id stringlengths 11 15 | title stringlengths 7 171 | problem stringlengths 9 4.33k | solution stringlengths 6 19k | problem_wikitext stringlengths 9 4.42k | solution_wikitext stringlengths 7 19.1k | proof_title stringlengths 9 171 | theorem_url stringlengths 34 198 | proof_url stringlengths 36 198 | categories listlengths 0 9 | theorem_references listlengths 0 36 | proof_references listlengths 0 253 |
|---|---|---|---|---|---|---|---|---|---|---|---|
proofwiki-7400 | Inversion Mapping Reverses Ordering in Ordered Group/Corollary/Proof 2 | Let $\struct {G, \circ, \preccurlyeq}$ be an ordered group with identity $e$.
Let $x \in G$.
Then the following equivalences hold:
{{:Inversion Mapping Reverses Ordering in Ordered Group/Corollary}} | By the definition of an ordered group, $\preccurlyeq$ is a relation compatible with $\circ$.
Thus by Inverses of Elements Related by Compatible Relation: Corollary:
{{begin-eqn}}
{{eqn | q = \forall x \in G
| l = x \preccurlyeq e
| o = \iff
| r = e \preccurlyeq x^{-1}
}}
{{eqn | l = e \preccurlyeq x
... | Let $\struct {G, \circ, \preccurlyeq}$ be an [[Definition:Ordered Group|ordered group]] with [[Definition:Identity Element|identity]] $e$.
Let $x \in G$.
Then the following equivalences hold:
{{:Inversion Mapping Reverses Ordering in Ordered Group/Corollary}} | By the definition of an [[Definition:Ordered Group|ordered group]], $\preccurlyeq$ is a relation [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
Thus by [[Inverses of Elements Related by Compatible Relation/Corollary|Inverses of Elements Related by Compatible Relation: Corollary]]:
{{begin-... | Inversion Mapping Reverses Ordering in Ordered Group/Corollary/Proof 2 | https://proofwiki.org/wiki/Inversion_Mapping_Reverses_Ordering_in_Ordered_Group/Corollary/Proof_2 | https://proofwiki.org/wiki/Inversion_Mapping_Reverses_Ordering_in_Ordered_Group/Corollary/Proof_2 | [
"Inversion Mapping Reverses Ordering in Ordered Group"
] | [
"Definition:Ordered Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [
"Definition:Ordered Group",
"Definition:Relation Compatible with Operation",
"Inverses of Elements Related by Compatible Relation/Corollary",
"Reflexive Reduction of Relation Compatible with Group Operation is Compatible",
"Inverses of Elements Related by Compatible Relation/Corollary"
] |
proofwiki-7401 | Mapping from Totally Ordered Set is Dual Order Embedding iff Strictly Decreasing | Let $\left({S, \preceq_1}\right)$ be a totally ordered set.
Let $\left({T, \preceq_2}\right)$ be an ordered set.
Let $\phi: S \to T$ be a mapping.
Then $\phi$ is a dual order embedding {{iff}} $\phi$ is strictly decreasing.
That is:
:$\forall x, y \in S: x \preceq_1 y \iff \phi \left({y}\right) \preceq_2 \phi \left({x}... | === 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, \p... | Let $\left({S, \preceq_1}\right)$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Let $\left({T, \preceq_2}\right)$ be an [[Definition:Ordered Set|ordered set]].
Let $\phi: S \to T$ be a [[Definition:Mapping|mapping]].
Then $\phi$ is a [[Definition:Dual Order Embedding|dual order embedding]] {{iff}} $\... | === Forward Implication ===
Let $\phi$ be a [[Definition:Dual Order Embedding|dual order embedding]].
Then $\phi$ is an [[Definition:Order Embedding|order embedding]] of $\left({S, \preceq_1}\right)$ into $\left({T, \succeq_2}\right)$, where $\succeq_2$ is the [[Definition:Dual Ordering|dual]] of $\preceq_2$.
Thus b... | Mapping from Totally Ordered Set is Dual Order Embedding iff Strictly Decreasing | https://proofwiki.org/wiki/Mapping_from_Totally_Ordered_Set_is_Dual_Order_Embedding_iff_Strictly_Decreasing | https://proofwiki.org/wiki/Mapping_from_Totally_Ordered_Set_is_Dual_Order_Embedding_iff_Strictly_Decreasing | [
"Total Orderings",
"Order Embeddings"
] | [
"Definition:Totally Ordered Set",
"Definition:Ordered Set",
"Definition:Mapping",
"Definition:Dual Order Embedding",
"Definition:Strictly Decreasing/Mapping"
] | [
"Definition:Dual Order Embedding",
"Definition:Order Embedding",
"Definition:Dual Ordering",
"Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing",
"Definition:Strictly Increasing/Mapping",
"Definition:Strictly Decreasing/Mapping",
"Definition:Strictly Decreasing/Mapping",
"Def... |
proofwiki-7402 | Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing/Reverse Implication | Let $\struct {S, \preceq_1}$ be a totally ordered set.
Let $\struct {T, \preceq_2}$ be an ordered set.
Let $\phi: S \to T$ be a strictly increasing mapping.
Then $\phi$ is an order embedding. | 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 $\prec_2$:
:$\map \phi x \preceq_2 \map \phi y$... | Let $\struct {S, \preceq_1}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Let $\struct {T, \preceq_2}$ be an [[Definition:Ordered Set|ordered set]].
Let $\phi: S \to T$ be a [[Definition:Strictly Increasing/Mapping|strictly increasing mapping]].
Then $\phi$ is an [[Definition:Order Embedding|order em... | 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 [[Definition:Strictly Increasing Mapping|strictly increasing mapping]]:
:$\map \phi x \prec_2 \map \phi y$
so by the definition ... | Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing/Reverse Implication/Proof 1 | https://proofwiki.org/wiki/Mapping_from_Totally_Ordered_Set_is_Order_Embedding_iff_Strictly_Increasing/Reverse_Implication | https://proofwiki.org/wiki/Mapping_from_Totally_Ordered_Set_is_Order_Embedding_iff_Strictly_Increasing/Reverse_Implication/Proof_1 | [
"Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing"
] | [
"Definition:Totally Ordered Set",
"Definition:Ordered Set",
"Definition:Strictly Increasing/Mapping",
"Definition:Order Embedding"
] | [
"Definition:Strictly Increasing/Mapping",
"Definition:Total Ordering",
"Definition:Strictly Increasing/Mapping",
"Rule of Transposition"
] |
proofwiki-7403 | Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing/Reverse Implication | Let $\struct {S, \preceq_1}$ be a totally ordered set.
Let $\struct {T, \preceq_2}$ be an ordered set.
Let $\phi: S \to T$ be a strictly increasing mapping.
Then $\phi$ is an order embedding. | 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 contr... | Let $\struct {S, \preceq_1}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Let $\struct {T, \preceq_2}$ be an [[Definition:Ordered Set|ordered set]].
Let $\phi: S \to T$ be a [[Definition:Strictly Increasing/Mapping|strictly increasing mapping]].
Then $\phi$ is an [[Definition:Order Embedding|order em... | Let $\phi$ be [[Definition:Strictly Increasing Mapping|strictly increasing]].
Let $\map \phi x \preceq_2 \map \phi y$.
As $\struct {S, \prec_1}$ is a [[Definition:Strictly Totally Ordered Set|strictly totally ordered set]]:
:Either $y \prec_1 x$, $y = x$, or $x \prec_1 y$.
{{AimForCont}} that $y \prec_1 x$.
By the... | Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing/Reverse Implication/Proof 2 | https://proofwiki.org/wiki/Mapping_from_Totally_Ordered_Set_is_Order_Embedding_iff_Strictly_Increasing/Reverse_Implication | https://proofwiki.org/wiki/Mapping_from_Totally_Ordered_Set_is_Order_Embedding_iff_Strictly_Increasing/Reverse_Implication/Proof_2 | [
"Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing"
] | [
"Definition:Totally Ordered Set",
"Definition:Ordered Set",
"Definition:Strictly Increasing/Mapping",
"Definition:Order Embedding"
] | [
"Definition:Strictly Increasing/Mapping",
"Definition:Strictly Totally Ordered Set",
"Definition:Strictly Increasing/Mapping",
"Definition:Contradiction",
"Definition:Order Embedding"
] |
proofwiki-7404 | Strictly Well-Founded Relation is Antireflexive | Let $\RR$ be a strictly well-founded relation on a set or class $A$.
Then $\RR$ is antireflexive. | Let $p \in A$.
Then $\set p \ne \O$ and $\set p \subseteq A$.
Thus, by the definition of strictly well-founded relation:
:$\exists x \in \set p: \forall y \in \set p: \neg \paren {y \mathrel \RR x}$
Since $x \in \set p$, it must be that $x = p$.
It follows that $p \not \mathrel \RR p$.
Since this holds for all $p \in A... | Let $\RR$ be a [[Definition:Strictly Well-Founded Relation|strictly well-founded relation]] on a [[Definition:Set|set]] or class $A$.
Then $\RR$ is [[Definition:Antireflexive Relation|antireflexive]]. | Let $p \in A$.
Then $\set p \ne \O$ and $\set p \subseteq A$.
Thus, by the definition of [[Definition:Strictly Well-Founded Relation|strictly well-founded relation]]:
:$\exists x \in \set p: \forall y \in \set p: \neg \paren {y \mathrel \RR x}$
Since $x \in \set p$, it must be that $x = p$.
It follows that $p \not ... | Strictly Well-Founded Relation is Antireflexive | https://proofwiki.org/wiki/Strictly_Well-Founded_Relation_is_Antireflexive | https://proofwiki.org/wiki/Strictly_Well-Founded_Relation_is_Antireflexive | [
"Well-Founded Relations",
"Antireflexive Relations"
] | [
"Definition:Strictly Well-Founded Relation",
"Definition:Set",
"Definition:Antireflexive Relation"
] | [
"Definition:Strictly Well-Founded Relation",
"Definition:Antireflexive Relation",
"Category:Well-Founded Relations",
"Category:Antireflexive Relations"
] |
proofwiki-7405 | Strictly Well-Founded Relation is Asymmetric | Let $\struct {S, \RR}$ be a relational structure, where $S$ is a set or a proper class.
Let $\RR$ be a strictly well-founded relation.
Then $\RR$ is asymmetric. | 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 strictly well-founded relation, $\set {p, q}$ has a strictly minimal element under $\RR$.
Since $p \mathrel \RR q$, $q$ is not an $\RR$-minimal element of $\set {p, q}$.
Thus $p$ is a stri... | Let $\struct {S, \RR}$ be a [[Definition:Relational Structure|relational structure]], where $S$ is a [[Definition:Set|set]] or a [[Definition:Proper Class|proper class]].
Let $\RR$ be a [[Definition:Strictly Well-Founded Relation|strictly well-founded relation]].
Then $\RR$ is [[Definition:Asymmetric Relation|asymme... | 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 [[Definition:Strictly Well-Founded Relation|strictly well-founded relation]], $\set {p, q}$ has a [[Definition:Strictly Minimal Element|strictly minimal element]] under $\RR$.
Since $p ... | Strictly Well-Founded Relation is Asymmetric | https://proofwiki.org/wiki/Strictly_Well-Founded_Relation_is_Asymmetric | https://proofwiki.org/wiki/Strictly_Well-Founded_Relation_is_Asymmetric | [
"Well-Founded Relations",
"Asymmetric Relations"
] | [
"Definition:Relational Structure",
"Definition:Set",
"Definition:Class (Class Theory)/Proper Class",
"Definition:Strictly Well-Founded Relation",
"Definition:Asymmetric Relation"
] | [
"Definition:Strictly Well-Founded Relation",
"Definition:Strictly Minimal Element",
"Definition:Strictly Minimal Element",
"Definition:Asymmetric Relation",
"Category:Well-Founded Relations",
"Category:Asymmetric Relations"
] |
proofwiki-7406 | Upper Section with no Smallest Element is Open in GO-Space | Let $\struct {S, \preceq, \tau}$ be a generalized ordered space.
Let $U$ be an upper section in $S$ with no smallest element.
Then $U$ is open in $\struct {S, \preceq, \tau}$. | By Minimal Element in Toset is Unique and Smallest, $U$ has no minimal element.
By Upper Section 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}} | Let $\struct {S, \preceq, \tau}$ be a [[Definition:Generalized Ordered Space|generalized ordered space]].
Let $U$ be an [[Definition:Upper Section|upper section]] in $S$ with no [[Definition:Smallest Element|smallest element]].
Then $U$ is [[Definition:Open Set (Topology)|open]] in $\struct {S, \preceq, \tau}$. | By [[Minimal Element in Toset is Unique and Smallest]], $U$ has no [[Definition:Minimal Element|minimal element]].
By [[Upper Section with no Minimal Element]]:
:$U = \bigcup \set {u^\succ: u \in U}$
where $u^\succ$ is the [[Definition:Strict Upper Closure of Element|strict upper closure]] of $u$.
By [[Open Ray is Op... | Upper Section with no Smallest Element is Open in GO-Space | https://proofwiki.org/wiki/Upper_Section_with_no_Smallest_Element_is_Open_in_GO-Space | https://proofwiki.org/wiki/Upper_Section_with_no_Smallest_Element_is_Open_in_GO-Space | [
"Generalized Ordered Spaces"
] | [
"Definition:Generalized Ordered Space",
"Definition:Upper Section",
"Definition:Smallest Element",
"Definition:Open Set/Topology"
] | [
"Minimal Element in Toset is Unique and Smallest",
"Definition:Minimal/Element",
"Upper Section with no Minimal Element",
"Definition:Strict Upper Closure/Element",
"Open Ray is Open in GO-Space",
"Definition:Set Union",
"Definition:Open Set/Topology",
"Definition:Open Set/Topology",
"Definition:Ope... |
proofwiki-7407 | Lower Section with no Maximal Element | Let $\struct {S, \preceq}$ be an ordered set.
Let $L \subseteq S$.
Then:
:$L$ is a lower section in $S$ with no maximal element
{{iff}}:
:$\ds L = \bigcup \set {l^\prec: l \in L}$
where $l^\prec$ is the strict lower closure of $l$. | By Dual Pairs (Order Theory):
* Lower section is dual to upper section.
* 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 Section with no Minimal Element.
{{qed}} | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $L \subseteq S$.
Then:
:$L$ is a [[Definition:Lower Section|lower section]] in $S$ with no [[Definition:Maximal Element|maximal element]]
{{iff}}:
:$\ds L = \bigcup \set {l^\prec: l \in L}$
where $l^\prec$ is the [[Definition:Strict Lower C... | By [[Dual Pairs (Order Theory)]]:
* [[Definition:Lower Section|Lower section]] is dual to [[Definition:Upper Section|upper section]].
* [[Definition:Maximal Element|Maximal element]] is dual to [[Definition:Minimal Element|minimal element]].
* [[Definition:Strict Lower Closure of Element|Strict lower closure]] is dual ... | Lower Section with no Maximal Element | https://proofwiki.org/wiki/Lower_Section_with_no_Maximal_Element | https://proofwiki.org/wiki/Lower_Section_with_no_Maximal_Element | [
"Lower Sections"
] | [
"Definition:Ordered Set",
"Definition:Lower Section",
"Definition:Maximal/Element",
"Definition:Strict Lower Closure/Element"
] | [
"Dual Pairs (Order Theory)",
"Definition:Lower Section",
"Definition:Upper Section",
"Definition:Maximal/Element",
"Definition:Minimal/Element",
"Definition:Strict Lower Closure/Element",
"Definition:Strict Upper Closure/Element",
"Duality Principle (Order Theory)/Global Duality",
"Upper Section wit... |
proofwiki-7408 | Lower Section with no Greatest Element is Open in GO-Space | Let $\struct {S, \preceq, \tau}$ be a generalized ordered space.
Let $L$ be a lower section in $S$ with no greatest element.
Then $L$ is open in $\struct {S, \preceq, \tau}$. | By Maximal Element in Toset is Unique and Greatest, $L$ has no maximal element.
By Lower Section with no Maximal Element:
:$\ds L = \bigcup \set {l^\prec: l \in L}$
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}} | Let $\struct {S, \preceq, \tau}$ be a [[Definition:Generalized Ordered Space|generalized ordered space]].
Let $L$ be a [[Definition:Lower Section|lower section]] in $S$ with no [[Definition:Greatest Element|greatest element]].
Then $L$ is [[Definition:Open Set (Topology)|open]] in $\struct {S, \preceq, \tau}$. | By [[Maximal Element in Toset is Unique and Greatest]], $L$ has no [[Definition:Maximal Element|maximal element]].
By [[Lower Section with no Maximal Element]]:
:$\ds L = \bigcup \set {l^\prec: l \in L}$
where $l^\prec$ is the [[Definition:Strict Lower Closure of Element|strict lower closure]] of $l$.
By [[Open Ray ... | Lower Section with no Greatest Element is Open in GO-Space | https://proofwiki.org/wiki/Lower_Section_with_no_Greatest_Element_is_Open_in_GO-Space | https://proofwiki.org/wiki/Lower_Section_with_no_Greatest_Element_is_Open_in_GO-Space | [
"Generalized Ordered Spaces"
] | [
"Definition:Generalized Ordered Space",
"Definition:Lower Section",
"Definition:Greatest Element",
"Definition:Open Set/Topology"
] | [
"Maximal Element in Toset is Unique and Greatest",
"Definition:Maximal/Element",
"Lower Section with no Maximal Element",
"Definition:Strict Lower Closure/Element",
"Open Ray is Open in GO-Space",
"Definition:Open Set/Topology"
] |
proofwiki-7409 | Lower Section is Dual to Upper Section | Let $\struct {S, \preceq}$ be an ordered set.
Let $T \subseteq S$.
The following are dual statements:
:$T$ is a lower section in $S$
:$T$ is an upper section in $S$ | By definition, $T$ is a lower section 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 section in $S$.
The converse fo... | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $T \subseteq S$.
The following are [[Definition:Dual Statement (Order Theory)|dual statements]]:
:$T$ is a [[Definition:Lower Section|lower section]] in $S$
:$T$ is an [[Definition:Upper Section|upper section]] in $S$ | By definition, $T$ is a [[Definition:Lower Section|lower section]] in $S$ {{iff}}:
:$\forall t \in T: \forall s \in S: s \preceq t \implies s \in T$
The [[Definition:Dual Statement (Order Theory)|dual]] of this statement is:
:$\forall t \in T: \forall s \in S: t \preceq s \implies s \in T$
by [[Dual Pairs (Order Th... | Lower Section is Dual to Upper Section | https://proofwiki.org/wiki/Lower_Section_is_Dual_to_Upper_Section | https://proofwiki.org/wiki/Lower_Section_is_Dual_to_Upper_Section | [
"Upper Sections",
"Lower Sections",
"Dual Pairs (Order Theory)"
] | [
"Definition:Ordered Set",
"Definition:Dual Statement (Order Theory)",
"Definition:Lower Section",
"Definition:Upper Section"
] | [
"Definition:Lower Section",
"Definition:Dual Statement (Order Theory)",
"Dual Pairs (Order Theory)",
"Definition:Upper Section",
"Dual of Dual Statement (Order Theory)"
] |
proofwiki-7410 | Order Topology equals Dual Order Topology | Let $\struct {S, \preceq}$ be a totally ordered set.
Let $\tau$ be the $\preceq$-order topology on $S$.
Let $\tau'$ be the $\succeq$-order topology on $S$, where $\succeq$ is the dual ordering of $\preceq$.
Then $\tau' = \tau$. | {{improve|recast in terms of dual statements}}
Let $U$ be an open ray in $\struct {S, \preceq}$.
By Open Ray is Dual to Open Ray, $U$ is an open ray in $\struct {S, \preceq}$.
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 ... | Let $\struct {S, \preceq}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Let $\tau$ be the $\preceq$-[[Definition:Order Topology|order topology]] on $S$.
Let $\tau'$ be the $\succeq$-order topology on $S$, where $\succeq$ is the [[Definition:Dual Ordering|dual ordering]] of $\preceq$.
Then $\tau' = \... | {{improve|recast in terms of dual statements}}
Let $U$ be an [[Definition:Open Ray|open ray]] in $\struct {S, \preceq}$.
By [[Open Ray is Dual to Open Ray]], $U$ is an [[Definition:Open Ray|open ray]] in $\struct {S, \preceq}$.
Since the [[Definition:Open Ray|open rays]] in a [[Definition:Totally Ordered Set|totally... | Order Topology equals Dual Order Topology | https://proofwiki.org/wiki/Order_Topology_equals_Dual_Order_Topology | https://proofwiki.org/wiki/Order_Topology_equals_Dual_Order_Topology | [
"Order Topologies"
] | [
"Definition:Totally Ordered Set",
"Definition:Order Topology",
"Definition:Dual Ordering"
] | [
"Definition:Ray (Order Theory)/Open",
"Open Ray is Dual to Open Ray",
"Definition:Ray (Order Theory)/Open",
"Definition:Ray (Order Theory)/Open",
"Definition:Totally Ordered Set",
"Definition:Sub-Basis",
"Definition:Finer Topology",
"Definition:Finer Topology",
"Definition:Set Equality/Definition 2"... |
proofwiki-7411 | Open Ray is Dual to Open Ray | Let $\struct {S, \preceq}$ be a totally ordered set.
