hcju/proofwikiqa huggingface dataset:
By Restriction of Ordering is Ordering, $\preceq \restriction_T$ is an ordering. Let $x, y \in T$. As $T \subseteq S$ it follows by definition of subset that: :$x, y \in S$ As $\preceq$ is a total ordering: :$\tuple {x, y} \in {\preceq}$ or: :$\tuple {y, x} \in {\preceq}$ Suppose $\tuple {x, y} \in {\preceq}$. As $x, y \in T$, it follows by definition of cartesian product that: :$\tuple {x, y} \in T \times T$ Thus: :$\tuple {x, y} \in \paren {T \times T} \cap {\preceq}$ By definition of the restriction of $\preceq$ to $T$: :$\paren {T \times T} \cap {\preceq} = {\preceq \restriction_T}$ That is: :$\tuple {x, y} \in {\preceq \restriction_T}$ A similar argument shows that: :$\tuple {y, x} \in {\preceq} \implies \tuple {y, x} \in {\preceq \restriction_T}$ Thus $\preceq \restriction_T$ is a total ordering of $T$. {{qed}}
Since $\preceq$ is an ordering it is transitive and antisymmetric. By Transitive Chaining, it follows from the first premise that for all $k$ with $0 \le k \le n$: :$x_0 \preceq x_k$ and also: :$x_k \preceq x_n$ The other premise states that $x_n \preceq x_0$. By transitivity of $\preceq$, this combines with the above to: :$x_k \preceq x_0$ Since $\preceq$ is antisymmetric, this means that $x_0 = x_k$ for $0 \le k \le n$. That is, $x_0 = x_1 = \dots = x_n$. {{qed}}
We will show that each finite subset of $S$ has a minimal element. The existence of a maximal element then follows from duality. Proof by induction: For all $n \in \Z_{\ge 1}$, let $P \left({n}\right)$ be the proposition: :Every set with $n$ elements has a minimal element. === Basis for the Induction === Let $T$ have exactly one element $x$. Since $x \not \prec x$ it follows that $x$ is minimal. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is the induction hypothesis: :Every set with $k$ elements has a minimal element. from which it is to be shown that: :Every set with $k + 1$ elements has a minimal element. === Induction Step === Suppose that every subset of $S$ with $k$ elements has a minimal element. Let $T$ have $k + 1$ elements. Then: : $T = T_0 \cup \left\{{x}\right\}$ where $T_0$ has $k$ elements and $x \in T \setminus T_0$. Then $T_0$ has a minimal element $m_0$. If $m_0$ is not a minimal element of $T$, then: : $x \prec m_0$ Thus $x$ is a minimal element of $T$. Thus either $m_0$ or $x$ is a minimal element of $T$. So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :For every finite, non-empty subset $T$ of $S$, $T$ has a maximal element and a minimal element  The result follows by the Principle of Mathematical Induction. {{qed}}
Let $x_1, x_2 \in A$. Let $a \in A$ be a star center of $A$. By definition of star convex set, it follows that for all $t \in \left[{0 \,.\,.\, 1}\right]$, we have $t x_1 + \left({1 - t}\right) a, t x_2 + \left({1 - t}\right) a  \in A$. Define two paths $\gamma_1, \gamma_2: t \in \left[{0 \,.\,.\, 1}\right] \to A$ by $\gamma_1 \left({t}\right) =  t x_1 + \left({1 - t}\right) a$, and $\gamma_2 \left({t}\right) =  t a + \left({1 - t}\right) x_2$. As $\gamma_2 \left({t}\right) = \left({1 - t}\right) x_2 + \left({1 - \left({1 - t}\right) }\right) a$, and $\left({1 - t}\right) \in \left[{0 \,.\,.\, 1}\right]$, it follows that $\gamma_2 \left({t}\right) \in A$. Note that $\gamma_1 \left({0}\right) = x_1$, $\gamma_1 \left({1}\right) = \gamma_2 \left({0}\right) = a$, and $\gamma_2 \left({1}\right) = x_2$. Define $\gamma: \left[{0 \,.\,.\, 1}\right] \to A$ as the concatenation $\gamma_1 * \gamma_2$. Then $\gamma$ is a path in $A$ joining $x_1$ and $x_2$, so $A$ is path-connected. {{qed}}
