# Common Point Criterion for Connectedness of Unions of Topological Subspaces

Recall from the Connected and Disconnected Sets in Topological Spaces page that if $X$ is a topological space then a set $A$ is said to be connected if the topological subspace $A$ is connected. Similarly, $A$ is said to be disconnected if the topological subspace $A$ is disconnected.

Let $X$ be any topological space and suppose that we have an arbitrary collection of connected topological subspaces $\{ A_i \}_{i \in I}$. Then the union $\bigcup_{i \in I} A_i$ need not be connected.

For example, consider the topological space $\mathbb{R}$ and let $A_n = (2n - 1, 2n)$ for all $n \in \mathbb{N}$. Consider the collection of topological spaces $\{ A_n \}_{n \in \mathbb{N}}$:

(1)All of the $A_n$ are clearly connected subspaces. However, $\displaystyle{\bigcup_{n = 1}^{\infty} (2n - 1, 2n)}$ is not a connected subspace since if we set $A = (1, 2)$ and $\displaystyle{B = \bigcup_{n=2}^{\infty} (2n - 1, 2n)}$ then $A$ and $B$ are open in $\displaystyle{\bigcup_{n=1}^{\infty} (2n - 1, 2n)}$ with the subspace topology, $A, B \neq \emptyset$, $A \cap B = \emptyset$, and $\displaystyle{\bigcup_{n=1}^{\infty} (2n - 1, 2n) = A \cup B}$, so $\{ A, B \}$ is a separation of $\displaystyle{\bigcup_{n=1}^{\infty} (2n - 1, 2n)}$.

So when exactly is an arbitrary union of connected subspaces also connected? The following theorem says that if they all share a common point then the union is connected.

Theorem 1: Let $X$ be a topological space and let $\{ A_i \}_{i \in I}$ be an arbitrary collection of connected topological subspaces. If $\displaystyle{\bigcap_{i \in I} A_i \neq \emptyset}$ then $\displaystyle{\bigcup_{i \in I} A_i}$ is connected. |

**Proof:**Suppose instead that $\displaystyle{\bigcup_{i \in I} A_i}$ is instead disconnected. Then by the Clopen Set Criterion for Connected Topological Spaces there exists a clopen set $\displaystyle{B \subset \bigcup_{i \in I} A_i}$ such that $B \neq \emptyset$ and $\displaystyle{B \neq \bigcup_{i \in I} A_i}$.

- Let $j \in I$. Suppose that $B \cap A_j \neq \emptyset$. We will show that then $A_j \subseteq B$.

- Suppose that $A_j \not \subseteq B$. Since $B$ is an open set in $\bigcup_{i \in I} A_i$, then this would imply that $B \cap A_j$ is a nonempty open set in $A_j$. But then since $B$ is also closed, $B^c$ is open and $B^c \cap A_j \neq \emptyset$ (otherwise $A_j \subseteq B$) so $B^c \cap A_j$ is also a nonempty open set in $A_j$. But then $(B \cap A_j) \cap (B^c \cap A_j) = \emptyset$ and $A_j = (B \cap A_j) \cup (B^c \cap A_j)$ so in fact $\{ B \cap A_j, B^c \cap A_j \}$ form a separation of $A_j$ which contradicts $A_j$ from being connected.

- Therefore if $B \cap A_j \neq \emptyset$ then $A_j \subseteq B$. So there exists a $J \subseteq I$ such that:

- Now since $\displaystyle{B \neq \bigcup_{i \in I} A_i }$ we have that there exists a $k \in I$ (more precisely, $k \in I \setminus J$ such that:

- We're given that $\displaystyle{\bigcap_{i \in I} A_i \neq \emptyset}$ though, and $B \cap A_k \neq \emptyset$. Hence from the same explanation above, $\{ B \cap A_k, B^c \cap A_k \}$ form a separation of $A_k$ which contradicts $A_k$ being connected.

- Regardless, the assumption that $\displaystyle{\bigcup_{i \in I} A_i}$ is disconnected leads to a contradiction. Therefore if $\{ A_i \}_{i \in I}$ is an arbitrary collection of connected subspaces such that $\displaystyle{\bigcap_{i \in I} A_i} \neq \emptyset$ then $\displaystyle{\bigcup_{i \in I} A_i}$ is a connected space. $\blacksquare$