Equal Images of Mappings to Hausdorff Space form Closed Set
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Theorem
Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $T_B$ be a Hausdorff space.
Let $f, g: T_A \to T_B$ be continuous mappings.
Let $W$ be the set defined as:
- $W = \set {x \in T_A: \map f x = \map g x}$
Then $W$ is closed in $T_A$.
Proof
Consider the set $V = S_A \setminus W$.
Hence:
- $V = \set {x \in T_A: \map f x \ne \map g x}$
Let $x \in V$.
Then:
- $\map f x \ne \map g x$
and as $T_B$ is Hausdorff:
- $\exists U_1, U_2 \in \tau_B: \map f x \in U_1, \map g x \in U_2, U_1 \cap U_2 = \O$
As $f$ and $g$ are continuous mappings:
- $f^{-1} \sqbrk {U_1}$ and $g^{-1} \sqbrk {U_2}$ are open in $T_B$.
Because $f^{-1} \sqbrk {U_1}$ and $g^{-1} \sqbrk {U_2}$ are open in $T_A$, $f^{-1} \sqbrk U_1 \cap g^{-1} \sqbrk U_2$ is also open in $T_A$.
Then we have that:
- $x \in f^{-1} \sqbrk {U_1} \cap g^{-1} \sqbrk {U_2}$
such that:
- $f^{-1} \sqbrk {U_1} \cap g^{-1} \sqbrk {U_2} \subseteq V$
Hence $V$ is open in $T_A$.
Hence $W = T_A \setminus V$ is closed in $T_A$.
$\blacksquare$
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $4$: The Hausdorff condition: Exercise $4.3: 6$