Let $R$ be an open ray in $\struct {S, \preceq}$.
Then $R$ is an open ray in $\struct {S, \succeq}$, where $\succeq$ is the dual ordering of $\preceq$. | 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 $R$... | Let $\struct {S, \preceq}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Let $R$ be an [[Definition:Open Ray|open ray]] in $\struct {S, \preceq}$.
Then $R$ is an [[Definition:Open Ray|open ray]] in $\struct {S, \succeq}$, where $\succeq$ is the [[Definition:Dual Ordering|dual ordering]] of $\preceq$. | By the definition of [[Definition:Open Ray|open ray]], there is some $p \in S$ such that:
:$R$ is the [[Definition:Strict Upper Closure of Element|strict upper]] or [[Definition:Strict Lower Closure of Element|strict lower closure]] of $p$ with respect to $\preceq$.
By [[Strict Lower Closure is Dual to Strict Upper C... | Open Ray is Dual to Open Ray | https://proofwiki.org/wiki/Open_Ray_is_Dual_to_Open_Ray | https://proofwiki.org/wiki/Open_Ray_is_Dual_to_Open_Ray | [
"Total Orderings"
] | [
"Definition:Totally Ordered Set",
"Definition:Ray (Order Theory)/Open",
"Definition:Ray (Order Theory)/Open",
"Definition:Dual Ordering"
] | [
"Definition:Ray (Order Theory)/Open",
"Definition:Strict Upper Closure/Element",
"Definition:Strict Lower Closure/Element",
"Strict Lower Closure is Dual to Strict Upper Closure",
"Definition:Dual Statement (Order Theory)",
"Definition:Strict Upper Closure/Element",
"Definition:Strict Lower Closure/Elem... |
proofwiki-7412 | Topologies on Set form Complete Lattice | Let $X$ be a non-empty set.
Let $\LL$ be the set of topologies on $X$.
Then $\struct {\LL, \subseteq}$ is a complete lattice. | Let $\KK \subseteq \LL$.
Then by Intersection of Topologies is Topology:
:$\bigcap \KK \in \LL$
By Intersection is Largest Subset, $\bigcap \LL$ is the infimum of $\KK$.
{{explain}}
Let $\tau$ be the topology generated by the sub-basis $\bigcup \KK$.
Then $\tau \in \LL$ and $\tau$ is the supremum of $\KK$.
We have that... | Let $X$ be a [[Definition:Non-Empty Set|non-empty set]].
Let $\LL$ be the [[Definition:Set|set]] of [[Definition:Topology|topologies]] on $X$.
Then $\struct {\LL, \subseteq}$ is a [[Definition:Complete Lattice|complete lattice]]. | Let $\KK \subseteq \LL$.
Then by [[Intersection of Topologies is Topology]]:
:$\bigcap \KK \in \LL$
By [[Intersection is Largest Subset]], $\bigcap \LL$ is the [[Definition:Infimum of Set|infimum]] of $\KK$.
{{explain}}
Let $\tau$ be the topology generated by the sub-basis $\bigcup \KK$.
Then $\tau \in \LL$ and $\... | Topologies on Set form Complete Lattice | https://proofwiki.org/wiki/Topologies_on_Set_form_Complete_Lattice | https://proofwiki.org/wiki/Topologies_on_Set_form_Complete_Lattice | [
"Topology",
"Complete Lattices"
] | [
"Definition:Non-Empty Set",
"Definition:Set",
"Definition:Topology",
"Definition:Complete Lattice"
] | [
"Intersection of Topologies is Topology",
"Intersection is Largest Subset",
"Definition:Infimum of Set",
"Definition:Supremum of Set",
"Definition:Complete Lattice",
"Category:Topology",
"Category:Complete Lattices"
] |
proofwiki-7413 | Complement of Lower Section is Upper Section | Let $\struct {S, \preceq}$ be an ordered set.
Let $L$ be a lower section.
Then $S \setminus L$ is an upper section. | Let $u \in S \setminus L$.
Let $s \in S$ such that $u \preceq s$.
{{AimForCont}} $s \notin S \setminus L$.
Then $s \in L$.
By definition of lower section, $u \in L$, a contradiction.
Hence $s \in S \setminus L$.
Since this holds for all such $u$ and $s$, $S \setminus L$ is an upper section.
{{qed}} | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $L$ be a [[Definition:Lower Section|lower section]].
Then $S \setminus L$ is an [[Definition:Upper Section|upper section]]. | Let $u \in S \setminus L$.
Let $s \in S$ such that $u \preceq s$.
{{AimForCont}} $s \notin S \setminus L$.
Then $s \in L$.
By definition of [[Definition:Lower Section|lower section]], $u \in L$, a contradiction.
Hence $s \in S \setminus L$.
Since this holds for all such $u$ and $s$, $S \setminus L$ is an [[Defini... | Complement of Lower Section is Upper Section | https://proofwiki.org/wiki/Complement_of_Lower_Section_is_Upper_Section | https://proofwiki.org/wiki/Complement_of_Lower_Section_is_Upper_Section | [
"Upper Sections",
"Lower Sections"
] | [
"Definition:Ordered Set",
"Definition:Lower Section",
"Definition:Upper Section"
] | [
"Definition:Lower Section",
"Definition:Upper Section"
] |
proofwiki-7414 | GO-Space Embeds as Closed Subspace of Linearly Ordered Space | Let $\struct {X, \preceq_X, \tau_X}$ be a generalized ordered space.
Then there is a linearly ordered space $\struct {Y, \preceq_Y, \tau_Y}$ and a mapping $\phi: X \to Y$ such that $\phi$ is a topological embedding and an order embedding, and $\phi \sqbrk {X}$ is closed in $Y$. | By GO-Space Embeds Densely into Linearly Ordered Space, there is a linearly ordered space $\struct {W, \preceq_W, \tau_W}$ and a mapping $\psi:X \to W$ which is an order embedding and a topological embedding.
{{WLOG}}, assume $X$ is a subspace of $W$.
Let $Y = \set {\tuple {x, 0}: x \in X} \cup \paren {W \setminus X} \... | Let $\struct {X, \preceq_X, \tau_X}$ be a [[Definition:Generalized Ordered Space|generalized ordered space]].
Then there is a [[Definition:Linearly Ordered Space|linearly ordered space]] $\struct {Y, \preceq_Y, \tau_Y}$ and a mapping $\phi: X \to Y$ such that $\phi$ is a topological embedding and an order embedding, ... | By [[GO-Space Embeds Densely into Linearly Ordered Space]], there is a linearly ordered space $\struct {W, \preceq_W, \tau_W}$ and a mapping $\psi:X \to W$ which is an order embedding and a topological embedding.
{{WLOG}}, assume $X$ is a subspace of $W$.
Let $Y = \set {\tuple {x, 0}: x \in X} \cup \paren {W \setminu... | GO-Space Embeds as Closed Subspace of Linearly Ordered Space | https://proofwiki.org/wiki/GO-Space_Embeds_as_Closed_Subspace_of_Linearly_Ordered_Space | https://proofwiki.org/wiki/GO-Space_Embeds_as_Closed_Subspace_of_Linearly_Ordered_Space | [
"Generalized Ordered Spaces",
"Linearly Ordered Spaces"
] | [
"Definition:Generalized Ordered Space",
"Definition:Linearly Ordered Space"
] | [
"GO-Space Embeds Densely into Linearly Ordered Space",
"Category:Generalized Ordered Spaces",
"Category:Linearly Ordered Spaces"
] |
proofwiki-7415 | Union of Total Ordering with Lower Sections is Total Ordering | Let $\struct {Y, \preceq}$ be a totally ordered set.
Let $X$ be the disjoint union of $Y$ with the set of lower sections of $Y$.
Define a relation $\preceq'$ on $X$ extending $\preceq$ by letting:
:$y_1 \preceq' y_2 \iff y_1 \preceq y_2$
:$y \preceq' L \iff y \in L$
:$L_1 \preceq' L_2 \iff L_1 \subseteq L_2$
:$L \prece... | First note that by Lower Sections in Totally Ordered Set form Chain:
:$\subseteq$ is a total ordering on the set of lower sections.
Also note that by Complement of Lower Section is Upper Section, the complement of each $\preceq$-lower section is a $\preceq$-upper section. | Let $\struct {Y, \preceq}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Let $X$ be the [[Definition:Disjoint Union (Set Theory)|disjoint union]] of $Y$ with the [[Definition:Set|set]] of [[Definition:Lower Section|lower sections]] of $Y$.
Define a [[Definition:Relation|relation]] $\preceq'$ on $X$ ext... | First note that by [[Lower Sections in Totally Ordered Set form Chain]]:
:$\subseteq$ is a [[Definition:Total Ordering|total ordering]] on the [[Definition:Set|set]] of [[Definition:Lower Section|lower sections]].
Also note that by [[Complement of Lower Section is Upper Section]], the [[Definition:Relative Complement|... | Union of Total Ordering with Lower Sections is Total Ordering | https://proofwiki.org/wiki/Union_of_Total_Ordering_with_Lower_Sections_is_Total_Ordering | https://proofwiki.org/wiki/Union_of_Total_Ordering_with_Lower_Sections_is_Total_Ordering | [
"Lower Sections",
"Total Orderings"
] | [
"Definition:Totally Ordered Set",
"Definition:Disjoint Union (Set Theory)",
"Definition:Set",
"Definition:Lower Section",
"Definition:Relation",
"Definition:Total Ordering"
] | [
"Lower Sections in Totally Ordered Set form Chain",
"Definition:Total Ordering",
"Definition:Set",
"Definition:Lower Section",
"Complement of Lower Section is Upper Section",
"Definition:Relative Complement",
"Definition:Lower Section",
"Definition:Upper Section",
"Definition:Lower Section",
"Defi... |
proofwiki-7416 | Lower Closure is Lower Section | Let $\struct {S, \preceq}$ be an ordered set.
Let $T$ be a subset of $S$.
Let $L$ be the lower closure of $T$.
Then $L$ is a lower section. | 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 section.
{{qed}} | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $T$ be a [[Definition:Subset|subset]] of $S$.
Let $L$ be the [[Definition:Lower Closure of Subset|lower closure]] of $T$.
Then $L$ is a [[Definition:Lower Section|lower section]]. | Let $a \in L$.
Let $b \in S$ with $b \preceq a$.
By the definition of [[Definition:Lower Closure of Subset|lower closure]], there is a $t \in T$ such that $a \preceq t$.
By [[Definition:Transitive Relation|transitivity]], $b \preceq t$.
Thus, again by the definition of [[Definition:Lower Closure of Subset|lower clo... | Lower Closure is Lower Section | https://proofwiki.org/wiki/Lower_Closure_is_Lower_Section | https://proofwiki.org/wiki/Lower_Closure_is_Lower_Section | [
"Lower Sections",
"Lower Closures"
] | [
"Definition:Ordered Set",
"Definition:Subset",
"Definition:Lower Closure/Set",
"Definition:Lower Section"
] | [
"Definition:Lower Closure/Set",
"Definition:Transitive Relation",
"Definition:Lower Closure/Set",
"Definition:Lower Section"
] |
proofwiki-7417 | Lower Closure is Lower Section | Let $\struct {S, \preceq}$ be an ordered set.
Let $T$ be a subset of $S$.
Let $L$ be the lower closure of $T$.
Then $L$ is a lower section. | 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 section.
{{qed}} | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $T$ be a [[Definition:Subset|subset]] of $S$.
Let $L$ be the [[Definition:Lower Closure of Subset|lower closure]] of $T$.
Then $L$ is a [[Definition:Lower Section|lower section]]. | Let $l \in p^\prec$.
Let $s \in S$ with $s \preceq l$.
Then by the definition of [[Definition:Strict Lower Closure of Element|strict lower closure]]:
:$l \prec p$
Thus by [[Extended Transitivity]]:
:$s \prec p$
So by the definition of [[Definition:Strict Lower Closure of Element|strict lower closure]]:
:$s \in p^\p... | Strict Lower Closure is Lower Section/Proof 1 | https://proofwiki.org/wiki/Lower_Closure_is_Lower_Section | https://proofwiki.org/wiki/Strict_Lower_Closure_is_Lower_Section/Proof_1 | [
"Lower Sections",
"Lower Closures"
] | [
"Definition:Ordered Set",
"Definition:Subset",
"Definition:Lower Closure/Set",
"Definition:Lower Section"
] | [
"Definition:Strict Lower Closure/Element",
"Extended Transitivity",
"Definition:Strict Lower Closure/Element",
"Definition:Lower Section"
] |
proofwiki-7418 | Lower Closure is Lower Section | Let $\struct {S, \preceq}$ be an ordered set.
Let $T$ be a subset of $S$.
Let $L$ be the lower closure of $T$.
Then $L$ is a lower section. | By Dual Pairs (Order Theory):
:strict upper closure is dual to strict lower closure
:Upper section is dual to lower section
Thus the theorem holds by Strict Upper Closure is Upper Section and the duality principle.
{{qed}} | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $T$ be a [[Definition:Subset|subset]] of $S$.
Let $L$ be the [[Definition:Lower Closure of Subset|lower closure]] of $T$.
Then $L$ is a [[Definition:Lower Section|lower section]]. | By [[Dual Pairs (Order Theory)]]:
:[[Definition:Strict Upper Closure of Element|strict upper closure]] is dual to [[Definition:Strict Lower Closure of Element|strict lower closure]]
:[[Definition:Upper Section|Upper section]] is dual to [[Definition:Lower Section|lower section]]
Thus the theorem holds by [[Strict Uppe... | Strict Lower Closure is Lower Section/Proof 2 | https://proofwiki.org/wiki/Lower_Closure_is_Lower_Section | https://proofwiki.org/wiki/Strict_Lower_Closure_is_Lower_Section/Proof_2 | [
"Lower Sections",
"Lower Closures"
] | [
"Definition:Ordered Set",
"Definition:Subset",
"Definition:Lower Closure/Set",
"Definition:Lower Section"
] | [
"Dual Pairs (Order Theory)",
"Definition:Strict Upper Closure/Element",
"Definition:Strict Lower Closure/Element",
"Definition:Upper Section",
"Definition:Lower Section",
"Strict Upper Closure is Upper Section",
"Duality Principle (Order Theory)/Global Duality"
] |
proofwiki-7419 | Ordered Set is Upper Section in Itself | Let $\struct {S, \preceq}$ be an ordered set.
Then $S$ is an upper section in $S$. | Follows immediately from the definition of upper section.
{{qed}}
Category:Upper Sections
fpxlqpe10b7pn56mo2o69xcv5qw295c | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Then $S$ is an [[Definition:Upper Section|upper section]] in $S$. | Follows immediately from the definition of [[Definition:Upper Section|upper section]].
{{qed}}
[[Category:Upper Sections]]
fpxlqpe10b7pn56mo2o69xcv5qw295c | Ordered Set is Upper Section in Itself | https://proofwiki.org/wiki/Ordered_Set_is_Upper_Section_in_Itself | https://proofwiki.org/wiki/Ordered_Set_is_Upper_Section_in_Itself | [
"Upper Sections"
] | [
"Definition:Ordered Set",
"Definition:Upper Section"
] | [
"Definition:Upper Section",
"Category:Upper Sections"
] |
proofwiki-7420 | Ordered Set is Lower Section in Itself | Let $\struct {S, \preceq}$ be an ordered set.
Then $S$ is a lower section in $S$. | Follows immediately from the definition of lower section.
{{qed}}
Category:Lower Sections
1t4xb95cwrx8blpmzfytmwfhz8wfd74 | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Then $S$ is a [[Definition:Lower Section|lower section]] in $S$. | Follows immediately from the definition of [[Definition:Lower Section|lower section]].
{{qed}}
[[Category:Lower Sections]]
1t4xb95cwrx8blpmzfytmwfhz8wfd74 | Ordered Set is Lower Section in Itself | https://proofwiki.org/wiki/Ordered_Set_is_Lower_Section_in_Itself | https://proofwiki.org/wiki/Ordered_Set_is_Lower_Section_in_Itself | [
"Lower Sections"
] | [
"Definition:Ordered Set",
"Definition:Lower Section"
] | [
"Definition:Lower Section",
"Category:Lower Sections"
] |
proofwiki-7421 | Ordered Set is Order-Convex in Itself | Let $\struct {S, \preceq}$ be an ordered set.
Then $S$ is an order-convex set in $S$. | Follows immediately from the definition of order-convex set.
{{qed}}
Category:Order-Convex Sets
Category:Order Theory
oimlxnt4fz28anw171f4x7tqdk3o8z0 | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Then $S$ is an [[Definition:Order-Convex Set|order-convex set]] in $S$. | Follows immediately from the definition of [[Definition:Order-Convex Set|order-convex set]].
{{qed}}
[[Category:Order-Convex Sets]]
[[Category:Order Theory]]
oimlxnt4fz28anw171f4x7tqdk3o8z0 | Ordered Set is Order-Convex in Itself | https://proofwiki.org/wiki/Ordered_Set_is_Order-Convex_in_Itself | https://proofwiki.org/wiki/Ordered_Set_is_Order-Convex_in_Itself | [
"Order-Convex Sets",
"Order Theory"
] | [
"Definition:Ordered Set",
"Definition:Order-Convex Set"
] | [
"Definition:Order-Convex Set",
"Category:Order-Convex Sets",
"Category:Order Theory"
] |
proofwiki-7422 | Equivalence of Definitions of Generalized Ordered Space/Definition 3 implies Definition 1 | Let $\struct {S, \preceq, \tau}$ be a generalized ordered space by Definition 3:
{{:Definition:Generalized Ordered Space/Definition 3}}
Then $\struct {S, \preceq, \tau}$ is a generalized ordered space by Definition 1:
{{:Definition:Generalized Ordered Space/Definition 1}} | Let $\SS$ be a sub-basis for $\tau$ consisting of upper sections and lower sections.
Let $\BB$ be the set of intersections of finite subsets of $\SS$.
By Upper Section is Order-Convex, Lower Section is Order-Convex and Intersection of Order-Convex Sets is Order-Convex:
:the elements of $\BB$ are order-convex.
From Synt... | Let $\struct {S, \preceq, \tau}$ be a [[Definition:Generalized Ordered Space/Definition 3|generalized ordered space by Definition 3]]:
{{:Definition:Generalized Ordered Space/Definition 3}}
Then $\struct {S, \preceq, \tau}$ is a [[Definition:Generalized Ordered Space/Definition 1|generalized ordered space by Definiti... | Let $\SS$ be a [[Definition:Sub-Basis|sub-basis]] for $\tau$ consisting of [[Definition:Upper Section|upper sections]] and [[Definition:Lower Section|lower sections]].
Let $\BB$ be the [[Definition:Set|set]] of [[Definition:Set Intersection|intersections]] of [[Definition:Finite Set|finite]] [[Definition:Subset|subset... | Equivalence of Definitions of Generalized Ordered Space/Definition 3 implies Definition 1 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Generalized_Ordered_Space/Definition_3_implies_Definition_1 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Generalized_Ordered_Space/Definition_3_implies_Definition_1 | [
"Equivalence of Definitions of Generalized Ordered Space"
] | [
"Definition:Generalized Ordered Space/Definition 3",
"Definition:Generalized Ordered Space/Definition 1"
] | [
"Definition:Sub-Basis",
"Definition:Upper Section",
"Definition:Lower Section",
"Definition:Set",
"Definition:Set Intersection",
"Definition:Finite Set",
"Definition:Subset",
"Upper Section is Order-Convex",
"Lower Section is Order-Convex",
"Intersection of Order-Convex Sets is Order-Convex",
"D... |
proofwiki-7423 | Strict Upper Closure is Upper Section | Let $\struct {S, \preceq}$ be an ordered set.
Let $p \in S$.
Let $p^\succ$ denote the strict upper closure of $p$.
Then $p^\succ$ is an upper section. | 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 section.
{{qed}} | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $p \in S$.
Let $p^\succ$ denote the [[Definition:Strict Upper Closure of Element|strict upper closure]] of $p$.
Then $p^\succ$ is an [[Definition:Upper Section|upper section]]. | Let $u \in p^\succ$.
Let $s \in S$ with $u \preceq s$.