Let $x \in C$. Since $m$ is minimal in $C$, $x \not\prec m$. Since $C$ is a chain, $x = m$ or $m \prec x$. Thus for each $x \in C$, $m \preceq x$. Therefore $m$ is the smallest element of $C$. {{qed}}
Let $x \in C$. Since $m$ is maximal in $C$, $m \not\prec x$. Since $C$ is a chain, $x = m$ or $x \prec m$. Thus for each $x \in C$, $x \preceq m$. Therefore $m$ is the greatest element of $C$. {{qed}}
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connective is true for all boolean interpretations. $\begin{array}{|cccc|c|cccc|} \hline p & \land & \neg & q & \iff & \neg & (p & \implies & q) \\ \hline F & F & T & F & T & F & F & T & F \\ F & F & F & T & T & F & F & T & T \\ T & T & T & F & T & T & T & F & F \\ T & F & F & T & T & F & T & T & T \\ \hline \end{array}$ {{qed}}
Let us use the following abbreviations {{begin-eqn}} {{eqn | l = \phi       | o = \text{ for }       | r = p \implies q       | c =  }} {{eqn | l = \psi       | o = \text{ for }       | r = q \implies r       | c =  }} {{eqn | l = \chi       | o = \text{ for }       | r = p \implies r       | c =  }} {{end-eqn}} {{BeginTableau|\paren {\paren {p \implies q} \land \paren {q \implies r} } \implies \paren {p \implies r} }} {{Assumption|1|\phi \land \psi}} {{Simplification|2|1|\phi|1|1}} {{Simplification|3|1|\psi|1|2}} {{SequentIntro|4|1|\chi|2, 3|Hypothetical Syllogism: Formulation 1}} {{Implication|5||\paren {\phi \land \psi} \implies \chi|1|4}} {{EndTableau}} Expanding the abbreviations leads us back to: :$\paren {\paren {p \implies q} \land \paren {q \implies r} } \implies \paren {p \implies r}$ {{qed}}
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connective is true for all boolean interpretations. :$\begin{array}{|ccccccc|c|ccc|} \hline ((p & \implies & q) & \land & (q & \implies & r)) & \implies & (p & \implies & r) \\ \hline F & T & F & T & F & T & F & T & F & T & F \\ F & T & F & T & F & T & T & T & F & T & T \\ F & T & T & T & T & F & F & T & F & T & F \\ F & T & T & T & T & T & T & T & F & T & T \\ T & F & F & F & F & T & F & T & T & F & F \\ T & F & F & T & F & T & T & T & T & T & T \\ T & T & T & F & T & F & F & T & T & F & F \\ T & T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$ {{qed}}
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connective are true for all boolean interpretations. $\begin{array}{|ccccc|c|c|} \hline ((p & \implies & q) & \land & p) & \implies & q \\ \hline F & T & F & F & F & T & F \\ F & T & T & F & F & T & T \\ T & F & F & F & T & T & F \\ T & T & T & T & T & T & T \\ \hline \end{array}$ {{qed}}
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|ccc|ccccccc|} \hline p & \iff & q & (p & \implies & q) & \land & (q & \implies & p) \\ \hline F & T & F & F & T & F & T & F & T & F \\ F & F & T & F & T & T & F & T & F & F \\ T & F & F & T & F & F & F & F & T & T \\ T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$ {{qed}}
