hcju/proofwikiqa huggingface dataset:
=== Forward Implication === Follows directly from Set in Discrete Topology is Clopen. {{qed|lemma}} === Reverse Implication === {{MissingLinks}} Suppose that: : For all $x \in X$: $\{ x \} \in \tau$ Let $S \subseteq X$. Then $S = \bigcup \left\{{ \{ s \}: s \in S }\right\}$. Then since each $\{ s \}$ is open, and a union of open sets is open, $S$ is open. Since this holds for all $S \subseteq X$, $\tau$ is the discrete topology. {{qed}} Category:Topology 8ze217dzd9ex1x6k6yqayam6ogag6xk
From Characteristic Function Determined by 1-Fiber, $\chi_S$ is the mapping determined by: :$\forall s \in S: \chi_S \left({s}\right) = 1 \iff s \in S$ Thus: :$\forall s \in S: \chi_S \left({s}\right) = 1$ By definition of constant mapping: :$\chi_S = f_1$ {{qed}} Category:Characteristic Functions gf9zrtuostpsz441vhfkigouidtz6u6
By Supremum and Infimum are Unique we need only show that $s$ is a supremum of $L$ {{iff}} it is a supremum of $T$.  === Forward Implication === Let $s$ be a supremum of $T$. $s$ is an upper bound of $L$: Let $l \in L$. Then by the definition of lower closure, there is a $t \in T$ such that $l \preceq t$. By the definition of supremum, $s$ is an upper bound of $T$. Thus $t \preceq s$. Since $\preceq$ is transitive, $l \preceq s$. Since this holds for all $l \in L$, $s$ is an upper bound of $L$. Suppose that $u$ is an upper bound of $L$. Then since $T \subseteq L$, $u$ is an upper bound of $T$. Thus by the definition of supremum, $s \preceq u$. So: : $s$ is an upper bound of $L$ : $s$ precedes all upper bounds of $L$ Thus it follows that $s$ is the supremum of $L$. {{qed|lemma}} === Reverse Implication === Let $s$ be a supremum of $L$. Then $s$ is an upper bound of $L$. Since $T \subseteq L$, $s$ is an upper bound of $T$. Let $u$ be any upper bound of $T$ and let $l \in L$. Then by the definition of lower closure, there is a $t \in T$ such that $l \preceq t$. Then since $u$ is an upper bound of $T$, $t \preceq u$. Thus since $\preceq$ is transitive, $l \preceq u$. Thus $u$ is an upper bound of $L$. By the definition of supremum, $s \preceq u$. Thus $s$ is an upper bound of $T$ which precedes every upper bound of $T$. Therefore $s$ is the supremum of $T$. {{qed}} Category:Lower Closures ksm1xyro52wfs8c899gn9ltzff1h9jw
Follows from Upper Closure is Closure Operator and Set Closure is Smallest Closed Set/Closure Operator. {{qed}}
=== Inflationary === Let $T \subseteq S$. Let $t \in T$. Then since $T \subseteq S$, $t \in S$ by the definition of subset. Since $\preceq$ is reflexive, $t \preceq t$. Thus by the definition of upper closure, $t \in T^\succeq$. Since this holds for all $t \in T$, $T \subseteq T^\succeq$. Since this holds for all $T \subseteq S$: : $\cdot^\succeq$ is inflationary. {{qed|lemma}} === Order-Preserving=== Let $T \subseteq U \subseteq S$. Let $x \in T^\succeq$. Then by the definition of upper closure: for some $t \in T$, $t \preceq x$. By the definition of subset: : $t \in U$ Thus by the definition of upper closure: : $x \in U^\succeq$ Since this holds for all $x \in T^\succeq$: : $T^\succeq \subseteq U^\succeq$ Since this holds for all $T$ and $U$: : $\cdot^\succeq$ is order-preserving. {{qed|lemma}} === Idempotent === Let $T \subseteq S$. By Upper Closure is Upper Set, $T^\succeq$ is an upper set. Thus by Equivalence of Definitions of Upper Set: : $\left({T^\succeq}\right)^\succeq = T^\succeq$ Since this holds for all $T$: : $\cdot^\succeq$ is idempotent. {{qed}}
