hcju/proofwikiqa huggingface dataset:
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connective on the {{LHS}} match those for $p$ on the {{RHS}} for all boolean interpretations: $\begin{array}{|ccccc||c|} \hline (p & \oplus & q) & \oplus & q & p \\ \hline \F & \F & \F & \F & \F & \F \\ \F & \T & \T & \F & \T & \F \\ \T & \T & \F & \T & \F & \T \\ \T & \F & \T & \T & \T & \T \\ \hline \end{array}$ {{qed}}
This will be proven by contradiction. Let such an operation $\circ$ exist. Let $f^\circ: \mathbb B^2 \to \mathbb B$ be the associated truth function. Suppose now that $q = F$, while $p$ remains unspecified. Then: :$p \land q = f^\land \left({p, F}\right) = F$ where $f^\land$ is the truth function of conjunction. It does not matter what $p$ is, for: :$f^\land \left({T, F}\right) = f^\land \left({F, F}\right) = F$ Hence, for $\left({p \land q}\right) \circ q = p$ to hold, $f^\circ$ must satisfy: :$f^\circ \left({F, F}\right) = p$ However, because $p$ could still be either $T$ or $F$, this identity cannot always hold. Therefore, $\circ$ cannot exist. {{qed}} Category:Conjunction iz8xrrjo16at5wz9wiqbjhnuzh6k4nl
Let $q$ be true. Then $p \lor q = T$, whatever truth value $p$ holds. Either $T \circ T = T$ or $T \circ T = F$, but not both. So if $q = T$ either: :$\left({p \land q}\right) \circ q = T$ or: :$\left({p \land q}\right) \circ q = F$ If the first, then $(1)$ does not hold when $p = F$. If the second, then $(1)$ does not hold when $p = T$. Hence there can be no such $\circ$. {{qed}} {{improve|I'm not proud of this - it needs to be couched in more rigorous language. Also possible to prove it by truth table.}} Category:Disjunction dnjqp2uu46fx7uxt68yrblp9sbhwmor
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connective on the {{LHS}} match those for $p$ on the {{RHS}} for all boolean interpretations: $\begin{array}{|ccccc||c|} \hline (p & \iff & q) & \iff & q & p \\ \hline F & T & F & F & F & F \\ F & F & T & F & T & F \\ T & F & F & T & F & T \\ T & T & T & T & T & T \\ \hline \end{array}$ {{qed}} Category:Biconditional 5ybdy25cskm1eyv6h2mol5sojt7fmiv
By the definition of finite set: :there exists an $n \in \N$ such that there exists a bijection $f: C \to \N_n$. This $n$ is unique by Equality of Natural Numbers and Set Equivalence is Equivalence Relation. Define a mapping $g: \N_n \to C$ recursively as: :$g \left({0}\right) = \min C$ :$g \left({k + 1}\right) = \min \left({ C \setminus g \left({N_k}\right) }\right)$ {{explain|How is $\min C$ is defined?}} {{finish}}
By Compactness from Basis, it is sufficient to prove that an open cover of $X$ consisting of open intervals and rays has a finite subcover. Let $\mathcal A$ be an open cover of $X$ consisting of open rays and open intervals. Let $m = \inf X$. This infimum exists because $\left({X, \preceq}\right)$ is complete. Let $C$ be the set of all $x \in X$ such that a finite subset of $\mathcal A$ covers $\left[{m \,.\,.\, x}\right]$. $C$ is non-empty because $m \in C$. Let $s = \sup C$. Since $\mathcal A$ covers $X$, there is a $U \in \mathcal A$ such that $s \in U$. Then we must have $U = \left({a \,.\,.\, b}\right)$, $U = {\dot\uparrow} a$, or $U = {\dot\downarrow} b$. Suppose that $U = \left({a \,.\,.\, b}\right)$. Let $V \in \mathcal U$ contain $b$. Then by the definition of supremum, there is an $x \succ a$ such that there is a finite $\mathcal F \subseteq \mathcal A$ that covers $\left[{m \,.