Path Connectivity of Countable Unions of Connected Sets

# Path Connectivity of Countable Unions of Connected Sets

Suppose that we have a countable collection $\{ A_i \}_{i=1}^{\infty}$ of path connected sets. Then $\displaystyle{\bigcup_{i=1}^{\infty} A_i}$ need not be path connected as the union itself may not connected. However, if $A_i \cap A_{i+1} \neq \emptyset$ for all $i \in I$ then it reasonable to suspect that $\displaystyle{\bigcup_{i=1}^{\infty} A_i}$ is connected. To prove this we will first need to get some notation out of the way.

Definition: Let $\alpha, \beta : [0, 1] \to X$ be paths such that $\alpha(1) = \beta(0)$. Then a new path denoted $\alpha * \beta : I \to X$ can be defined for all $x \in [0, 1]$ by $\alpha * \beta(x) = \left\{\begin{matrix} \alpha (2x) & \mathrm{if} \: 0 \leq x \leq \frac{1}{2} \\ \beta(2x - 1) & \mathrm{if} \: \frac{1}{2} \leq x \leq 1 \end{matrix}\right.$ |

We now prove the result stated above.

Theorem 1: Let $X$ be a topological space and let $\{ A_i \}_{i = 1}^{\infty}$ be a countable collection of path connected subsets of $X$. If $A_i \cap A_{i+1} \neq \emptyset$ for all $i \in \{1, 2, … \}$ then $\displaystyle{\bigcup_{i=1}^{\infty} A_i}$ is also path connected. |

**Proof:**Let $\{ A_i \}_{i =1}^{\infty}$ be a countable collection of path connected subsets of $X$. For each $i \in \{ 1, 2, …, \infty \}$ take $a_i \in A_i \cap A_{i+1}$, which is possible since $A_i \cap A_{i+1} \neq \emptyset$ for all $i \in \{ 1, 2, … \}$.

- Now since $A_i$ is connected for each $i \in \{ 1, 2, … \}$ there exists paths $\alpha_i : I \to X$ such that $\alpha_i(0) = a_i$ and $\alpha_i(1) = a_{i+1}$.

- Now, let $\displaystyle{x, y \in \bigcup_{i=1}^{\infty} A_i}$ be distinct points. Then there exists a $j, k \in \{ 1, 2, … \}$ such that $x \in A_j$ and $y \in A_k$. Assume without loss of generality that $j < k$. Then there exists paths $\beta_1, \beta_2 : I \to X$ such that $\beta_1(0) = x$ and $\beta_1(1) = a_j$, while $\beta_2(0) = a_k$ and $\beta_2(1) = y$.

- Consider the following path:

\begin{align} \beta_1 * a_j * a_{j+1} * … * a_{k-1} * \beta_2 \end{align}

- Then this is a path from $x$ to $y$. This shows that $\bigcup_{i=1}^{\infty} A_i$ is path connected.