=== Definition 1 implies Definition 2 === Suppose that: :$\forall u \in U: \forall s \in S: u \preceq s \implies s \in U$ Let $k \in U^\succeq$. Then by the definition of upper closure, there is some $u \in U$ such that $u \preceq k$. Since $k \in U^\succeq \subseteq S$, the premise proves that $k \in U$. Since this holds for all $k \in U^\succeq$, it follows that: : $U^\succeq \subseteq U$ {{qed|lemma}} === Definition 2 implies Definition 3 === Suppose that $U^\succeq \subseteq U$. Let $u \in U$. Then since $U \subseteq S$, $u \in S$ by the definition of subset. Since $\preceq$ is reflexive: : $u \preceq u$ Thus by the definition of upper closure: : $u \in U^\succeq$. Since this holds for all $u \in U$: : $U \subseteq U^\succeq$ Thus by definition of set equality: : $U^\succeq = U$ {{qed|lemma}} === Definition 3 implies Definition 1 === Suppose that $U^\succeq = U$. Let $u \in U$. Let $s \in S$. Let $u \preceq s$. Then by the definition of upper closure, $s \in U$. Thus we have shown that: :$\forall u \in U: \forall s \in S: u \preceq s \implies s \in U$ {{qed}} Category:Upper Sets pfk2cix6ah4phtbuf7v3g63yku3xwvj
=== Extensive === Follows from Set is Subset of its Topological Closure. === Increasing === Follows immediately from Topological Closure of Subset is Subset of Topological Closure. === Idempotent === Follows immediately from Closure of Topological Closure equals Closure. {{MissingLinks|the three concepts in the headers (link in the ''text'')}} {{qed}} Category:Set Closures Category:Examples of Closure Operators 7j6hyd3oqs9bvg7unfcg2u4munbralz
Let $\mathcal Q$ be the set of reflexive relations on $S$. By Intersection of Reflexive Relations is Reflexive, the intersection of any subset of $\mathcal Q$ is in $Q$. By the definition of reflexive closure as the intersection of reflexive supersets: :The reflexive closure of a relation $\mathcal R$ on $S$ is the intersection of elements of $\mathcal Q$ that contain $S$. From Closure Operator from Closed Sets we conclude that reflexive closure is a closure operator. {{qed}}
=== Reflexive Closure is Inflationary === {{:Reflexive Closure is Inflationary}}{{qed|lemma}} === Reflexive Closure is Order Preserving === {{:Reflexive Closure is Order Preserving}}{{qed|lemma}} === Reflexive Closure is Idempotent === {{:Reflexive Closure is Idempotent}}{{qed|lemma}} Thus by the definition of closure operator, reflexive closure is a closure operator. {{qed}}
By definition, $\map \cl T$ is closed. Let $C$ be closed. Let $T \subseteq C$. By the definition of closure operator, $\cl$ is $\subseteq$-increasing. So: :$\map \cl T \subseteq \map \cl C$ Since $C$ is closed, $\map \cl C = C$. So: :$\map \cl T \subseteq C$ Thus $\map \cl T$ is the smallest closed set containing $T$ as a subset. {{qed}} Category:Closure Operators Category:Set Closure is Smallest Closed Set rgb5hevh3a910j1evzkrqqrzjppg3ey
By the Duality Principle, it suffices to prove that: :$(1^*)$, $(2^*)$ and $(3^*)$ are equivalent where these are the dual statements of $(1)$, $(2)$ and $(3)$, respectively. By Dual Pairs, it can be seen that these dual statements are as follows: {{begin-axiom}} {{axiom | n = 1^*         | m = \forall u \in U: \forall s \in S: u \preceq s \implies s \in U }} {{axiom | n = 2^*         | m = U^\succeq \subseteq U }} {{axiom | n = 3^*         | m = U^\succeq = U }} {{end-axiom}} Their equivalence is proved on Equivalence of Definitions of Upper Set. {{qed}} Category:Lower Sets f08gllkp30urt8cig57kk7zesq4jvog