Then by the definition of [[Definition:Strict Upper Closure of Element|strict upper closure]]:
:$p \prec u$
Thus by [[Extended Transitivity]]:
:$p \prec s$
So by the definition of [[Definition:Strict Upper Closure of Element|strict upper closure]]:
:$s \in p^\s... | Strict Upper Closure is Upper Section | https://proofwiki.org/wiki/Strict_Upper_Closure_is_Upper_Section | https://proofwiki.org/wiki/Strict_Upper_Closure_is_Upper_Section | [
"Upper Sections",
"Upper Closures"
] | [
"Definition:Ordered Set",
"Definition:Strict Upper Closure/Element",
"Definition:Upper Section"
] | [
"Definition:Strict Upper Closure/Element",
"Extended Transitivity",
"Definition:Strict Upper Closure/Element",
"Definition:Upper Section"
] |
proofwiki-7424 | Strict Lower Closure is Lower Section | Let $\struct {S, \preceq}$ be an ordered set.
Let $p \in S$.
Let $p^\prec$ denote the strict lower closure of $p$.
Then $p^\prec$ is a lower section. | 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 section.
{{qed}} | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $p \in S$.
Let $p^\prec$ denote the [[Definition:Strict Lower Closure of Element|strict lower closure]] of $p$.
Then $p^\prec$ is a [[Definition:Lower Section|lower section]]. | Let $l \in p^\prec$.
Let $s \in S$ with $s \preceq l$.
Then by the definition of [[Definition:Strict Lower Closure of Element|strict lower closure]]:
:$l \prec p$
Thus by [[Extended Transitivity]]:
:$s \prec p$
So by the definition of [[Definition:Strict Lower Closure of Element|strict lower closure]]:
:$s \in p^\p... | Strict Lower Closure is Lower Section/Proof 1 | https://proofwiki.org/wiki/Strict_Lower_Closure_is_Lower_Section | https://proofwiki.org/wiki/Strict_Lower_Closure_is_Lower_Section/Proof_1 | [
"Lower Sections",
"Lower Closures",
"Strict Lower Closure is Lower Section"
] | [
"Definition:Ordered Set",
"Definition:Strict Lower Closure/Element",
"Definition:Lower Section"
] | [
"Definition:Strict Lower Closure/Element",
"Extended Transitivity",
"Definition:Strict Lower Closure/Element",
"Definition:Lower Section"
] |
proofwiki-7425 | Strict Lower Closure is Lower Section | Let $\struct {S, \preceq}$ be an ordered set.
Let $p \in S$.
Let $p^\prec$ denote the strict lower closure of $p$.
Then $p^\prec$ is a lower section. | By Dual Pairs (Order Theory):
:strict upper closure is dual to strict lower closure
:Upper section is dual to lower section
Thus the theorem holds by Strict Upper Closure is Upper Section and the duality principle.
{{qed}} | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $p \in S$.
Let $p^\prec$ denote the [[Definition:Strict Lower Closure of Element|strict lower closure]] of $p$.
Then $p^\prec$ is a [[Definition:Lower Section|lower section]]. | By [[Dual Pairs (Order Theory)]]:
:[[Definition:Strict Upper Closure of Element|strict upper closure]] is dual to [[Definition:Strict Lower Closure of Element|strict lower closure]]
:[[Definition:Upper Section|Upper section]] is dual to [[Definition:Lower Section|lower section]]
Thus the theorem holds by [[Strict Uppe... | Strict Lower Closure is Lower Section/Proof 2 | https://proofwiki.org/wiki/Strict_Lower_Closure_is_Lower_Section | https://proofwiki.org/wiki/Strict_Lower_Closure_is_Lower_Section/Proof_2 | [
"Lower Sections",
"Lower Closures",
"Strict Lower Closure is Lower Section"
] | [
"Definition:Ordered Set",
"Definition:Strict Lower Closure/Element",
"Definition:Lower Section"
] | [
"Dual Pairs (Order Theory)",
"Definition:Strict Upper Closure/Element",
"Definition:Strict Lower Closure/Element",
"Definition:Upper Section",
"Definition:Lower Section",
"Strict Upper Closure is Upper Section",
"Duality Principle (Order Theory)/Global Duality"
] |
proofwiki-7426 | Topology is Discrete iff All Singletons are Open | Let $\struct {S, \tau}$ be a topological space.
Then:
:$\tau$ is the discrete topology on $S$
{{iff}}:
:$\forall x \in S: \set x \in \tau$
That is, {{iff}} every singleton of $S$ is $\tau$-open. | === Sufficient Condition ===
Let $\tau$ be the discrete topology on $S$.
Let $x \in S$ be arbitrary.
Then from Set in Discrete Topology is Clopen it follows directly that $\set x$ is open in $\struct {S, \tau}$.
{{qed|lemma}} | Let $\struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Then:
:$\tau$ is the [[Definition:Discrete Topology|discrete topology]] on $S$
{{iff}}:
:$\forall x \in S: \set x \in \tau$
That is, {{iff}} every [[Definition:Singleton|singleton]] of $S$ is [[Definition:Open Set (Topology)|$\tau$-open... | === Sufficient Condition ===
Let $\tau$ be the [[Definition:Discrete Topology|discrete topology]] on $S$.
Let $x \in S$ be [[Definition:Arbitrary|arbitrary]].
Then from [[Set in Discrete Topology is Clopen]] it follows directly that $\set x$ is [[Definition:Open Set (Topology)|open]] in $\struct {S, \tau}$.
{{qed|le... | Topology is Discrete iff All Singletons are Open | https://proofwiki.org/wiki/Topology_is_Discrete_iff_All_Singletons_are_Open | https://proofwiki.org/wiki/Topology_is_Discrete_iff_All_Singletons_are_Open | [
"Topology is Discrete iff All Singletons are Open",
"Discrete Topologies",
"Singletons"
] | [
"Definition:Topological Space",
"Definition:Discrete Topology",
"Definition:Singleton",
"Definition:Open Set/Topology"
] | [
"Definition:Discrete Topology",
"Definition:Arbitrary",
"Set in Discrete Topology is Clopen",
"Definition:Open Set/Topology",
"Definition:Arbitrary",
"Definition:Open Set/Topology",
"Definition:Open Set/Topology",
"Definition:Open Set/Topology",
"Definition:Open Set/Topology",
"Definition:Arbitrar... |
proofwiki-7427 | Characteristic Function of Universe | Let $S$ be a set.
Let $\chi_S: S \to \set {0, 1}$ be its characteristic function (in itself).
Then:
:$\chi_S = f_1$
where $f_1: S \to \set {0, 1}$ is the constant mapping with value $1$. | From Characteristic Function Determined by 1-Fiber, $\chi_S$ is the mapping determined by:
:$\forall s \in S: \map {\chi_S} s = 1 \iff s \in S$
Thus:
:$\forall s \in S: \map {\chi_S} s = 1$
By definition of constant mapping:
:$\chi_S = f_1$
{{qed}}
Category:Characteristic Functions
1wa2giegor01nb3o0133fgh93fanx4u | Let $S$ be a [[Definition:Set|set]].
Let $\chi_S: S \to \set {0, 1}$ be its [[Definition:Characteristic Function of Set|characteristic function]] (in itself).
Then:
:$\chi_S = f_1$
where $f_1: S \to \set {0, 1}$ is the [[Definition:Constant Mapping|constant mapping]] with value $1$. | From [[Characteristic Function Determined by 1-Fiber]], $\chi_S$ is the [[Definition:Mapping|mapping]] determined by:
:$\forall s \in S: \map {\chi_S} s = 1 \iff s \in S$
Thus:
:$\forall s \in S: \map {\chi_S} s = 1$
By definition of [[Definition:Constant Mapping|constant mapping]]:
:$\chi_S = f_1$
{{qed}}
[[Cate... | Characteristic Function of Universe | https://proofwiki.org/wiki/Characteristic_Function_of_Universe | https://proofwiki.org/wiki/Characteristic_Function_of_Universe | [
"Characteristic Functions"
] | [
"Definition:Set",
"Definition:Characteristic Function (Set Theory)/Set",
"Definition:Constant Mapping"
] | [
"Characteristic Function Determined by 1-Fiber",
"Definition:Mapping",
"Definition:Constant Mapping",
"Category:Characteristic Functions"
] |
proofwiki-7428 | Supremum of Lower Closure of Set | Let $\left({S, \preceq}\right)$ be an ordered set.
Let $T \subseteq S$.
Let $L = T^\preceq$ be the lower closure of $T$ in $S$.
Let $s \in S$
Then $s$ is the supremum of $T$ {{iff}} it is the supremum of $L$. | By Supremum and Infimum are Unique we need only show that $s$ is a supremum of $L$ {{iff}} it is a supremum of $T$. | Let $\left({S, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
Let $T \subseteq S$.
Let $L = T^\preceq$ be the [[Definition:Lower Closure of Subset|lower closure]] of $T$ in $S$.
Let $s \in S$
Then $s$ is the [[Definition:Supremum of Set|supremum]] of $T$ {{iff}} it is the [[Definition:Supremum of Se... | By [[Supremum and Infimum are Unique]] we need only show that $s$ is a [[Definition:Supremum of Set|supremum]] of $L$ {{iff}} it is a [[Definition:Supremum of Set|supremum]] of $T$. | Supremum of Lower Closure of Set | https://proofwiki.org/wiki/Supremum_of_Lower_Closure_of_Set | https://proofwiki.org/wiki/Supremum_of_Lower_Closure_of_Set | [
"Lower Closures"
] | [
"Definition:Ordered Set",
"Definition:Lower Closure/Set",
"Definition:Supremum of Set",
"Definition:Supremum of Set"
] | [
"Supremum and Infimum are Unique",
"Definition:Supremum of Set",
"Definition:Supremum of Set",
"Definition:Supremum of Set",
"Definition:Supremum of Set",
"Definition:Supremum of Set",
"Definition:Supremum of Set",
"Definition:Supremum of Set",
"Definition:Supremum of Set",
"Definition:Supremum of... |
proofwiki-7429 | Upper Closure is Smallest Containing Upper Section | Let $\struct {S, \preceq}$ be an ordered set.
Let $T \subseteq S$.
Let $U = T^\succeq$ be the upper closure of $T$.
Then $U$ is the smallest upper section containing $T$ as a subset. | Follows from Upper Closure is Closure Operator and Set Closure is Smallest Closed Set/Closure Operator.
{{qed}} | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $T \subseteq S$.
Let $U = T^\succeq$ be the [[Definition:Upper Closure of Subset|upper closure]] of $T$.
Then $U$ is the smallest [[Definition:Upper Section|upper section]] containing $T$ as a [[Definition:Subset|subset]]. | Follows from [[Upper Closure is Closure Operator]] and [[Set Closure is Smallest Closed Set/Closure Operator]].
{{qed}} | Upper Closure is Smallest Containing Upper Section | https://proofwiki.org/wiki/Upper_Closure_is_Smallest_Containing_Upper_Section | https://proofwiki.org/wiki/Upper_Closure_is_Smallest_Containing_Upper_Section | [
"Upper Closures",
"Upper Sections"
] | [
"Definition:Ordered Set",
"Definition:Upper Closure/Set",
"Definition:Upper Section",
"Definition:Subset"
] | [
"Upper Closure is Closure Operator",
"Set Closure is Smallest Closed Set/Closure Operator"
] |
proofwiki-7430 | Upper Closure is Closure Operator | Let $\struct {S, \preceq}$ be an ordered set.
Let $T^\succeq$ be the upper closure of $T$ for each $T \subseteq S$.
Then $\cdot^\succeq$ is a closure operator. | === 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 \s... | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $T^\succeq$ be the [[Definition:Upper Closure of Subset|upper closure]] of $T$ for each $T \subseteq S$.
Then $\cdot^\succeq$ is a [[Definition:Closure Operator|closure operator]]. | === Inflationary ===
Let $T \subseteq S$.
Let $t \in T$.
Then since $T \subseteq S$, $t \in S$ by the definition of [[Definition:subset|subset]].
Since $\preceq$ is [[Definition:Reflexive Relation|reflexive]], $t \preceq t$.
Thus by the definition of [[Definition:Upper Closure of Subset|upper closure]], $t \in T^\... | Upper Closure is Closure Operator | https://proofwiki.org/wiki/Upper_Closure_is_Closure_Operator | https://proofwiki.org/wiki/Upper_Closure_is_Closure_Operator | [
"Upper Closures",
"Closure Operators"
] | [
"Definition:Ordered Set",
"Definition:Upper Closure/Set",
"Definition:Closure Operator"
] | [
"Definition:subset",
"Definition:Reflexive Relation",
"Definition:Upper Closure/Set",
"Definition:Inflationary Mapping",
"Definition:Upper Closure/Set",
"Definition:Upper Closure/Set"
] |
proofwiki-7431 | Equivalence of Definitions of Upper Section | Let $\struct {S, \preceq}$ be an ordered set.
Let $U \subseteq S$.
{{TFAE|def = Upper Section}} | === 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 ho... | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $U \subseteq S$.
{{TFAE|def = Upper Section}} | === 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 [[Definition:Upper Closure of Subset|upper closure]], there is some $u \in U$ such that $u \preceq k$.
Since $k \in U^\succeq \subseteq S$, the ... | Equivalence of Definitions of Upper Section | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Upper_Section | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Upper_Section | [
"Upper Sections"
] | [
"Definition:Ordered Set"
] | [
"Definition:Upper Closure/Set",
"Definition:Upper Closure/Set",
"Definition:Upper Closure/Set"
] |
proofwiki-7432 | Topological Closure is Closure Operator | The topological closure operator is a closure operator. | === Extensive ===
Follows from Set is Subset of its Topological Closure. | The [[Definition:Closure (Topology)|topological closure]] operator is a [[Definition:Closure Operator|closure operator]]. | === Extensive ===
Follows from [[Set is Subset of its Topological Closure]]. | Topological Closure is Closure Operator | https://proofwiki.org/wiki/Topological_Closure_is_Closure_Operator | https://proofwiki.org/wiki/Topological_Closure_is_Closure_Operator | [
"Set Closures",
"Examples of Closure Operators"
] | [
"Definition:Closure (Topology)",
"Definition:Closure Operator"
] | [
"Set is Subset of its Topological Closure"
] |
proofwiki-7433 | Reflexive Closure is Closure Operator | Let $S$ be a set.
Let $R$ be the set of all endorelations on $S$.
Then the reflexive closure operator on $R$ is a closure operator. | Let $\QQ$ be the set of reflexive relations on $S$.
By Intersection of Reflexive Relations is Reflexive, the intersection of any subset of $\QQ$ is in $Q$.
By the definition of reflexive closure as the intersection of reflexive supersets:
:The reflexive closure of a relation $\RR$ on $S$ is the intersection of elements... | Let $S$ be a [[Definition:set|set]].
Let $R$ be the set of all [[Definition:Endorelation|endorelations]] on $S$.
Then the [[Definition:Reflexive Closure|reflexive closure]] operator on $R$ is a [[Definition:Closure Operator|closure operator]]. | Let $\QQ$ be the [[Definition:Set of Sets|set]] of [[Definition:Reflexive Relation|reflexive relations]] on $S$.
By [[Intersection of Reflexive Relations is Reflexive]], the [[Definition:Set Intersection|intersection]] of any [[Definition:subset|subset]] of $\QQ$ is in $Q$.
By the definition of [[Definition:Reflexive... | Reflexive Closure is Closure Operator/Proof 1 | https://proofwiki.org/wiki/Reflexive_Closure_is_Closure_Operator | https://proofwiki.org/wiki/Reflexive_Closure_is_Closure_Operator/Proof_1 | [
"Reflexive Closures",
"Closure Operators",
"Reflexive Closure is Closure Operator"
] | [
"Definition:set",
"Definition:Endorelation",
"Definition:Reflexive Closure",
"Definition:Closure Operator"
] | [
"Definition:Set of Sets",
"Definition:Reflexive Relation",
"Intersection of Reflexive Relations is Reflexive",
"Definition:Set Intersection",
"Definition:subset",
"Definition:Reflexive Closure/Intersection of Reflexive Supersets",
"Definition:Endorelation",
"Definition:Set Intersection",
"Closure Op... |
proofwiki-7434 | Reflexive Closure is Closure Operator | Let $S$ be a set.
Let $R$ be the set of all endorelations on $S$.
Then the reflexive closure operator on $R$ is a closure operator. | === 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 ... | Let $S$ be a [[Definition:set|set]].
Let $R$ be the set of all [[Definition:Endorelation|endorelations]] on $S$.
Then the [[Definition:Reflexive Closure|reflexive closure]] operator on $R$ is a [[Definition:Closure Operator|closure operator]]. | === [[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 defi... | Reflexive Closure is Closure Operator/Proof 2 | https://proofwiki.org/wiki/Reflexive_Closure_is_Closure_Operator | https://proofwiki.org/wiki/Reflexive_Closure_is_Closure_Operator/Proof_2 | [
"Reflexive Closures",
"Closure Operators",
"Reflexive Closure is Closure Operator"
] | [
"Definition:set",
"Definition:Endorelation",
"Definition:Reflexive Closure",
"Definition:Closure Operator"
] | [
"Reflexive Closure is Inflationary",
"Reflexive Closure is Order Preserving",
"Reflexive Closure is Idempotent",
"Definition:Closure Operator",
"Definition:Reflexive Closure"
] |
proofwiki-7435 | Set Closure is Smallest Closed Set/Closure Operator | Let $S$ be a set.
Let $\cl: \powerset S \to \powerset S$ be a closure operator.
Let $T \subseteq S$.
Then $\map \cl T$ is the smallest closed set (with respect to $\cl$) containing $T$ as a subset. | 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 $T$ a... | Let $S$ be a [[Definition:Set|set]].
Let $\cl: \powerset S \to \powerset S$ be a [[Definition:Closure Operator|closure operator]].
Let $T \subseteq S$.
Then $\map \cl T$ is the smallest [[Definition:Closed Set under Closure Operator|closed set]] (with respect to $\cl$) containing $T$ as a [[Definition:Subset|subset]... | By definition, $\map \cl T$ is [[Definition:Closed Set under Closure Operator|closed]].
Let $C$ be closed.
Let $T \subseteq C$.
By the definition of [[Definition:Closure Operator|closure operator]], $\cl$ is $\subseteq$-[[Definition:Increasing Mapping|increasing]].
So:
:$\map \cl T \subseteq \map \cl C$
Since $C$ ... | Set Closure is Smallest Closed Set/Closure Operator | https://proofwiki.org/wiki/Set_Closure_is_Smallest_Closed_Set/Closure_Operator | https://proofwiki.org/wiki/Set_Closure_is_Smallest_Closed_Set/Closure_Operator | [
"Closure Operators",
"Set Closure is Smallest Closed Set"
] | [
"Definition:Set",
"Definition:Closure Operator",
"Definition:Closed Set/Closure Operator",
"Definition:Subset"
] | [
"Definition:Closed Set/Closure Operator",
"Definition:Closure Operator",
"Definition:Increasing/Mapping",
"Definition:Closed Set/Closure Operator",
"Definition:Closed Set/Closure Operator",
"Definition:Subset",
"Category:Closure Operators",
"Category:Set Closure is Smallest Closed Set"
] |
proofwiki-7436 | Equivalence of Definitions of Lower Section | Let $\struct {S, \preceq}$ be an ordered set.
Let $U \subseteq S$.
{{TFAE|def = Lower Section}} | We are required to show that the following are equivalent:
{{begin-axiom}}
{{axiom | n = 1
| m = \forall l \in L: \forall s \in S: s \preceq l \implies s \in L
}}
{{axiom | n = 2
| m = L^\preceq \subseteq L
}}
{{axiom | n = 3
| m = L^\preceq = L
}}
{{end-axiom}}
By the Duality Principle, it suff... | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $U \subseteq S$.
{{TFAE|def = Lower Section}} | We are required to show that the following are [[Definition:Logically Equivalent|equivalent]]:
{{begin-axiom}}
{{axiom | n = 1
| m = \forall l \in L: \forall s \in S: s \preceq l \implies s \in L
}}
{{axiom | n = 2
| m = L^\preceq \subseteq L
}}
{{axiom | n = 3
| m = L^\preceq = L
}}
{{end-axio... | Equivalence of Definitions of Lower Section | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Lower_Section | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Lower_Section | [
"Lower Sections"
] | [
"Definition:Ordered Set"
] | [
"Definition:Logical Equivalence",
"Duality Principle (Order Theory)/Global Duality",
"Definition:Logical Equivalence",
"Definition:Dual Statement (Order Theory)",
"Dual Pairs (Order Theory)",
"Definition:Logical Equivalence",
"Equivalence of Definitions of Upper Section",
"Category:Lower Sections"
] |
proofwiki-7437 | Convergent Series of Natural Numbers | Let $\sequence {a_n}_{n \mathop \in \N}$ be a sequence of natural numbers.
Then the following are equivalent:
$(1): \quad \ds \sum_{n \mathop = 1}^\infty a_n$ converges
$(2): \quad \exists N \in \N: \forall n \ge N: a_n = 0$
That is, $\ds \sum_{n \mathop = 1}^\infty a_n$ converges {{iff}} only finitely many of the $a_n... | $(1) \implies (2)$:
Suppose that there is an infinite subsequence $\sequence {a_{n_k} }_{k \mathop \in \N}$ such that for each $k$, $a_{n_k} \ne 0$.