{{BeginTableau|\vdash \left({p \iff q}\right) \iff \left({\left({p \implies q}\right) \land \left({q \implies p}\right)}\right)}} {{Assumption|1|p \iff q}} {{SequentIntro|2|1|\left({\left({p \implies q}\right) \land \left({q \implies p}\right)}\right)|1|Rule of Material Equivalence: Formulation 1}} {{Implication|3||\left({p \iff q}\right) \implies \left({\left({p \implies q}\right) \land \left({q \implies p}\right)}\right)|1|2}} {{Assumption|4|\left({p \implies q}\right) \land \left({q \implies p}\right)}} {{SequentIntro|5|4|p \iff q|4|Rule of Material Equivalence: Formulation 1}} {{Implication|6||\left({\left({p \implies q}\right) \land \left({q \implies p}\right)}\right) \implies \left({p \iff q}\right)|4|5}} {{BiconditionalIntro|7||\left({p \iff q}\right) \iff \left({\left({p \implies q}\right) \land \left({q \implies p}\right)}\right)|3|6}} {{EndTableau}} {{qed}}
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connective are true for all boolean interpretations. $\begin{array}{|ccc|c|cccc|} \hline (p & \implies & q) & \iff & (\neg & p & \lor & q) \\ \hline F & T & F & T & T & F & T & F \\ F & T & T & T & T & F & T & T \\ T & F & F & F & T & T & F & F \\ T & T & T & F & T & T & T & T \\ \hline \end{array}$ {{qed}}
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connective are true for all boolean interpretations. $\begin{array}{|ccc|c|ccccccc|} \hline (p & \iff & q) & \iff & (p & \implies & q) & \land & (q & \implies & p) \\ \hline F & T & F & T & F & T & F & T & F & T & F \\ F & F & T & T & F & T & T & F & T & F & F \\ T & F & F & T & T & F & F & F & F & T & T \\ T & T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$ {{qed}}
{{BeginTableau|\vdash \paren {p \land q} \iff \paren {\neg \paren {\neg p \lor \neg q} } }} {{Assumption|1|p \land q}} {{SequentIntro|2|1|\neg \paren {\neg p \lor \neg q}|1|De Morgan's Laws (Logic): Conjunction: Formulation 1}} {{Implication|3||\paren {p \land q} \implies \paren {\neg \paren {\neg p \lor \neg q} }|1|2}} {{Assumption|4|\neg \paren {\neg p \lor \neg q} }} {{SequentIntro|5|4|p \land q|4|De Morgan's Laws (Logic): Conjunction: Formulation 1}} {{Implication|6||\paren {\neg \paren {\neg p \lor \neg q} } \implies \paren {p \land q}|4|5}} {{BiconditionalIntro|7||\paren {p \land q} \iff \paren {\neg \paren {\neg p \lor \neg q} }|3|6}} {{EndTableau}} {{qed}}
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connective are true for all boolean interpretations. $\begin{array}{|ccc|c|cccccc|} \hline (p & \land & q) & \iff & (\neg & (\neg & p & \lor & \neg & q))  \\ \hline F & F & F & T & F & T & F & T & T & F \\ F & F & T & T & F & T & F & T & F & T \\ T & F & F & T & F & F & T & T & T & F \\ T & T & T & T & T & F & T & F & F & T \\ \hline \end{array}$ {{qed}}
{{BeginTableau|\vdash \paren {p \land q} \iff \paren {\neg \paren {p \implies \neg q} } }} {{Assumption|1|p \land q}} {{SequentIntro|2|1|\neg \paren {p \implies \neg q}|1|Conjunction Equivalent to Negation of Implication of Negative: Formulation 1: Forward Implication}} {{Implication|3||\paren {p \land q} \implies \paren {\neg \paren {p \implies \neg q} }|1|2}} {{Assumption|4|\neg \paren {p \implies \neg q} }} {{SequentIntro|5|4|p \land q|4|Conjunction Equivalent to Negation of Implication of Negative: Formulation 1: Reverse Implication}} {{Implication|6||\paren {\neg \paren {p \implies \neg q} \implies \paren {p \land q} }|4|5}} {{BiconditionalIntro|7||\paren {p \land q} \iff \paren {\neg \paren {p \implies \neg q} }|3|6}} {{EndTableau}} {{qed}}
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connective are true for all boolean interpretations. $\begin{array}{|ccc|c|ccccc|} \hline (p & \land & q) & \iff & (\neg & (p & \implies & \neg & q)) \\ \hline F & F & F & T & F & F & T & T & F \\ F & F & T & T & F & F & T & F & T \\ T & F & F & T & F & T & T & T & F \\ T & T & T & T & T & T & F & F & T \\ \hline \end{array}$ {{qed}}
