[proofplan] We prove both directions of the equivalence. For the forward direction, we express $R = (\pi \times \pi)^{-1}(\Delta_{X/\sim})$ and use the [Equivalent Characterisations of the Hausdorff Property](/theorems/1025) to conclude that $\Delta_{X/\sim}$ is closed, making $R$ the preimage of a closed set under a continuous map. For the reverse direction, given $R$ closed and distinct classes $[x] \neq [y]$, we find open $U \ni x$, $V \ni y$ with $U \times V \subset (X \times X) \setminus R$, show $\pi(U) \cap \pi(V) = \varnothing$ by a transitivity argument, and verify openness of $\pi(U)$ and $\pi(V)$ using the fact that $\pi$ is an open map. [/proofplan] [step:Express $R$ as the preimage of the diagonal under $\pi \times \pi$] Define the map \begin{align*} \pi \times \pi: X \times X &\to (X/{\sim}) \times (X/{\sim}), \\ (x, y) &\mapsto (\pi(x), \pi(y)). \end{align*} A map into a product is continuous if and only if each component is continuous. Since $\pi$ is continuous (as every quotient map is), $\pi \times \pi$ is continuous with respect to the product topologies. Let $\Delta_{X/\sim} = \{(q, q) : q \in X/{\sim}\}$ denote the diagonal. Then \begin{align*} (\pi \times \pi)^{-1}(\Delta_{X/\sim}) = \{(x, y) \in X \times X : \pi(x) = \pi(y)\} = \{(x, y) \in X \times X : x \sim y\} = R, \end{align*} since $\pi(x) = \pi(y)$ if and only if $x$ and $y$ belong to the same equivalence class, which is the definition of $x \sim y$. [/step] [step:Prove that $X/{\sim}$ Hausdorff implies $R$ is closed] Assume $X/{\sim}$ is Hausdorff. By the [Equivalent Characterisations of the Hausdorff Property](/theorems/1025) (implication $(1) \Rightarrow (2)$), the diagonal $\Delta_{X/\sim}$ is closed in $(X/{\sim}) \times (X/{\sim})$. Since $\pi \times \pi$ is continuous and $R = (\pi \times \pi)^{-1}(\Delta_{X/\sim})$, the set $R$ is closed in $X \times X$ as the preimage of a closed set under a continuous map. [guided] Assume $X/{\sim}$ is Hausdorff. The [Equivalent Characterisations of the Hausdorff Property](/theorems/1025) provides, among other equivalences, the characterisation that a space is Hausdorff if and only if its diagonal is closed. Applying the implication $(1) \Rightarrow (2)$ to $X/{\sim}$, the diagonal $\Delta_{X/\sim}$ is closed in $(X/{\sim}) \times (X/{\sim})$. The map $\pi \times \pi$ is continuous (established in the previous step). A basic property of continuous maps is that the preimage of every closed set is closed. Since $\Delta_{X/\sim}$ is closed and $R = (\pi \times \pi)^{-1}(\Delta_{X/\sim})$, the set $R$ is closed in $X \times X$. This direction is clean and requires no special properties of $\pi$ beyond continuity — in particular, $\pi$ need not be open. The converse direction, however, requires more structure. [/guided] [/step] [step:Prove that $R$ closed implies $X/{\sim}$ is Hausdorff when the quotient map is open] Assume $R$ is closed in $X \times X$ and that $\pi: X \to X/{\sim}$ is an open map (equivalently, for every open set $U \subset X$, the saturation $\pi^{-1}(\pi(U))$ is open in $X$). Let $[x], [y] \in X/{\sim}$ with $[x] \neq [y]$, so $x \not\sim y$ and $(x, y) \notin R$. Since $(X \times X) \setminus R$ is open, there exist open sets $U, V \subset X$ with \begin{align*} (x, y) \in U \times V \subset (X \times X) \setminus R. \end{align*} The inclusion $U \times V \subset (X \times X) \setminus R$ means: for all $a \in U$ and $b \in V$, $(a, b) \notin R$, i.e., $a \not\sim b$. **Disjointness.** We show $\pi(U) \cap \pi(V) = \varnothing$. Suppose $[z] \in \pi(U) \cap \pi(V)$. Then $z \sim a$ for some $a \in U$ and $z \sim b$ for some $b \in V$. By transitivity of $\sim$, $a \sim b$, so $(a, b) \in R \cap (U \times V) = \varnothing$, a contradiction. **Openness.** Since $\pi$ is an open map and $U$ is open in $X$, the image $\pi(U)$ is open in $X/{\sim}$. Similarly, $\pi(V)$ is open in $X/{\sim}$. Therefore $\pi(U)$ and $\pi(V)$ are disjoint open subsets of $X/{\sim}$ with $[x] \in \pi(U)$ and $[y] \in \pi(V)$, establishing the Hausdorff property. [guided] Assume $R$ is closed in $X \times X$ and that $\pi$ is an open map. Let $[x], [y] \in X/{\sim}$ with $[x] \neq [y]$. Then $x \not\sim y$, so $(x, y) \notin R$. Since $R$ is closed, the complement $(X \times X) \setminus R$ is open, and by the definition of the product topology there exist open sets $U \ni x$, $V \ni y$ in $X$ with $U \times V \subset (X \times X) \setminus R$. The condition $U \times V \subset (X \times X) \setminus R$ says: no element of $U$ is equivalent to any element of $V$. This immediately yields disjointness of the images: if $[z] \in \pi(U) \cap \pi(V)$, then $z \sim a$ for some $a \in U$ and $z \sim b$ for some $b \in V$. By transitivity, $a \sim b$, giving $(a, b) \in R$. But $(a, b) \in U \times V \subset (X \times X) \setminus R$, a contradiction. So $\pi(U) \cap \pi(V) = \varnothing$. The key condition is openness of $\pi(U)$ and $\pi(V)$. By the definition of the quotient topology, $\pi(U)$ is open in $X/{\sim}$ if and only if $\pi^{-1}(\pi(U))$ is open in $X$. The set $\pi^{-1}(\pi(U))$ is the saturation of $U$ — the union of all equivalence classes that meet $U$. In general, the saturation of an open set need not be open (for example, if $X = \mathbb{R}$ with $0 \sim 1$ and $U = (-0.1, 0.1)$, then $\pi^{-1}(\pi(U)) = (-0.1, 0.1) \cup \{1\}$, which is not open). The hypothesis that $\pi$ is an open map is precisely the condition ensuring that the saturation of every open set is open. Under this hypothesis, $\pi(U)$ and $\pi(V)$ are open, disjoint, and contain $[x]$ and $[y]$ respectively, establishing the Hausdorff property. The condition that $\pi$ is open is equivalent to requiring that for every open $W \subset X$, the set $\{z \in X : z \sim w \text{ for some } w \in W\}$ is open. This holds automatically when equivalence classes are finite, when $X$ is a topological group and $\sim$ is defined by a closed normal subgroup, or more generally when the equivalence relation is "open" in the sense that the saturation map preserves open sets. [/guided] [/step]