For $N \in \N$ let
:$\ds s_N = \sum_{n \mathop = 1}^N a_n$
To show that $s_N$ diverges it suffices to show that:
:$\forall M > 0: \exists N \in \N : \forall n > N : \size... | Let $\sequence {a_n}_{n \mathop \in \N}$ be a [[Definition:Sequence|sequence]] of [[Definition:Natural Number|natural numbers]].
Then the following are [[Definition:Logical Equivalence|equivalent]]:
$(1): \quad \ds \sum_{n \mathop = 1}^\infty a_n$ [[Definition:Convergent Series|converges]]
$(2): \quad \exists N \in... | $(1) \implies (2)$:
Suppose that there is an [[Definition:Infinity|infinite]] [[Definition:Subsequence|subsequence]] $\sequence {a_{n_k} }_{k \mathop \in \N}$ such that for each $k$, $a_{n_k} \ne 0$.
For $N \in \N$ let
:$\ds s_N = \sum_{n \mathop = 1}^N a_n$
To show that $s_N$ [[Definition:Divergent Sequence|diver... | Convergent Series of Natural Numbers | https://proofwiki.org/wiki/Convergent_Series_of_Natural_Numbers | https://proofwiki.org/wiki/Convergent_Series_of_Natural_Numbers | [
"Natural Numbers",
"Series"
] | [
"Definition:Sequence",
"Definition:Natural Numbers",
"Definition:Logical Equivalence",
"Definition:Convergent Series",
"Definition:Finite Set"
] | [
"Definition:Infinity",
"Definition:Subsequence",
"Definition:Divergent Sequence",
"Definition:Positive",
"Definition:Increasing/Sequence",
"Category:Natural Numbers",
"Category:Series"
] |
proofwiki-7438 | Closure Operator from Closed Sets | Let $S$ be a set.
Let $\CC$ be a set of subsets of $S$.
Let $\CC$ be closed under arbitrary intersections:
:$\forall \KK \in \powerset \CC: \bigcap \KK \in \CC$
where $\bigcap \O$ is taken to be $S$.
Define $\cl: \powerset S \to \CC$ by letting:
:$\map \cl T = \bigcap \set {C \in \CC: T \subseteq C}$
Then $\cl$ is a cl... | First we will show that $\cl$ is a closure operator. | Let $S$ be a [[Definition:set|set]].
Let $\CC$ be a set of [[Definition:Subset|subsets]] of $S$.
Let $\CC$ be closed under arbitrary [[Definition:Set Intersection|intersections]]:
:$\forall \KK \in \powerset \CC: \bigcap \KK \in \CC$
where $\bigcap \O$ is taken to be $S$.
Define $\cl: \powerset S \to \CC$ by lettin... | First we will show that $\cl$ is a [[Definition:Closure Operator|closure operator]]. | Closure Operator from Closed Sets | https://proofwiki.org/wiki/Closure_Operator_from_Closed_Sets | https://proofwiki.org/wiki/Closure_Operator_from_Closed_Sets | [
"Closure Operators"
] | [
"Definition:set",
"Definition:Subset",
"Definition:Set Intersection",
"Definition:Closure Operator",
"Definition:Closed Set/Closure Operator"
] | [
"Definition:Closure Operator"
] |
proofwiki-7439 | Intersection is Decreasing | Let $U$ be a set.
Let $\FF$ and $\GG$ be sets of subsets of $U$.
Then $\FF \subseteq \GG \implies \bigcap \GG \subseteq \bigcap \FF$, where by convention $\bigcap \O = U$.
That is, $\bigcap$ is a decreasing mapping from $\struct {\powerset {\powerset U}, \subseteq}$ to $\struct {\powerset U, \subseteq}$, where $\powers... | Let $\FF \subseteq \GG$.
Let $x \in \bigcap \GG$.
Then for each $S \in \FF$, $S \in \GG$.
By the definition of intersection, $x \in S$.
Since this holds for all $S \in \FF$, $x \in \bigcap \FF$.
Since this holds for all $ x \in \bigcap \GG$:
:$\bigcap \GG \subseteq \bigcap \FF$
{{qed}}
Category:Set Intersection
sb5cyeo... | Let $U$ be a [[Definition:Set|set]].
Let $\FF$ and $\GG$ be [[Definition:Set of Sets|sets]] of [[Definition:Subset|subsets]] of $U$.
Then $\FF \subseteq \GG \implies \bigcap \GG \subseteq \bigcap \FF$, where by convention $\bigcap \O = U$.
That is, $\bigcap$ is a [[Definition:Decreasing Mapping|decreasing mapping]]... | Let $\FF \subseteq \GG$.
Let $x \in \bigcap \GG$.
Then for each $S \in \FF$, $S \in \GG$.
By the definition of [[Definition:Set Intersection|intersection]], $x \in S$.
Since this holds for all $S \in \FF$, $x \in \bigcap \FF$.
Since this holds for all $ x \in \bigcap \GG$:
:$\bigcap \GG \subseteq \bigcap \FF$
{{q... | Intersection is Decreasing | https://proofwiki.org/wiki/Intersection_is_Decreasing | https://proofwiki.org/wiki/Intersection_is_Decreasing | [
"Set Intersection"
] | [
"Definition:Set",
"Definition:Set of Sets",
"Definition:Subset",
"Definition:Decreasing/Mapping",
"Definition:Power Set"
] | [
"Definition:Set Intersection",
"Category:Set Intersection"
] |
proofwiki-7440 | Open Ray is Open in GO-Space/Definition 2 | Let $\struct {S, \preceq, \tau}$ be a generalized ordered space by Definition 2.
That is:
:Let $\struct {S, \preceq}$ be a totally ordered set.
:Let $\struct {S, \tau}$ be a topological space.
Let there be:
:a linearly ordered space $\struct {S', \preceq', \tau'}$
and:
:a mapping $\phi: S \to S'$ which is both:
::a $\p... | 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 ... | Let $\struct {S, \preceq, \tau}$ be a [[Definition:Generalized Ordered Space/Definition 2|generalized ordered space by Definition 2]].
That is:
:Let $\struct {S, \preceq}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
:Let $\struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
... | We will prove that $p^\succ$ is [[Definition:Open Set (Topology)|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 ... | Open Ray is Open in GO-Space/Definition 2 | https://proofwiki.org/wiki/Open_Ray_is_Open_in_GO-Space/Definition_2 | https://proofwiki.org/wiki/Open_Ray_is_Open_in_GO-Space/Definition_2 | [
"Generalized Ordered Spaces"
] | [
"Definition:Generalized Ordered Space/Definition 2",
"Definition:Totally Ordered Set",
"Definition:Topological Space",
"Definition:Linearly Ordered Space",
"Definition:Mapping",
"Definition:Order Embedding",
"Definition:Embedding (Topology)",
"Definition:Open Set/Topology",
"Definition:Strict Lower ... | [
"Definition:Open Set/Topology",
"Inverse Image under Order Embedding of Strict Upper Closure of Image of Point",
"Definition:Ray (Order Theory)/Open",
"Definition:Open Set/Topology",
"Definition:Order Topology",
"Definition:Open Set/Topology",
"Definition:Continuous Mapping (Topology)",
"Definition:Op... |
proofwiki-7441 | Union is Increasing | Let $U$ be a set.
Let $\FF$ and $\GG$ be sets of subsets of $U$.
Then $\FF \subseteq \GG \implies \bigcup \FF \subseteq \bigcup \GG$.
That is, $\bigcup$ is an increasing mapping from $\struct {\powerset {\powerset U}, \subseteq}$ to $\struct {\powerset U, \subseteq}$, where $\powerset U$ is the power set of $U$. | Let $\FF \subseteq \GG$.
Let $x \in \bigcup \FF$.
Then by the definition of union:
:$\exists S \in \FF: x \in S$
By the definition of subset:
:$S \in \GG$
Thus by the definition of union:
:$x \in \bigcup \GG$
Since this holds for all $x \in \bigcup \FF$:
:$\bigcup \FF \subseteq \bigcup \GG$
{{qed}}
Category:Set Union
m... | Let $U$ be a [[Definition:Set|set]].
Let $\FF$ and $\GG$ be [[Definition:Set of Sets|sets]] of [[Definition:Subset|subsets]] of $U$.
Then $\FF \subseteq \GG \implies \bigcup \FF \subseteq \bigcup \GG$.
That is, $\bigcup$ is an [[Definition:Increasing Mapping|increasing mapping]] from $\struct {\powerset {\powerset ... | Let $\FF \subseteq \GG$.
Let $x \in \bigcup \FF$.
Then by the definition of [[Definition:Set Union|union]]:
:$\exists S \in \FF: x \in S$
By the definition of [[Definition:Subset|subset]]:
:$S \in \GG$
Thus by the definition of [[Definition:Set Union|union]]:
:$x \in \bigcup \GG$
Since this holds for all $x \in \b... | Union is Increasing | https://proofwiki.org/wiki/Union_is_Increasing | https://proofwiki.org/wiki/Union_is_Increasing | [
"Set Union"
] | [
"Definition:Set",
"Definition:Set of Sets",
"Definition:Subset",
"Definition:Increasing/Mapping",
"Definition:Power Set"
] | [
"Definition:Set Union",
"Definition:Subset",
"Definition:Set Union",
"Category:Set Union"
] |
proofwiki-7442 | Set Intersection is Idempotent/Indexed Family | Let $\family {F_i}_{i \mathop \in I}$ be a non-empty indexed family of sets.
Suppose that all the sets in the $\family {F_i}_{i \mathop \in I}$ are the same.
That is, suppose that for some set $S$:
:$\forall i \in I: F_i = S$
Then:
:$\ds \bigcap_{i \mathop \in I} F_i = S$
where $\ds \bigcap_{i \mathop \in I} F_i$ is th... | First we show that:
:$\ds \bigcap_{i \mathop \in I} F_i \subseteq S$
Let $x \in \ds \bigcap_{i \mathop \in I} F_i$.
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 $x \in \ds \bigcap_{i \mathop \in I} F_i$:
... | Let $\family {F_i}_{i \mathop \in I}$ be a non-empty [[Definition:Indexed Family of Sets|indexed family of sets]].
Suppose that all the [[Definition:Set|sets]] in the $\family {F_i}_{i \mathop \in I}$ are the same.
That is, suppose that for some [[Definition:Set|set]] $S$:
:$\forall i \in I: F_i = S$
Then:
:$\ds ... | First we show that:
:$\ds \bigcap_{i \mathop \in I} F_i \subseteq S$
Let $x \in \ds \bigcap_{i \mathop \in I} F_i$.
Since $I$ is [[Definition:Non-Empty Set|non-empty]], it has an [[Definition:Element|element]] $k$.
By the definition of [[Definition:Intersection of Family|intersection]], $x \in F_k$.
By the premise... | Set Intersection is Idempotent/Indexed Family | https://proofwiki.org/wiki/Set_Intersection_is_Idempotent/Indexed_Family | https://proofwiki.org/wiki/Set_Intersection_is_Idempotent/Indexed_Family | [
"Set Intersection",
"Indexed Families",
"Examples of Idempotence"
] | [
"Definition:Indexing Set/Family of Sets",
"Definition:Set",
"Definition:Set",
"Definition:Set Intersection/Family of Sets"
] | [
"Definition:Non-Empty Set",
"Definition:Element",
"Definition:Set Intersection/Family of Sets",
"Definition:Set Intersection/Family of Sets",
"Definition:Set Equality",
"Category:Set Intersection",
"Category:Indexed Families",
"Category:Examples of Idempotence"
] |
proofwiki-7443 | Factor Principles/Disjunction on Left/Formulation 1/Proof 1 | :$p \implies q \vdash \paren {r \lor p} \implies \paren {r \lor q}$ | {{BeginTableau|p \implies q \vdash \paren {r \lor p} \implies \paren {r \lor q} }}
{{Premise|1|p \implies q}}
{{TheoremIntro|2|r \implies r|Law of Identity: Formulation 2}}
{{SequentIntro|3|1|\paren {r \lor p} \implies \paren {r \lor q}|2, 1|Constructive Dilemma}}
{{EndTableau}}
{{qed}} | :$p \implies q \vdash \paren {r \lor p} \implies \paren {r \lor q}$ | {{BeginTableau|p \implies q \vdash \paren {r \lor p} \implies \paren {r \lor q} }}
{{Premise|1|p \implies q}}
{{TheoremIntro|2|r \implies r|[[Law of Identity/Formulation 2|Law of Identity: Formulation 2]]}}
{{SequentIntro|3|1|\paren {r \lor p} \implies \paren {r \lor q}|2, 1|[[Constructive Dilemma/Formulation 1|Constru... | Factor Principles/Disjunction on Left/Formulation 1/Proof 1 | https://proofwiki.org/wiki/Factor_Principles/Disjunction_on_Left/Formulation_1/Proof_1 | https://proofwiki.org/wiki/Factor_Principles/Disjunction_on_Left/Formulation_1/Proof_1 | [
"Factor Principles"
] | [] | [
"Law of Identity/Formulation 2",
"Constructive Dilemma/Formulation 1"
] |
proofwiki-7444 | Constructive Dilemma for Join Semilattices | Let $\struct {S, \vee, \preceq}$ be a join semilattice.
Let $a, b, c, d \in S$.
Let $a \preceq b$.
Let $c \preceq d$.
Then $\paren {a \vee c} \preceq \paren {b \vee d}$. | 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 | Let $\struct {S, \vee, \preceq}$ be a [[Definition:Join Semilattice|join semilattice]].
Let $a, b, c, d \in S$.
Let $a \preceq b$.
Let $c \preceq d$.
Then $\paren {a \vee c} \preceq \paren {b \vee d}$. | By [[Join Semilattice is Ordered Structure]], $\preceq$ is [[Definition:Relation Compatible with Operation|compatible]] with $\vee$.
By the definition of [[Definition:ordering|ordering]], $\preceq$ is [[Definition:Transitive Relation|transitive]].
Thus the theorem holds by [[Operating on Transitive Relationships Comp... | Constructive Dilemma for Join Semilattices | https://proofwiki.org/wiki/Constructive_Dilemma_for_Join_Semilattices | https://proofwiki.org/wiki/Constructive_Dilemma_for_Join_Semilattices | [
"Lattice Theory"
] | [
"Definition:Join Semilattice"
] | [
"Join Semilattice is Ordered Structure",
"Definition:Relation Compatible with Operation",
"Definition:ordering",
"Definition:Transitive Relation",
"Operating on Transitive Relationships Compatible with Operation",
"Category:Lattice Theory"
] |
proofwiki-7445 | Praeclarum Theorema for Meet Semilattices | Let $(S, \wedge, \preceq)$ be a meet semilattice.
Let $a, b, c, d \in S$.
Let $a \preceq b$.
Let $c \preceq d$.
Then $(a \wedge c) \preceq (b \wedge d)$. | 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 $(S, \wedge, \preceq)$ be a [[Definition:Meet Semilattice|meet semilattice]].
Let $a, b, c, d \in S$.
Let $a \preceq b$.
Let $c \preceq d$.
Then $(a \wedge c) \preceq (b \wedge d)$. | By [[Meet Semilattice is Ordered Structure]], $\preceq$ is [[Definition:Relation Compatible with Operation|compatible]] with $\wedge$.
By the definition of [[Definition:ordering|ordering]], $\preceq$ is [[Definition:Transitive Relation|transitive]].
Thus the theorem holds by [[Operating on Transitive Relationships Co... | Praeclarum Theorema for Meet Semilattices | https://proofwiki.org/wiki/Praeclarum_Theorema_for_Meet_Semilattices | https://proofwiki.org/wiki/Praeclarum_Theorema_for_Meet_Semilattices | [
"Lattice Theory"
] | [
"Definition:Meet Semilattice"
] | [
"Meet Semilattice is Ordered Structure",
"Definition:Relation Compatible with Operation",
"Definition:ordering",
"Definition:Transitive Relation",
"Operating on Transitive Relationships Compatible with Operation"
] |
proofwiki-7446 | Supremum is Increasing relative to Product Ordering | Let $\struct {S, \preceq}$ be an ordered set.
Let $I$ be a set.
Let $f, g: I \to S$.
Let $f \sqbrk I$ denote the image of $I$ under $f$.
Let:
:$\forall i \in I: \map f i \preceq \map g i$
That is, let $f \preceq g$ in the product ordering.
Let $f \sqbrk I$ and $g \sqbrk I$ admit suprema.
Then:
:$\sup f \sqbrk I \preceq... | Let $x \in f \sqbrk I$.
Then:
:$\exists j \in I: \map f j = x$
Then:
:$\map f j \prec \map g j$
By the definition of supremum:
:$\sup g \sqbrk I$ is an upper bound of $g \sqbrk I$
Thus:
:$\map g j \preceq \sup g \sqbrk I$
Since $\preceq$ is transitive:
:$x = \map f j \preceq \sup g \sqbrk I$
Since this holds for all $x... | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $I$ be a [[Definition:Set|set]].
Let $f, g: I \to S$.
Let $f \sqbrk I$ denote the [[Definition:Image of Subset under Mapping|image of $I$ under $f$]].
Let:
:$\forall i \in I: \map f i \preceq \map g i$
That is, let $f \preceq g$ in the p... | Let $x \in f \sqbrk I$.
Then:
:$\exists j \in I: \map f j = x$
Then:
:$\map f j \prec \map g j$
By the definition of [[Definition:Supremum of Set|supremum]]:
:$\sup g \sqbrk I$ is an [[Definition:Upper Bound of Set|upper bound]] of $g \sqbrk I$
Thus:
:$\map g j \preceq \sup g \sqbrk I$
Since $\preceq$ is [[Definit... | Supremum is Increasing relative to Product Ordering | https://proofwiki.org/wiki/Supremum_is_Increasing_relative_to_Product_Ordering | https://proofwiki.org/wiki/Supremum_is_Increasing_relative_to_Product_Ordering | [
"Order Theory",
"Increasing Mappings"
] | [
"Definition:Ordered Set",
"Definition:Set",
"Definition:Image (Set Theory)/Mapping/Subset",
"Definition:Supremum of Set"
] | [
"Definition:Supremum of Set",
"Definition:Upper Bound of Set",
"Definition:Transitive Relation",
"Definition:Upper Bound of Set",
"Definition:Supremum of Set",
"Category:Order Theory",
"Category:Increasing Mappings"
] |
proofwiki-7447 | Reflexive Closure of Strict Ordering is Ordering | Let $S$ be a set.
Let $\prec$ be a strict ordering on $S$.
Let $\preceq$ be the reflexive closure of $\prec$.
Then $\preceq$ is an ordering. | 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}} | Let $S$ be a [[Definition:set|set]].
Let $\prec$ be a [[Definition:Strict Ordering|strict ordering]] on $S$.
Let $\preceq$ be the [[Definition:Reflexive Closure|reflexive closure]] of $\prec$.
Then $\preceq$ is an [[Definition:ordering|ordering]]. | Since $\prec$ is a [[Definition:Strict Ordering|strict ordering]], it is by definition [[Definition:Transitive Relation|transitive]] and [[Definition:Asymmetric Relation|asymmetric]].
By [[Asymmetric Relation is Antisymmetric]], $\prec$ is [[Definition:Antisymmetric Relation|antisymmetric]].
Thus by [[Reflexive Closu... | Reflexive Closure of Strict Ordering is Ordering | https://proofwiki.org/wiki/Reflexive_Closure_of_Strict_Ordering_is_Ordering | https://proofwiki.org/wiki/Reflexive_Closure_of_Strict_Ordering_is_Ordering | [
"Strict Orderings",
"Reflexive Closures"
] | [
"Definition:set",
"Definition:Strict Ordering",
"Definition:Reflexive Closure",
"Definition:ordering"
] | [
"Definition:Strict Ordering",
"Definition:Transitive Relation",
"Definition:Asymmetric Relation",
"Asymmetric Relation is Antisymmetric",
"Definition:Antisymmetric Relation",
"Reflexive Closure of Transitive Antisymmetric Relation is Ordering",
"Definition:Ordering"
] |
proofwiki-7448 | Reflexive Closure is Reflexive | Let $\RR$ be a relation on a set $S$.
Then $\RR^=$, the reflexive closure of $\RR$, is reflexive. | Recall the definition of reflexive closure:
:$\RR^= := \RR \cup \Delta_S$
From Set is Subset of Union:
:$\Delta_S \subseteq \RR^=$
The result follows directly from Relation Contains Diagonal Relation iff Reflexive.
{{qed}} | Let $\RR$ be a [[Definition:Binary Relation|relation]] on a [[Definition:Set|set]] $S$.