\,.\, x}\right]$. Then $\mathcal F \cup \left\{{U, V}\right\}$ covers $\left[{m \,.\,.\, b}\right]$, contradicting the fact that $s$ is an upper bound of $C$. Suppose next that $U = \dot\downarrow b$. Then for some $V \in \mathcal A$, $b \in V$. Then $\left[{m \,.\,.\, b}\right]$ is covered by $\left\{{U, V}\right\}$, contradicting the fact that $s$ is the supremum of $C$. Thus $U = \dot\uparrow a$. By the definition of supremum, $a$ is not an upper bound of $C$. So there is an $x \succ a$ such that there is a finite subset $\mathcal F$ of $\mathcal A$ that covers $\left[{m \,.\,.\, x}\right]$. Thus $\mathcal F \cup \left\{{U}\right\}$ is a finite subcover of $A$. {{qed}}
Suppose $\struct {S, \preceq}$ is not well-founded. So by definition there exists a non-empty subset $T$ of $S$ which has no minimal element. Let $a \in T$. Since $a$ is not minimal in $T$, we can find $b \in T: b \prec a$. Since this holds for all $a \in T$, $\prec \restriction_{T \times T}$, the restriction of $\prec$ to $T \times T$, is a right-total endorelation on $T$. So, by the Axiom of Dependent Choice, it follows that there is an infinite sequence $\sequence {a_n}$ in $T$ such that $\forall n \in \N: a_{n + 1} \prec a_n$. {{qed|lemma}}
Suppose $\struct {S, \preceq}$ is not well-founded. Let $T \subseteq S$ have no minimal element. Let $a_0 \in T$. We have that $a_0$ is not minimal in $T$. So: :$\exists a_1 \in T: a_1 \prec a_0$ Similarly, $a_1$ is not minimal in $T$. So: :$\exists a_2 \in T: a_2 \prec a_1$ Let $a_{k + 1}$ be an arbitrary element for which $a_{k + 1} \prec a_k$. In order to allow this to be possible in the infinite case, it is necessary to invoke the Axiom of Dependent Choice as follows: Let $a_k \in T$. Then as $a_k$ is not minimal in $T$: :$\exists a_{k + 1} \in T: a_{k + 1} \prec a_k$ Hence by definition $\prec$ is a right-total relation. So, by the Axiom of Dependent Choice, it follows that: :$\forall n \in \N: \exists a_n \in T: a_{n + 1} \prec a_n$ Thus we have been able to construct an infinite sequence $\sequence {a_n}$ in $T$ such that: :$\forall n \in \N: a_{n + 1} \prec a_n$. It follows by the Rule of Transposition that if there is no infinite sequence $\sequence {a_n}$ of elements of $S$ such that: :$\forall n \in \N: a_{n + 1} \prec a_n$ then $\struct {S, \preceq}$ is well-founded. {{qed|lemma}}
By Inversion Mapping is Involution and Involution is Permutation, $\iota$ is a permutation and so by definition bijective. Let $x, y \in G$ such that $x \prec y$. Then $y^{-1} \prec x^{-1}$ by Inversion Mapping Reverses Ordering in Ordered Group. Thus $\map \iota y \prec \map \iota x$. Since this holds for all $x$ and $y$ with $x \prec y$, $\iota$ is strictly decreasing. If $\map \iota x \prec \map \iota y$, then $\map \iota {\map \iota y} \prec \map \iota {\map \iota x}$ by the above. Thus by Inverse of Group Inverse: $y \prec x$. Therefore, $\iota$ reverses ordering in both directions, and is thus a dual isomorphism. {{qed}} Category:Ordered Groups Category:Inversion Mappings r4yxtzyk6j10d5wzp88nv29h4wf9ga9
By the definition of an ordered group, $\preceq$ is a relation compatible with $\circ$. Thus by Properties of Relation Compatible with Group Operation/CRG2 , we obtain the first four results. By Reflexive Reduction of Relation Compatible with Group Operation is Compatible, $\prec$ is compatible with $\circ$. Again by Properties of Relation Compatible with Group Operation/CRG2, we obtain the remaining results. {{qed}}
