Common Point Criterion for Connectedness of Unions of Topo. Subs.

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)
\begin{align} \quad \{ A_n \}_{n \in \mathbb{N}} = \{(1, 2), (3, 4), (5, 6), (7, 8), ... \} \end{align}

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:
(2)
\begin{align} \quad B = \bigcup_{i \in J} A_i \end{align}
• 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:
(3)
\begin{align} \quad A_k \not \subseteq B \end{align}
• 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. $\blacksquare$