[proofplan] We prove both directions. For the forward direction ($\Leftarrow$), given distinct points $x \neq y$ in the product, we find an index where their coordinates differ, separate them in that factor using the Hausdorff property, and pull back via the projection to obtain disjoint open sets in the product. For the reverse direction ($\Rightarrow$), we embed each factor into the product by fixing all other coordinates, use the Hausdorff property of the product to separate two points differing only in the $\beta$-th coordinate, and extract separating open sets in $X_\beta$ from basic product neighbourhoods. [/proofplan] [step:Show that the Hausdorff property of each factor implies the product is Hausdorff] Assume each $(X_\alpha, \tau_\alpha)$ is Hausdorff. Let $x, y \in \prod_{\alpha \in A} X_\alpha$ with $x \neq y$. Since $x \neq y$, there exists an index $\beta \in A$ with $x_\beta \neq y_\beta$. Since $X_\beta$ is Hausdorff, there exist disjoint open sets $U_\beta, V_\beta \in \tau_\beta$ with $x_\beta \in U_\beta$ and $y_\beta \in V_\beta$. Define \begin{align*} U = \pi_\beta^{-1}(U_\beta), \qquad V = \pi_\beta^{-1}(V_\beta), \end{align*} where $\pi_\beta: \prod_{\alpha \in A} X_\alpha \to X_\beta$ is the canonical projection. Both $U$ and $V$ are open in the product topology as preimages of open sets under the continuous projection $\pi_\beta$. We have $x \in U$ (since $x_\beta \in U_\beta$) and $y \in V$ (since $y_\beta \in V_\beta$). If $p \in U \cap V$, then $p_\beta \in U_\beta \cap V_\beta = \varnothing$, a contradiction. Therefore $U \cap V = \varnothing$. [guided] Assume each $(X_\alpha, \tau_\alpha)$ is Hausdorff. Let $x, y \in \prod_{\alpha \in A} X_\alpha$ with $x \neq y$. Two elements of a Cartesian product are equal if and only if they agree at every index, so $x \neq y$ guarantees the existence of an index $\beta \in A$ where $x_\beta \neq y_\beta$. Since $X_\beta$ is Hausdorff, there exist disjoint open sets $U_\beta, V_\beta \in \tau_\beta$ with $x_\beta \in U_\beta$ and $y_\beta \in V_\beta$. We lift these to the product by taking preimages under the projection $\pi_\beta: \prod_{\alpha \in A} X_\alpha \to X_\beta$: \begin{align*} U = \pi_\beta^{-1}(U_\beta), \qquad V = \pi_\beta^{-1}(V_\beta). \end{align*} By the definition of the product topology, $\pi_\beta$ is continuous, so $U$ and $V$ are open. These are subbasic open sets in the product topology. Concretely, $U$ consists of all points in the product whose $\beta$-th coordinate lies in $U_\beta$, with no constraint on any other coordinate. The containments $x \in U$ and $y \in V$ follow from $x_\beta \in U_\beta$ and $y_\beta \in V_\beta$. Disjointness: if $p \in U \cap V$, then $p_\beta \in U_\beta \cap V_\beta = \varnothing$, a contradiction. Therefore $U$ and $V$ are disjoint open sets separating $x$ and $y$. This direction is straightforward because we need only *one* index where $x$ and $y$ differ — the product topology is fine enough to separate points using information from a single factor. [/guided] [/step] [step:Show that the product being Hausdorff implies each factor is Hausdorff] Assume $\prod_{\alpha \in A} X_\alpha$ is Hausdorff. Fix $\beta \in A$ and let $a, b \in X_\beta$ with $a \neq b$. For each $\alpha \neq \beta$, choose a point $c_\alpha \in X_\alpha$. Define $x, y \in \prod_{\alpha \in A} X_\alpha$ by \begin{align*} x_\alpha = \begin{cases} a & \text{if } \alpha = \beta, \\ c_\alpha & \text{if } \alpha \neq \beta, \end{cases} \qquad y_\alpha = \begin{cases} b & \text{if } \alpha = \beta, \\ c_\alpha & \text{if } \alpha \neq \beta. \end{cases} \end{align*} Since $a \neq b$, we have $x \neq y$. By the Hausdorff property of the product, there exist disjoint open sets $U, V$ in the product topology with $x \in U$ and $y \in V$. Since $U$ is open and contains $x$, there exists a basic open set $\prod_{\alpha \in A} W_\alpha \subset U$ containing $x$, where $W_\alpha \in \tau_\alpha$ for each $\alpha$ and $W_\alpha = X_\alpha$ for all but finitely many $\alpha$. In particular, $a = x_\beta \in W_\beta$. Similarly, there exists a basic open set $\prod_{\alpha \in A} W'_\alpha \subset V$ containing $y$, with $b = y_\beta \in W'_\beta$. We show $W_\beta \cap W'_\beta = \varnothing$. Suppose $z \in W_\beta \cap W'_\beta$. Define $p \in \prod_{\alpha \in A} X_\alpha$ by $p_\beta = z$ and $p_\alpha = c_\alpha$ for all $\alpha \neq \beta$. For each $\alpha \neq \beta$, $p_\alpha = c_\alpha = x_\alpha \in W_\alpha$ (since $x \in \prod_\alpha W_\alpha$), and $p_\beta = z \in W_\beta$. So $p \in \prod_\alpha W_\alpha \subset U$. By the same reasoning with $W'_\alpha$ in place of $W_\alpha$ (using $p_\alpha = c_\alpha = y_\alpha \in W'_\alpha$ for $\alpha \neq \beta$ and $p_\beta = z \in W'_\beta$), $p \in \prod_\alpha W'_\alpha \subset V$. This gives $p \in U \cap V$, contradicting $U \cap V = \varnothing$. Therefore $W_\beta$ and $W'_\beta$ are disjoint open sets in $X_\beta$ separating $a$ and $b$, and $X_\beta$ is Hausdorff. [guided] Assume $\prod_{\alpha \in A} X_\alpha$ is Hausdorff. We show each factor $X_\beta$ is Hausdorff by embedding the separation problem in $X_\beta$ into the product. Fix $\beta \in A$ and let $a, b \in X_\beta$ with $a \neq b$. Choose arbitrary points $c_\alpha \in X_\alpha$ for each $\alpha \neq \beta$ (the product is non-empty since it contains points, so each factor is non-empty). Define $x, y \in \prod_{\alpha} X_\alpha$ by $x_\beta = a$, $y_\beta = b$, and $x_\alpha = y_\alpha = c_\alpha$ for $\alpha \neq \beta$. These points differ only in the $\beta$-th coordinate, so $x \neq y$. By the Hausdorff property of the product, there exist disjoint open sets $U, V$ with $x \in U$, $y \in V$. To extract open sets in $X_\beta$, we pass to basic open sets. Since $U$ is open and $x \in U$, there is a basic open set $\prod_\alpha W_\alpha \subset U$ containing $x$, where $W_\alpha = X_\alpha$ for all but finitely many $\alpha$ and $W_\alpha \in \tau_\alpha$ for all $\alpha$. In particular, $a \in W_\beta$. Similarly, there is a basic open set $\prod_\alpha W'_\alpha \subset V$ containing $y$, with $b \in W'_\beta$. Why must $W_\beta$ and $W'_\beta$ be disjoint? Suppose $z \in W_\beta \cap W'_\beta$. We construct a point $p$ that lies in both $\prod_\alpha W_\alpha$ and $\prod_\alpha W'_\alpha$: set $p_\beta = z$ and $p_\alpha = c_\alpha$ for $\alpha \neq \beta$. At each index $\alpha \neq \beta$, $p_\alpha = c_\alpha = x_\alpha \in W_\alpha$ (since $x \in \prod_\alpha W_\alpha$) and $p_\alpha = c_\alpha = y_\alpha \in W'_\alpha$ (since $y \in \prod_\alpha W'_\alpha$). At index $\beta$, $p_\beta = z \in W_\beta \cap W'_\beta$. So $p \in \prod_\alpha W_\alpha \subset U$ and $p \in \prod_\alpha W'_\alpha \subset V$, giving $p \in U \cap V = \varnothing$ — a contradiction. The construction works because the points $x$ and $y$ share all coordinates except the $\beta$-th. This forces any basic open sets containing $x$ and $y$ to have "room" at every non-$\beta$ coordinate for the same point $c_\alpha$, so the contradiction point $p$ can be assembled from $z$ at index $\beta$ and $c_\alpha$ elsewhere. [/guided] [/step]