Each result follows from Properties of Ordered Group/OG1.  For example, by Properties of Ordered Group/OG1, :$x \preceq y \iff x \circ x^{-1} \preceq y \circ x^{-1}$ Since $x \circ x^{-1} = e$: :$(\operatorname{OG}2.1):\quad x \preceq y \iff e \preceq y \circ x^{-1}$ {{qed}}
By Inversion Mapping Reverses Ordering in Ordered Group: {{begin-axiom}} {{axiom | m = x \preceq e \iff e^{-1} \preceq x^{-1} }} {{axiom | m = e \preceq x \iff x^{-1} \preceq e^{-1} }} {{axiom | m = x \prec e \iff e^{-1} \prec x^{-1} }} {{axiom | m = e \prec x \iff x^{-1} \prec e^{-1} }} {{end-axiom}} Since $e^{-1} = e$, the theorem holds. {{qed}}
By the definition of an ordered group, $\preceq$ is a relation compatible with $\circ$. Thus by Properties of Relation Compatible with Group Operation/CRG4, we obtain the first two results: :$(\operatorname{OG}4.1):\quad x \preceq e \iff e \preceq x^{-1}$ :$(\operatorname{OG}4.2):\quad e \preceq x \iff x^{-1} \preceq e$ By Reflexive Reduction of Relation Compatible with Group Operation is Compatible, $\prec$ is also compatible with $\circ$. Thus by again Properties of Relation Compatible with Group Operation/CRG4, we obtain the remaining results: :$(\operatorname{OG}4.1'):\quad x \prec e \iff e \prec x^{-1}$ :$(\operatorname{OG}4.2'):\quad e \prec x \iff x^{-1} \prec e$ {{qed}}
By Power Function Strictly Preserves Ordering in Ordered Group: :$x \preceq e \implies x^n \preceq e^n$ :$e \preceq x \implies e^n \preceq x^n$ :$x \prec e \implies x^n \prec e^n$ :$e \prec x \implies e^n \prec x^n$ By Identity Element is Idempotent, $e$ is idempotent with respect to $\circ$. Therefore by the definition of an idempotent element, $e^n = e$. Thus the theorem holds. {{qed}}
By the definition of an ordered group, $\preceq$ is a transitive relation compatible with $\circ$. By Transitive Relation Compatible with Semigroup Operation Relates Powers of Related Elements: :$x \preceq e \implies x^n \preceq e^n$ :$e \preceq x \implies e^n \preceq x^n$ By Identity Element is Idempotent, $e$ is idempotent with respect to $\circ$. Thus we obtain the first two results: :$x \preceq e \implies x^n \preceq e$ :$e \preceq x \implies e \preceq x^n$ By Reflexive Reduction of Relation Compatible with Group Operation is Compatible, $\prec$ is compatible with $\circ$. By Reflexive Reduction of Transitive Antisymmetric Relation is Strict Ordering, $\prec$ is transitive. Thus by the same method as above, we obtain the remaining results: :$x \prec e \implies x^n \prec e$ :$e \prec x \implies e \prec x^n$ {{qed}}
=== Forward Implication === Let $\phi$ be a dual order embedding. Then $\phi$ is an order embedding of $\left({S, \preceq_1}\right)$ into $\left({T, \succeq_2}\right)$, where $\succeq_2$ is the dual of $\preceq_2$. Thus by Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing: :$\phi: \left({S, \preceq_1}\right) \to \left({T, \succeq_2}\right)$ is strictly increasing. Thus: :$\forall x, y \in S: x \prec_1 y \implies \phi \left({x}\right) \succ_2 \phi \left({y}\right)$ so: :$\forall x, y \in S: x \prec_1 y \implies \phi \left({y}\right) \prec_2 \phi \left({x}\right)$  Thus $\phi: \left({S, \preceq_1}\right) \to \left({T, \preceq_2}\right)$ is strictly decreasing. {{qed|lemma}} === Reverse Implication === Suppose that $\phi: \left({S, \preceq_1}\right) \to \left({T, \preceq_2}\right)$ is strictly decreasing. Then by the same argument as above: :$\phi: \left({S, \preceq_1}\right) \to \left({T, \preceq_2}\right)$ is strictly increasing. Thus by Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing, $\phi$ is an order embedding of $\left({S, \preceq_1}\right)$ into $\left({T, \succeq_2}\right)$. So $\phi$ is a dual order embedding of $\left({S, \preceq_1}\right)$ into $\left({T, \preceq_2}\right)$. {{qed}} Category:Total Orderings Category:Order Embeddings cvbcsnw2k9bzz4617102zbbql23f35t
