User:Leigh.Samphier/Topology/Homeomorphism Preserves Sobriety

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Theorem

Let $T_1 = \struct{S_1, \tau_1}$ be a sober space.

Let $T_2 = \struct{S_2, \tau_2}$ be a topological space.

Let $f : T_1 \to T_2$ be a homeomorphism.

Then $T_2$ is a sober space.

Proof

From Sober Space iff Completely Prime Filter is Unique System of Open Neighborhoods it is sufficent to show:

for each completely prime filter $\FF$ in the complete lattice $\struct{\tau_2, \subseteq}$:

$\exists ! y \in S_2 : \FF = \map \UU y$

Let $g$ denote the inverse of $f$.

From Inverse of Homeomorphism is Homeomorphism:

$g : T_2 \to T_1$ is a homeomorphism

Let $\FF$ be a completely prime filter in $\struct{\tau_2, \subseteq}$.

From Frame Homomorphism of Continuous Mapping is Frame Homomorphism:

the mapping $\map \Omega g : \tau_1 \to \tau_2$ defined by: $\map {\map \Omega g} U = \map {g^\gets} U$

is a frame homomorphism, where:

$g^\gets : \powerset {S_1} \to \powerset {S_2}$ denotes the inverse image mapping of $g$

From Inverse Image Mapping of Frame Homomorphism Preserves Completely Prime Filter:

$\map {\map \Omega g^\gets} \FF$ is a completely prime filter of $\struct{\tau_1, \subseteq}$

We have by hypothesis:

$T_1$ is a sober space

From Sober Space iff Completely Prime Filter is Unique System of Open Neighborhoods:

$\exists ! x \in S_1 : \map {\map \Omega g^\gets} \FF = \map \UU x$

$\blacksquare$