In order to prove equivalence, it is to be shown that if a subset of a topological space is not connected by one of the given definitions above, then it will likewise not be connected by one of the other definitions. Equivalence of definitions for connected set will then follow by the Rule of Transposition. === Definition 1 implies Definition 2=== Suppose that $H$ is not connected by Definition 1. Then: :$H = A \cup B$ where $A$ and $B$ are non-empty separated sets of $T$. By Separated Sets are Clopen in Union, $A$ and $B$ are open in $H$, considered with the subspace topology $\tau_H$. Thus there exist open sets $U$ and $V$ in $S$ such that: :$A = U \cap H$ and: :$B = V \cap H$ Thus $U \cap H \ne \O$ and $V \cap H \ne \O$ such that: {{begin-eqn}} {{eqn | l = H       | r = A \cup B }} {{eqn | r = \paren {U \cap H} \cup \paren {V \cap H} }} {{eqn | r = \paren {U \cup V} \cap H }} {{eqn | o = \subseteq       | r = U \cup V }} {{end-eqn}} It remains to be shown that: :$U \cap V \cap H = \O$ Thus: {{begin-eqn}} {{eqn | l = U \cap V \cap H       | r = \paren {U \cap H} \cap \paren {V \cap H} }} {{eqn | r = A \cap B }} {{eqn | r = \O }} {{end-eqn}} So $H$ is not connected by Definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Suppose that $H$ is not connected by Definition 2. Then there exist open sets $U$ and $V$ in $S$ such that: :$U \cap H \ne \O$ :$V \cap H \ne \O$ :$H \subseteq U \cup V$ :$U \cap V \cap H = \O$ Let $A = U \cap H$ and let $B = V \cap H$. Then by hypothesis: :$A \ne \O$ and: :$B \ne \O$ Since $H \subseteq U \cup V$, it follows that: :$H = H \cap \paren {U \cup V} = \paren {H \cap U} \cup \paren {H \cap V} = A \cup B$ It remains to be shown that $A$ and $B$ are separated sets of $T$. Since $H = A \cup B$ and $U \cap V \cap H = \O$, it follows that: :$A = H \setminus B$ Since $V$ is open in $\struct {S, \tau}$ it follows that $B$ is open in $\struct {H, \tau_H}$, where $\tau_H$ is the subspace topology. Thus by definition $A$ is closed in $\struct {H, \tau_H}$. By Closed Set equals its Closure: :$A = A^- \cap H$ We have that $B \subseteq H$. Therefore: {{begin-eqn}} {{eqn | l = A^- \cap B       | o = \subseteq       | r = A^- \cap B \cap H }} {{eqn | r = A \cap B }} {{eqn | r = \paren ({U \cap H} \cap \paren {V \cap H} }} {{eqn | r = U \cap V \cap H }} {{eqn | r = \O }} {{end-eqn}} A similar argument shows that $A \cap B^- = \O$. Thus, by definition, $A$ and $B$ are separated sets of $T$. That is, $H$ is not connected by Definition 1. {{qed|lemma}} === Definition 2 implies Definition 3 === Suppose that $H$ is not connected by Definition 2. Then there exist open sets $U$ and $V$ in $S$ such that: :$U \cap H \ne \O$ :$V \cap H \ne \O$ :$H \subseteq U \cup V$ :$U \cap V \cap H = \O$ Let $U' = U \cap H$ and $V' = V \cap H$. By assumption, $U'$ and $V'$ are non-empty. By the definition of the subspace topology, $U'$ and $V'$ are open in $\struct {H, \tau_H}$. Since $H \subseteq U \cup V$, it follows that: {{begin-eqn}} {{eqn | l = H       | r = H \cap \paren {U \cup V} }} {{eqn | r = A \cap B }} {{eqn | r = \paren {H \cap U} \cup \paren {H \cap V} }} {{eqn | r = U' \cup V' }} {{end-eqn}} Thus $U'$ and $V'$ form a separation of $H$. Thus $H$ is not a connected topological space. That is, $H$ is not connected by Definition 3. {{qed|lemma}} === Definition 3 implies Definition 2 === Let $\tau_H$ be the subspace topology induced on $H$ by $\tau$. Suppose that $H$ is not connected by Definition 3. Thus, by definition of connected topological space, $\struct {H, \tau_H}$ admits a separation. That is, there exist non-empty disjoint open sets $U'$ and $V'$ of $\struct {H, \tau_H}$ such that: :$H = U' \cup V'$. By the definition of the subspace topology, there exist open sets $U$ and $V$ of $\struct {S, \tau}$ such that: :$U' = U \cap H$ and: :$V' = V \cap H$ Then: {{begin-eqn}} {{eqn | l = U \cap V \cap H       | r = \paren {U \cap H} \cap \paren {V \cap H} }} {{eqn | r = U' \cap V' }} {{eqn | r = \O }} {{end-eqn}} Since $U'$ and $V'$ are non-empty: :$U' = U \cap H$ :$V' = V \cap H$ :$U \cap H \ne \O$ :$V \cap H \ne \O$ Then because: :$H = U' \cup V'$ :$U' \subseteq U$ :$V' \subseteq V$ it follows that: :$H \subseteq U \cup V$ Thus we have: :$U \cap H \ne \O$ :$V \cap H \ne \O$ :$H \subseteq U \cup V$ :$U \cap V \cap H = \O$ That is, $H$ is not connected by Definition 2. {{qed|lemma}} Hence it follows that all the definitions of connected set are equivalent. {{qed}}