Then $\RR^=$, the [[Definition:Reflexive Closure|reflexive closure]] of $\RR$, is [[Definition:Reflexive Relation|reflexive]]. | Recall the definition of [[Definition:Reflexive Closure/Union with Diagonal|reflexive closure]]:
:$\RR^= := \RR \cup \Delta_S$
From [[Set is Subset of Union]]:
:$\Delta_S \subseteq \RR^=$
The result follows directly from [[Relation Contains Diagonal Relation iff Reflexive]].
{{qed}} | Reflexive Closure is Reflexive | https://proofwiki.org/wiki/Reflexive_Closure_is_Reflexive | https://proofwiki.org/wiki/Reflexive_Closure_is_Reflexive | [
"Reflexive Closures"
] | [
"Definition:Relation",
"Definition:Set",
"Definition:Reflexive Closure",
"Definition:Reflexive Relation"
] | [
"Definition:Reflexive Closure/Union with Diagonal",
"Set is Subset of Union",
"Equivalence of Definitions of Reflexive Relation"
] |
proofwiki-7449 | Equivalence of Definitions of Reflexive Closure | {{TFAE|def = Reflexive Closure}}
Let $\RR$ be a relation on a set $S$. | Let $\RR$ be a relation on a set $S$. | {{TFAE|def = Reflexive Closure}}
Let $\RR$ be a [[Definition:Binary Relation|relation]] on a [[Definition:Set|set]] $S$. | Let $\RR$ be a [[Definition:Endorelation|relation]] on a [[Definition:set|set]] $S$. | Equivalence of Definitions of Reflexive Closure | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Reflexive_Closure | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Reflexive_Closure | [
"Reflexive Closures"
] | [
"Definition:Relation",
"Definition:Set"
] | [
"Definition:Endorelation",
"Definition:set",
"Definition:Endorelation",
"Definition:set"
] |
proofwiki-7450 | Reflexive Closure is Inflationary | Let $S$ be a set.
Let $R$ denote the set of all endorelations on $S$.
Then the reflexive closure operator is an inflationary mapping on $R$. | Let $\RR \in R$.
The reflexive closure $\RR^=$ of $\RR$ is defined as:
:$\RR^= := \RR \cup \Delta_S$
From Set is Subset of Union:
:$\RR \subseteq \RR^=$
Hence the reflexive closure operator is an inflationary mapping. | Let $S$ be a [[Definition:Set|set]].
Let $R$ denote the set of all [[Definition:Endorelation|endorelations]] on $S$.
Then the [[Definition:Reflexive Closure|reflexive closure]] operator is an [[Definition:Inflationary Mapping|inflationary mapping]] on $R$. | Let $\RR \in R$.
The [[Definition:Reflexive Closure/Union with Diagonal|reflexive closure]] $\RR^=$ of $\RR$ is defined as:
:$\RR^= := \RR \cup \Delta_S$
From [[Set is Subset of Union]]:
:$\RR \subseteq \RR^=$
Hence the [[Definition:Reflexive Closure|reflexive closure]] operator is an [[Definition:Inflationary Map... | Reflexive Closure is Inflationary | https://proofwiki.org/wiki/Reflexive_Closure_is_Inflationary | https://proofwiki.org/wiki/Reflexive_Closure_is_Inflationary | [
"Reflexive Closures"
] | [
"Definition:Set",
"Definition:Endorelation",
"Definition:Reflexive Closure",
"Definition:Inflationary Mapping"
] | [
"Definition:Reflexive Closure/Union with Diagonal",
"Set is Subset of Union",
"Definition:Reflexive Closure",
"Definition:Inflationary Mapping"
] |
proofwiki-7451 | Reflexive Closure is Order Preserving | Let $S$ be a set.
Let $R$ denote the set of all endorelations on $S$.
Then the reflexive closure operator is an order preserving mapping on $R$.
That is:
:$\forall \RR, \SS \in R: \RR \subseteq \SS \implies \RR^= \subseteq \SS^=$
where $\RR^=$ and $\SS^=$ denote the reflexive closure of $\RR$ and $\SS$ respectively. | 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|Set Union Preserves Subsets}}:
:$\RR^= \subseteq \SS^=$ | Let $S$ be a [[Definition:Set|set]].
Let $R$ denote the set of all [[Definition:Endorelation|endorelations]] on $S$.
Then the [[Definition:Reflexive Closure|reflexive closure]] operator is an [[Definition:Increasing Mapping|order preserving mapping]] on $R$.
That is:
:$\forall \RR, \SS \in R: \RR \subseteq \SS \imp... | Let $\RR, \SS \in R$.
Suppose:
:$\RR \subseteq \SS$
Their respective [[Definition:Reflexive Closure/Union with Diagonal|reflexive closures]] $\RR^=$ and $\SS^=$ are defined as:
:$\RR^= := \RR \cup \Delta_S$
:$\SS^= := \SS \cup \Delta_S$
Hence by {{Corollary|Set Union Preserves Subsets}}:
:$\RR^= \subseteq \SS^=$ | Reflexive Closure is Order Preserving | https://proofwiki.org/wiki/Reflexive_Closure_is_Order_Preserving | https://proofwiki.org/wiki/Reflexive_Closure_is_Order_Preserving | [
"Reflexive Closures"
] | [
"Definition:Set",
"Definition:Endorelation",
"Definition:Reflexive Closure",
"Definition:Increasing/Mapping"
] | [
"Definition:Reflexive Closure/Union with Diagonal"
] |
proofwiki-7452 | Reflexive Closure is Idempotent | Let $S$ be a set.
Let $R$ denote the set of all endorelations on $S$.
Then the reflexive closure operator is an idempotent mapping on $R$.
That is:
:$\forall \RR \in R: \RR^= = \paren {\RR^=}^=$
where $\RR^=$ denotes the reflexive closure of $\RR$. | 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 Set Union is Idempotent:
:$\paren {\RR^=}^= = \RR \cup \Delta_S$
Hence:
:$\foral... | Let $S$ be a [[Definition:Set|set]].
Let $R$ denote the set of all [[Definition:Endorelation|endorelations]] on $S$.
Then the [[Definition:Reflexive Closure|reflexive closure]] operator is an [[Definition:Idempotent Mapping|idempotent mapping]] on $R$.
That is:
:$\forall \RR \in R: \RR^= = \paren {\RR^=}^=$
where ... | Let $\RR \in R$.
By the definition of [[Definition:Reflexive Closure/Union with Diagonal|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 [[Set Union is Id... | Reflexive Closure is Idempotent | https://proofwiki.org/wiki/Reflexive_Closure_is_Idempotent | https://proofwiki.org/wiki/Reflexive_Closure_is_Idempotent | [
"Reflexive Closures"
] | [
"Definition:Set",
"Definition:Endorelation",
"Definition:Reflexive Closure",
"Definition:Idempotence/Mapping",
"Definition:Reflexive Closure"
] | [
"Definition:Reflexive Closure/Union with Diagonal",
"Union is Associative",
"Set Union is Idempotent"
] |
proofwiki-7453 | Transitive Closure of Relation Always Exists | Let $\RR$ be a relation on a set $S$.
Then the transitive closure $\RR^+$ of $\RR$ always exists. | === Outline ===
{{:Transitive Closure of Relation Always Exists/Outline}}{{qed|lemma}}
By definition, the trivial relation $\TT = S \times S$ on $S$ contains $\RR$ as a subset.
Further, $\TT$ is transitive by Trivial Relation is Equivalence.
Thus there exists at least one transitive relation on $S$ that contains $\RR$.... | Let $\RR$ be a [[Definition:Relation|relation]] on a [[Definition:Set|set]] $S$.
Then the [[Definition:Transitive Closure of Relation|transitive closure]] $\RR^+$ of $\RR$ always exists. | === [[Transitive Closure of Relation Always Exists/Outline|Outline]] ===
{{:Transitive Closure of Relation Always Exists/Outline}}{{qed|lemma}}
By definition, the [[Definition:Trivial Relation|trivial relation]] $\TT = S \times S$ on $S$ contains $\RR$ as a [[Definition:Subset|subset]].
Further, $\TT$ is [[Definitio... | Transitive Closure of Relation Always Exists/Proof | https://proofwiki.org/wiki/Transitive_Closure_of_Relation_Always_Exists | https://proofwiki.org/wiki/Transitive_Closure_of_Relation_Always_Exists/Proof | [
"Transitive Closure of Relation Always Exists",
"Transitive Closures"
] | [
"Definition:Relation",
"Definition:Set",
"Definition:Transitive Closure of Relation"
] | [
"Transitive Closure of Relation Always Exists/Outline",
"Definition:Trivial Relation",
"Definition:Subset",
"Definition:Transitive Relation",
"Trivial Relation is Equivalence",
"Definition:Transitive Relation",
"Definition:Set Intersection",
"Definition:Transitive Relation",
"Intersection of Transit... |
proofwiki-7454 | Equivalence of Definitions of Transitive Closure of Relation/Union of Compositions is Smallest | Let $\RR$ be a relation on a set $S$.
Let:
:$\RR^n := \begin {cases} \RR & : n = 0 \\ \RR^{n - 1} \circ \RR & : n > 0 \end {cases}$
where $\circ$ denotes composition of relations.
{{explain|Really? I would have thought $\RR^1 {{=}} \RR$, not $\RR^0 {{=}} \RR$. If anything, the diagonal relation $\Delta_S$ should be $\R... | ==== $\RR^+$ 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:
:$\e... | Let $\RR$ be a [[Definition:Endorelation|relation]] on a [[Definition:Set|set]] $S$.
Let:
:$\RR^n := \begin {cases} \RR & : n = 0 \\ \RR^{n - 1} \circ \RR & : n > 0 \end {cases}$
where $\circ$ denotes [[Definition:Composition of Relations|composition of relations]].
{{explain|Really? I would have thought $\RR^1 {{=}... | ==== $\RR^+$ is Transitive ====
By [[Relation contains Composite with Self iff Transitive]], we can prove that $\RR^+$ is [[Definition:Transitive Relation|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... | Equivalence of Definitions of Transitive Closure of Relation/Union of Compositions is Smallest | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Transitive_Closure_of_Relation/Union_of_Compositions_is_Smallest | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Transitive_Closure_of_Relation/Union_of_Compositions_is_Smallest | [
"Equivalence of Definitions of Transitive Closure of Relation"
] | [
"Definition:Endorelation",
"Definition:Set",
"Definition:Composition of Relations",
"Definition:Diagonal Relation",
"Definition:Smallest Set by Set Inclusion",
"Definition:Transitive Relation",
"Definition:Subset"
] | [
"Equivalence of Definitions of Transitive Relation",
"Definition:Transitive Relation",
"Composition of Relations is Associative",
"Definition:Transitive Relation",
"Set is Subset of Union/Family of Sets",
"Definition:Transitive Relation",
"Definition:Composition of Relations",
"Definition:Transitive R... |
proofwiki-7455 | Equivalence of Definitions of Transitive Closure of Relation/Finite Chain Equivalent to Union of Compositions | The finite chain and union of compositions definitions of '''transitive closure''' are equivalent. | {{explain|more detail required}}
Follows from the definition of composition of relations.
{{qed}} | The [[Definition:Transitive Closure of Relation/Finite Chain|finite chain]] and [[Definition:Transitive Closure of Relation/Union of Compositions|union of compositions]] definitions of '''[[Definition:Transitive Closure of Relation|transitive closure]]''' are equivalent. | {{explain|more detail required}}
Follows from the definition of [[Definition:Composition of Relations|composition of relations]].
{{qed}} | Equivalence of Definitions of Transitive Closure of Relation/Finite Chain Equivalent to Union of Compositions | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Transitive_Closure_of_Relation/Finite_Chain_Equivalent_to_Union_of_Compositions | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Transitive_Closure_of_Relation/Finite_Chain_Equivalent_to_Union_of_Compositions | [
"Equivalence of Definitions of Transitive Closure of Relation"
] | [
"Definition:Transitive Closure of Relation/Finite Chain",
"Definition:Transitive Closure of Relation/Union of Compositions",
"Definition:Transitive Closure of Relation"
] | [
"Definition:Composition of Relations"
] |
proofwiki-7456 | Chasles' Relation | Let $\EE$ be an affine space.
Let $p, q, r \in \EE$ be points.
Then:
:$\vec {p q} = \vec {p r} + \vec {r q}$ | We have:
{{begin-eqn}}
{{eqn | l = \vec {p r} + \vec {r q}
| r = \paren {r - p} + \paren {q - r}
| c = {{Defof|Vector (Affine Geometry)|Vector in Affine Space}}
}}
{{eqn | r = \paren {r + \paren {q - r} } - p
| c = {{Defof|Affine Space}}: axiom $(\text A 3)$
}}
{{eqn | r = q - p
| c = {{Defof|Af... | Let $\EE$ be an [[Definition:Affine Space|affine space]].
Let $p, q, r \in \EE$ be [[Definition:Point (Affine Geometry)|points]].
Then:
:$\vec {p q} = \vec {p r} + \vec {r q}$ | We have:
{{begin-eqn}}
{{eqn | l = \vec {p r} + \vec {r q}
| r = \paren {r - p} + \paren {q - r}
| c = {{Defof|Vector (Affine Geometry)|Vector in Affine Space}}
}}
{{eqn | r = \paren {r + \paren {q - r} } - p
| c = {{Defof|Affine Space}}: axiom $(\text A 3)$
}}
{{eqn | r = q - p
| c = {{Defof|Af... | Chasles' Relation | https://proofwiki.org/wiki/Chasles'_Relation | https://proofwiki.org/wiki/Chasles'_Relation | [
"Affine Geometry"
] | [
"Definition:Affine Space",
"Definition:Point (Affine Geometry)"
] | [
"Category:Affine Geometry"
] |
proofwiki-7457 | Affine Coordinates are Well-Defined | Let $\EE$ be an affine space with difference space $V$ over a field $k$.
Let $\RR = \left({p_0, e_1, \ldots, e_n}\right)$ be an affine frame in $\EE$.
Define a mapping $\Theta_\RR : k^n \to \EE$ by:
:$\ds \map {\Theta_\RR} {\lambda_1, \ldots, \lambda_n} = p_0 + \sum_{i \mathop = 1}^n \lambda_i e_i$
Then $\Theta_\RR$ is... | === Proof of Surjection ===
Let $p \in \EE$.
Let $v = p - p_0 \in V$.
Let $\tuple {\lambda_1, \ldots, \lambda_n}$ be coordinates of $v$ in the basis $\tuple {e_1, \ldots, e_n}$.
Then:
{{begin-eqn}}
{{eqn | l = p_0 + \sum_{i \mathop = 1}^n \lambda_ie_i
| r = p_0 + v
}}
{{eqn | r = p_0 + \paren {p - p_0}
}}
{{eqn |... | Let $\EE$ be an [[Definition:Affine Space|affine space]] with [[Definition:Difference Space|difference space]] $V$ over a [[Definition:Field (Abstract Algebra)|field]] $k$.
Let $\RR = \left({p_0, e_1, \ldots, e_n}\right)$ be an [[Definition:Affine Frame|affine frame]] in $\EE$.
Define a mapping $\Theta_\RR : k^n \to ... | === Proof of Surjection ===
Let $p \in \EE$.
Let $v = p - p_0 \in V$.
Let $\tuple {\lambda_1, \ldots, \lambda_n}$ be [[Definition:Coordinate|coordinates]] of $v$ in the [[Definition:Ordered Basis|basis]] $\tuple {e_1, \ldots, e_n}$.
Then:
{{begin-eqn}}
{{eqn | l = p_0 + \sum_{i \mathop = 1}^n \lambda_ie_i
| r... | Affine Coordinates are Well-Defined | https://proofwiki.org/wiki/Affine_Coordinates_are_Well-Defined | https://proofwiki.org/wiki/Affine_Coordinates_are_Well-Defined | [
"Affine Geometry"
] | [
"Definition:Affine Space",
"Definition:Tangent Space (Affine Geometry)",
"Definition:Field (Abstract Algebra)",
"Definition:Affine Frame",
"Definition:Bijection"
] | [
"Definition:Coordinate",
"Definition:Ordered Basis",
"Definition:Surjection"
] |
proofwiki-7458 | Barycenter Exists and is Well Defined | Let $\EE$ be an affine space over a field $k$.
Let $p_1, \ldots, p_n \in \EE$ be points.
Let $\lambda_1, \ldots, \lambda_n \in k$ such that $\ds \sum_{i \mathop = 1}^n \lambda_i = 1$.
Then the barycentre of $p_1, \ldots, p_n$ with weights $\lambda_1, \ldots, \lambda_n$ exists and is unique. | Let $r$ be any point in $\EE$.
Set:
:$\ds q = r + \sum_{i \mathop = 1}^n \lambda_i \vec{r p_i}$
We are required to prove that for any other point $m \in \EE$:
:$\ds q = m + \sum_{i \mathop = 1}^n \lambda_i \vec{m p_i}$
So:
{{begin-eqn}}
{{eqn | l = m + \sum_{i \mathop = 1}^n \lambda_i \vec{m p_i}
| r = m + \sum_{... | Let $\EE$ be an [[Definition:Affine Space|affine space]] over a [[Definition:Field (Abstract Algebra)|field]] $k$.
Let $p_1, \ldots, p_n \in \EE$ be points.
Let $\lambda_1, \ldots, \lambda_n \in k$ such that $\ds \sum_{i \mathop = 1}^n \lambda_i = 1$.
Then the [[Definition:Barycentre|barycentre]] of $p_1, \ldots, p... | Let $r$ be any point in $\EE$.
Set:
:$\ds q = r + \sum_{i \mathop = 1}^n \lambda_i \vec{r p_i}$
We are required to prove that for any other point $m \in \EE$:
:$\ds q = m + \sum_{i \mathop = 1}^n \lambda_i \vec{m p_i}$
So:
{{begin-eqn}}
{{eqn | l = m + \sum_{i \mathop = 1}^n \lambda_i \vec{m p_i}
| r = m + \s... | Barycenter Exists and is Well Defined | https://proofwiki.org/wiki/Barycenter_Exists_and_is_Well_Defined | https://proofwiki.org/wiki/Barycenter_Exists_and_is_Well_Defined | [
"Affine Geometry"
] | [
"Definition:Affine Space",
"Definition:Field (Abstract Algebra)",
"Definition:Barycenter"
] | [
"Chasles' Relation",
"Definition:Affine Space",
"Category:Affine Geometry"
] |
proofwiki-7459 | Transitive Chaining | Let $\RR$ be a transitive relation on a set $S$.
Let $n \in \N$ be a natural number.
Let $n \ge 2$.
Let $\sequence {x_k}_{k \mathop \in \set {1, 2, \dots, n} }$ be a sequence of $n$ terms.
For each $k \in \set {1, 2, \dots, n - 1}$, let $x_k \mathrel \RR x_{k + 1}$.
That is, let $x_1 \mathrel \RR x_2$, $x_2 \mathrel \R... | The proof proceeds by induction on $n$, the number of terms in the sequence.
We first define a propositional function, $P$, as follows:
For each $n \in \N$ such that $n \ge 2$, let $\map P n$ be the proposition that if both of the following hold:
:$\sequence {x_k}_{k \mathop \in \set {1, 2, \dots, n} }$ is a sequence o... | Let $\RR$ be a [[Definition:Transitive Relation|transitive relation]] on a [[Definition:Set|set]] $S$.
Let $n \in \N$ be a [[Definition:Natural Number|natural number]].
Let $n \ge 2$.
Let $\sequence {x_k}_{k \mathop \in \set {1, 2, \dots, n} }$ be a [[Definition:Sequence of n Terms|sequence of $n$ terms]].
For each... | The proof proceeds by [[Principle of Mathematical Induction|induction]] on $n$, the number of terms in the [[Definition:Finite Sequence|sequence]].
We first define a [[Definition:Propositional Function|propositional function]], $P$, as follows:
For each $n \in \N$ such that $n \ge 2$, let $\map P n$ be the [[Definiti... | Transitive Chaining | https://proofwiki.org/wiki/Transitive_Chaining | https://proofwiki.org/wiki/Transitive_Chaining | [
"Transitive Relations"
] | [
"Definition:Transitive Relation",
"Definition:Set",
"Definition:Natural Numbers",
"Definition:Sequence of n Terms"
] | [
"Principle of Mathematical Induction",
"Definition:Finite Sequence",
"Definition:Propositional Function",
"Definition:Proposition",
"Definition:Sequence of n Terms",
"Definition:Finite Sequence",
"Definition:Sequence of n Terms",
"Principle of Mathematical Induction"
] |
proofwiki-7460 | Szpilrajn Extension Theorem | Let $\struct {S, \prec}$ be a strictly ordered set.
{{Disambiguate|Definition:Strictly Ordered Set}}
Then there is a strict total ordering on $S$ of which $\prec$ is a subset. | {{proof wanted}}
{{Namedfor|Edward Szpilrajn|cat = Marczewski}}
Category:Order Theory
lji5i6b8gp30a7jge1hz1i76m5xmudk | Let $\struct {S, \prec}$ be a [[Definition:Strictly Ordered Set|strictly ordered set]].