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connective are true for all boolean interpretations. $\begin{array}{|ccc|c|cccc|} \hline p & \lor & q & \iff & \neg & p & \implies & q \\ \hline F & F & F & T & T & F & F & F \\ F & T & T & T & T & F & T & T \\ T & T & F & T & F & T & T & F \\ T & T & T & T & F & T & T & T \\ \hline \end{array}$ {{qed}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|ccc|c|ccccc|} \hline p & \implies & q) & \iff & (\neg & q & \implies & \neg & p) \\ \hline F & T & F & T & T & F & T & T & F \\ F & T & T & T & F & T & T & T & F \\ T & F & F & T & T & F & F & F & T \\ T & T & T & T & F & T & T & F & T \\ \hline \end{array}$ {{qed}}
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connective are true for all boolean interpretations. $\begin{array}{|ccc|c|cccccc|} \hline (p & \lor & q) & \iff & (\neg & (\neg & p & \land & \neg & q)) \\ \hline F & F & F & T & F & T & F & T & T & F \\ F & T & T & T & T & T & F & F & F & T \\ T & T & F & T & T & F & T & F & T & F \\ T & T & T & T & T & F & T & F & F & T \\ \hline \end{array}$ {{qed}}
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connective is true for all boolean interpretations. $\begin{array}{|ccccc|c|cccc|} \hline \neg & p & \lor & \neg & q & \iff & \neg & (p & \land & q) \\ \hline T & F & T & T & F & T & T & F & F & F \\ T & F & T & F & T & T & T & F & F & T \\ F & T & T & T & F & T & T & T & F & F \\ F & T & F & F & T & T & F & T & T & T \\ \hline \end{array}$ {{qed}}
{{BeginTableau|\vdash \paren {p \iff q} \iff \paren {\paren {p \land q} \lor \paren {\neg p \land \neg q} } }} {{Assumption        |1|p \iff q}} {{SequentIntro      |2|1|\paren {p \land q} \lor \paren {\neg p \land \neg q}|1|Biconditional as Disjunction of Conjunctions: Formulation 1}} {{Implication       |3| |\paren {p \iff q} \implies \paren {\paren {p \land q} \lor \paren {\neg p \land \neg q} }|1|2}} {{Assumption        |4|\paren {p \land q} \lor \paren {\neg p \land \neg q} }} {{SequentIntro      |5|4|p \iff q|4|Biconditional as Disjunction of Conjunctions: Formulation 1}} {{Implication       |6| |\paren {\paren {p \land q} \lor \paren {\neg p \land \neg q} } \implies \paren {p \iff q}|4|5}} {{BiconditionalIntro|7| |\paren {p \iff q} \iff \paren {\paren {p \land q} \lor \paren {\neg p \land \neg q} }|3|6}} {{EndTableau}} {{qed}}
We apply the Method of Truth Tables. As can be seen by inspection, in all cases the truth values under the main connective is true for all boolean interpretations. $\begin{array}{|ccc|c|ccccccccc|} \hline (p & \iff & q) & \iff & ((p & \land & q) & \lor & (\neg & p & \land & \neg & q)) \\ \hline F & T & F & T & F & F & F & T & T & F & T & T & F \\ F & F & T & T & F & F & T & F & T & F & F & F & T \\ T & F & F & T & T & F & F & F & F & T & F & T & F \\ T & T & T & T & T & T & T & T & F & T & F & F & T \\ \hline \end{array}$ {{qed}}
