After seeing that closures preserve connectedness, we now learn how connected sets may be joined. Unions of connected subspaces are connected when the pieces overlap in a suitable way. The simplest version says that a family of connected subsets with a common point has connected union. A finite chain version is also useful: each set need not meet every other set, but consecutive overlap is enough. The focus keyword unions of connected subspaces is important because it will later support path-connectedness, components, and connected open sets in the plane. :::theorem[Union of Connected Sets with Common Intersection] Let $(X,d)$ be a metric space and let $\{Y_\lambda:\lambda\in\Lambda\}$ be a family of connected subsets of $X$ such that $\bigcap_{\lambda\in\Lambda}Y_\lambda\neq\varnothing$. Then $Y=\bigcup_{\lambda\in\Lambda}Y_\lambda$ is connected. ::: :::proof Given that each $Y_\lambda$ is connected and that all the sets have a common point. To prove that their union is connected. Choose $p\in\bigcap_{\lambda\in\Lambda}Y_\lambda$. Suppose, if possible, that $Y=\bigcup_{\lambda\in\Lambda}Y_\lambda$ is disconnected. Then there exist nonempty separated subsets $A$ and $B$ of $Y$ such that $Y=A\cup B$. Since $p\in Y$, either $p\in A$ or $p\in B$. Assume, without loss of generality, that $p\in A$. Since $B$ is nonempty, choose $b\in B$. Then $b\in Y_{\lambda_0}$ for some $\lambda_0\in\Lambda$. Also $p\in Y_{\lambda_0}$. Thus $Y_{\lambda_0}\cap A\neq\varnothing$ and $Y_{\lambda_0}\cap B\neq\varnothing$. These two sets form a separation of $Y_{\lambda_0}$. This contradicts the connectedness of $Y_{\lambda_0}$. Hence $Y$ is connected. $\square$ ::: :::corollary Let $Y_1,Y_2,\ldots,Y_n$ be connected subsets of a metric space $(X,d)$ such that $Y_i\cap Y_{i+1}\neq\varnothing$ for $1\leq i\leq n-1$. Then $\bigcup_{i=1}^{n}Y_i$ is connected. ::: :::proof Given that $Y_1,Y_2,\ldots,Y_n$ are connected and consecutive sets intersect. To prove that their union is connected. For $n=1$, the result holds. Assume that $\bigcup_{i=1}^{k}Y_i$ is connected. Since $Y_k\cap Y_{k+1}\neq\varnothing$, we get $\left(\bigcup_{i=1}^{k}Y_i\right)\cap Y_{k+1}\neq\varnothing$. Therefore $\left(\bigcup_{i=1}^{k}Y_i\right)\cup Y_{k+1}$ is connected as a union of two connected sets with common intersection. Hence $\bigcup_{i=1}^{k+1}Y_i$ is connected. Thus, by induction, $\bigcup_{i=1}^{n}Y_i$ is connected. $\square$ ::: :::theorem[Connected Subsets Through Every Pair of Points] Let $X$ be a metric space. If every two points of $X$ are contained in some connected subset of $X$, then $X$ is connected. ::: :::proof Given that every two points of $X$ are contained in a connected subset of $X$. To prove that $X$ is connected. Suppose, if possible, that $X$ is disconnected. Then $X=A\cup B$, where $A$ and $B$ are nonempty separated subsets of $X$. Choose $a\in A$ and $b\in B$. By hypothesis, there exists a connected subset $C$ of $X$ such that $a,b\in C$. Since $A$ and $B$ separate $X$, a connected subset of $X$ must lie entirely in $A$ or entirely in $B$. Therefore either $C\subseteq A$ or $C\subseteq B$. Both alternatives contradict the fact that $a$ and $b$ lie in different separated sets. Hence $X$ is connected. $\square$ ::: :::example[Union of Intersecting Line Segments] In $\mathbb{R}^2$, let $Y_\theta$ be the line segment from the origin to the point $(\cos\theta,\sin\theta)$ for $0\leq\theta\leq\pi$. Each $Y_\theta$ is connected, and every $Y_\theta$ contains the origin. Therefore $\bigcup_{0\leq\theta\leq\pi}Y_\theta$ is connected. $\square$ ::: :::solved-problem Show that the union of the coordinate axes in $\mathbb{R}^2$ is connected. ::: :::solution Given that $Y=(\mathbb{R}\times\{0\})\cup(\{0\}\times\mathbb{R})$. To prove that $Y$ is connected. The $x$-axis $\mathbb{R}\times\{0\}$ is connected because it is homeomorphic to $\mathbb{R}$. The $y$-axis $\{0\}\times\mathbb{R}$ is also connected. Their intersection contains $(0,0)$. Therefore their union is connected by the union theorem. $\square$ ::: :::exercise [1] State the common-intersection theorem for connected unions. [2] State the finite chain corollary. [3] Give an example of two connected subsets with disconnected union. [4] Prove that a space is connected if every two points lie in a connected subset. ::: :::answer [1] A family of connected subsets with nonempty total intersection has connected union. [2] A finite union of connected sets is connected if consecutive sets intersect. [3] In $\mathbb{R}$, the sets $[0,1]$ and $[2,3]$ are connected but their union is disconnected. [4] Any separation would put two points in different pieces, contradicting the connected subset containing both. ::: :::faq[Frequently Asked Questions] Q: Must all sets intersect pairwise? A: No. A common point is sufficient, and a finite chain of intersections is also sufficient. Q: Can the union of connected sets be disconnected? A: Yes. Disjoint connected sets may have disconnected union. Q: Why is the pair-of-points criterion useful? A: It lets us prove connectedness by building connected bridges between arbitrary points. ::: :::call-to-action[Continue to Components] subtitle: Use unions of connected subspaces to define maximal connected pieces called components. button: Next Lesson | /functional-analysis/metric-spaces/connectedness-and-path-structure-in-metric-spaces/components-of-metric-spaces button-ghost: Previous Lesson | /functional-analysis/metric-spaces/connectedness-and-path-structure-in-metric-spaces/closures-and-enlargements-of-connected-sets :::