{{Disambiguate|Definition:Strictly Ordered Set}}
Then there is a [[Definition:Strict Total Ordering|strict total ordering]] on $S$ of which $\prec$ is a [[Definition:Subset|subset]]. | {{proof wanted}}
{{Namedfor|Edward Szpilrajn|cat = Marczewski}}
[[Category:Order Theory]]
lji5i6b8gp30a7jge1hz1i76m5xmudk | Szpilrajn Extension Theorem | https://proofwiki.org/wiki/Szpilrajn_Extension_Theorem | https://proofwiki.org/wiki/Szpilrajn_Extension_Theorem | [
"Order Theory"
] | [
"Definition:Strictly Ordered Set",
"Definition:Strict Total Ordering",
"Definition:Subset"
] | [
"Category:Order Theory"
] |
proofwiki-7461 | Strict Ordering can be Expanded to Compare Additional Pair | Let $\struct {S, \prec}$ be an ordered set.
Let $a$ and $b$ be distinct, $\prec$-incomparable elements of $S$.
That is, let:
:$a \nprec b$ and $b \nprec a$.
Let $\prec' = {\prec} \cup \set {\tuple {a, b} }$.
Define a relation $\prec'^+$ by letting $p \prec'^+ q$ {{iff}}:
:$p \prec q$
or:
:$p \preceq a$ and $b \preceq q... | First, note that since $\prec$ is a strict ordering, $\preceq$ is an ordering by Reflexive Closure of Strict Ordering is Ordering.
=== $a$ and $b$ are $\preceq$-incomparable ===
Suppose that $a \preceq b$.
By the definition of reflexive closure, either $a \prec b$ or $a = b$.
Each possibility contradicts one of the pre... | Let $\struct {S, \prec}$ be an [[Definition:Ordered Set|ordered set]].
Let $a$ and $b$ be distinct, [[Definition:Non-Comparable Elements|$\prec$-incomparable]] elements of $S$.
That is, let:
:$a \nprec b$ and $b \nprec a$.
Let $\prec' = {\prec} \cup \set {\tuple {a, b} }$.
Define a [[Definition:Endorelation|relatio... | First, note that since $\prec$ is a [[Definition:Strict Ordering|strict ordering]], $\preceq$ is an [[Definition:Ordering|ordering]] by [[Reflexive Closure of Strict Ordering is Ordering]].
=== $a$ and $b$ are $\preceq$-incomparable ===
Suppose that $a \preceq b$.
By the definition of [[Definition:Reflexive Closure|... | Strict Ordering can be Expanded to Compare Additional Pair/Proof 1 | https://proofwiki.org/wiki/Strict_Ordering_can_be_Expanded_to_Compare_Additional_Pair | https://proofwiki.org/wiki/Strict_Ordering_can_be_Expanded_to_Compare_Additional_Pair/Proof_1 | [
"Strict Orderings",
"Strict Ordering can be Expanded to Compare Additional Pair"
] | [
"Definition:Ordered Set",
"Definition:Non-Comparable Elements",
"Definition:Endorelation",
"Definition:Reflexive Closure",
"Definition:Strict Ordering",
"Definition:Transitive Closure of Relation"
] | [
"Definition:Strict Ordering",
"Definition:Ordering",
"Reflexive Closure of Strict Ordering is Ordering",
"Definition:Reflexive Closure",
"Definition:Strict Ordering",
"Definition:Transitive Relation",
"Definition:Non-Comparable Elements",
"Definition:Antireflexive Relation",
"Definition:Transitive R... |
proofwiki-7462 | Characterization of Interior of Triangle | Let $\triangle$ be a triangle embedded in $\R^2$.
Denote the vertices of $\triangle$ as $A_1, A_2, A_3$.
For $i \in \set {1, 2, 3}$, put $j = i \bmod 3 + 1$, $k = \paren {i + 1} \bmod 3 + 1$, and:
:$U_i = \set {A_i + s t \paren {A_j - A_i} + \paren {1 - s} t \paren {A_k - A_i} : s \in \openint 0 1, t \in \R_{>0} }$
The... | From Boundary of Polygon is Jordan Curve, it follows that the boundary of $\triangle$ is equal to the image of a Jordan curve, so $\Int \triangle$ is well-defined. | Let $\triangle$ be a [[Definition:Triangle (Geometry)|triangle]] embedded in $\R^2$.
Denote the [[Definition:Vertex of Polygon|vertices]] of $\triangle$ as $A_1, A_2, A_3$.
For $i \in \set {1, 2, 3}$, put $j = i \bmod 3 + 1$, $k = \paren {i + 1} \bmod 3 + 1$, and:
:$U_i = \set {A_i + s t \paren {A_j - A_i} + \paren... | From [[Boundary of Polygon is Jordan Curve]], it follows that the [[Definition:Boundary (Geometry)|boundary]] of $\triangle$ is equal to the [[Definition:Image of Mapping|image]] of a [[Definition:Jordan Curve|Jordan curve]], so $\Int \triangle$ is well-defined. | Characterization of Interior of Triangle | https://proofwiki.org/wiki/Characterization_of_Interior_of_Triangle | https://proofwiki.org/wiki/Characterization_of_Interior_of_Triangle | [
"Topology"
] | [
"Definition:Triangle (Geometry)",
"Definition:Polygon/Vertex",
"Definition:Jordan Curve/Interior",
"Definition:Boundary (Geometry)"
] | [
"Boundary of Polygon is Jordan Curve",
"Definition:Boundary (Geometry)",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Jordan Curve",
"Definition:Boundary (Geometry)"
] |
proofwiki-7463 | Equivalence of Definitions of Reflexive Transitive Closure | Let $\RR$ be a relation on a set $S$.
{{TFAE|def = Reflexive Transitive Closure}} | The result follows from:
:Transitive Closure of Reflexive Relation is Reflexive
:Reflexive Closure of Transitive Relation is Transitive
:Composition of Compatible Closure Operators
{{qed}}
Category:Reflexive Closures
Category:Transitive Closures
3zuw09718neju3nn5iksghyv70vs1p2 | Let $\RR$ be a [[Definition:Endorelation|relation]] on a [[Definition:set|set]] $S$.
{{TFAE|def = Reflexive Transitive Closure}} | The result follows from:
:[[Transitive Closure of Reflexive Relation is Reflexive]]
:[[Reflexive Closure of Transitive Relation is Transitive]]
:[[Composition of Compatible Closure Operators]]
{{qed}}
[[Category:Reflexive Closures]]
[[Category:Transitive Closures]]
3zuw09718neju3nn5iksghyv70vs1p2 | Equivalence of Definitions of Reflexive Transitive Closure | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Reflexive_Transitive_Closure | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Reflexive_Transitive_Closure | [
"Reflexive Closures",
"Transitive Closures"
] | [
"Definition:Endorelation",
"Definition:set"
] | [
"Transitive Closure of Reflexive Relation is Reflexive",
"Reflexive Closure of Transitive Relation is Transitive",
"Composition of Compatible Closure Operators",
"Category:Reflexive Closures",
"Category:Transitive Closures"
] |
proofwiki-7464 | Intersection of Relation with Inverse is Symmetric Relation | Let $\RR$ be a relation on a set $S$.
Then $\RR \cap \RR^{-1}$, the intersection of $\RR$ with its inverse, is symmetric. | Let $\tuple {x, y} \in \RR \cap \RR^{-1}$
By definition of intersection:
:$\tuple {x, y} \in \RR$
:$\tuple {x, y} \in \RR^{-1}$
By definition of inverse relation:
:$\tuple {x, y} \in \RR \implies \tuple {y, x} \in \RR^{-1}$
:$\tuple {x, y} \in \RR^{-1} \implies \tuple {y, x} \in \paren {\RR^{-1} }^{-1}$
By Inverse of I... | Let $\RR$ be a [[Definition:Binary Relation|relation]] on a [[Definition:Set|set]] $S$.
Then $\RR \cap \RR^{-1}$, the [[Definition:Set Intersection|intersection]] of $\RR$ with its [[Definition:Inverse Relation|inverse]], is [[Definition:Symmetric Relation|symmetric]]. | Let $\tuple {x, y} \in \RR \cap \RR^{-1}$
By definition of [[Definition:Set Intersection|intersection]]:
:$\tuple {x, y} \in \RR$
:$\tuple {x, y} \in \RR^{-1}$
By definition of [[Definition:Inverse Relation|inverse relation]]:
:$\tuple {x, y} \in \RR \implies \tuple {y, x} \in \RR^{-1}$
:$\tuple {x, y} \in \RR^... | Intersection of Relation with Inverse is Symmetric Relation | https://proofwiki.org/wiki/Intersection_of_Relation_with_Inverse_is_Symmetric_Relation | https://proofwiki.org/wiki/Intersection_of_Relation_with_Inverse_is_Symmetric_Relation | [
"Set Intersection",
"Inverse Relations",
"Symmetric Relations"
] | [
"Definition:Relation",
"Definition:Set",
"Definition:Set Intersection",
"Definition:Inverse Relation",
"Definition:Symmetric Relation"
] | [
"Definition:Set Intersection",
"Definition:Inverse Relation",
"Inverse of Inverse Relation",
"Definition:Set Intersection",
"Definition:Symmetric Relation"
] |
proofwiki-7465 | Intersection of Closed Sets is Closed/Closure Operator | Let $S$ be a set.
Let $f: \powerset S \to \powerset S$ be a closure operator on $S$.
Let $\CC$ be the set of all subsets of $S$ that are closed with respect to $f$.
Let $\AA \subseteq \CC$.
Then $\bigcap \AA \in \CC$. | Let $Q = \bigcap \AA$.
By the definition of closure operator, $f$ is inflationary, order-preserving, and idempotent.
Let $A \in \AA$.
By Intersection is Largest Subset, $Q \subseteq A$.
Since $f$ is order-preserving, $\map f Q \subseteq \map f A$.
By the definition of closed set, $\map f A = A$
Thus $\map f Q \subseteq... | Let $S$ be a [[Definition:Set|set]].
Let $f: \powerset S \to \powerset S$ be a [[Definition:Closure Operator|closure operator]] on $S$.
Let $\CC$ be the [[Definition:Set|set]] of all [[Definition:Subset|subsets]] of $S$ that are [[Definition:Closed Set under Closure Operator|closed]] with respect to $f$.
Let $\AA \s... | Let $Q = \bigcap \AA$.
By the definition of [[Definition:Closure Operator|closure operator]], $f$ is [[Definition:Inflationary Mapping|inflationary]], [[Definition:Order-Preserving Mapping|order-preserving]], and [[Definition:Idempotent Mapping|idempotent]].
Let $A \in \AA$.
By [[Intersection is Largest Subset/Set o... | Intersection of Closed Sets is Closed/Closure Operator | https://proofwiki.org/wiki/Intersection_of_Closed_Sets_is_Closed/Closure_Operator | https://proofwiki.org/wiki/Intersection_of_Closed_Sets_is_Closed/Closure_Operator | [
"Closure Operators",
"Set Intersection"
] | [
"Definition:Set",
"Definition:Closure Operator",
"Definition:Set",
"Definition:Subset",
"Definition:Closed Set/Closure Operator"
] | [
"Definition:Closure Operator",
"Definition:Inflationary Mapping",
"Definition:Increasing/Mapping",
"Definition:Idempotence/Mapping",
"Intersection is Largest Subset/Set of Sets",
"Definition:Increasing/Mapping",
"Definition:Closed Set/Closure Operator",
"Intersection is Largest Subset/Set of Sets",
... |
proofwiki-7466 | Closure is Closed/Power Set | Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Let $\cl: \powerset S \to \powerset S$ be a closure operator.
Let $T \subseteq S$.
Then $\map \cl T$ is a closed set with respect to $\cl$. | By the definition of closure operator, $\cl$ is idempotent.
Therefore $\map \cl {\map \cl T} = \map \cl T$, so $\map \cl T$ is closed.
{{qed}}
Category:Closure Operators
84pspzef6qd7pt11nc0mnk94pg71ies | Let $S$ be a [[Definition:Set|set]].
Let $\powerset S$ be the [[Definition:Power Set|power set]] of $S$.
Let $\cl: \powerset S \to \powerset S$ be a [[Definition:Closure Operator on Set|closure operator]].
Let $T \subseteq S$.
Then $\map \cl T$ is a [[Definition:Closed Set under Closure Operator|closed set]] with r... | By the definition of [[Definition:Closure Operator on Set|closure operator]], $\cl$ is [[Definition:Idempotent Mapping|idempotent]].
Therefore $\map \cl {\map \cl T} = \map \cl T$, so $\map \cl T$ is [[Definition:Closed Set under Closure Operator|closed]].
{{qed}}
[[Category:Closure Operators]]
84pspzef6qd7pt11nc0mnk... | Closure is Closed/Power Set | https://proofwiki.org/wiki/Closure_is_Closed/Power_Set | https://proofwiki.org/wiki/Closure_is_Closed/Power_Set | [
"Closure Operators"
] | [
"Definition:Set",
"Definition:Power Set",
"Definition:Closure Operator/Power Set",
"Definition:Closed Set/Closure Operator"
] | [
"Definition:Closure Operator/Power Set",
"Definition:Idempotence/Mapping",
"Definition:Closed Set/Closure Operator",
"Category:Closure Operators"
] |
proofwiki-7467 | Relation Intersection Inverse is Greatest Symmetric Subset of Relation | Let $\RR$ be a relation on a set $S$.
Let $\powerset \RR$ be the power set of $\RR$.
By definition, $\powerset \RR$ is the set of all relations on $S$ that are subsets of $\RR$.
Then the greatest element of $\powerset \RR$ that is symmetric is:
:$\RR \cap \RR^{-1}$ | By Intersection of Relation with Inverse is Symmetric Relation:
:$\RR \cap \RR^{-1}$ is a symmetric relation.
Suppose for some $\SS \in \powerset \RR$ that $S$ is symmetric and not equal to $\RR \cap \RR^{-1}$.
We will show that it is a proper subset of $\RR \cap \RR^{-1}$.
Suppose $\tuple {x, y} \in \SS$.
Then as $\SS... | Let $\RR$ be a [[Definition:Binary Relation|relation]] on a [[Definition:Set|set]] $S$.
Let $\powerset \RR$ be the [[Definition:Power Set|power set]] of $\RR$.
By definition, $\powerset \RR$ is the [[Definition:Set|set]] of all [[Definition:Binary Relation|relation]]s on $S$ that are [[Definition:Subset|subsets]] of ... | By [[Intersection of Relation with Inverse is Symmetric Relation]]:
:$\RR \cap \RR^{-1}$ is a [[Definition:Symmetric Relation|symmetric relation]].
Suppose for some $\SS \in \powerset \RR$ that $S$ is [[Definition:Symmetric Relation|symmetric]] and not equal to $\RR \cap \RR^{-1}$.
We will show that it is a [[Defini... | Relation Intersection Inverse is Greatest Symmetric Subset of Relation | https://proofwiki.org/wiki/Relation_Intersection_Inverse_is_Greatest_Symmetric_Subset_of_Relation | https://proofwiki.org/wiki/Relation_Intersection_Inverse_is_Greatest_Symmetric_Subset_of_Relation | [
"Relation Theory",
"Symmetric Relations"
] | [
"Definition:Relation",
"Definition:Set",
"Definition:Power Set",
"Definition:Set",
"Definition:Relation",
"Definition:Subset",
"Definition:Greatest Element",
"Definition:Symmetric Relation"
] | [
"Intersection of Relation with Inverse is Symmetric Relation",
"Definition:Symmetric Relation",
"Definition:Symmetric Relation",
"Definition:Proper Subset",
"Definition:Symmetric Relation",
"Definition:Inverse Relation",
"Definition:Set Intersection",
"Definition:Subset",
"Category:Relation Theory",... |
proofwiki-7468 | Composition of Compatible Closure Operators | Let $S$ be a set.
Let $f, g: \powerset S \to \powerset S$ be closure operators on $S$.
Let $\CC_f$ and $\CC_g$ be the sets of closed sets of $S$ with respect to $f$ and $g$ respectively.
For each subset $T$ of $S$, let the following hold:
:$(1): \quad$ If $T$ is closed with respect to $g$, then $\map f T$ is closed wit... | First we show that $\CC_h$ induces a closure operator on $S$.
Let $\AA \subseteq \CC_h$.
By Intersection is Largest Subset:
:$\AA \subseteq \CC_f$
and:
:$\AA \subseteq \CC_g$
Thus by Intersection of Closed Sets is Closed/Closure Operator:
:$\ds \bigcap \AA \in \CC_f$
and
:$\ds \bigcap \AA \in \CC_g$
Thus by the definit... | Let $S$ be a [[Definition:Set|set]].
Let $f, g: \powerset S \to \powerset S$ be [[Definition:Closure Operator|closure operators]] on $S$.
Let $\CC_f$ and $\CC_g$ be the sets of [[Definition:Closed Set under Closure Operator|closed sets]] of $S$ with respect to $f$ and $g$ respectively.
For each [[Definition:Subset|... | First we show that $\CC_h$ [[Closure Operator from Closed Sets|induces a closure operator]] on $S$.
Let $\AA \subseteq \CC_h$.
By [[Intersection is Largest Subset]]:
:$\AA \subseteq \CC_f$
and:
:$\AA \subseteq \CC_g$
Thus by [[Intersection of Closed Sets is Closed/Closure Operator]]:
:$\ds \bigcap \AA \in \CC_f$
and... | Composition of Compatible Closure Operators | https://proofwiki.org/wiki/Composition_of_Compatible_Closure_Operators | https://proofwiki.org/wiki/Composition_of_Compatible_Closure_Operators | [
"Closure Operators"
] | [
"Definition:Set",
"Definition:Closure Operator",
"Definition:Closed Set/Closure Operator",
"Definition:Subset",
"Definition:Closed Set/Closure Operator",
"Closure Operator from Closed Sets",
"Definition:Composition of Mappings"
] | [
"Closure Operator from Closed Sets",
"Intersection is Largest Subset",
"Intersection of Closed Sets is Closed/Closure Operator",
"Definition:Set Intersection",
"Closure Operator from Closed Sets",
"Closure Operator from Closed Sets",
"Definition:Closed Set/Closure Operator",
"Definition:Set Intersecti... |
proofwiki-7469 | Closure is Closed | Let $\struct {S, \preceq}$ be an ordered set.
Let $\cl: S \to S$ be a closure operator.
Let $x \in S$.
Then $\map \cl x$ is a closed element of $S$ with respect to $\cl$. | By the definition of closure operator, $\cl$ is idempotent.
Therefore:
:$\map \cl {\map \cl x} = \map \cl x$
It follows by definition that $\map \cl x$ is a closed element.
{{qed}} | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $\cl: S \to S$ be a [[Definition:Closure Operator (Order Theory)|closure operator]].
Let $x \in S$.
Then $\map \cl x$ is a [[Definition:Closed Element|closed element]] of $S$ with respect to $\cl$. | By the definition of [[Definition:Closure Operator (Order Theory)|closure operator]], $\cl$ is [[Definition:Idempotent Mapping|idempotent]].
Therefore:
:$\map \cl {\map \cl x} = \map \cl x$
It follows by definition that $\map \cl x$ is a [[Definition:Closed Element|closed element]].
{{qed}} | Closure is Closed | https://proofwiki.org/wiki/Closure_is_Closed | https://proofwiki.org/wiki/Closure_is_Closed | [
"Closure Operators"
] | [
"Definition:Ordered Set",
"Definition:Closure Operator/Ordering",
"Definition:Closed Element"
] | [
"Definition:Closure Operator/Ordering",
"Definition:Idempotence/Mapping",
"Definition:Closed Element"
] |
proofwiki-7470 | Closure Operator from Closed Elements | Let $\struct {S, \preceq}$ be an ordered set.
Let $C \subseteq S$.
Suppose that $C$ is a subset of $S$ with the property that every element of $S$ has a smallest successor in $C$.
Let $\cl: S \to S$ be defined as follows:
For $x \in S$:
:$\map \cl x = \map \min {C \cap x^\succeq}$
where $x^\succeq$ is the upper closure... | === Inflationary ===
$x$ is a lower bound of $x^\succeq$.
Hence by Lower Bound for Subset, $x$ is also a lower bound of $C \cap x^\succeq$.
By the definition of smallest element, $x \preceq \map \cl x$.
{{qed|lemma}} | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $C \subseteq S$.
Suppose that $C$ is a [[Definition:Subset|subset]] of $S$ with the property that every [[Definition:Element|element]] of $S$ has a [[Definition:Smallest Element|smallest]] [[Definition:Successor Element|successor]] in $C$.
... | === Inflationary ===
$x$ is a [[Definition:Lower Bound of Set|lower bound]] of $x^\succeq$.
Hence by [[Lower Bound for Subset]], $x$ is also a [[Definition:Lower Bound of Set|lower bound]] of $C \cap x^\succeq$.
By the definition of [[Definition:Smallest Element|smallest element]], $x \preceq \map \cl x$.