$(1) \implies (2)$: Suppose that there is an infinite subsequence $\left({ a_{n_k} }\right)_{k \in \N}$ such that for each $k$, $a_{n_k} \neq 0$. For $N \in \N$ let  :$\displaystyle s_N = \sum_{n \mathop = 1}^N a_n$ To show that $s_N$ diverges it suffices to show that: :$\forall M > 0\ \exists N \in \N : \forall n > N : \left\vert{ s_n }\right\vert > M$ Since for each $n$, $a_n \ge 0$, $s_N$ is a positive increasing sequence in $N$. Therefore it suffices to show that: :$\forall M > 0\ \exists N \in \N :  s_N  > M$ Fix $M > 0$. Let $k$ be any positive integer such that $n_k > M$. Then we have: {{begin-eqn}} {{eqn | l = s_{n_k}       | r = \sum_{n \mathop = 1}^{n_k} a_n }} {{eqn | r =\sum_{n \mathop = 1}^{n_k} 1       | o = \ge       | c = as the $a_n$ are positive and non-zero }} {{eqn | r = n_k       | c = }} {{eqn | r = M       | c = By the choice of $n_k$ }} {{end-eqn}} Therefore the sequence $s_N$ diverges. $(2) \implies (1)$: Suppose there exists $N > 0$ such that $a_n = 0$ for all $n > N$. Then we have, for all $L > N$: :$\displaystyle s_L = \sum_{n \mathop = 1}^L a_n = \sum_{n \mathop = 1}^N a_n = s_N$ In particular, for any $\epsilon > 0$ and all $L > N$: :$\left\vert{s_L - s_N}\right\vert = 0 < \epsilon$ Therefore the sequence converges to $s_N$. {{Qed}} Category:Natural Numbers Category:Series p7ollayrl8lk9jknqx3lw4nfcmlpv5f
First we will show that $\cl$ is a closure operator. === Inflationary === Let $T \subseteq S$. By Set Intersection Preserves Subsets/General Result/Corollary, $T \subseteq \map \cl T$. Since this holds for all such $T$, $\cl$ is inflationary. {{qed|lemma}} === Increasing === Let $T \subseteq U \subseteq S$. Let $\TT$ and $\UU$ be the sets of elements of $\CC$ containing $T$ and $U$, respectively. Since Subset Relation is Transitive, every set containing $U$ contains $T$, so $\UU \subseteq \TT$. By Intersection is Decreasing, $\bigcap \TT \subseteq \bigcap \UU$. Thus $\map \cl T \subseteq \map \cl U$. {{qed|lemma}} === Idempotent === Let $T \subseteq S$. By the premise, the intersection of a subset of $\CC$ is in $\CC$. Thus in particular $\map \cl T \in \CC$. Therefore: :$\map \cl {\map \cl T} \subseteq \map \cl T$ Since $\cl$ is inflationary: :$\map \cl T \subseteq \map \cl {\map \cl T}$ By definition of set equality: :$\map \cl {\map \cl T} = \map \cl T$ Since this holds for all $T \subseteq S$, $\cl$ is idempotent. {{qed|lemma}} Finally, we need to show that the elements of $\CC$ are the closed sets with respect to $\cl$. If $C \in \CC$, then since $\cl$ is inflationary: :$C \subseteq \map \cl C$ But since $C \subseteq C$, $\map \cl C \subseteq C$. Thus by definition of set equality: :$\map \cl C = C$ so $C$ is closed with respect to $\cl$. Suppose instead that $C$ is closed with respect to $\cl$. Then $\map \cl C = C$. Since $\CC$ is closed under intersections, $C \in \CC$.  {{qed}} Category:Closure Operators 8l27ak827j0rbdlqdhr4a1qkp4tt57z
Let $\mathcal F \subseteq \mathcal G$. Let $x \in \bigcap \mathcal G$. Then for each $S \in \mathcal F$, $S \in \mathcal G$. By the definition of intersection, $x \in S$. Since this holds for all $S \in \mathcal F$, $x \in \bigcap \mathcal F$. Since this holds for all $ x \in \bigcap \mathcal G$: : $\bigcap \mathcal G \subseteq \bigcap \mathcal F$ {{qed}} Category:Set Intersection n0i62gr2ponbnrgzwsyc1rt36hg0id4