Let $\phi$ be strictly increasing. Let $\map \phi x \preceq_2 \map \phi y$. As $\struct {S, \prec_1}$ is a strictly totally ordered set: :Either $y \prec_1 x$, $y = x$, or $x \prec_1 y$. {{AimForCont}} that $y \prec_1 x$. By the definition of a strictly increasing mapping: :$\map \phi y \prec_2 \map \phi x$ which contradicts the fact that $\map \phi x \preceq_2 \map \phi y$. Therefore $y \nprec_1 x$. Thus $y = x$, or $x \prec_1 y$, so $x \preceq_1 y$. Hence: :$\map \phi x \preceq_2 \map \phi y \iff x \preceq_1 y$ and $\phi$ has been proved to be an order embedding. {{qed}}
Let $x \preceq_1 y$. Then $x = y$ or $x \prec_1 y$. Let $x = y$. Then :$\map \phi x = \map \phi y$ so: :$\map \phi x \preceq_2 \map \phi y$ Let $x \prec_1 y$. Then by the definition of strictly increasing mapping: :$\map \phi x \prec_2 \map \phi y$ so by the definition of $\prec_2$: :$\map \phi x \preceq_2 \map \phi y$ Thus: :$x \preceq_1 y \implies \map \phi x \preceq_2 \map \phi y$ It remains to be shown that: :$\map \phi x \preceq_2 \map \phi y \implies x \preceq_1 y$ Suppose that $x \npreceq_1 y$. Since $\preceq_1$ is a total ordering: :$y \prec_1 x$ Thus since $\phi$ is strictly increasing: :$\map \phi y \prec_1 \map \phi x$ Thus: :$\map \phi x \npreceq_1 \map \phi y$ Therefore: :$x \npreceq_1 y \implies \map \phi x \npreceq_2 \map \phi y$ By the Rule of Transposition: :$\map \phi x \preceq_2 \map \phi y \implies x \preceq y$ {{qed}}
Let $p \in A$. Then $\left\{{p}\right\} \ne \varnothing$ and $\left\{{p}\right\} \subseteq A$. Thus, by the definition of foundational relation: :$\exists x \in \left\{{p}\right\}: \forall y \in \left\{{p}\right\}: \neg \left({y \mathrel{\mathcal R} x}\right)$ Since $x \in \left\{{p}\right\}$, it must be that $x = p$. It follows that $p \not\mathrel{\mathcal R} p$. Since this holds for all $p \in A$, $\mathcal R$ is antireflexive. {{qed}} Category:Foundational Relations Category:Reflexive Relations 5un4m8mr37k753e9fnq83rdai1bc2zb
Let $p, q \in S$ and suppose that $p \mathrel \RR q$. Then $\set {p, q} \ne \O$ and $\set {p, q} \subseteq S$. By the definition of foundational relation, $\set {p, q}$ has an $\RR$-minimal element. Since $p \mathrel \RR q$, $q$ is not an $\RR$-minimal element of $\set {p, q}$. Thus $p$ is an $\RR$-minimal element of $\set {p, q}$. Thus $q \not \mathrel \RR p$. Since for all $p, q \in S$, $p \mathrel \RR q \implies q \not \mathrel \RR p$, $\RR$ is asymmetric. {{qed}} Category:Foundational Relations Category:Symmetric Relations dxlx9qoocb3xu2h7wv4ce074awnbyzr
