Graph of Continuous Mapping to Hausdorff Space is Closed in Product
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
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $4$: The Hausdorff condition: Exercise $4.3: 3$