=== Definition 2 implies Definition 1 === A closure operator, by definition, is inflationary and idempotent. Thus it follows immediately that Definition 2 implies Definition 1. === Definition 1 implies Definition 2 === Let $X$ be a set. Let $\cl$ be a Kuratowski closure operator on $X$ by Definition 1. By definition of closure operator, it remains to be proved that $\cl$ is increasing. Let $A \subseteq B \subseteq X$. Then by Definition 1 and Union with Superset is Superset: :$\map \cl B = \map \cl {A \cup B} = \map \cl A \cup \map \cl B$ By Set is Subset of Union: :$\map \cl A \subseteq \map \cl A \cup \map \cl B = \map \cl B$ {{qed}} Category:Topology Category:Closure Operators 96izmdzooovx1zcwt4nhvtnw8glo56r
We will prove that $p^\succ$ is open. {{explain|follow by duality how?}} That $p^\prec$ is open will follow by duality. By Inverse Image under Order Embedding of Strict Upper Closure of Image of Point: :$\map {\phi^{-1} } {\map \phi p^\succ} = p^\succ$ :$\map \phi p^\succ$ is an open ray in $S'$ Therefore $\tau'$-open by the definition of the order topology. {{explain|What exactly is $\tau'$-open here?}} Since $\phi$ is a topological embedding, it is continuous. Thus $p^\succ$ is $\tau$-open. {{qed}}
Let $\mathcal F \subseteq \mathcal G$. Let $x \in \bigcup \mathcal F$. Then by the definition of union, for some $S \in \mathcal F$, $x \in S$. By the definition of subset, $S \in \mathcal G$. Thus by the definition of union, $x \in \bigcup \mathcal G$. Since this holds for all $x \in \bigcup \mathcal F$: : $\bigcup \mathcal F \subseteq \bigcup \mathcal G$ {{qed}} Category:Set Union mebmzub0fzw8n5o285nhwbfmgikasrm
First we show that: :$\displaystyle \bigcap_{i \mathop \in I} F_i \subseteq S$ Let $x \in \displaystyle \bigcap_{i \mathop \in I} F_i$. Since $I$ is non-empty, it has an element $k$. By the definition of intersection, $x \in F_k$. By the premise, $F_k = S$, so $x \in S$. Since this holds for all $x \in \displaystyle \bigcap_{i \mathop \in I} F_i$: :$\displaystyle \bigcap_{i \mathop \in I} F_i \subseteq S$ Next we show that: :$\displaystyle S \subseteq \bigcap_{i \mathop \in I} F_i$ Let $x \in S$. Then for all $i \in I$, $F_i = S$, so $x \in F_i$. Thus by the definition of intersection: :$x \in \displaystyle \bigcap_{i \mathop \in I} F_i$ Since this holds for all $x \in S$: :$S \subseteq \displaystyle \bigcap_{i \mathop \in I} F_i$ By definition of set equality: :$\displaystyle \bigcap_{i \mathop \in I} F_i = S$ {{qed}} Category:Set Intersection Category:Indexed Families Category:Idempotence h4ptw6dbvab2024vremmg8oo427n5xt
{{BeginTableau   |\paren {p \implies q} \land \paren {r \implies s} \vdash \paren {p \land r} \implies \paren {q \land s} }} {{Premise        |1|\paren {p \implies q} \land \paren {r \implies s} }} {{Simplification |2|1|p \implies q|1|1}} {{Simplification |3|1|r \implies s|1|2}} {{SequentIntro   |4|1|p \land r \implies q \land r|2|Factor Principles/Conjunction on Right/Formulation 1/Proof 2}} {{SequentIntro   |5|1|q \land r \implies q \land s|3|Factor Principles/Conjunction on Left/Formulation 1/Proof 2}} {{SequentIntro   |6|1|p \land r \implies q \land s|4, 5|Hypothetical Syllogism}} {{EndTableau|qed}}
{{BeginTableau|p \implies q \vdash \left({p \land r}\right) \implies \left ({q \land r}\right)}} {{Premise|1|p \implies q}} {{IdentityLaw|2||r \implies r|(None)|This is a theorem so depends on nothing}} {{Conjunction|3|1|\left({p \implies q}\right) \land \left({r \implies r}\right)|1|2}} {{SequentIntro|4|1|\left({p \land r}\right) \implies \left ({q \land r}\right)|3|Praeclarum Theorema}} {{EndTableau}} {{qed}}
{{BeginTableau|p \implies q \vdash \paren {r \land p} \implies \paren {r \land q} }} {{Premise|1|p \implies q}} {{IdentityLaw|2||r \implies r|(None)|This is a theorem so depends on nothing}} {{Conjunction|3|1|\paren {r \implies r} \land \paren {p \implies q}|2|1}} {{SequentIntro|4|1|\paren {r \land p} \implies \paren {r \land q}|3|Praeclarum Theorema}} {{EndTableau}} {{qed}}