=== $(2)$ implies $(1)$ === Follows from Upper Set is Convex, Lower Set is Convex, and Intersection of Convex Sets is Convex Set (Order Theory). {{qed|lemma}} === $(1)$ implies $(3)$ === {{MissingLinks}} Let $C$ be a convex set in $S$. Let $U$ and $L$ be the upper and lower closures of $C$, respectively. Since $C \subseteq U$ and $C \subseteq L$: :$C \subseteq U \cap L$. Let $p \in U \cap L$. Then $a \preceq p \preceq b$ for some $a, b \in C$. Since $C$ is convex, $p \in C$. Since this holds for all $p \in U \cap L$: :$U \cap L \subseteq C$. Since we know that $C \subseteq U \cap L$, $C = U \cap L$. {{qed|lemma}} === $(3)$ implies $(2)$ === Follows from Upper Closure is Upper Set and Lower Closure is Lower Set. {{qed}} Category:Convex Sets 2rudyxkfvqqz8vkpr8bjamysr5o11lu
{{BeginTableau|\paren {p \implies q} \land \paren {r \implies s}, p \lor r \vdash q \lor s}} {{Premise|1|\paren {p \implies q} \land \paren {r \implies s} }} {{Premise|2|p \lor r}} {{Conjunction|3|1, 2|\paren {p \lor r} \land \paren {p \implies q} \land \paren {r \implies s}|2|1|Associativity is implicit}} {{TheoremIntro|4|\paren {\paren {p \lor r} \land \paren {p \implies q} \land \paren {r \implies s} } \implies \paren {q \lor s}|Constructive Dilemma: Formulation 2}} {{ModusPonens|5|1, 2|q \lor s|4|3}} {{EndTableau}} {{qed}}
Let $a \in U$. Let $b \in S$ with $a \preceq b$. By the definition of upper closure, there is a $t \in T$ such that $t \preceq a$. By transitivity, $t \preceq b$. Thus, agin by the definition of upper closure, $b \in U$. Since this holds for all such $a$ and $b$, $U$ is an upper set. {{qed}}
{{BeginTableau|p \implies q, r \implies s \vdash \neg q \lor \neg s \implies \neg p \lor \neg r}} {{Premise|1|p \implies q}} {{Premise|2|r \implies s}} {{SequentIntro|3|1, 2|\paren {p \land r} \implies \paren {q \land s}|1, 2|Praeclarum Theorema}} {{Assumption|4|\neg q \lor \neg s}} {{DeMorgan|5|4|\neg \paren {q \land s}|4|Disjunction of Negations}} {{ModusTollens|6|1, 2, 4|\neg \paren {p \land r}|3|5}} {{DeMorgan|7|1, 2|\neg p \lor \neg r|6|Disjunction of Negations}} {{Implication|8|1, 2|\neg q \lor \neg s \implies \neg p \lor \neg r|4|7}} {{EndTableau}} {{qed}}
We apply the Method of Truth Tables to the proposition. 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 $T$: : $\begin{array}{|ccccccc||ccccccccccc|} \hline (p & \implies & q) & \land & (r & \implies & s) & (\neg & q & \lor & \neg & s) & \implies & (\neg & p & \lor & \neg & r) \\ \hline F & T & F & T & F & T & F & T & F & T & T & F & T & T & F & T & T & F \\ F & T & F & T & F & T & T & T & F & T & F & T & T & T & F & T & T & F \\ F & T & F & F & T & F & F & T & F & T & T & F & T & T & F & T & F & T \\ F & T & F & T & T & T & T & T & F & T & F & T & T & T & F & T & F & T \\ F & T & T & T & F & T & F & F & T & T & T & F & T & T & F & T & T & F \\ F & T & T & T & F & T & T & F & T & F & F & T & T & T & F & T & T & F \\ F & T & T & F & T & F & F & F & T & T & T & F & T & T & F & T & F & T \\ F & T & T & T & T & T & T & F & T & F & F & T & T & T & F & T & F & T \\ T & F & F & F & F & T & F & T & F & T & T & F & T & F & T & T & T & F \\ T & F & F & F & F & T & T & T & F & T & F & T & T & F & T & T & T & F \\ T & F & F & F & T & F & F & T & F & T & T & F & F & F & T & F & F & T \\ T & F & F & F & T & T & T & T & F & T & F & T & F & F & T & F & F & T \\ T & T & T & T & F & T & F & F & T & T & T & F & T & F & T & T & T & F \\ T & T & T & T & F & T & T & F & T & F & F & T & T & F & T & T & T & F \\ T & T & T & F & T & F & F & F & T & T & T & F & F & F & T & F & F & T \\ T & T & T & T & T & T & T & F & T & F & F & T & T & F & T & F & F & T \\ \hline \end{array}$ Hence the result. {{qed}} Note that the two formulas are not equivalent, as the relevant columns do not match exactly.
