Path Connectedness of Arbitrary Topological Products

# Path Connectedness of Arbitrary Topological Products

One very nice property of an arbitrary collection of path connected topological spaces is that the resulting topological product is also path connected as we will prove in the following theorem.

Theorem 1: Let $\{ X_i \}_{i \in I}$ be an arbitrary collection of path connected topological spaces. Then the topological product $\displaystyle{\prod_{i \in I} X_i}$ is path connected. |

**Proof:**For each $j \in I$, let $\displaystyle{p_j : \prod_{i \in I} X_i \to X_j}$ denote the projection maps defined for all $\displaystyle{(x_i)_{i \in I} \in \prod_{i \in I} X_i}$ by:

\begin{align} \quad p_j((x_i)_{i \in I}) = x_j \end{align}

- Now, let $\displaystyle{(x_i)_{i \in I}, (y_i)_{i \in I} \in \prod_{i \in I} X_i}$. Since each $X_j$ is connected, there exists paths $\alpha_j : [0, 1] \to X_j$ such that $\alpha_j(0) = p_j((x_i)_{i \in I}) = x_j$ and $\alpha_j(1) = p_j((y_i)_{i \in I}) = y_j$.

- We now define a path $\displaystyle{\alpha : [0, 1] \to \prod_{i \in I} X_i}$ for all $x \in [0, 1]$ by:

\begin{align} \quad \alpha(x) = (\alpha_j(x))_{j \in I} \end{align}

- Note that $\alpha$ is continuous since each component $\alpha_j$ for all $j \in I$ is continuous. Moreover we see that:

\begin{align} \quad \alpha(0) = (\alpha_j(0))_{j \in I} = (x_j)_{j \in J} = (x_i)_{i \in I} \end{align}

(4)
\begin{align} \quad \alpha(1) = (\alpha_j(1))_{j \in I} = (y_j)_{j \in J} = (y_i)_{i \in I} \end{align}

- So $\displaystyle{\prod_{i \in I} X_i}$ is path connected. $\blacksquare$