[proofplan] We construct a path between any two points $x, y \in \prod_{\alpha \in A} X_\alpha$ by building it coordinatewise: for each $\alpha$, use path-connectedness of $X_\alpha$ to obtain a path $\gamma_\alpha$ from $x_\alpha$ to $y_\alpha$, then assemble the map $\gamma: [0,1] \to \prod_{\alpha \in A} X_\alpha$ by $\gamma(t) = (\gamma_\alpha(t))_{\alpha \in A}$. Continuity of $\gamma$ follows from the [Universal Property of the Product Topology](/theorems/962): a map into a product is continuous if and only if each coordinate projection is continuous. [/proofplan] [step:Choose two arbitrary points and construct coordinatewise paths] Let $x = (x_\alpha)_{\alpha \in A}$ and $y = (y_\alpha)_{\alpha \in A}$ be two points in $\prod_{\alpha \in A} X_\alpha$. For each $\alpha \in A$, the space $X_\alpha$ is path-connected by hypothesis, so there exists a continuous map \begin{align*} \gamma_\alpha: [0,1] &\to X_\alpha \end{align*} with $\gamma_\alpha(0) = x_\alpha$ and $\gamma_\alpha(1) = y_\alpha$. [guided] Each factor $X_\alpha$ contributes one path $\gamma_\alpha$, connecting the $\alpha$-th coordinates of $x$ and $y$. Note that we invoke the Axiom of Choice here: we are choosing, for each $\alpha \in A$, a path $\gamma_\alpha$ from a non-empty set of paths (guaranteed to be non-empty by path-connectedness of $X_\alpha$). When $A$ is uncountable, this choice requires the full Axiom of Choice. This mirrors the situation in the proof of [Arbitrary Products of Connected Spaces](/theorems/963), where the Axiom of Choice is similarly needed to select elements from each factor. [/guided] [/step] [step:Assemble the coordinatewise paths into a single map into the product] Define the map \begin{align*} \gamma: [0,1] &\to \prod_{\alpha \in A} X_\alpha \\ t &\mapsto (\gamma_\alpha(t))_{\alpha \in A}. \end{align*} This is well-defined: for each $t \in [0,1]$ and each $\alpha \in A$, the value $\gamma_\alpha(t) \in X_\alpha$, so $\gamma(t)$ is an element of $\prod_{\alpha \in A} X_\alpha$. By construction, \begin{align*} \gamma(0) &= (\gamma_\alpha(0))_{\alpha \in A} = (x_\alpha)_{\alpha \in A} = x, \\ \gamma(1) &= (\gamma_\alpha(1))_{\alpha \in A} = (y_\alpha)_{\alpha \in A} = y. \end{align*} [/step] [step:Verify continuity of $\gamma$ using the universal property of the product topology] We apply the [Universal Property of the Product Topology](/theorems/962). That theorem states: a map $f: Z \to \prod_{\alpha \in A} X_\alpha$ from a topological space $Z$ into a product (equipped with the product topology) is continuous if and only if $\pi_\alpha \circ f: Z \to X_\alpha$ is continuous for every $\alpha \in A$, where $\pi_\alpha: \prod_{\alpha \in A} X_\alpha \to X_\alpha$ is the projection. We apply this with $Z = [0,1]$ and $f = \gamma$. For each $\alpha \in A$, \begin{align*} \pi_\alpha \circ \gamma: [0,1] &\to X_\alpha \\ t &\mapsto \pi_\alpha((\gamma_\beta(t))_{\beta \in A}) = \gamma_\alpha(t). \end{align*} Thus $\pi_\alpha \circ \gamma = \gamma_\alpha$, which is continuous by construction. Since this holds for every $\alpha \in A$, the universal property gives that $\gamma: [0,1] \to \prod_{\alpha \in A} X_\alpha$ is continuous. Therefore $\gamma$ is a path from $x$ to $y$ in $\prod_{\alpha \in A} X_\alpha$. Since $x$ and $y$ were arbitrary, the product space is path-connected. [guided] The universal property of the product topology is the engine of this proof. It reduces the continuity of a map into an (arbitrarily large) product to the continuity of each coordinate function — a condition we have by construction. Without the universal property, one would need to work directly with the product topology basis: a basic open set in $\prod_{\alpha \in A} X_\alpha$ has the form $\bigcap_{i=1}^{n} \pi_{\alpha_i}^{-1}(U_{\alpha_i})$ for finitely many indices $\alpha_1, \ldots, \alpha_n$ and open sets $U_{\alpha_i} \subset X_{\alpha_i}$. Then \begin{align*} \gamma^{-1}\Bigl(\bigcap_{i=1}^{n} \pi_{\alpha_i}^{-1}(U_{\alpha_i})\Bigr) = \bigcap_{i=1}^{n} \gamma^{-1}(\pi_{\alpha_i}^{-1}(U_{\alpha_i})) = \bigcap_{i=1}^{n} (\pi_{\alpha_i} \circ \gamma)^{-1}(U_{\alpha_i}) = \bigcap_{i=1}^{n} \gamma_{\alpha_i}^{-1}(U_{\alpha_i}), \end{align*} which is a finite intersection of open subsets of $[0,1]$ (since each $\gamma_{\alpha_i}$ is continuous), hence open. This is exactly what the universal property encapsulates. It is worth noting the contrast with connectedness: the proof that [Arbitrary Products of Connected Spaces](/theorems/963) are connected is substantially more difficult and requires a delicate argument (often via finite subproducts and a limiting process). Path-connectedness, by contrast, is *easier* to verify in products because paths are maps from $[0,1]$, and the universal property immediately reduces the problem to individual coordinates. This is one of the rare instances where path-connectedness is more tractable than connectedness. Finally, observe that the box topology would also make this proof work: the construction of $\gamma$ and the verification that each $\pi_\alpha \circ \gamma$ is continuous are unchanged. However, the box topology is strictly finer than the product topology (when $A$ is infinite), so a map continuous into the product is also continuous into the box topology — but not conversely. The universal property we invoked is specific to the product topology. [/guided] [/step]