[proofplan] Fix a basepoint $a = (a_\alpha)_{\alpha \in A}$ in the product. For each finite subset $F \subset A$, the "slice" of points agreeing with $a$ outside $F$ is homeomorphic to the finite product $\prod_{\alpha \in F} X_\alpha$, which is connected (by induction from the two-factor case). Every such slice contains $a$, so their union $Y$ is connected (as a union of connected sets with a common point). We then show that $Y$ is [dense](/page/Dense%20Subset) in $\prod_{\alpha \in A} X_\alpha$: every non-empty basic [open set](/page/Open%20Set) intersects $Y$. Since the [closure of a connected set is connected](/theorems/297), $\overline{Y} = \prod_{\alpha \in A} X_\alpha$ is connected. [/proofplan] [step:Fix a basepoint and define the finite-coordinate slices] Since each $X_\alpha \neq \varnothing$, the Axiom of Choice provides a point $a = (a_\alpha)_{\alpha \in A} \in \prod_{\alpha \in A} X_\alpha$. For each finite subset $F \subset A$, define \begin{align*} Y_F = \Bigl\{x \in \prod_{\alpha \in A} X_\alpha : x_\alpha = a_\alpha \text{ for all } \alpha \in A \setminus F\Bigr\}. \end{align*} The set $Y_F$ consists of all points in the product that agree with $a$ on every coordinate outside $F$, with the coordinates in $F$ free to vary over the corresponding factor spaces. Note that $a \in Y_F$ for every finite $F \subset A$. [guided] The idea behind the slice $Y_F$ is to "unfreeze" finitely many coordinates while keeping the rest pinned at the basepoint $a$. Points in $Y_F$ can differ from $a$ only in the coordinates indexed by $F$. This makes $Y_F$ essentially a copy of the finite product $\prod_{\alpha \in F} X_\alpha$ embedded in the full product — and finite products of connected spaces are connected. The basepoint $a$ serves as the glue: it belongs to every $Y_F$, which will let us apply the union theorem for connected sets sharing a common point. [/guided] [/step] [step:Show each $Y_F$ is connected by identifying it with a finite product] Fix a finite subset $F = \{\alpha_1, \ldots, \alpha_n\} \subset A$. Define the map \begin{align*} \iota_F: \prod_{\alpha \in F} X_\alpha &\to \prod_{\alpha \in A} X_\alpha \\ (x_\alpha)_{\alpha \in F} &\mapsto y, \quad \text{where } y_\alpha = \begin{cases} x_\alpha & \text{if } \alpha \in F, \\ a_\alpha & \text{if } \alpha \in A \setminus F. \end{cases} \end{align*} The image of $\iota_F$ is exactly $Y_F$. We verify that $\iota_F$ is [continuous](/page/Continuity): for each $\beta \in A$, the composition $\pi_\beta \circ \iota_F: \prod_{\alpha \in F} X_\alpha \to X_\beta$ is given by \begin{align*} \pi_\beta \circ \iota_F = \begin{cases} \pi_\beta|_{\prod_{\alpha \in F} X_\alpha} & \text{if } \beta \in F, \\ c_{a_\beta} & \text{if } \beta \in A \setminus F, \end{cases} \end{align*} where $c_{a_\beta}: \prod_{\alpha \in F} X_\alpha \to X_\beta$ is the constant map sending every point to $a_\beta$. Projection maps on finite products are continuous, and constant maps are continuous, so each $\pi_\beta \circ \iota_F$ is continuous. By the [Universal Property of the Product Topology](/theorems/962), $\iota_F$ is continuous. The finite product $\prod_{\alpha \in F} X_\alpha = X_{\alpha_1} \times \cdots \times X_{\alpha_n}$ is connected: each $X_{\alpha_i}$ is connected by hypothesis, and the [product of two connected spaces is connected](/theorems/299), so by induction on $|F|$, the finite product is connected. (The base case $|F| = 1$ holds since $X_{\alpha_1}$ is connected, and the inductive step uses the fact that $\prod_{i=1}^{m+1} X_{\alpha_i} \cong \bigl(\prod_{i=1}^m X_{\alpha_i}\bigr) \times X_{\alpha_{m+1}}$.) Since $\iota_F$ is continuous and $\prod_{\alpha \in F} X_\alpha$ is connected, the [continuous image of a connected space is connected](/theorems/296), so $Y_F = \iota_F\!\bigl(\prod_{\alpha \in F} X_\alpha\bigr)$ is connected. [guided] We need to confirm that the hypotheses of the [Universal Property of the Product Topology](/theorems/962) are satisfied: we require each component $\pi_\beta \circ \iota_F$ to be continuous. When $\beta \in F$, this component is a projection from the finite product $\prod_{\alpha \in F} X_\alpha$ onto the factor $X_\beta$, which is continuous in the product [topology](/page/Topology). When $\beta \notin F$, it is the constant map $c_{a_\beta}$, which is continuous for any topology on the domain. So the universal property applies, and $\iota_F$ is continuous. The connectedness of the finite product follows by induction. The base case is that a single factor $X_\alpha$ is connected (given by hypothesis). For the inductive step, if $\prod_{i=1}^m X_{\alpha_i}$ is connected and $X_{\alpha_{m+1}}$ is connected, then the [product of two connected spaces is connected](/theorems/299) gives that $\prod_{i=1}^{m+1} X_{\alpha_i}$ is connected. Why not just cite the theorem we are proving for the finite case? Because that would be circular. We rely on the two-factor result (theorem 299), which has an independent proof, and extend it by induction. The passage from finite to arbitrary products is the content of this theorem. [/guided] [/step] [step:Show the union $Y = \bigcup_F Y_F$ is connected] Define \begin{align*} Y = \bigcup_{\substack{F \subset A \\ F \text{ finite}}} Y_F. \end{align*} Each $Y_F$ is connected (as shown above), and $a \in Y_F$ for every finite $F \subset A$. In particular, any two sets $Y_F$ and $Y_{F'}$ in this union both contain $a$, so $Y_F \cap Y_{F'} \neq \varnothing$. By the theorem on [unions of overlapping connected sets](/theorems/298), $Y$ is connected. More precisely, we verify the hypothesis of [theorem 298](/theorems/298): the collection $\{Y_F\}_{F \subset A, F \text{ finite}}$ consists of connected subsets of $\prod_{\alpha \in A} X_\alpha$, and for any two members $Y_F, Y_{F'}$, the point $a$ lies in $Y_F \cap Y_{F'}$, so the intersection is non-empty. The theorem concludes that $\bigcup_F Y_F = Y$ is connected. [guided] The union theorem for connected sets (theorem 298) requires a family of connected sets such that every pair has non-empty intersection. Our family is $\{Y_F\}$ indexed by finite subsets $F$ of $A$. The basepoint $a$ belongs to every $Y_F$ (since $a$ agrees with itself on all coordinates), so any two sets in the family share the point $a$. The theorem then gives connectedness of the union $Y$. Note that the pairwise intersection condition is stronger than merely requiring a common point — but here we have the strongest possible version: a single point $a$ that lies in *every* member of the family. [/guided] [/step] [step:Show $Y$ is dense in $\prod_{\alpha \in A} X_\alpha$] We show $\overline{Y} = \prod_{\alpha \in A} X_\alpha$ by proving that every non-empty basic [open set](/page/Open%20Set) in the product [topology](/page/Topology) intersects $Y$. Let $U = \prod_{\alpha \in A} U_\alpha$ be a non-empty basic open set, where $U_\alpha \in \tau_\alpha$ for all $\alpha$ and $U_\alpha = X_\alpha$ for all $\alpha$ outside some finite set $F_0 \subset A$. Since $U$ is non-empty, each $U_\alpha \neq \varnothing$; in particular, for each $\alpha \in F_0$, choose $x_\alpha \in U_\alpha$. Define the point $z \in \prod_{\alpha \in A} X_\alpha$ by \begin{align*} z_\alpha = \begin{cases} x_\alpha & \text{if } \alpha \in F_0, \\ a_\alpha & \text{if } \alpha \in A \setminus F_0. \end{cases} \end{align*} Then $z \in U$ because $z_\alpha \in U_\alpha$ for all $\alpha$ (for $\alpha \in F_0$ by construction, and for $\alpha \notin F_0$ because $U_\alpha = X_\alpha$). Also $z \in Y_{F_0} \subset Y$ because $z_\alpha = a_\alpha$ for all $\alpha \in A \setminus F_0$. Therefore $z \in U \cap Y$, so $U \cap Y \neq \varnothing$. Since every non-empty basic open set intersects $Y$, the set $Y$ is [dense](/page/Dense%20Subset): $\overline{Y} = \prod_{\alpha \in A} X_\alpha$. [guided] Density of $Y$ is the step that bridges finite-coordinate slices and the full product. The argument is constructive: given a basic open set $U$ that constrains coordinates in a finite set $F_0$, we build a point $z$ that satisfies those constraints (by picking $x_\alpha \in U_\alpha$ for $\alpha \in F_0$) and agrees with $a$ elsewhere (ensuring $z \in Y_{F_0}$). This works because basic open sets in the product topology constrain only *finitely many* coordinates. This is the essential feature of the product topology (as opposed to the box topology). In the box topology, a basic open set could constrain all coordinates, and we would not be able to place $z$ in any finite-coordinate slice $Y_F$. Indeed, arbitrary products of connected spaces need *not* be connected in the box topology. [/guided] [/step] [step:Conclude connectedness via the closure of a connected set] We have shown that $Y$ is a connected subset of $\prod_{\alpha \in A} X_\alpha$ and that $\overline{Y} = \prod_{\alpha \in A} X_\alpha$. By the theorem that the [closure of a connected set is connected](/theorems/297), applied with $Y$ connected and $Y \subset \prod_{\alpha \in A} X_\alpha \subset \overline{Y}$, we conclude that $\prod_{\alpha \in A} X_\alpha = \overline{Y}$ is connected. The hypothesis of [theorem 297](/theorems/297) requires $Y$ to be a connected subspace and $\overline{Y}$ to be its closure — both of which have been established. The theorem gives that any set $Z$ with $Y \subset Z \subset \overline{Y}$ is connected; taking $Z = \overline{Y} = \prod_{\alpha \in A} X_\alpha$ yields the result. [guided] The passage from $Y$ to its closure is a standard technique in connectivity arguments: connectedness is preserved under closure because any clopen partition of $\overline{Y}$ restricts to a clopen partition of $Y$, and since $Y$ is connected, one part must be empty, forcing the corresponding part of $\overline{Y}$ to have empty interior in $\overline{Y}$ — hence to be empty by density. To summarise the proof structure: we built $Y$ from finite-coordinate slices (each connected by the finite product theorem), glued them via a common basepoint (applying the union theorem for connected sets), showed $Y$ is [dense](/page/Dense%20Subset) (using the finite-coordinate character of product-[topology](/page/Topology) basic [open sets](/page/Open%20Set)), and passed to the closure (using the closure theorem for connected sets). Each of these ingredients is independently motivated and necessary. [/guided] [/step]