{{BeginTableau|p \implies q \vdash \paren {p \lor r} \implies \paren {q \lor r} }} {{Premise|1|p \implies q}} {{Assumption|2|p \lor r}} {{Commutation|3|2|r \lor p|2|Disjunction}} {{SequentIntro|4|1|\paren {r \lor p} \implies \paren {r \lor q}|1|Factor Principles/Disjunction on Left/Formulation 1}} {{ModusPonens|5|1,2|r \lor q|4|3}} {{Commutation|6|1,2|q \lor r|5|Disjunction}} {{Implication|7|1|\paren {p \lor r} \implies \paren {q \lor r}|2|6}} {{EndTableau}} {{qed}}
{{BeginTableau|p \implies q \vdash \paren {p \lor r} \implies \paren {q \lor r} }} {{Premise|1|p \implies q}} {{TheoremIntro|2|r \implies r|Law of Identity: Formulation 2}} {{SequentIntro|3|1|\paren {p \lor r} \implies \paren {q \lor r}|1, 2|Constructive Dilemma}} {{EndTableau}} {{qed}}
As can be seen for all boolean interpretations by inspection, where the truth value under the main connective on the {{LHS}} is $T$, that under the one on the {{RHS}} is also $T$: :$\begin{array}{|ccc||ccc||ccccccc|} \hline p & q & r & (p & \implies & q) & (p & \lor & r) & \implies & (q & \lor & r) \\ \hline F & F & F & F & T & F & F & F & F & T & F & F & F \\ F & F & T & F & T & F & F & T & T & T & F & T & T \\  F & T & F & F & T & T & F & F & F & T & T & T & F \\ F & T & T & F & T & T & F & T & T & T & T & T & T \\ T & F & F & T & F & F & T & T & F & F & F & F & F \\ T & F & T & T & F & F & T & T & T & T & F & T & T \\ T & T & F & T & T & T & T & T & F & T & T & T & F \\ T & T & T & T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$ {{qed}}
By Join Semilattice is Ordered Structure, $\preceq$ is compatible with $\vee$. By the definition of ordering, $\preceq$ is transitive. Thus the theorem holds by Operating on Transitive Relationships Compatible with Operation. {{qed}} Category:Lattice Theory gsg75kiusqg9sajkk7sl9l8iitilbgy
By Meet Semilattice is Ordered Structure, $\preceq$ is compatible with $\wedge$. By the definition of ordering, $\preceq$ is transitive. Thus the theorem holds by Operating on Transitive Relationships Compatible with Operation. {{qed}}
Let $x \in f \left[{I}\right]$. Then: :$\exists j \in I: f \left({j}\right) = x$ Then: :$f \left({j}\right) \prec g \left({j}\right)$ By the definition of supremum: :$\sup g \left[{I}\right]$ is an upper bound of $g \left[{I}\right]$ Thus: :$g \left({j}\right) \preceq \sup g \left[{I}\right]$ Since $\preceq$ is transitive: :$x = f \left({j}\right) \preceq \sup g \left[{I}\right]$ Since this holds for all $x \in f \left[{I}\right]$, $\sup g \left[{I}\right]$ is an upper bound of $f \left[{I}\right]$. Thus by the definition of supremum: :$\sup f \left[{I}\right] \preceq \sup g \left[{I}\right]$ {{qed}} Category:Order Theory Category:Increasing Mappings t4ucd6b4ofbzzdyz2idqo0fxee3o15s
Since $\prec$ is a strict ordering, it is by definition transitive and asymmetric. By Asymmetric Relation is Antisymmetric, $\prec$ is antisymmetric. Thus by Reflexive Closure of Transitive Antisymmetric Relation is Ordering, $\preceq$ is an ordering. {{qed}}
Recall the definition of reflexive closure: :$\mathcal R^= := \mathcal R \cup \Delta_S$ From Set is Subset of Union: :$\Delta_S \subseteq \mathcal R^=$ The result follows directly from Relation Contains Diagonal Relation iff Reflexive. {{qed}}
