Graph of Continuous Mapping to Hausdorff Space is Closed in Product

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Theorem

Let $T_A = \struct {A, \tau_A}$ and $T_B = \struct {B, \tau_B}$ be topological spaces.

Let $T_B$ be a Hausdorff space.

Let $f: T_A \to T_B$ be a continuous mapping.


Then the graph of $f$ is a closed subset of $T_A \times T_B$ under the product topology.


Proof 1

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:

\(\ds (\paren {f \times I_B}^{-1} \sqbrk {\Delta_B}\) \(=\) \(\ds \set {\tuple {x, y} \in A \times B: \map {f \times I_B} {x, y} \in \Delta_B}\)
\(\ds \) \(=\) \(\ds \set {\tuple {x, y} \in A \times B : \tuple {\map f x, y} \in \Delta_B}\)
\(\ds \) \(=\) \(\ds \set {\tuple {x, y} \in A \times B : \map f x = y}\)
\(\ds \) \(=\) \(\ds G_f\)

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

$\blacksquare$


Proof 2

Let $G_f$ be the graph of $f$:

$G_f = \set {\tuple {x, y} \in A \times B: \map f x = y}$

Fix a point $\tuple {x_0, y_0}$ in $\paren {A \times B} \setminus G_f = \set {\tuple {x, y} \in A \times B : \map f x \ne y}$.

As $\map f {x_0} \ne y_0$ and $T_B$ is Hausdorff, there exists disjoint, non-empty, open sets $V_{x_0}$ and $V_{y_0}$ of $T_B$ containing $\map f {x_0}$ and $y_0$ respectively.

From the definition of continuous mapping, $f^{-1} \sqbrk {V_{x_0} }$ is an open set of $T_A$.

So we have that:

$f^{-1} \sqbrk {V_{x_0} } \times V_{y_0}$ is open in $T_A \times T_B$
$f^{-1} \sqbrk {V_{x_0} } \times V_{y_0}$ is a subset of $\paren {A \times B} \setminus G_f$ containing $\tuple {x_0, y_0}$

Hence it follows that $\paren {A \times B} \setminus G_f$ is open in $T_A \times T_B$.

Hence $G_f$ is closed in $T_A \times T_B$.

$\blacksquare$


Sources

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