{{qed|lemm... | Closure Operator from Closed Elements | https://proofwiki.org/wiki/Closure_Operator_from_Closed_Elements | https://proofwiki.org/wiki/Closure_Operator_from_Closed_Elements | [
"Closure Operators",
"Order Theory"
] | [
"Definition:Ordered Set",
"Definition:Subset",
"Definition:Element",
"Definition:Smallest Element",
"Definition:Succeed",
"Definition:Upper Closure/Set",
"Definition:Smallest Element",
"Definition:Succeed",
"Definition:Closure Operator/Ordering",
"Definition:Closed Element"
] | [
"Definition:Lower Bound of Set",
"Lower Bound for Subset",
"Definition:Lower Bound of Set",
"Definition:Smallest Element",
"Definition:Smallest Element",
"Definition:Smallest Element"
] |
proofwiki-7471 | Product of Affine Spaces is Affine Space | Let $\EE, \FF$ be affine spaces.
Let $\GG = \EE \times \FF$ be the product of $\EE$ and $\FF$.
Then $\GG$ is an affine space. | Let $G = \vec \GG$ be the difference space of $\GG$.
We are required to show that the following axioms are satisfied:
{{begin-axiom}}
{{axiom | n = 1
| q = \forall p, q \in \GG
| m = p + \paren {q - p} = q
}}
{{axiom | n = 2
| q = \forall p \in \GG: \forall u, v \in G
| m = \paren {p + u... | Let $\EE, \FF$ be [[Definition:Affine Space|affine spaces]].
Let $\GG = \EE \times \FF$ be the [[Definition:Product of Affine Spaces|product]] of $\EE$ and $\FF$.
Then $\GG$ is an [[Definition:Affine Space|affine space]]. | Let $G = \vec \GG$ be the [[Definition:Difference Space|difference space]] of $\GG$.
We are required to show that the following axioms are satisfied:
{{begin-axiom}}
{{axiom | n = 1
| q = \forall p, q \in \GG
| m = p + \paren {q - p} = q
}}
{{axiom | n = 2
| q = \forall p \in \GG: \forall u, v ... | Product of Affine Spaces is Affine Space | https://proofwiki.org/wiki/Product_of_Affine_Spaces_is_Affine_Space | https://proofwiki.org/wiki/Product_of_Affine_Spaces_is_Affine_Space | [
"Affine Geometry"
] | [
"Definition:Affine Space",
"Definition:Product of Affine Spaces",
"Definition:Affine Space"
] | [
"Definition:Tangent Space (Affine Geometry)",
"Definition:Product of Affine Spaces",
"Definition:Product of Affine Spaces",
"Definition:Affine Space",
"Definition:Product of Affine Spaces",
"Definition:Product of Affine Spaces",
"Definition:Affine Space",
"Definition:Product of Affine Spaces",
"Defi... |
proofwiki-7472 | Intersection of Complete Meet Subsemilattices induces Closure Operator | Let $\struct {S, \preccurlyeq}$ be an ordered set.
Let $I$ be an indexing set.
Let $\family {f_i}_{i \mathop \in I}$ be an indexed family of closure operators on $S$.
Let $C_i = \map {f_i} S$ be the set of closed elements with respect to $f_i$ for each $i \in I$.
Suppose that for each $i \in I$, $C_i$ is a '''complete ... | === Lemma ===
{{:Intersection of Complete Meet Subsemilattices induces Closure Operator/Lemma}}{{qed|lemma}}
By the lemma, $C$ is a '''complete meet semilattice'''.
Let $x \in S$.
Then $C \cap x^\succcurlyeq$ has an infimum in $S$ which lies in $C$, where $x^\succcurlyeq$ is the upper closure of $x$.
By the definition ... | Let $\struct {S, \preccurlyeq}$ be an [[Definition:Ordered Set|ordered set]].
Let $I$ be an [[Definition:Indexing Set|indexing set]].
Let $\family {f_i}_{i \mathop \in I}$ be an [[Definition:Indexed Family|indexed family]] of [[Definition:Closure Operator (Order Theory)|closure operators]] on $S$.
Let $C_i = \map {f... | === [[Intersection of Complete Meet Subsemilattices induces Closure Operator/Lemma|Lemma]] ===
{{:Intersection of Complete Meet Subsemilattices induces Closure Operator/Lemma}}{{qed|lemma}}
By the [[Intersection of Complete Meet Subsemilattices induces Closure Operator/Lemma|lemma]], $C$ is a '''[[Definition:Complete... | Intersection of Complete Meet Subsemilattices induces Closure Operator | https://proofwiki.org/wiki/Intersection_of_Complete_Meet_Subsemilattices_induces_Closure_Operator | https://proofwiki.org/wiki/Intersection_of_Complete_Meet_Subsemilattices_induces_Closure_Operator | [
"Closure Operators",
"Intersection of Complete Meet Subsemilattices induces Closure Operator"
] | [
"Definition:Ordered Set",
"Definition:Indexing Set",
"Definition:Indexing Set/Family",
"Definition:Closure Operator/Ordering",
"Definition:Set",
"Definition:Closed Element",
"Definition:Complete Meet Subsemilattice",
"Definition:Infimum of Set",
"Closure Operator from Closed Elements",
"Definition... | [
"Intersection of Complete Meet Subsemilattices induces Closure Operator/Lemma",
"Intersection of Complete Meet Subsemilattices induces Closure Operator/Lemma",
"Definition:Complete Meet Semilattice",
"Definition:Infimum of Set",
"Definition:Upper Closure/Element",
"Definition:Infimum of Set",
"Definitio... |
proofwiki-7473 | Vector Space with Standard Affine Structure is Affine Space | Let $E$ be a vector space.
Let $\struct {\EE, E, +, -}$ be the standard affine structure on $E$.
Then with this structure, $\EE$ is an affine space. | We are required to show that:
{{begin-axiom}}
{{axiom|n = 1
|q = \forall p, q \in \EE
|m = p + \paren {q - p} = q
}}
{{axiom|n = 2
|q = \forall p \in \EE: \forall u, v \in E
|m = \paren {p + u} + v = p + \paren {u + v}
}}
{{axiom|n = 3
|q = \forall p, q \in \EE: \forall u \in E
... | Let $E$ be a [[Definition:Vector Space|vector space]].
Let $\struct {\EE, E, +, -}$ be the [[Definition:Standard Affine Structure on Vector Space|standard affine structure]] on $E$.
Then with this structure, $\EE$ is an [[Definition:Affine Space|affine space]]. | We are required to show that:
{{begin-axiom}}
{{axiom|n = 1
|q = \forall p, q \in \EE
|m = p + \paren {q - p} = q
}}
{{axiom|n = 2
|q = \forall p \in \EE: \forall u, v \in E
|m = \paren {p + u} + v = p + \paren {u + v}
}}
{{axiom|n = 3
|q = \forall p, q \in \EE: \forall u \in E
... | Vector Space with Standard Affine Structure is Affine Space | https://proofwiki.org/wiki/Vector_Space_with_Standard_Affine_Structure_is_Affine_Space | https://proofwiki.org/wiki/Vector_Space_with_Standard_Affine_Structure_is_Affine_Space | [
"Affine Geometry"
] | [
"Definition:Vector Space",
"Definition:Standard Affine Structure on Vector Space",
"Definition:Affine Space"
] | [
"Definition:Standard Affine Structure on Vector Space",
"Definition:Vector Space",
"Definition:Vector Space",
"Definition:Commutative/Operation",
"Definition:Associative Operation",
"Definition:Commutative/Operation",
"Definition:Associative Operation",
"Category:Affine Geometry"
] |
proofwiki-7474 | Reflexive Reduction of Ordering is Strict Ordering | Let $\RR$ be an ordering on a set $S$.
Let $\RR^\ne$ be the reflexive reduction of $\RR$.
Then $\RR^\ne$ is a strict ordering on $S$. | === Antireflexivity ===
Follows from Reflexive Reduction is Antireflexive.
{{qed|lemma}}
=== Transitivity ===
Suppose $\tuple {x, y}, \tuple {y, z} \in \RR^\ne$.
By antireflexivity $x \ne y$ and $y \ne z$.
We consider the two remaining cases.
==== Case 1: $x = z$ ====
If $x = z$ then:
:$\tuple {x, y}, \tuple {y, x} \in... | Let $\RR$ be an [[Definition:Ordering|ordering]] on a [[Definition:Set|set]] $S$.
Let $\RR^\ne$ be the [[Definition:Reflexive Reduction|reflexive reduction]] of $\RR$.
Then $\RR^\ne$ is a [[Definition:Strict Ordering|strict ordering]] on $S$. | === Antireflexivity ===
Follows from [[Reflexive Reduction is Antireflexive]].
{{qed|lemma}}
=== Transitivity ===
Suppose $\tuple {x, y}, \tuple {y, z} \in \RR^\ne$.
By [[Definition:Antireflexive Relation|antireflexivity]] $x \ne y$ and $y \ne z$.
We consider the two remaining cases.
==== Case 1: $x = z$ ====
... | Reflexive Reduction of Ordering is Strict Ordering/Proof 1 | https://proofwiki.org/wiki/Reflexive_Reduction_of_Ordering_is_Strict_Ordering | https://proofwiki.org/wiki/Reflexive_Reduction_of_Ordering_is_Strict_Ordering/Proof_1 | [
"Strict Orderings",
"Reflexive Reductions",
"Reflexive Reduction of Ordering is Strict Ordering"
] | [
"Definition:Ordering",
"Definition:Set",
"Definition:Reflexive Reduction",
"Definition:Strict Ordering"
] | [
"Reflexive Reduction is Antireflexive",
"Definition:Antireflexive Relation",
"Definition:Antisymmetric Relation",
"Definition:Antireflexive Relation",
"Definition:Transitive Relation",
"Definition:Reflexive Reduction",
"Definition:Transitive Relation"
] |
proofwiki-7475 | Reflexive Reduction of Ordering is Strict Ordering | Let $\RR$ be an ordering on a set $S$.
Let $\RR^\ne$ be the reflexive reduction of $\RR$.
Then $\RR^\ne$ is a strict ordering on $S$. | By definition, an ordering is both antisymmetric and transitive.
The result then follows from Reflexive Reduction of Transitive Antisymmetric Relation is Strict Ordering.
{{qed}} | Let $\RR$ be an [[Definition:Ordering|ordering]] on a [[Definition:Set|set]] $S$.
Let $\RR^\ne$ be the [[Definition:Reflexive Reduction|reflexive reduction]] of $\RR$.
Then $\RR^\ne$ is a [[Definition:Strict Ordering|strict ordering]] on $S$. | By definition, an [[Definition:Ordering|ordering]] is both [[Definition:Antisymmetric Relation|antisymmetric]] and [[Definition:Transitive Relation|transitive]].
The result then follows from [[Reflexive Reduction of Transitive Antisymmetric Relation is Strict Ordering]].
{{qed}} | Reflexive Reduction of Ordering is Strict Ordering/Proof 2 | https://proofwiki.org/wiki/Reflexive_Reduction_of_Ordering_is_Strict_Ordering | https://proofwiki.org/wiki/Reflexive_Reduction_of_Ordering_is_Strict_Ordering/Proof_2 | [
"Strict Orderings",
"Reflexive Reductions",
"Reflexive Reduction of Ordering is Strict Ordering"
] | [
"Definition:Ordering",
"Definition:Set",
"Definition:Reflexive Reduction",
"Definition:Strict Ordering"
] | [
"Definition:Ordering",
"Definition:Antisymmetric Relation",
"Definition:Transitive Relation",
"Reflexive Reduction of Transitive Antisymmetric Relation is Strict Ordering"
] |
proofwiki-7476 | Vectorialization of Affine Space is Vector Space | Let $\EE$ be an affine space over a field $K$ with difference space $E$.
Let $\RR = \tuple {p_0, e_1, \ldots, e_n}$ be an affine frame in $\EE$.
Let $\struct {\EE, +, \cdot}$ be the vectorialization of $\EE$.
Then $\struct {\EE, +, \cdot}$ is a vector space. | By the definition of the vectorialization of an affine space, the mapping $\Theta_\RR : K^n \to \EE$ defined by:
:$\ds \map {\Theta_\RR} {\lambda_1, \ldots, \lambda_n} = p_0 + \sum_{i \mathop = 1}^n \lambda_i e_i$
is a bijection from $K^n$ to $\EE$.
Therefore, by Homomorphic Image of Vector Space, it suffices to prove ... | Let $\EE$ be an [[Definition:Affine Space|affine space]] over a [[Definition:Field (Abstract Algebra)|field]] $K$ with [[Definition:Difference Space|difference space]] $E$.
Let $\RR = \tuple {p_0, e_1, \ldots, e_n}$ be an [[Definition:Affine Frame|affine frame]] in $\EE$.
Let $\struct {\EE, +, \cdot}$ be the [[Defini... | By the definition of the [[Definition:Vectorialization of Affine Space|vectorialization of an affine space]], the [[Definition:Mapping|mapping]] $\Theta_\RR : K^n \to \EE$ defined by:
:$\ds \map {\Theta_\RR} {\lambda_1, \ldots, \lambda_n} = p_0 + \sum_{i \mathop = 1}^n \lambda_i e_i$
is a [[Definition:Bijection|bijecti... | Vectorialization of Affine Space is Vector Space | https://proofwiki.org/wiki/Vectorialization_of_Affine_Space_is_Vector_Space | https://proofwiki.org/wiki/Vectorialization_of_Affine_Space_is_Vector_Space | [
"Affine Geometry"
] | [
"Definition:Affine Space",
"Definition:Field (Abstract Algebra)",
"Definition:Tangent Space (Affine Geometry)",
"Definition:Affine Frame",
"Definition:Vectorialization of Affine Space",
"Definition:Vector Space"
] | [
"Definition:Vectorialization of Affine Space",
"Definition:Mapping",
"Definition:Bijection",
"Homomorphic Image of Vector Space",
"Definition:Linear Transformation/Vector Space",
"General Linear Group is Group",
"Definition:Linear Transformation/Vector Space",
"Definition:Inverse Mapping",
"Definiti... |
proofwiki-7477 | Reflexive Reduction is Antireflexive | Let $\RR$ be a relation on a set $S$.
Let $\RR^\ne$ denote the reflexive reduction of $\RR$.
Then $\RR^\ne$ is antireflexive. | By the definition of reflexive reduction:
:$\RR^\ne = \RR \setminus \Delta_S$
where $\Delta_S$ denotes the diagonal relation on $S$.
By Set Difference Intersection with Second Set is Empty Set:
:$\paren {\RR \setminus \Delta_S} \cap \Delta_S = \O$
Hence by Relation is Antireflexive iff Disjoint from Diagonal Relation, ... | Let $\RR$ be a [[Definition:Endorelation|relation]] on a [[Definition:Set|set]] $S$.
Let $\RR^\ne$ denote the [[Definition:Reflexive Reduction|reflexive reduction]] of $\RR$.
Then $\RR^\ne$ is [[Definition:Antireflexive Relation|antireflexive]]. | By the definition of [[Definition:Reflexive Reduction|reflexive reduction]]:
:$\RR^\ne = \RR \setminus \Delta_S$
where $\Delta_S$ denotes the [[Definition:Diagonal Relation|diagonal relation]] on $S$.
By [[Set Difference Intersection with Second Set is Empty Set]]:
:$\paren {\RR \setminus \Delta_S} \cap \Delta_S = ... | Reflexive Reduction is Antireflexive | https://proofwiki.org/wiki/Reflexive_Reduction_is_Antireflexive | https://proofwiki.org/wiki/Reflexive_Reduction_is_Antireflexive | [
"Reflexive Reductions"
] | [
"Definition:Endorelation",
"Definition:Set",
"Definition:Reflexive Reduction",
"Definition:Antireflexive Relation"
] | [
"Definition:Reflexive Reduction",
"Definition:Diagonal Relation",
"Set Difference Intersection with Second Set is Empty Set",
"Relation is Antireflexive iff Disjoint from Diagonal Relation",
"Definition:Antireflexive Relation"
] |
proofwiki-7478 | Equivalence of Definitions of Strict Ordering | Let $S$ be a set.
Let $\RR$ be a relation on $S$.
{{TFAE|def = Strict Ordering}} | Let $\RR$ be transitive.
Then by Transitive Relation is Antireflexive iff Asymmetric it follows directly that:
:$(1): \quad$ If $\RR$ is antireflexive then it is asymmetric
:$(2): \quad$ If $\RR$ is asymmetric then it is antireflexive.
{{qed}}
Category:Strict Orderings
5qtldkiz3xr0f1zgffx5gc1m0m0a7fr | Let $S$ be a [[Definition:Set|set]].
Let $\RR$ be a [[Definition:Endorelation|relation]] on $S$.
{{TFAE|def = Strict Ordering}} | Let $\RR$ be [[Definition:Transitive|transitive]].
Then by [[Transitive Relation is Antireflexive iff Asymmetric]] it follows directly that:
:$(1): \quad$ If $\RR$ is [[Definition:Antireflexive Relation|antireflexive]] then it is [[Definition:Asymmetric Relation|asymmetric]]
:$(2): \quad$ If $\RR$ is [[Definition:Asy... | Equivalence of Definitions of Strict Ordering | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Strict_Ordering | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Strict_Ordering | [
"Strict Orderings"
] | [
"Definition:Set",
"Definition:Endorelation"
] | [
"Definition:Transitive",
"Transitive Relation is Antireflexive iff Asymmetric",
"Definition:Antireflexive Relation",
"Definition:Asymmetric Relation",
"Definition:Asymmetric Relation",
"Definition:Antireflexive Relation",
"Category:Strict Orderings"
] |
proofwiki-7479 | Subband iff Idempotent under Induced Operation | Let $\struct {S, \circ}$ be a band.
Let $\struct {\powerset S, \circ_\PP}$ be the algebraic structure consisting of the power set of $S$ and the operation induced on $\powerset S$ by $\circ$.
Let $X \in \powerset S$.
Then $X$ is idempotent {{iff}} $\struct {X, \circ}$ is a subband of $\struct {S, \circ}$. | === Subbandhood implies Idempotency === | Let $\struct {S, \circ}$ be a [[Definition:Band|band]].
Let $\struct {\powerset S, \circ_\PP}$ be the [[Definition:Algebraic Structure|algebraic structure]] consisting of the [[Definition:Power Set|power set]] of $S$ and the [[Definition:Subset Product|operation induced on $\powerset S$ by $\circ$]].
Let $X \in \powe... | === Subbandhood implies Idempotency === | Subband iff Idempotent under Induced Operation | https://proofwiki.org/wiki/Subband_iff_Idempotent_under_Induced_Operation | https://proofwiki.org/wiki/Subband_iff_Idempotent_under_Induced_Operation | [
"Idempotent Semigroups"
] | [
"Definition:Idempotent Semigroup",
"Definition:Algebraic Structure",
"Definition:Power Set",
"Definition:Subset Product",
"Definition:Idempotence/Element",
"Definition:Subband"
] | [] |
proofwiki-7480 | Restriction of Idempotent Operation is Idempotent | Let $\struct {S, \circ}$ be an algebraic structure.
Let $T \subseteq S$.
Let the operation $\circ$ be idempotent.
Then $\circ$ is also idempotent upon restriction to $\struct {T, \circ \restriction_T}$. | {{begin-eqn}}
{{eqn | l = T
| o = \subseteq
| r = S
| c = {{hypothesis}}
}}
{{eqn | ll= \leadsto
| q = \forall a \in T
| l = a
| o = \in
| r = S
| c = {{Defof|Subset}}
}}
{{eqn | ll= \leadsto
| l = a \mathop {\circ \restriction_T} a
| r = a \circ a
| c =... | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]].
Let $T \subseteq S$.
Let the [[Definition:Binary Operation|operation]] $\circ$ be [[Definition:Idempotent Operation|idempotent]].
Then $\circ$ is also [[Definition:Idempotent Operation|idempotent]] upon [[Definition:Restriction o... | {{begin-eqn}}
{{eqn | l = T
| o = \subseteq
| r = S
| c = {{hypothesis}}
}}
{{eqn | ll= \leadsto
| q = \forall a \in T
| l = a
| o = \in
| r = S
| c = {{Defof|Subset}}
}}
{{eqn | ll= \leadsto
| l = a \mathop {\circ \restriction_T} a
| r = a \circ a
| c =... | Restriction of Idempotent Operation is Idempotent | https://proofwiki.org/wiki/Restriction_of_Idempotent_Operation_is_Idempotent | https://proofwiki.org/wiki/Restriction_of_Idempotent_Operation_is_Idempotent | [
"Abstract Algebra",
"Idempotence"
] | [
"Definition:Algebraic Structure",
"Definition:Operation/Binary Operation",
"Definition:Idempotence/Operation",
"Definition:Idempotence/Operation",
"Definition:Restriction/Operation"
] | [
"Category:Abstract Algebra",
"Category:Idempotence"
] |
proofwiki-7481 | Subband of Induced Operation is Set of Subbands | Let $\struct {S, \circ}$ be a band.