Let $\RR$ be a relation on a set $S$. === Union with Diagonal is Smallest Reflexive Superset === Let $\Delta_S$ be the diagonal relation on $S$. Let $\RR^= = \RR \cup \Delta_S$ By Smallest Element is Unique, at most one relation on $S$ can be the smallest reflexive superset of $\RR$. From Subset of Union: :$\RR \subseteq \RR^=$ :$\Delta_S \subseteq \RR^=$ By Relation Contains Diagonal Relation iff Reflexive, $\RR^=$ is reflexive. Thus $\RR^=$ is a reflexive relation containing $\RR$. Again by Relation Contains Diagonal Relation iff Reflexive, ''every'' reflexive relation containing $\RR$ must also contain $\Delta_S$. From Union is Smallest Superset, it follows that $\RR^=$ is the smallest reflexive relation on $S$ which contains $\RR$. {{qed|lemma}} === Intersection of Reflexive Supersets is Union with Diagonal === Let $\QQ$ be the set of all reflexive relations containing $\RR$ as a subset. Let $\RR^= = \bigcap \QQ$. By the above proof that $\RR \cup \Delta_S$ is a reflexive relation containing $\RR$: :$\RR \cup \Delta_S \in \QQ$ By Intersection is Subset: :$\RR^= \subseteq \RR \cup \Delta_S$ By the above proof that $\RR \cup \Delta_S$ is the smallest reflexive relation containing $\RR$: :$\forall \PP \in \QQ: \RR \cup \Delta_S \subseteq \PP$ By Intersection is Largest Subset: :$\RR \cup \Delta_S \subseteq \RR^=$ Thus by definition of set equality: :$\RR^= = \RR \cup \Delta_S$ {{qed}} Category:Reflexive Closures fu65mekifuk8slb72c8z8wsj4zcft5m
Let $\mathcal R \in R$. The reflexive closure $\mathcal R^=$ of $\mathcal R$ is defined as: :$\mathcal R^= := \mathcal R \cup \Delta_S$ From Set is Subset of Union: :$\mathcal R \subseteq \mathcal R^=$ Hence the reflexive closure operator is an inflationary mapping.
Let $\RR, \SS \in R$. Suppose: :$\RR \subseteq \SS$ Their respective reflexive closures $\RR^=$ and $\SS^=$ are defined as: :$\RR^= := \RR \cup \Delta_S$ :$\SS^= := \SS \cup \Delta_S$ Hence by Corollary to Set Union Preserves Subsets: :$\RR^= \subseteq \SS^=$
Let $\RR \in R$. By the definition of reflexive closure: :$\RR^= = \RR \cup \Delta_S$ :$\paren {\RR^=}^= = \paren {\RR \cup \Delta_S} \cup \Delta_S$ By Union is Associative: :$\paren {\RR^=}^= = \RR \cup \paren {\Delta_S \cup \Delta_S}$ By Union is Idempotent: :$\paren {\RR^=}^= = \RR \cup \Delta_S$ Hence: :$\forall \RR \in R: \RR^= = \paren {\RR^=}^=$
Let $\RR$ be a relation on a set $S$. First note that by Smallest Element is Unique, once it has been shown that some relation, $\QQ$, is the smallest transitive superset of $\RR$, it is the only such. Thus we need only prove that each of the other definitions lead to relations with this property. First we have: === Intersection of Transitive Supersets is Smallest Transitive Superset === {{:Equivalence of Definitions of Transitive Closure (Relation Theory)/Intersection is Smallest}} === The Finite Chain Definition is Equivalent to the Union of Compositions Definition === {{:Equivalence of Definitions of Transitive Closure (Relation Theory)/Finite Chain Equivalent to Union of Compositions}} === Union of Compositions is Smallest Transitive Superset === {{:Equivalence of Definitions of Transitive Closure (Relation Theory)/Union of Compositions is Smallest}} === Finite Chain Definition gives Smallest Transitive Superset === {{:Equivalence of Definitions of Transitive Closure (Relation Theory)/Finite Chain is Smallest}}
First, note that there exists at least one transitive relation containing $\mathcal R$. That is, the trivial relation $S \times S$, which is an equivalence and therefore transitive by definition. Next, note that the Intersection of Transitive Relations is Transitive. Hence the transitive closure of $\mathcal R$ is the intersection of all transitive relations containing $\mathcal R$.