Let $\FF$ be an ultrafilter on $Y$. For $S \in \FF$, let $\map f S = \inf S$. Let $p = \sup \map f \FF$. {{improve|perhaps directly use $p :{{=}} \ds \sup_{S \mathop \in \FF} \inf S$?}} Then $\FF$ converges to $p$: === Upward rays === Let $a \in X$ with $a \prec p$. Since $\FF$ is an ultrafilter, either $Y \cap {\uparrow} a \in \FF$ or $Y \cap {\bar \downarrow} a \in \FF$. {{AimForCont}} that $Y \cap {\bar \downarrow} a \in \FF$. For each $S \in \FF$: :$S \cap {\bar \downarrow} a \in \FF$ because an ultrafilter is a filter. {{explain|Mention that $S \cap (Y \cap {\bar \downarrow} a) {{=}} S \cap {\bar \downarrow} a$ is used}} :$S \cap {\bar \downarrow} a \ne \O$ because a filter on a set is proper. By applying the definition of supremum to $p$, it follows that there exists an $S \in \FF$ such that $a \prec \inf S$. By the definition of infimum: :$S \cap {\bar \downarrow} a = \O$ which is a contradiction. Thus $Y \cap {\uparrow} a \in \FF$. === Downward rays === Let $b \in X$ with $p \prec b$. Either $Y \cap {\downarrow} b \in \FF$ or $Y \cap {\bar \uparrow} b \in \FF$. {{AimForCont}} that $Y \cap {\bar \uparrow} b \in \FF$. Let $b' = \map \inf {Y \cap {\bar \uparrow} b}$. We have that $b$ is a lower bound of $Y \cap {\bar \uparrow} b$ So by the definition of infimum: :$b \preceq b'$ Since $p \prec b$ and $b \preceq b'$, $p \prec b'$ by Extended Transitivity. By the definition of $b'$ and the definition of $f$: :$b' \in \map f \FF$ But this contradicts the fact that $p$ is the supremum, and hence an upper bound, of $\map f \FF$. {{qed|lemma}} By the definition of the order topology, the upward and downward rays containing each point form a neighborhood sub-basis for that point. {{wtd|an invocation of "Sub-Basis is Nbhd Sub-Basis" would be appropriate}} Thus by the Neighborhood Sub-Basis Criterion for Filter Convergence, $\FF$ converges. Since every ultrafilter on $Y$ converges, $Y$ is compact by Equivalent Definitions of Compactness.
By Minimal Element in Toset is Unique and Smallest, $U$ has no minimal element. By Upper Set with no Minimal Element: :$U = \bigcup \set {u^\succ: u \in U}$ where $u^\succ$ is the strict upper closure of $u$. By Open Ray is Open in GO-Space and the fact that a union of open sets is open, $U$ is open. {{qed}}
By Dual Pairs (Order Theory): * Lower set is dual to upper set. * Maximal element is dual to minimal element. * Strict lower closure is dual to strict upper closure. Thus the theorem holds by the duality principle applied to Upper Set with no Minimal Element. {{qed}}
By Maximal Element in Toset is Unique and Greatest, $L$ has no maximal element. By Lower Set with no Maximal Element: :$\displaystyle L = \bigcup \set {l^\prec: l \in L}$ where $l^\prec$ is the strict lower closure of $l$. By Open Ray is Open in GO-Space and the fact that a union of open sets is open, $L$ is open. {{qed}}
By definition, $T$ is a lower set in $S$ iff: :$\forall t \in T: \forall s \in S: s \preceq t \implies s \in T$ The dual of this statement is: :$\forall t \in T: \forall s \in S: t \preceq s \implies s \in T$ by Dual Pairs (Order Theory).   By definition, this means $T$ is an upper set in $S$. The converse follows from Dual of Dual Statement (Order Theory). {{qed}}
{{improve|recast in terms of dual statements}} Let $U$ be an open ray in $\left({S, \preceq}\right)$. By Open Ray is Dual to Open Ray, $U$ is an open ray in $\left({S, \preceq}\right)$. Since the open rays in a totally ordered set form a sub-basis for the topology on that set, $\tau'$ is finer than $\tau$. {{explain|Invoke some other duality principle?}} By the same argument, $\tau$ is finer than $\tau'$. Thus by definition of set equality: : $\tau' = \tau$ {{qed}} Category:Order Topology 4o99v8qf2um7xjidwbxm35zjftnjqcc
