# Common Connector Criterion for Connectedness of Unions of Topological Subspaces

Recall from the Common Point Criterion for Connectedness of Unions of Topological Subspaces page that if $X$ is a topological space and $\{ A_i \}_{i \in I}$ is an arbitrary collection of connected topological subspaces then if $\displaystyle{\bigcap_{i \in I} A_i \neq \emptyset}$, i.e., there exists a point in common between all of the topological subspaces $A_i$, then the union $\displaystyle{\bigcap_{i \in I} A_i}$ is also a connected space.

The result stated above is sufficient to show that the union of topological subspaces is connected, however, it is not necessary. For example, if we consider the topological space $\mathbb{R}$ with the usual topology and the subspaces $A_n = [n, n + 1]$, then the collection $\{ A_n \}_{n \in \mathbb{N}} = \{ [1, 2], [2, 3], [3, 4], ... \}$ is such that $\displaystyle{\bigcap_{i \in I} A_n = \emptyset}$. However, the union $\displaystyle{\bigcup_{n = 1}^{\infty} A_n = [0, \infty)}$ is connected.

We now develop a similar result to determine whether the union of a collection of topological spaces is connected or not.

Theorem 1: Let $X$ be a topological space and let $\{ A_i \}_{i \in I}$ be an arbitrary collection of connected topological subspaces. If $A_0$ is another connected topological subspace and $A_0 \cap A_i \neq \emptyset$ for all $i \in I$ then $\displaystyle{A_0 \cup \bigcup_{i \in I} A_i}$ is connected. |

**Proof:**Suppose instead that $\displaystyle{A_0 \cup \bigcup_{i \in I} A_i}$ is disconnected. Then there exists a clopen set $\displaystyle{B \subset A_0 \cup \bigcup_{i \in I} A_i}$ such that $B \neq \emptyset$ and $\displaystyle{B \neq A_0 \cup \bigcup_{i \in I} A_i}$.

- Let $j \in I \cup \{ 0 \}$ and suppose that $B \cap A_j \neq \emptyset$ (and/or $B \cap A_0 \neq \emptyset$. We have already established that then $A_j \subseteq B$ (and/or $A_0 \subseteq B$) otherwise we could form a separation of $A_j$ (and/or $A_0$) which contradicts $A_j$ (or $A_0$) from being connected.

- Hence we see that for some $J \subseteq I \cup \{ 0 \}$ that then:

- There are two cases to consider:

- If $A_0 \subseteq B$, then $B \cap A_i \neq \emptyset$ for all $i \in I$. So $\displaystyle{B = A_0 \cup \bigcup_{i \in I} A_i}$ which contradicts $B$ being a proper subset of $\displaystyle{A_0 \cup \bigcup_{i \in I} A_i}$.

- If $A_0 \not \subseteq B$ then this implies that $A_0 \cap B = \emptyset$. But $A_0 \cap A_i \neq \emptyset$ for all $i \in I$, and since $B$ is a union of $A_i$s we have that this is a contradiction.

- In both cases we see that the assumption that $\displaystyle{A_0 \cup \bigcup_{i \in I} A_i}$ is disconnected leads to contradictions. Therefore, if $X$ is a topological space, $\{ A_i \}_{i \in I}$ is an arbitrary collection of connected topological subspaces, and $A_0$ is another connected topological subspace such that $A_0 \cap A_i \neq \emptyset$ for all $i \in I$, then $\displaystyle{A_0 \cup \bigcup_{i \in I} A_i}$ is also connected. $\blacksquare$