---- > [!definition] Definition. ([[connected component]]) > Given a [[topological space]] $X$, define an [[equivalence relation]] on $X$ by setting $x \sim y$ if there is a [[connected]] [[subspace topology|subspace]] containing $x$ and $y$. The [[equivalence class|equivalence classes]] [[partitions are always determined uniquely by equivalence relations|partition]] $X$ into disjoint constituent connected sets, called the **connected components** of $X$. > \ > Each nonempty connected subspace of $X$ intersects exactly one connected component nontrivially. > [!justification] > Reflexivity and symmetry of $\sim$ are obvious; for transitivity note that if $x \sim y$ as witnessed by a connected subspace $A$ and $y \sim z$ as witnessed by a connected subspace $B$, then $x \sim z$ as witnessed by the [[arbitrary union of nontrivially-intersecting connected subspaces is connected|the connected subapce]] $A \cup B$ containing them. > > From [[partitions are always determined uniquely by equivalence relations]] it is clear that the connected components disjoint-union to $X$. If a connected subspace $Y \subset X$, $Y \neq \emptyset$, is [[connected]], then $Y$ intersects exactly one CC nontrivially. For if $Y$ intersects the components $C_{1}$ and $C_{2}$ of $X$, say in points $x_{1}$ and $x_{2}$, then $Y$ witnesses $x_{1} \sim x_{2}$, which in turn mandates $C_{1}=C_{2}$. > Finally, we show the component $C$ must be connected. Choose a point $x_{0}$ of $C$. For each $x \in C$, we know that $x_{0} \sim x$, so there is a connected subspace $A_{x}$ containing $x_{0}$ and $x$. By the result just proved, $A_{x} \subset C$. Hence $C=\bigcup_{x \in C}^{}A_{x}$ and because each ([[connected]]) $A_{x}$ shares the point $x_{0}$ [[arbitrary union of nontrivially-intersecting connected subspaces is connected|their union]], $C$, is [[connected]]. > [!basicproperties] > - The [[connected component]]s of $X$ are [[closed set|closed]] in $X$, since [[adjoining limit points preserves connectedness|closure preserves connectedness]]. (Explicitly, if $C$ is the [[connected component]] of $x \in X$ and $\overline{x} \in \overline{C}$, then $\overline{x}$ belongs to $C$ because there exists a connected subset containing both $x$ and $\overline{x}$, namely $\overline{C}$. So $C=\overline{C}$.) ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```