Define $B = \left\{ {\complement_X\left({x^{\succeq'}}\right): x \in X}\right\}$ We will prove that :$\forall A \in B: f^{-1}\left[{\complement_X\left({A}\right)}\right]$ is closed. Let $A \in B$. By definition of $B$: :$\exists x \in X: A = \complement_X\left({x^{\succeq'} }\right)$ By Relative Complement of Relative Complement: :$\complement_X\left({A}\right) = x^{\succeq'}$ By Infimum of Upper Closure of Element: :$\inf \left({\complement_X\left({A}\right)}\right) = x$ Suppose that the case: $f^{-1}\left[{\complement_X\left({A}\right)}\right] = \varnothing$ holds. Thus by Empty Set is Closed in Topological Space: :$f^{-1}\left[{\complement_X\left({A}\right)}\right]$ is closed. {{qed|lemma}} Suppose that the case: $f^{-1}\left[{\complement_X\left({A}\right)}\right] \ne \varnothing$ holds. By assumption: :$f$ preserves the infimum of $f^{-1}\left[{\complement_X\left({A}\right)}\right]$ By definitions of mapping preserves the infimum and complete lattice: :$f\left({\inf \left({f^{-1}\left[{x^{\succeq'} }\right]}\right)}\right) = \inf\left({f\left[{f^{-1}\left[{\complement_X\left({A}\right)}\right]}\right]}\right)$ By Image of Preimage under Mapping: :$f\left[{f^{-1}\left[{\complement_X\left({A}\right)}\right]}\right] \subseteq x^{\succeq'}$ By Infimum of Subset and definition of complete lattice: :$x \preceq' f\left({\inf \left({f^{-1}\left[{x^{\succeq'} }\right]}\right)}\right)$ We will prove that :$f^{-1}\left[{\complement_X\left({A}\right)}\right] = {\left({\inf \left({f^{-1}\left[{x^{\succeq'} }\right]}\right)}\right)}^\succeq$ Let $a \in f^{-1}\left[{\complement_X\left({A}\right)}\right]$ By definitions of infimum and lower bound: :$\inf \left({f^{-1}\left[{x^{\succeq'} }\right]}\right) \preceq a$ By definition of upper closure of element: :$a \in {\left({\inf \left({f^{-1}\left[{x^{\succeq'} }\right]}\right)}\right)}^\succeq$ {{qed|lemma}} Let $a \in {\left({\inf \left({f^{-1}\left[{x^{\succeq'} }\right]}\right)}\right)}^\succeq$ By assumption: :$f$ preserves the infimum of $\left\{ {\inf \left({f^{-1}\left[{x^{\succeq'} }\right]}\right), a}\right\}$ By definitions of mapping preserves the infimum and  complete lattice: :$f\left({\left({\inf \left({f^{-1}\left[{x^{\succeq'} }\right]}\right)}\right) \wedge a}\right) = f\left({\inf \left({f^{-1}\left[{x^{\succeq'} }\right]}\right)}\right) \wedge f\left({a}\right)$ By definition of upper closure of element: :$\inf \left({f^{-1}\left[{x^{\succeq'} }\right]}\right) \preceq a$ By Meet Precedes Operands: :$f\left({\inf \left({f^{-1}\left[{x^{\succeq'} }\right]}\right)}\right) = f\left({\inf \left({f^{-1}\left[{x^{\succeq'} }\right]}\right)}\right) \wedge f\left({a}\right)$ By Preceding iff Meet equals Less Operand: :$f\left({\inf \left({f^{-1}\left[{x^{\succeq'} }\right]}\right)}\right) \preceq' f\left({a}\right)$ By definition of transitivity: :$x \preceq' f\left({a}\right)$ By definition of upper closure of element: :$f\left({a}\right) \in x^{\succeq'}$ Thus by definition of preimage of set: :$a \in f^{-1}\left[{\complement_X\left({A}\right)}\right]$ {{qed|lemma}} Thus by Complement of Upper Closure of Element is Open in Lower Topology: :$f^{-1}\left[{\complement_X\left({A}\right)}\right]$ is closed. {{qed|lemma}} We will prove that :$\forall A \in B: f^{-1}\left[{A}\right] \in \tau$ Let $A \in B$. Then by previous: :$f^{-1}\left[{\complement_X\left({A}\right)}\right]$ is closed. By Complement of Preimage equals Preimage of Complement :$f^{-1}\left[{\complement_X\left({A}\right)}\right] = \complement_S\left({f^{-1}\left[{A}\right]}\right)$ Thus by definition of closed set :$f^{-1}\left[{A}\right] \in \tau$ {{qed|lemma}} Thus Continuity Test using Sub-Basis: :$f$ is continuous mapping. {{qed}}