=== Necessary Condition === Suppose that $T \times D$ is a $T_5$ space. Let $A, B \subseteq T$ be two separated sets. First, we will show that $A \times D$ and $B \times D$ are also separated. To this end, observe that: {{begin-eqn}} {{eqn | l = \paren {A \times D}^- \cap \paren {B \times D}       | r = \paren {A^- \times D} \cap \paren {B \times D}       | c = Closure of Subset of Double Pointed Topological Space }} {{eqn | r = \paren {A^- \cap B} \times D       | c = Cartesian Product Distributes over Intersection }} {{eqn | r = \O \times D       | c = $A$ and $B$ are separated }} {{eqn | r = \varnothing       | c = Cartesian Product is Empty iff Factor is Empty }} {{end-eqn}} Mutatis mutandis, also: :$\paren {B \times D}^- \cap \paren {A \times D} = \O$ Thus $A \times D$ and $B \times D$ are separated. Since $T \times D$ is assumed to be a $T_5$ space, find open sets $U, V$ such that: :$U \cap V = \O$ :$A^- \times D \subseteq U$ :$B^- \times D \subseteq V$ By Open Sets of Double Pointed Topology, there are open sets $U', V'$ of $\tau$ such that: :$U = U' \times D$ :$V = V' \times D$ We have, furthermore: {{begin-eqn}} {{eqn | l = \varnothing       | r = U \cap V }} {{eqn | r = \paren {U' \times D} \cap \paren {V' \times D} }} {{eqn | r = \paren {U' \cap V'} \times D       | c = Cartesian Product Distributes over Intersection }} {{end-eqn}} From Cartesian Product is Empty iff Factor is Empty, $U' \cap V' = \O$ since $D$ is non-empty. By Cartesian Product of Subsets, we also infer: :$A^- \subseteq U'$ :$B^- \subseteq V'$ It follows that $T$ is a $T_5$ space. {{qed|lemma}} === Sufficient Condition === Suppose that $T$ is a $T_5$ space. Let $X, Y \subseteq T \times D$ be separated sets. By Closure of Subset of Double Pointed Topological Space, we know that: :$X^- = {\pr_1 \sqbrk X}^- \times D$ :$Y^- = {\pr_1 \sqbrk Y}^- \times D$ where $\pr_1$ is the first projection. Suppose now that for some $s \in S$, one has: :$s \in {\pr_1 \sqbrk X}^- \cap \pr_1 \sqbrk Y$ Then it would be that for some $d \in D$: :$\tuple {s, d} \in Y$ However, by the expression for $X^-$ above, also $\tuple {s, d} \in X^-$. This is a contradiction, for $X$ and $Y$ were assumed separated. Thus: :${\pr_1 \sqbrk X}^- \cap \pr_1 \sqbrk Y = \O$ and analogously, we derive: :$\pr_1 \sqbrk X \cap {\pr_1 \sqbrk Y}^- = \O$ that is, $\pr_1 \sqbrk X$ and $\pr_1 \sqbrk Y$ are separated. Since $T$ is a $T_5$ space, we find open sets $U, V$ of $\tau$ such that: :$U \cap V = \varnothing$ :${\pr_1 \sqbrk X}^- \subseteq U$ :${\pr_1 \sqbrk Y}^- \subseteq V$ By Cartesian Product of Subsets, we have: :$X^- = {\pr_1 \sqbrk X}^- \times D \subseteq U \times D$ :$Y^- = {\pr_1 \sqbrk Y}^- \times D \subseteq V \times D$ By Open Sets of Double Pointed Topology, $U \times D$ and $V \times D$ are also open sets in $S \times D$. Finally, we compute: {{begin-eqn}} {{eqn | l = \paren {U \times D} \cap \paren {V \times D}       | r = \paren {U \cap V} \times D       | c = Cartesian Product Distributes over Intersection }} {{eqn | r = \O \times D       | c = by choice of $U$ and $V$ }} {{eqn | r = \O       | c = Cartesian Product is Empty iff Factor is Empty }} {{end-eqn}} That is, $T \times D$ is a $T_5$ space. {{qed}}