[proofplan] We prove the three preservation results (Hausdorff, regular, completely regular) by reducing separation in the product to separation in the factors via projections. For the Hausdorff property, two distinct points differ in some coordinate, and we pull back disjoint [open sets](/page/Open%20Set) from that factor. For regularity, we separate a point from a [closed set](/page/Closed%20Set) by expressing the complement as a union of basic open sets, identifying a finite collection of coordinates where separation occurs, and intersecting the resulting open sets to construct the required neighbourhoods. For complete regularity, we construct a separating function on the product from separating functions on the factors. [/proofplan] [step:Separate distinct points in the product by pulling back disjoint open sets from a single factor] Assume each $X_\alpha$ is Hausdorff. Let $x, y \in X = \prod_{\alpha \in A} X_\alpha$ with $x \neq y$. Since $x \neq y$ as elements of the product, there exists $\beta \in A$ with $\pi_\beta(x) \neq \pi_\beta(y)$. Since $X_\beta$ is Hausdorff, there exist disjoint [open sets](/page/Open%20Set) $U_\beta, V_\beta \subset X_\beta$ with $\pi_\beta(x) \in U_\beta$ and $\pi_\beta(y) \in V_\beta$. Define \begin{align*} U &:= \pi_\beta^{-1}(U_\beta), \\ V &:= \pi_\beta^{-1}(V_\beta). \end{align*} By the [Universal Property of the Product Topology](/theorems/962), $\pi_\beta$ is [continuous](/page/Continuity), so $U$ and $V$ are open in $X$. We have $x \in U$, $y \in V$, and $U \cap V = \pi_\beta^{-1}(U_\beta \cap V_\beta) = \pi_\beta^{-1}(\varnothing) = \varnothing$. Hence $X$ is Hausdorff. [guided] Assume each $X_\alpha$ is Hausdorff. Let $x, y \in X = \prod_{\alpha \in A} X_\alpha$ with $x \neq y$. Two elements of a product are equal if and only if they agree in every coordinate, so $x \neq y$ means there exists at least one index $\beta \in A$ with $\pi_\beta(x) \neq \pi_\beta(y)$. We now use the Hausdorff property of the single factor $X_\beta$. Since $\pi_\beta(x) \neq \pi_\beta(y)$ in $X_\beta$ and $X_\beta$ is Hausdorff, there exist disjoint open sets $U_\beta, V_\beta \subset X_\beta$ with $\pi_\beta(x) \in U_\beta$ and $\pi_\beta(y) \in V_\beta$. Define \begin{align*} U &:= \pi_\beta^{-1}(U_\beta), \\ V &:= \pi_\beta^{-1}(V_\beta). \end{align*} By the [Universal Property of the Product Topology](/theorems/962), every projection $\pi_\beta$ is continuous, so $U$ and $V$ are open in the product [topology](/page/Topology). Moreover, $x \in U$ and $y \in V$ because $\pi_\beta(x) \in U_\beta$ and $\pi_\beta(y) \in V_\beta$. Finally, \begin{align*} U \cap V = \pi_\beta^{-1}(U_\beta) \cap \pi_\beta^{-1}(V_\beta) = \pi_\beta^{-1}(U_\beta \cap V_\beta) = \pi_\beta^{-1}(\varnothing) = \varnothing. \end{align*} Hence $X$ is Hausdorff. The key observation is that separation in a *single* coordinate suffices to separate in the product. We do not need to separate in every coordinate simultaneously, which would require the box topology. [/guided] [/step] [step:Separate a point from a closed set by applying regularity in finitely many factors] Assume each $X_\alpha$ is regular (hence also Hausdorff, since regularity