First we will prove that :$\partial A \subseteq \partial \left({A \cup B}\right) \cup \partial \left({A \cap B}\right) \cup \left({\partial A \cap \partial B}\right)$ Let $x \in \partial A$. Aiming for a contradiction suppose that :$x \notin \partial \left({A \cup B}\right) \cup \partial \left({A \cap B}\right) \cup \left({\partial A \cap \partial B}\right)$ Then by definition of union: :$x \notin \partial \left({A \cup B}\right) \land x \notin \partial \left({A \cap B}\right) \land x \notin \left({\partial A \cap \partial B}\right)$ By Characterization of Boundary by Open Sets: :$\exists Q \in \tau: x \in Q \land \left({\left({A \cup B}\right) \cap Q = \varnothing \lor \complement_S \left({A \cup B}\right) \cap Q = \varnothing}\right)$ By Intersection Distributes over Union. Complement of Union: :$\left({A \cap Q}\right) \cup \left({B \cap Q}\right) = \varnothing \lor \complement_S \left({ A }\right) \cap \complement_S \left({B}\right) \cap Q = \varnothing$ Because $x \notin \left({\partial A \cap \partial B}\right)$ therefore by definition of intersection: :$x \notin \partial B$ By Characterization of Boundary by Open Sets: :$\exists U \in \tau: x \in U \land \left({B \cap U = \varnothing \lor \complement_S \left({B}\right) \cap U = \varnothing}\right)$ As $x \in \partial A$ by Characterization of Boundary by Open Sets: :$A \cap Q \ne \varnothing$ Then by Union is Empty iff Sets are Empty: :$\left({A \cap Q}\right) \cup \left({B \cap Q}\right) \ne \varnothing$ Hence :$\complement_S \left({ A }\right) \cap \complement_S \left({B}\right) \cap Q = \varnothing$ We will show that :$B \cap U = \varnothing \implies \complement_S \left({ A }\right) \cap Q \cap U = \varnothing$ Assume :$B \cap U = \varnothing$ Then: :$U \subseteq \complement_S \left({B}\right)$ By Intersection with Empty Set: :$\complement_S \left({ A }\right) \cap Q \cap \complement_S \left({B}\right) \cap U = \varnothing \cap U = \varnothing$ Thus by Intersection with Subset is Subset: :$\complement_S \left({ A }\right) \cap Q \cap U = \varnothing$ By definition of intersection: :$x \in Q \cap U$ By definition of topological space: :$Q \cap U$ is open. Then by Characterization of Boundary by Open Sets: :$\complement_S \left({A}\right) \cap Q \cap U \ne \varnothing$ Hence: :$B \cap U \ne \varnothing$ Then: :$\complement_S \left({B}\right) \cap U = \varnothing$ Therefore: :$U \subseteq B$ Because $x \notin \partial \left({A \cap B}\right)$ by Characterization of Boundary by Open Sets: :$\exists V \in \tau: x \in V \land \left({A \cap B \cap V = \varnothing \lor \complement_S \left({A \cap B}\right) \cap V = \varnothing}\right)$ By Complement of Intersection: :$A \cap B \cap V = \varnothing \lor \left({\complement_S \left({A}\right) \cup \complement_S \left({B}\right)}\right) \cap V = \varnothing$ By Intersection Distributes over Union: :$A \cap V \cap B = \varnothing \lor \left({\complement_S \left({A}\right) \cap V}\right) \cup \left({\complement_S \left({B}\right) \cap V}\right) = \varnothing$ Because $x \in \partial A$ therefore by Characterization of Boundary by Open Sets: :$\complement_S \left({A}\right) \cap V \ne \varnothing$ Then by Union is Empty iff Sets are Empty: :$\left({\complement_S \left({A}\right) \cap V}\right) \cup \left({\complement_S \left({B}\right) \cap V}\right) \ne \varnothing$ Hence: :$A \cap V \cap B = \varnothing$ By Intersection with Empty Set: :$A \cap V \cap B \cap U = \varnothing \cap U = \varnothing$ By Intersection with Subset is Subset: :$A \cap V \cap U = \varnothing$ By definition of intersection: :$x \in V \cap U$ By definition of topological space: :$V \cap U$ is open. Then by Characterization of Boundary by Open Sets: :$A \cap Q \cap V \ne \varnothing$ This contradicts with $A \cap V \cap U = \varnothing$ Thus the inclusion is proved. {{qed|lemma}} Analogically  :$\partial B \subseteq \partial \left({A \cup B}\right) \cup \partial \left({A \cap B}\right) \cup \left({\partial A \cap \partial B}\right)$ By Union of Subsets is Subset: :$\partial A \cup \partial B \subseteq \partial \left({A \cup B}\right) \cup \partial \left({A \cap B}\right) \cup \left({\partial A \cap \partial B}\right)$ By Boundary of Union is Subset of Union of Boundaries: :$\partial \left({A \cup B}\right) \subseteq \partial A \cup \partial B$ By Boundary of Intersection is Subset of Union of Boundaries: :$\partial \left({A \cap B}\right) \subseteq \partial A \cup \partial B$ By Intersection is Subset of Union: :$\partial A \cap \partial B \subseteq \partial A \cup \partial B$ Hence by Union of Subsets is Subset :$\partial \left({A \cup B}\right) \cup \partial \left({A \cap B}\right) \cup \left({\partial A \cap \partial B}\right) \subseteq \partial A \cup \partial B$ Thus by definition of set equality the result follows: :$\partial A \cup \partial B = \partial \left({A \cup B}\right) \cup \partial \left({A \cap B}\right) \cup \left({\partial A \cap \partial B}\right)$ {{qed}}