Inverse of Open Set in Product Space is Open in Inverse Product Space
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Theorem
Let $\struct{S_1, \tau_1}$ and $\struct{S_2, \tau_2}$ be topological spaces.
Let $\tau_{S_1 \times S_2}$ be the product topology on the Cartesian product $S_1 \times S_2$.
Let $\tau_{S_2 \times S_1}$ be the product topology on the Cartesian product $S_2 \times S_1$.
Let $V \subseteq S_1 \times S_2$.
Let $V^{-1} \subseteq S_2 \times S_1$ denote the inverse of $V$ viewed as a relation on $S_1 \times S_2$.
Then:
- $V$ is open in $\struct{S_1 \times S_2, \tau_{S_1 \times S_2}}$ if and only if $V^{-1}$ is open in $\struct{S_2 \times S_1, \tau_{S_2 \times S_1}}$
Proof
Necessary Condition
Let $V$ be open in $\struct{S_1 \times S_2, \tau_{S_1 \times S_2}}$.
Let $\tuple{s_2, s_1} \in V^{-1}$
By definition of inverse:
- $\tuple{s_1, s_2} \in V$
By definition of product topology:
- $\BB_{12} = \set{U \times W : U \in \tau_1, W \in \tau_2}$ is a basis for $\tau_{S_1 \times S_2}$
By definition of basis:
- $\exists U \in \tau_1, W \in \tau_2 : \tuple{s_1, s_2} \in U \times W \subseteq V$
By definition of inverse:
- $\tuple{s_2, s_1} \in W \times U \subseteq V^{-1}$
By definition of product topology:
- $\BB_{21} = \set{W \times U : W \in \tau_2, U \in \tau_1}$ is a basis for $\tau_{S_2 \times S_1}$
Since $\tuple{s_2, s_1}$ was arbitrary, we have:
- $\forall \tuple{s_2, s_1} \in V^{-1} : \exists W \times U \in \BB_{21} : \tuple{s_2, s_1} \in W \times U \subseteq V^{-1}$
From Characterization of Set Equals Union of Sets:
- $V^{-1} = \bigcup \set{W \times U \in \BB_{21} : W \times U \subseteq V^{-1}}$
By Open Set Axiom $\paren {\text O 1 }$: Union of Open Sets:
- $V^{-1}$ is open in $\struct{S_2 \times S_1, \tau_{S_2 \times S_1}}$
$\Box$
Sufficient Condition
Let $V^{-1}$ be open in $\struct{S_2 \times S_1, \tau_{S_2 \times S_1}}$.
From Necessary Condition above:
- $\paren{V^{-1}}^{-1}$ is open in $\struct{S_1 \times S_2, \tau_{S_1 \times S_2}}$
From Inverse of Inverse Relation:
- $\paren{V^{-1}}^{-1} = V$
Hence:
- $V$ is open in $\struct{S_1 \times S_2, \tau_{S_1 \times S_2}}$
$\blacksquare$