includes the $T_1$ axiom). We must show that for every $x \in X$ and every [closed set](/page/Closed%20Set) $C \subset X$ with $x \notin C$, there exist disjoint [open sets](/page/Open%20Set) $U, V \subset X$ with $x \in U$ and $C \subset V$. Since $X \setminus C$ is open and contains $x$, there exists a basic open neighbourhood of $x$ contained in $X \setminus C$. This basic open set has the form \begin{align*} B = \bigcap_{j=1}^{m} \pi_{\beta_j}^{-1}(W_{\beta_j}) \end{align*} where $\beta_1, \ldots, \beta_m \in A$ and each $W_{\beta_j} \subset X_{\beta_j}$ is open with $\pi_{\beta_j}(x) \in W_{\beta_j}$, and $B \subset X \setminus C$. For each $j \in \{1, \ldots, m\}$, the set $X_{\beta_j} \setminus W_{\beta_j}$ is closed in $X_{\beta_j}$ and does not contain $\pi_{\beta_j}(x)$. Since $X_{\beta_j}$ is regular, there exist disjoint open sets $U_{\beta_j}, V_{\beta_j} \subset X_{\beta_j}$ with \begin{align*} \pi_{\beta_j}(x) \in U_{\beta_j} \quad \text{and} \quad X_{\beta_j} \setminus W_{\beta_j} \subset V_{\beta_j}. \end{align*} Define \begin{align*} U &:= \bigcap_{j=1}^{m} \pi_{\beta_j}^{-1}(U_{\beta_j}), \\ V &:= \bigcup_{j=1}^{m} \pi_{\beta_j}^{-1}(V_{\beta_j}). \end{align*} Then $U$ is open (finite intersection of open sets), $V$ is open (union of open sets), and $x \in U$ since $\pi_{\beta_j}(x) \in U_{\beta_j}$ for all $j$. [guided] Assume each $X_\alpha$ is regular. We must show that for every $x \in X$ and every closed set $C \subset X$ with $x \notin C$, there exist disjoint open sets separating $x$ from $C$. Since $x \notin C$ and $C$ is closed, $X \setminus C$ is an open neighbourhood of $x$. The product [topology](/page/Topology) has a basis of sets of the form $\bigcap_{j=1}^{m} \pi_{\beta_j}^{-1}(W_{\beta_j})$ (finite intersections of subbasis elements), so there exist finitely many indices $\beta_1, \ldots, \beta_m \in A$ and open sets $W_{\beta_j} \subset X_{\beta_j}$ with $\pi_{\beta_j}(x) \in W_{\beta_j}$ such that \begin{align*} x \in B := \bigcap_{j=1}^{m} \pi_{\beta_j}^{-1}(W_{\beta_j}) \subset X \setminus C. \end{align*} The inclusion $B \subset X \setminus C$ means $C \subset X \setminus B$. We rewrite this complement: \begin{align*} X \setminus B = X \setminus \bigcap_{j=1}^{m} \pi_{\beta_j}^{-1}(W_{\beta_j}) = \bigcup_{j=1}^{m} \pi_{\beta_j}^{-1}(X_{\beta_j} \setminus W_{\beta_j}). \end{align*} So every point of $C$ fails to lie in $W_{\beta_j}$ for at least one $j$. Now we use regularity of the individual factors. For each $j$, the point $\pi_{\beta_j}(x) \in W_{\beta_j}$ and the closed set $X_{\beta_j} \setminus W_{\beta_j}$ are disjoint in $X_{\beta_j}$. Since $X_{\beta_j}$ is regular, there exist disjoint open sets $U_{\beta_j}$ and $V_{\beta_j}$ in $X_{\beta_j}$ with \begin{align*} \pi_{\beta_j}(x) \in U_{\beta_j} \quad \text{and} \quad X_{\beta_j} \setminus W_{\beta_j} \subset V_{\beta_j}. \end{align*} We define the separating open sets in the product: \begin{align*} U &:= \bigcap_{j=1}^{m} \pi_{\beta_j}^{-1}(U_{\beta_j}), \\ V &:= \bigcup_{j=1}^{m} \pi_{\beta_j}^{-1}(V_{\beta_j}). \end{align*} The set $U$ is open as a finite intersection of preimages of open sets under [continuous](/page/Continuity) maps. The