We will prove that $U = p^\succ$ is $\tau$-open. That $p^\prec$ is $\tau$-open will follow by duality. Let $u \in U$. Since $p \notin U$, $p \ne u$. By the definition of GO-space, $\tau$ is Hausdorff, and therefore $T_1$. Thus by the definition of GO-space, there is an open, convex set $M$ such that $u \in M$ and $p \notin M$. Next we will show that $M \subseteq U$: Let $x \in S \setminus U$. Then $x \preceq p \preceq u$. Suppose for the sake of contradiction that $x \in M$. Since $x, u \in M$, $p \in M$ because $M$ is convex, contradicting the choice of $M$. Thus $x \notin M$. Since this hold for all $x \in S \setminus U$, $M \subseteq U$. Thus $U$ contains a neighborhood of each of its points, so it is open. {{qed}}
We will show that the upper closure $U$ of $A$ is open. The lower closure can be proven open by the same method. By the definition of upper closure: :$U = \left\{ {u \in X: \exists a \in A: a \preceq u}\right\}$ But then: {{begin-eqn}} {{eqn | l = U       | r = \left\{ {u \in X: \left({u \in A}\right) \lor \left({\exists a \in A: a \prec u}\right) }\right\} }} {{eqn | r = A \cup \bigcup \left\{ {a^\succeq: a \in A }\right\} }} {{end-eqn}} where $a^\preceq$ denotes the upper closure of $a$. By Open Ray is Open in GO-Space/Definition 1, each $a^\succeq$ is open. Thus $U$ is a union of open sets. Thus $U$ is open by the definition of a topology. {{qed}} Category:Topology Category:Total Orderings 39ofm2hzen1z5lbb1anoktvpftjwa7g
<section notitle="True" name="proof"> Let $\BB$ be a basis for $\tau$ consisting of convex sets. Let: :$\SS = \set {U^\succeq: U \in \BB} \cup \set {U^\preceq: U \in \BB}$ where $U^\succeq$ and $U^\preceq$ denote the upper closure and lower closure respectively of $U$. By Upper Closure is Upper Set and Lower Closure is Lower Set, the elements of $\SS$ are upper and lower sets. It is to be shown that $\SS$ is a sub-basis for $\tau$. By Upper and Lower Closures of Open Set in GO-Space are Open: :$\SS \subseteq \tau$ By Convex Set Characterization (Order Theory), each element of $\BB$ is the intersection of its upper closure with its lower closure. Thus each element of $\BB$ is generated by $\SS$. {{Disambiguate|Definition:Generated}} Thus $\SS$ is a sub-basis for $\tau$. {{qed}} </section> Category:Equivalence of Definitions of Generalized Ordered Space bctoewd1tp2ms06qszp7bg85bfvzjyv
Let $a, c \in T$. Let $b \in S$. Let $a \preceq b \preceq c$. Since: :$a \in T$ :$a \preceq b$ :$T$ is an upper set it follows that: :$b \in T$ This holds for all such $a$, $b$, and $c$. Therefore, by definition, $T$ is convex in $S$. {{qed}}
Let $a, c \in T$. Let $b \in S$. Let $a \preceq b \preceq c$. Since: :$c \in T$ :$b \preceq c$ :$T$ is a lower set it follows that: :$b \in T$ This holds for all such $a$, $b$, and $c$. Hence, by definition, $T$ is convex in $S$. {{qed}}