By the definition of open ray, there is some $p \in S$ such that: :$R$ is the strict upper or strict lower closure of $p$ with respect to $\preceq$. By Strict Lower Closure is Dual to Strict Upper Closure, the dual statement is: :$R$ is the strict upper or strict lower closure of $p$ with respect to $\succeq$. Thus $R$ is an open ray in $\struct {S, \succeq}$. {{qed}} Category:Total Orderings gvmyntzwkf86e0y22eoizxb1pgju6ic
Let $\mathcal K \subseteq \mathcal L$. Then by Intersection of Topologies is Topology: :$\bigcap \mathcal K \in \mathcal L$ By Intersection is Largest Subset, $\bigcap \mathcal L$ is the infimum of $\mathcal K$. {{explain}} Let $\tau$ be the topology generated by the sub-basis $\bigcup \mathcal K$. Then $\tau \in \mathcal L$ and $\tau$ is the supremum of $\mathcal K$. We have that each subset of $\mathcal L$ has a supremum and an infimum in $\mathcal L$. Thus it follows that $\left({\mathcal L, \subseteq}\right)$ is a complete lattice. {{qed}} Category:Topology Category:Complete Lattices 6s1khlyr62d80t7yamqkq7fgmssgpd7
Let $u \in S \setminus L$. Let $s \in S$ such that $u \preceq s$. Suppose for the sake of contradiction that $s \notin S \setminus L$. Then $s \in L$. By the definition of lower set, $u \in L$, a contradiction. Hence $s \in S \setminus L$. Since this holds for all such $u$ and $s$, $S \setminus L$ is an upper set. {{qed}}
By GO-Space Embeds Densely into Linearly Ordered Space, there is a linearly ordered space $(W, \preceq_W, \tau_W)$ and a mapping $\psi:X \to W$ which is an order embedding and a topological embedding. Assume without loss of generality that $X$ is a subspace of $W$. Let $Y = \left\{{  (x, 0): x \in X }\right\} \cup (W \setminus X) \times \Z$. Let $\preceq_Y$ be the restriction to $Y$ of the lexicographic ordering on $W \times \Z$. Let $\tau_Y$ be the $\preceq_Y$-order topology on $Y$. Let $\phi:X \to Y$ be given by $\phi(x) = (x,0)$. $\phi$ is clearly an order embedding. Next, we show that it is a topological embedding: If $$ Finally, we show that $\phi(X)$ is closed in $Y$: {{finish}}
First note that by Lower Sets in Totally Ordered Set form Nest: :$\subseteq$ is a total ordering on the set of lower sets. Also note that by Complement of Lower Set is Upper Set, the complement of each $\preceq$-lower set is a $\preceq$-upper set. === Reflexivity === This follows immediately from the fact that $\preceq$ and $\subseteq$ are reflexive. Thus $\preceq'$ is reflexive. {{qed|lemma}} === Transitivity === There are eight possibilities to consider. If $y_1 \preceq' y_2$ and $y_2 \preceq' y_3$, then $y_1 \preceq' y_3$ because $\preceq$ is transitive. If $L_1 \preceq' L_2$ and $L_2 \preceq' L_3$, then $L_1 \preceq' L_3$ because $\subseteq$ is transitive. If $y_1 \preceq' y_2$ and $y_2 \preceq' L$, then: :$y_1 \preceq y_2$ and $y_2 \in L$ Since $L$ is a lower set in $Y$: :$y_1 \in L$ so: :$y_1 \preceq' L$ If $L \preceq' y_1$ and $y_1 \preceq' y_2$, then: :$y_1 \in Y \setminus L$ and $y_1 \preceq y_2$ Since $Y \setminus L$ is an upper set in $Y$: :$y_2 \in Y \setminus L$ so: :$L \preceq' y_2$ If $y \preceq' L_1$ and $L_1 \preceq' L_2$, then: :$y \in L_1$ and $L_1 \subseteq L_2$ By the definition of subset: :$y \in L_2$ so: :$y \preceq' L_2$ If $L_1 \preceq' L_2$ and $L_2 \preceq' y$, then: :$y \in Y \setminus L_2$ and $L_2 \supseteq L_1$ so: :$y \in Y \setminus L_1$ so: :$L_1 \preceq' y$ If $y_1 \preceq' L$ and $L \preceq' y_2$ then :$y_1 \in L$ and $y_2 \in Y \setminus L$. Since $L$ is a lower set: : $y_2 \not\preceq y_1$ Since $\preceq$ is a total ordering: :$y_1 \preceq y_2$ so: :$y_1 \preceq' y_2$ If $L_1 \preceq' y$ and $y \preceq' L_2$, then: :$y \in L_2$ but $y \notin L_1$ Thus: :$L_2 \not\subseteq L_1$ Since $\subseteq$ is a total ordering on the lower sets: :$L_1 \subseteq L_2$ so: :$L_1 \preceq' L_2$ Thus $\preceq'$ is transitive. {{qed|lemma}} === Antisymmetry === There are three cases. If $y_1 \preceq' y_2$ and $y_2 \preceq' y_1$ then: :$y_1 \preceq y_2$ and $y_2 \preceq y_1$ Since $\preceq$ is antisymmetric: :$y_1 = y_2$ If $L_1 \preceq' L_2$ and $L_2 \preceq' L_1$ then: :$L_1 \subseteq L_2$ and $L_2 \subseteq L_1$ Since $\subseteq$ is antisymmetric: :$L_1 = L_2$ By the definition of $\preceq'$, it is impossible for $y \preceq' L$ and $L \preceq' y$, so the third case cannot occur. Thus$\preceq'$ is antisymmetric {{qed|lemma}} Since $\preceq'$ is reflexive, transitive, and antisymmetric, it is an ordering. $\preceq'$ is a total ordering of $X$ because: :$\preceq$ is a total ordering :the set of lower sets is a nest and: :for any $y$ and $L$ either $y \in L$ or $y \in Y\setminus L$. {{explain|Specify exactly why it follows from the above that $\preceq'$ is a total ordering}} {{qed}} Category:Lower Sets Category:Total Orderings 8p7dffqm112z0q5kjcc20o60fxg6c58
Let $a \in L$. Let $b \in S$ with $b \preceq a$. By the definition of lower closure, there is a $t \in T$ such that $a \preceq t$. By transitivity, $b \preceq t$. Thus, again by the definition of lower closure, $b \in L$. Since this holds for all such $a$ and $b$, $L$ is a lower set. {{qed}}
Let $l \in p^\prec$. Let $s \in S$ with $s \preceq l$. Then by the definition of strict lower closure: :$l \prec p$ Thus by Extended Transitivity: :$s \prec p$ So by the definition of strict lower closure: :$s \in p^\prec$ Since this holds for all such $l$ and $s$, $p^\prec$ is a lower set. {{qed}}
By Dual Pairs (Order Theory): :strict upper closure is dual to strict lower closure :Upper set is dual to lower set Thus the theorem holds by Strict Upper Closure is Upper Set and the duality principle. {{qed}}
Follows immediately from the definition of upper set. {{qed}} Category:Upper Sets 3b2tr3jpcfkuapt8nr2dx0ad887se19
Follows immediately from the definition of lower set. {{qed}} Category:Lower Sets 57868spvvmikxz5nls23916ev9xoxg8
Follows immediately from the definition of convex set. {{qed}} Category:Order Theory 6j6iuiv0lnnhfp51h069w7inoa9hvwq
<section  notitle="True" name="proof"> Let $\mathcal S$ be a sub-basis for $\tau$ consisting of upper sets and lower sets. Let $\mathcal B$ be the set of intersections of finite subsets of $\mathcal S$. By Upper Set is Convex, Lower Set is Convex and Intersection of Convex Sets is Convex Set (Order Theory) : :the elements of $\mathcal B$ are convex. {{explain|Link to theorem on why this is a basis.}} But $\mathcal B$ is a basis for $\tau$. Therefore $\tau$ has a basis consisting of convex sets. {{qed}} </section> Category:Equivalence of Definitions of Generalized Ordered Space et9g81yhxxur5f9ub0xyqbvfl2lbdbl
Let $u \in p^\succ$. Let $s \in S$ with $u \preceq s$. Then by the definition of strict upper closure: : $p \prec u$ Thus by Extended Transitivity: : $p \prec s$ So by the definition of strict upper closure: : $s \in p^\succ$ Since this holds for all such $u$ and $s$, $p^\succ$ is an upper set. {{qed}}