set $V$ is open as a union of such preimages. The point $x$ lies in $U$ because $\pi_{\beta_j}(x) \in U_{\beta_j}$ for each $j$. [/guided] [/step] [step:Verify that $C \subset V$ and $U \cap V = \varnothing$] **Containment:** Let $z \in C$. Since $C \subset X \setminus B$, there exists $j_0 \in \{1, \ldots, m\}$ with $\pi_{\beta_{j_0}}(z) \notin W_{\beta_{j_0}}$, i.e., $\pi_{\beta_{j_0}}(z) \in X_{\beta_{j_0}} \setminus W_{\beta_{j_0}} \subset V_{\beta_{j_0}}$. Hence $z \in \pi_{\beta_{j_0}}^{-1}(V_{\beta_{j_0}}) \subset V$. **Disjointness:** Suppose $z \in U \cap V$. From $z \in U$, we have $\pi_{\beta_j}(z) \in U_{\beta_j}$ for all $j = 1, \ldots, m$. From $z \in V$, there exists $j_0$ with $\pi_{\beta_{j_0}}(z) \in V_{\beta_{j_0}}$. But $U_{\beta_{j_0}} \cap V_{\beta_{j_0}} = \varnothing$, contradicting $\pi_{\beta_{j_0}}(z) \in U_{\beta_{j_0}} \cap V_{\beta_{j_0}}$. Therefore $U$ and $V$ are disjoint [open sets](/page/Open%20Set) with $x \in U$ and $C \subset V$, so $X$ is regular. [guided] **Containment: $C \subset V$.** Let $z \in C$. Since $B \subset X \setminus C$, we have $z \notin B$, so $z \in X \setminus B = \bigcup_{j=1}^{m} \pi_{\beta_j}^{-1}(X_{\beta_j} \setminus W_{\beta_j})$. This means there exists some $j_0 \in \{1, \ldots, m\}$ with $\pi_{\beta_{j_0}}(z) \in X_{\beta_{j_0}} \setminus W_{\beta_{j_0}}$. Since $X_{\beta_{j_0}} \setminus W_{\beta_{j_0}} \subset V_{\beta_{j_0}}$, we get $\pi_{\beta_{j_0}}(z) \in V_{\beta_{j_0}}$, so $z \in \pi_{\beta_{j_0}}^{-1}(V_{\beta_{j_0}}) \subset V$. **Disjointness: $U \cap V = \varnothing$.** Suppose for contradiction that some $z \in U \cap V$. From $z \in U = \bigcap_{j=1}^{m} \pi_{\beta_j}^{-1}(U_{\beta_j})$, we have $\pi_{\beta_j}(z) \in U_{\beta_j}$ for every $j = 1, \ldots, m$. From $z \in V = \bigcup_{j=1}^{m} \pi_{\beta_j}^{-1}(V_{\beta_j})$, there exists some $j_0$ with $\pi_{\beta_{j_0}}(z) \in V_{\beta_{j_0}}$. Combining these, $\pi_{\beta_{j_0}}(z) \in U_{\beta_{j_0}} \cap V_{\beta_{j_0}}$. But $U_{\beta_{j_0}}$ and $V_{\beta_{j_0}}$ were chosen to be disjoint in $X_{\beta_{j_0}}$, a contradiction. Notice the asymmetry in the construction: $U$ is an *intersection* (ensuring every coordinate of $x$ is controlled), while $V$ is a *union* (ensuring every point of $C$ is captured through whichever coordinate forces it out of $B$). This asymmetry is what makes the argument work — an intersection for $V$ would not guarantee $C \subset V$, since different points of $C$ may fail to lie in $B$ at different coordinates. [/guided] [/step] [step:Construct a separating function on the product as the maximum of factor separating functions] Assume each $X_\alpha$ is completely regular. Let $x \in X$ and let $C \subset X$ be closed with $x \notin C$. As in the regularity argument, choose a basic open neighbourhood $B = \bigcap_{j=1}^{m} \pi_{\beta_j}^{-1}(W_{\beta_j})$ with $x \in B \subset X \setminus C$. For each $j \in \{1, \ldots, m\}$, the point $\pi_{\beta_j}(x) \notin X_{\beta_j} \setminus W_{\beta_j}$ and $X_{\beta_j} \setminus W_{\beta_j}$ is closed in $X_{\beta_j}$. Since $X_{\beta_j}$ is completely regular, there exists a [continuous function](/page/Continuity%20(Real%20Analysis)) \begin{align*} g_j: X_{\beta_j} &\to [0, 1] \end{align*} with $g_j(\pi_{\beta_j}(x)) = 0$ and $g_j \equiv 1$ on $X_{\beta_j} \setminus W_{\beta_j}$. Define \begin{align*} f: X &\to [0, 1] \\ z &\mapsto \max_{1 \le j \le m} \, g_j(\pi_{\beta_j}(z)). \end{align*} Each map $g_j \circ \pi_{\beta_j}: X \to [0,1]$ is continuous as a composition of continuous functions ($\pi_{\beta_j}$ is continuous by the [Universal Property of the Product Topology](/theorems/962) and $g_j$ is continuous by construction). The maximum of finitely many continuous real-valued functions is continuous, so $f$ is continuous. We verify: $f(x) = \max_j g_j(\pi_{\beta_j}(x)) = \max_j 0 = 0$. For $z \in C$, since $C \subset X \setminus B$, there exists $j_0$ with $\pi_{\beta_{j_0}}(z) \notin W_{\beta_{j_0}}$, hence $g_{j_0}(\pi_{\beta_{j_0}}(z)) = 1$, so $f(z) = 1$. Therefore $X$ is completely regular. [guided] Assume each $X_\alpha$ is completely regular. Let $x \in X$ and let $C \subset X$ be closed with $x \notin C$. Complete regularity asks for a continuous function $f: X \to [0,1]$ with $f(x) = 0$ and $f \equiv 1$ on $C$. As before, choose a basic open neighbourhood $B = \bigcap_{j=1}^{m} \pi_{\beta_j}^{-1}(W_{\beta_j})$ with $x \in B \subset X \setminus C$. For each $j \in \{1, \ldots, m\}$, the point $\pi_{\beta_j}(x)$ lies in the [open set](/page/Open%20Set) $W_{\beta_j}$ and is therefore separated from the [closed set](/page/Closed%20Set) $X_{\beta_j} \setminus W_{\beta_j}$. Since $X_{\beta_j}$ is completely regular, there exists a continuous function \begin{align*} g_j: X_{\beta_j} &\to [0, 1] \end{align*} with $g_j(\pi_{\beta_j}(x)) = 0$ and $g_j(t) = 1$ for all $t \in X_{\beta_j} \setminus W_{\beta_j}$. We combine these $m$ separating functions into a single separating function on the product. Define \begin{align*} f: X &\to [0, 1] \\ z &\mapsto \max_{1 \le j \le m} \, g_j(\pi_{\beta_j}(z)). \end{align*} Why the maximum and not, say, a sum or product? The maximum achieves exactly the right behaviour: $f(x) = 0$ requires each $g_j$ to vanish at $x$ (which holds by construction), while $f(z) = 1$ for $z \in C$ requires only *one* $g_j$ to equal $1$ at $z$ (which is guaranteed because $z \notin B$ forces $\pi_{\beta_{j_0}}(z) \notin W_{\beta_{j_0}}$ for some $j_0$, giving $g_{j_0}(\pi_{\beta_{j_0}}(z)) = 1$). **Continuity of $f$.** Each composition $g_j \circ \pi_{\beta_j}: X \to [0,1]$ is continuous: $\pi_{\beta_j}$ is continuous by the [Universal Property of the Product Topology](/theorems/962), and $g_j$ is continuous by the complete regularity of $X_{\beta_j}$. The pointwise maximum of finitely many continuous real-valued functions is continuous (this follows from the identity $\max(a,b) = \frac{a + b + |a - b|}{2}$ applied inductively). Hence $f$ is continuous. **Verification.** At $x$: $f(x) = \max_j g_j(\pi_{\beta_j}(x)) = \max_j 0 = 0$. For $z \in C$: since $C \subset X \setminus B$, there exists $j_0$ with $\pi_{\beta_{j_0}}(z) \in X_{\beta_{j_0}} \setminus W_{\beta_{j_0}}$, so $g_{j_0}(\pi_{\beta_{j_0}}(z)) = 1$, giving $f(z) \ge 1$. Since $f$ takes values in $[0,1]$, we conclude $f(z) = 1$. Therefore $X$ is completely regular. [/guided] [/step]