the same subspace topology, the three connected components are its singleton sets.

Every point $x\in X$ lies in exactly one connected component, constructed by taking the unionAll members contain $x$, so their union is connected by connected union with a connected hub. The union includes every connected subset through $x$, so it is maximal. Equivalently, defineThis is an equivalence relation: singleton sets give reflexivity, symmetry is immediate, and two connected subsets sharing the middle point have a connected union, giving transitivity. The equivalence classes are precisely the connected components. They form a partition of a set: they cover $X$ and any two are either equal or disjoint.^[ref]

Every nonempty connected subset lies within one connected component. Moreover, every connected component is closed in $X$: closure of a connected set makes its closure connected, and maximality forces that closure to equal the original connected component. A connected component need not be open. For example, in $S=\{0\}\cup \{1/n:n=1,2,\dots \}$ with its usual subspace topology from the real line, all connected components are singleton sets, but $\{0\}$ is not open: every neighbourhood of $0$ in $S$ contains some $1/n$.^[ref]

Suppose the union $W$ has a separation $W=A\cup B$ into disjoint, nonempty, relatively open subsets. Each connected $U_j$ must lie entirely in one of them: otherwise its intersections with $A$ and $B$ would separate $U_j$ in its subspace topology. In particular $U_1$ lies on one side, say $A$. Every other $U_j$ intersects $U_1$, so it cannot lie in $B$ and must also lie in $A$. Then $W\subseteq A$, contradicting $B\ne \varnothing$. The union is connected. This is the connected union with a connected hub argument.

For path-connectedness, take any $x\in U_i$ and $y\in U_j$. Choose $a\in U_i\cap U_1$ and $b\in U_j\cap U_1$. There is a path from $x$ to $a$ inside $U_i$, a path from $a$ to $b$ inside $U_1$, and a path from $b$ to $y$ inside $U_j$. Concatenate the three paths on successive thirds of $[0,1]$. They agree at the joining points, so the resulting map is continuous and remains in $W$. The union is path-connected. There need not be a single point common to all the subsets.