Note that the trivial relation $\mathcal T = S \times S$ on $S$ contains $\mathcal R$, by definition. Further, $\mathcal T$ is transitive by Trivial Relation is Equivalence. Thus there is at least one transitive relation on $S$ that contains $\mathcal R$. Now define $\mathcal R^\cap$ as the intersection of all transitive relations on $S$ that contain $\mathcal R$: :$\displaystyle \mathcal R^\cap := \bigcap \left\{{\mathcal R': \text{$\mathcal R'$ is transitive and $\mathcal R \subseteq \mathcal R'$}}\right\}$ By Intersection of Transitive Relations is Transitive, $\mathcal R^\cap$ is also a transitive relation on $S$. By Set Intersection Preserves Subsets, it also holds that $\mathcal R \subseteq \mathcal R^\cap$. Lastly, by Intersection is Subset, for any transitive relation $\mathcal R'$ containing $\mathcal R$, it must be that $\mathcal R^\cap \subseteq \mathcal R'$. Thus $\mathcal R^\cap$ is indeed the minimal transitive relation on $S$ containing $\mathcal R$. That is, $\mathcal R^+ = \mathcal R^\cap$, and thence the transitive closure of $\mathcal R$ exists. {{qed}}
==== $\RR^+$ is Transitive ==== By Relation contains Composite with Self iff Transitive, we can prove that $\RR^+$ is transitive by proving the following: :$\RR^+ \circ \RR^+ \subseteq \RR^+$ Let $\tuple {a, c} \in \RR^+ \circ \RR^+$. Then: :$\exists b \in S: \tuple {a, b} \in \RR^+, \tuple {b, c} \in \RR^+$ Thus: :$\exists n \in \N: \tuple {a, b} \in \RR^n$ :$\exists m \in \N: \tuple {b, c} \in \RR^m$ From Composition of Relations is Associative: :$\RR^{n + m} = \RR^n \circ \RR^m$ so: :$\tuple {a, c} \in \RR^{n + m} \subseteq \RR^+$ Since this holds for all $\tuple {a, c} \in \RR^+ \circ \RR^+$: :$\RR^+ \circ \RR^+ \subseteq \RR^+$ Thus $\RR^+$ is transitive. {{qed|lemma}} ==== $\RR^+$ contains $\RR$ ==== $\RR \subseteq \RR^+$ by Set is Subset of Union. ==== $\RR^+$ is Smallest ==== Let $\RR'$ be a transitive relation on $S$ such that $\RR \subseteq \RR'$. We must show that $\RR^+ \subseteq \RR'$. Let $\tuple {a, b} \in \RR^+$. That is: :$a \mathrel \RR b$ Then: :$\exists n \in \N: \tuple {a, b} \in \RR^n$ Thus by the definition of composition of relations, there exists $x_{n-1} \in S$ such that: :$a \mathrel {\RR^{n - 1} } x_{n - 1} \land x_{n - 1} \mathrel \RR b$ Likewise there exists $x_{n-2} \in S$ such that: :$a \mathrel {\RR^{n - 2} } x_{n - 2} \land x_{n - 2} \mathrel \RR x_{n - 1}$   And so forth there exist elements $x_0, \dots, x_n \in S$ such that: :$x_0 = a$ :$x_n = b$ :$\forall k \in \N_n: x_k \mathrel \RR x_{k + 1}$ Since $\RR \subseteq \RR'$: :$\forall k \in \N_n: x_k \mathrel {\RR'} x_{k + 1}$ Since $\RR'$ is transitive: :$a \mathrel {\RR'} b$ That is: :$\tuple {a, b} \in \RR'$ Since this holds for all $\tuple {a, b} \in \RR^+$: :$\RR^+ \subseteq \RR'$ Since this holds for all transitive relations $\RR'$ that contain $\RR$: $\RR^+$ is the smallest transitive relation containing $\RR$. {{qed}}
{{explain|more detail required}} Follows from the definition of composition of relations. {{qed}}