Graph of Continuous Mapping to Hausdorff Space is Closed in Product/Proof 1

Proof
Let $G_f$ be the graph of $f$:
 * $G_f = \set {\tuple {x, y} \in A \times B: \map f x = y}$

Let $I_B: T_B \to T_B$ be the identity mapping on $B$:
 * $\forall y \in B: \map {I_B} y = y$

From Identity Mapping is Continuous, $I_B$ is continuous on $T_B$.

Let $f \times I_B: T_A \times T_B \to T_B \times T_B$ be the product map:
 * $\map {f \times I_B} {x, y} = \tuple {\map f x, y}$

From Continuous Mapping to Product Space, $f \times I_B$ is continuous.

Let $\Delta_B$ be the diagonal relation on $B$.

From Hausdorff Space iff Diagonal Set on Product is Closed, $\Delta_B$ is a closed set of $T_B \times T_B$ under the product topology.

From Continuity Defined from Closed Sets:

is closed in $T_X \times T_B$ endowed with the product topology.