Let $\tau'$ be the $\tau$-relative subspace topology on $Y$. Let $\preceq'$ be the restriction of $\preceq$ to $Y$. === Lemma === $\struct {Y, \preceq', \tau'}$ is a linearly ordered space. === Proof === By definition of a generalized ordered space: :$\tau'$ has a sub-basis consisting of upper and lower sets in $Y$. To prove that $\struct {Y, \preceq', \tau'}$ is a linearly ordered space, we need to show that each element of this sub-basis is open in the $\preceq'$ order topology. Let $U$ be a non-empty, $\tau'$-open upper set in $Y$. If $U = \O$ or $U = Y$, then $U$ is open in the $\preceq'$-order topology by the definition of topology. Suppose then that $\O \subsetneqq U \subsetneqq Y$. Let $C = Y \setminus U$. Then $C$ is non-empty and $\tau'$-closed. $Y$ is $\tau'$-closed by Closed Set in Linearly Ordered Space. Thus $C$ is also $\tau'$-closed. $C$ has a supremum in $X$ by the premise. Let $c = \sup_X C$ Since $C$ is $\tau'$-closed, Closed Set in Linearly Ordered Space shows that $c \in C$. Since $c \in C = Y \setminus U$ and $U$ is an upper set in $Y$: :$c \prec u$ for all $u \in U$. Furthermore, if $y \in Y$ and $c \prec y$, then by the definition of supremum, $y \notin C$, so $y \in U$. Thus: :$U = {c^\succeq}_Y$ where ${c^\succeq}_Y$ denotes the upper closure of $c$ in $Y$. So $U$ is open in the $\preceq'$-order topology for $Y$. The same argument shows that $\tau'$-open lower sets in $Y$ are open in the $\preceq'$-order topology. Thus $\tau'$ is the $\preceq'$-order topology. {{qed|lemma}} The premises immediately show that $\struct {Y, \preceq'}$ is a complete lattice. By Complete Linearly Ordered Space is Compact, $Y$ is a compact subspace of $X$. {{qed}}
=== Forward implication === Let $U$ be an upper set in $S$ with no minimal element. Then by the definition of upper set: :$\displaystyle \bigcup \set {u^\succ: u \in U} \subseteq U$ Let $x \in U$. Since $U$ has no minimal element, $x$ is not minimal. Thus there is a $u \in U$ such that $u \prec x$. Then $x \in u^\succ$, so: :$\displaystyle x \in \bigcup \set {u^\succ: u \in U }$ Since this holds for all $x \in U$: :$\displaystyle U \subseteq \bigcup \set {u^\succ: u \in U}$ Thus the theorem holds by definition of set equality. {{qed|lemma}} === Reverse implication === Let: : $\displaystyle U = \bigcup \set {u^\succ: u \in U}$ Then: :$\forall u \in U: u^\succ \subseteq U$ so $U$ is an upper set. Furthermore: :$\forall x \in U: \exists u \in U: x \in u^\succ$ But then: :$u \prec x$ so $x$ is not minimal. Since this holds for all $x \in U$, $U$ has no minimal element. {{qed}}
Let $L, M \in \mathcal L$. Suppose that $M \not\subseteq L$. Then for some $x \in M$: $x \notin L$. Let $y \in L$. Then since $\preceq$ is a total ordering, $x \preceq y$ or $y \preceq x$. If $x \preceq y$, then since $L$ is a lower set: $x \in L$, a contradiction. Thus $y \preceq x$. Since $M$ is a lower set, $y \in M$. Since this holds for all $y \in L$, $L \subseteq M$. Hence, for all $L, M \in \mathcal L$: :$M \subseteq L$ or $L \subseteq M$ That is, $\mathcal L$ is a nest. {{qed}} Category:Lower Sets Category:Total Orderings Category:Nests sxailfjx37zyldndgzq5xmy6y0j176n