Let $\struct {\powerset S, \circ_\PP}$ be the algebraic structure consisting of:
: the power set $\powerset S$ of $S$
and
: the operation $\circ_\PP$ induced on $\powerset S$ by $\circ$.
Let $T \subseteq \powerset S$.
Let $\struct {T, \circ_\PP}$ be a subband of $\struct {\powerset S... | {{improve|See talk page}} | Let $\struct {S, \circ}$ be a [[Definition:Band|band]].
Let $\struct {\powerset S, \circ_\PP}$ be the [[Definition:Algebraic Structure|algebraic structure]] consisting of:
: the [[Definition:Power Set|power set]] $\powerset S$ of $S$
and
: the [[Definition:Operation Induced on Power Set|operation $\circ_\PP$ induced o... | {{improve|See talk page}} | Subband of Induced Operation is Set of Subbands | https://proofwiki.org/wiki/Subband_of_Induced_Operation_is_Set_of_Subbands | https://proofwiki.org/wiki/Subband_of_Induced_Operation_is_Set_of_Subbands | [
"Idempotent Semigroups"
] | [
"Definition:Idempotent Semigroup",
"Definition:Algebraic Structure",
"Definition:Power Set",
"Definition:Subset Product",
"Definition:Subband",
"Definition:Element",
"Definition:Subband"
] | [] |
proofwiki-7482 | Composition of Commuting Idempotent Mappings is Idempotent | Let $S$ be a set.
Let $f, g: S \to S$ be idempotent mappings from $S$ to $S$.
Let:
:$f \circ g = g \circ f$
where $\circ$ denotes composition.
Then $f \circ g$ is idempotent. | {{begin-eqn}}
{{eqn | l = \paren {f \circ g} \circ \paren {f \circ g}
| r = f \circ \paren {g \circ f} \circ g
| c = Composition of Mappings is Associative
}}
{{eqn | r = f \circ \paren {f \circ g} \circ g
| c = by hypothesis
}}
{{eqn | r = \paren {f \circ f} \circ \paren {g \circ g}
| c = Comp... | Let $S$ be a [[Definition:Set|set]].
Let $f, g: S \to S$ be [[Definition:Idempotent Mapping|idempotent mappings]] from $S$ to $S$.
Let:
:$f \circ g = g \circ f$
where $\circ$ denotes [[Definition:Composition of Mappings|composition]].
Then $f \circ g$ is [[Definition:Idempotent Mapping|idempotent]]. | {{begin-eqn}}
{{eqn | l = \paren {f \circ g} \circ \paren {f \circ g}
| r = f \circ \paren {g \circ f} \circ g
| c = [[Composition of Mappings is Associative]]
}}
{{eqn | r = f \circ \paren {f \circ g} \circ g
| c = [[Definition:By Hypothesis|by hypothesis]]
}}
{{eqn | r = \paren {f \circ f} \circ \p... | Composition of Commuting Idempotent Mappings is Idempotent/Proof 1 | https://proofwiki.org/wiki/Composition_of_Commuting_Idempotent_Mappings_is_Idempotent | https://proofwiki.org/wiki/Composition_of_Commuting_Idempotent_Mappings_is_Idempotent/Proof_1 | [
"Idempotence",
"Mapping Theory",
"Composite Mappings",
"Composition of Commuting Idempotent Mappings is Idempotent"
] | [
"Definition:Set",
"Definition:Idempotence/Mapping",
"Definition:Composition of Mappings",
"Definition:Idempotence/Mapping"
] | [
"Composition of Mappings is Associative",
"Definition:By Hypothesis",
"Composition of Mappings is Associative",
"Definition:Idempotence/Mapping",
"Definition:By Hypothesis"
] |
proofwiki-7483 | Composition of Commuting Idempotent Mappings is Idempotent | Let $S$ be a set.
Let $f, g: S \to S$ be idempotent mappings from $S$ to $S$.
Let:
:$f \circ g = g \circ f$
where $\circ$ denotes composition.
Then $f \circ g$ is idempotent. | By Set of all Self-Maps under Composition forms Semigroup, the set of all self-maps on $S$ forms a semigroup under composition.
The result follows from Product of Commuting Idempotent Elements is Idempotent.
{{qed}} | Let $S$ be a [[Definition:Set|set]].
Let $f, g: S \to S$ be [[Definition:Idempotent Mapping|idempotent mappings]] from $S$ to $S$.
Let:
:$f \circ g = g \circ f$
where $\circ$ denotes [[Definition:Composition of Mappings|composition]].
Then $f \circ g$ is [[Definition:Idempotent Mapping|idempotent]]. | By [[Set of all Self-Maps under Composition forms Semigroup]], the [[Definition:Set|set]] of all [[Definition:Self-Map|self-maps]] on $S$ forms a [[Definition:Semigroup|semigroup]] under [[Definition:Composition of Mappings|composition]].
The result follows from [[Product of Commuting Idempotent Elements is Idempotent... | Composition of Commuting Idempotent Mappings is Idempotent/Proof 2 | https://proofwiki.org/wiki/Composition_of_Commuting_Idempotent_Mappings_is_Idempotent | https://proofwiki.org/wiki/Composition_of_Commuting_Idempotent_Mappings_is_Idempotent/Proof_2 | [
"Idempotence",
"Mapping Theory",
"Composite Mappings",
"Composition of Commuting Idempotent Mappings is Idempotent"
] | [
"Definition:Set",
"Definition:Idempotence/Mapping",
"Definition:Composition of Mappings",
"Definition:Idempotence/Mapping"
] | [
"Set of all Self-Maps under Composition forms Semigroup",
"Definition:Set",
"Definition:Self-Map",
"Definition:Semigroup",
"Definition:Composition of Mappings",
"Product of Commuting Idempotent Elements is Idempotent"
] |
proofwiki-7484 | Compositions of Closure Operators are both Closure Operators iff Operators Commute | Let $\struct {S, \preceq}$ be an ordered set.
Let $f$ and $g$ be closure operators on $S$.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$f \circ g$ and $g \circ f$ are both closure operators}}
{{item|(2):|$f$ and $g$ commute (that is, $f \circ g {{=}} g \circ f$)}}
{{item|(3):|$\Img {f \circ g} {{=}} \Img {g \circ f}$}}
{{en... | By Composition of Inflationary Mappings is Inflationary:
:$f \circ g$ and $g \circ f$ are inflationary.
By Composite of Increasing Mappings is Increasing:
:$f \circ g$ and $g \circ f$ are increasing.
Thus each of the two composite mappings will be a closure operator {{iff}} it is idempotent.
Therefore the equivalences ... | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $f$ and $g$ be [[Definition:Closure Operator (Order Theory)|closure operators]] on $S$.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$f \circ g$ and $g \circ f$ are both [[Definition:Closure Operator (Order Theory)|closure operators]]}}
{{item|(2... | By [[Composition of Inflationary Mappings is Inflationary]]:
:$f \circ g$ and $g \circ f$ are [[Definition:Inflationary Mapping|inflationary]].
By [[Composite of Increasing Mappings is Increasing]]:
:$f \circ g$ and $g \circ f$ are [[Definition:Increasing Mapping|increasing]].
Thus each of the two [[Definition:Comp... | Compositions of Closure Operators are both Closure Operators iff Operators Commute | https://proofwiki.org/wiki/Compositions_of_Closure_Operators_are_both_Closure_Operators_iff_Operators_Commute | https://proofwiki.org/wiki/Compositions_of_Closure_Operators_are_both_Closure_Operators_iff_Operators_Commute | [
"Closure Operators"
] | [
"Definition:Ordered Set",
"Definition:Closure Operator/Ordering",
"Definition:Closure Operator/Ordering",
"Definition:Commutative/Elements",
"Definition:Image (Set Theory)/Mapping/Mapping"
] | [
"Composition of Inflationary Mappings is Inflationary",
"Definition:Inflationary Mapping",
"Composite of Increasing Mappings is Increasing",
"Definition:Increasing/Mapping",
"Definition:Composition of Mappings",
"Definition:Closure Operator/Ordering",
"Definition:Idempotence/Mapping",
"Composition of ... |
proofwiki-7485 | Composition of Inflationary Mappings is Inflationary | Let $\struct {S, \preceq}$ be an ordered set.
Let $f, g: S \to S$ be inflationary mappings.
Then $f \circ g$, the composition of $f$ and $g$, is also inflationary. | Let $x \in S$.
{{begin-eqn}}
{{eqn | n = 1
| l = x
| o = \preceq
| r = \map g x
| c = $g$ is inflationary
}}
{{eqn | n = 2
| l = \map g x
| o = \preceq
| r = \map f {\map g x}
| c = $f$ is inflationary
}}
{{eqn | l = x
| o = \preceq
| r = \map f {\map g x}
... | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $f, g: S \to S$ be [[Definition:Inflationary Mapping|inflationary mappings]].
Then $f \circ g$, the [[Definition:Composition of Mappings|composition]] of $f$ and $g$, is also [[Definition:Inflationary Mapping|inflationary]]. | Let $x \in S$.
{{begin-eqn}}
{{eqn | n = 1
| l = x
| o = \preceq
| r = \map g x
| c = $g$ is [[Definition:Inflationary Mapping|inflationary]]
}}
{{eqn | n = 2
| l = \map g x
| o = \preceq
| r = \map f {\map g x}
| c = $f$ is [[Definition:Inflationary Mapping|inflationary... | Composition of Inflationary Mappings is Inflationary | https://proofwiki.org/wiki/Composition_of_Inflationary_Mappings_is_Inflationary | https://proofwiki.org/wiki/Composition_of_Inflationary_Mappings_is_Inflationary | [
"Order Theory"
] | [
"Definition:Ordered Set",
"Definition:Inflationary Mapping",
"Definition:Composition of Mappings",
"Definition:Inflationary Mapping"
] | [
"Definition:Inflationary Mapping",
"Definition:Inflationary Mapping",
"Definition:Ordering",
"Definition:Transitive Relation",
"Definition:Inflationary Mapping",
"Category:Order Theory"
] |
proofwiki-7486 | Fixed Point of Idempotent Mapping | Let $S$ be a set.
Let $f: S \to S$ be an idempotent mapping.
Let $\Img f$ be the image of $f$.
Let $x \in S$.
Then $x$ is a fixed point of $f$ {{iff}} $x \in \Img f$. | === Necessary Condition ===
Let $x$ be a fixed point of $f$.
Then:
:$\map f x = x$
and so by definition of image of mapping:
:$x \in \Img f$
{{qed|lemma}} | Let $S$ be a [[Definition:Set|set]].
Let $f: S \to S$ be an [[Definition:Idempotent Mapping|idempotent mapping]].
Let $\Img f$ be the [[Definition:Image of Mapping|image]] of $f$.
Let $x \in S$.
Then $x$ is a [[Definition:Fixed Point|fixed point]] of $f$ {{iff}} $x \in \Img f$. | === Necessary Condition ===
Let $x$ be a [[Definition:Fixed Point|fixed point]] of $f$.
Then:
:$\map f x = x$
and so by definition of [[Definition:Image of Mapping|image of mapping]]:
:$x \in \Img f$
{{qed|lemma}} | Fixed Point of Idempotent Mapping | https://proofwiki.org/wiki/Fixed_Point_of_Idempotent_Mapping | https://proofwiki.org/wiki/Fixed_Point_of_Idempotent_Mapping | [
"Idempotent Mappings"
] | [
"Definition:Set",
"Definition:Idempotence/Mapping",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Fixed Point"
] | [
"Definition:Fixed Point",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Fixed Point"
] |
proofwiki-7487 | Symmetric Closure of Ordering may not be Transitive | Let $\struct {S, \preceq}$ be an ordered set.
Let $\preceq^\leftrightarrow$ be the symmetric closure of $\preceq$.
Then it is not necessarily the case that $\preceq^\leftrightarrow$ is transitive. | Proof by Counterexample:
Let $S = \set {a, b, c}$ where $a$, $b$, and $c$ are distinct.
Let:
:${\preceq} = \set {\tuple {a, a}, \tuple {b, b}, \tuple {c, c}, \tuple {a, c}, \tuple {b, c} }$:
Then $\preceq$ is an ordering, but $\preceq^\leftrightarrow$ is not transitive, as follows:
$\preceq$ is reflexive because it con... | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $\preceq^\leftrightarrow$ be the [[Definition:Symmetric Closure|symmetric closure]] of $\preceq$.
Then it is not necessarily the case that $\preceq^\leftrightarrow$ is [[Definition:Transitive Relation|transitive]]. | [[Proof by Counterexample]]:
Let $S = \set {a, b, c}$ where $a$, $b$, and $c$ are [[Definition:Distinct|distinct]].
Let:
:${\preceq} = \set {\tuple {a, a}, \tuple {b, b}, \tuple {c, c}, \tuple {a, c}, \tuple {b, c} }$:
Then $\preceq$ is an [[Definition:Ordering|ordering]], but $\preceq^\leftrightarrow$ is not [[Def... | Symmetric Closure of Ordering may not be Transitive | https://proofwiki.org/wiki/Symmetric_Closure_of_Ordering_may_not_be_Transitive | https://proofwiki.org/wiki/Symmetric_Closure_of_Ordering_may_not_be_Transitive | [
"Orderings",
"Symmetric Closures",
"Transitive Relations"
] | [
"Definition:Ordered Set",
"Definition:Symmetric Closure",
"Definition:Transitive Relation"
] | [
"Proof by Counterexample",
"Definition:Distinct",
"Definition:Ordering",
"Definition:Transitive Relation",
"Definition:Reflexive Relation",
"Definition:Subset",
"Definition:Diagonal Relation",
"Definition:Transitive Relation",
"Definition:Antisymmetric Relation",
"Definition:Ordered Pair",
"Defi... |
proofwiki-7488 | Composition of Idempotent Mappings | Let $S$ be a set.
Let $f, g: S \to S$ be idempotent mappings.
Suppose that $f \circ g$ and $g \circ f$ have the same images.
That is, suppose that $f \sqbrk {g \sqbrk S} = g \sqbrk {f \sqbrk S}$.
Then $f \circ g$ and $g \circ f$ are idempotent. | Let $x \in S$.
By the premise:
:$\map f {\map g x} \in g \sqbrk {f \sqbrk S}$
Since $f \sqbrk S \subseteq S$:
:$\map f {\map g x} \in g \sqbrk S$
Thus for some $y \in S$:
:$\map f {\map g x} = \map g y$
Since $g$ is idempotent:
:$\map g {\map g y} = \map g y$
By the choice of $y$:
:$\map g {\map f {\map g x} } = \map g... | Let $S$ be a [[Definition:Set|set]].
Let $f, g: S \to S$ be [[Definition:Idempotent Mapping|idempotent mappings]].
Suppose that $f \circ g$ and $g \circ f$ have the same [[Definition:Image of Mapping|images]].
That is, suppose that $f \sqbrk {g \sqbrk S} = g \sqbrk {f \sqbrk S}$.
Then $f \circ g$ and $g \circ f$ a... | Let $x \in S$.
By the premise:
:$\map f {\map g x} \in g \sqbrk {f \sqbrk S}$
Since $f \sqbrk S \subseteq S$:
:$\map f {\map g x} \in g \sqbrk S$
Thus for some $y \in S$:
:$\map f {\map g x} = \map g y$
Since $g$ is [[Definition:Idempotent Mapping|idempotent]]:
:$\map g {\map g y} = \map g y$
By the choice of $y... | Composition of Idempotent Mappings | https://proofwiki.org/wiki/Composition_of_Idempotent_Mappings | https://proofwiki.org/wiki/Composition_of_Idempotent_Mappings | [
"Idempotent Mappings",
"Composite Mappings"
] | [
"Definition:Set",
"Definition:Idempotence/Mapping",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Idempotence/Mapping"
] | [
"Definition:Idempotence/Mapping",
"Definition:Idempotence/Mapping",
"Category:Idempotent Mappings",
"Category:Composite Mappings"
] |
proofwiki-7489 | Composition of Inflationary and Idempotent Mappings | Let $\struct {S, \preceq}$ be an ordered set.
Let $f$ and $g$ be inflationary and idempotent mappings on $S$.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$f \circ g$ and $g \circ f$ are both idempotent}}
{{item|(2):|$f$ and $g$ commute (that is, $f \circ g {{=}} g \circ f$)}}
{{item|(3):|$\Img {f \circ g} {{=}} \Img {g \cir... | === $(2)$ implies $(1)$ ===
Follows from Composition of Commuting Idempotent Mappings is Idempotent.
{{qed|lemma}} | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $f$ and $g$ be [[Definition:Inflationary Mapping|inflationary]] and [[Definition:Idempotent Mapping|idempotent]] [[Definition:Mapping|mappings]] on $S$.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$f \circ g$ and $g \circ f$ are both [[Definitio... | === $(2)$ implies $(1)$ ===
Follows from [[Composition of Commuting Idempotent Mappings is Idempotent]].
{{qed|lemma}} | Composition of Inflationary and Idempotent Mappings | https://proofwiki.org/wiki/Composition_of_Inflationary_and_Idempotent_Mappings | https://proofwiki.org/wiki/Composition_of_Inflationary_and_Idempotent_Mappings | [
"Mapping Theory"
] | [
"Definition:Ordered Set",
"Definition:Inflationary Mapping",
"Definition:Idempotence/Mapping",
"Definition:Mapping",
"Definition:Idempotence/Mapping",
"Definition:Commutative/Elements",
"Definition:Composition of Mappings",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Mapping"
] | [
"Composition of Commuting Idempotent Mappings is Idempotent"
] |
proofwiki-7490 | Closure is Smallest Closed Successor | Let $\struct {S, \preceq}$ be an ordered set.
Let $f: S \to S$ be a closure operator on $S$.
Let $x \in S$.
Then $\map f x$ is the smallest closed element that succeeds $x$. | By the definition of closure operator, $f$ is inflationary.
Thus $x \preceq \map f x$.
By definition, $\map f x$ is closed.
So $\map f x$ is a closed element that succeeds $x$.
We will now show that it is the smallest such.
Let $k$ be a closed element of $S$ such that $x \preceq k$.
Since $f$ is a closure operator, it ... | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $f: S \to S$ be a [[Definition:Closure Operator (Order Theory)|closure operator]] on $S$.
Let $x \in S$.
Then $\map f x$ is the [[Definition:Smallest Element|smallest]] [[Definition:Closed Element|closed element]] that [[Definition:Succeed... | By the definition of [[Definition:Closure Operator (Order Theory)|closure operator]], $f$ is [[Definition:Inflationary Mapping|inflationary]].
Thus $x \preceq \map f x$.
By definition, $\map f x$ is [[Definition:Closed Element|closed]].
So $\map f x$ is a closed element that [[Definition:Succeed|succeeds]] $x$.
We ... | Closure is Smallest Closed Successor | https://proofwiki.org/wiki/Closure_is_Smallest_Closed_Successor | https://proofwiki.org/wiki/Closure_is_Smallest_Closed_Successor | [
"Closure Operators"
] | [
"Definition:Ordered Set",
"Definition:Closure Operator/Ordering",
"Definition:Smallest Element",
"Definition:Closed Element",
"Definition:Succeed"
] | [
"Definition:Closure Operator/Ordering",
"Definition:Inflationary Mapping",
"Definition:Closed Element",
"Definition:Succeed",
"Definition:Smallest Element",
"Definition:Closure Operator/Ordering",
"Definition:Increasing/Mapping",
"Definition:Closed Element",
"Category:Closure Operators"
] |
proofwiki-7491 | Closed Elements Uniquely Determine Closure Operator | Let $\struct {S, \preceq}$ be an ordered set.
Let $f, g: S \to S$ be closure operators on $S$.
Suppose that $f$ and $g$ have the same closed elements.
Then $f = g$. | Let $x \in S$.
Let $C$ be the set of closed elements of $S$ ({{WRT}} either $f$ or $g$) that succeed $x$.
By Closure is Smallest Closed Successor, $\map f x$ and $\map g x$ are smallest closed successors of $x$.
That is, $\map f x$ and $\map g x$ are smallest elements of $C \cap x^\succeq$, where $x^\succeq$ denotes th... | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $f, g: S \to S$ be [[Definition:Closure Operator (Order Theory)|closure operators]] on $S$.
Suppose that $f$ and $g$ have the same [[Definition:Closed Element|closed elements]].
Then $f = g$. | Let $x \in S$.
Let $C$ be the [[Definition:Set|set]] of [[Definition:Closed Element|closed elements]] of $S$ ({{WRT}} either $f$ or $g$) that [[Definition:Succeed|succeed]] $x$.
By [[Closure is Smallest Closed Successor]], $\map f x$ and $\map g x$ are [[Definition:Smallest Element|smallest]] [[Definition:Closed Elem... | Closed Elements Uniquely Determine Closure Operator | https://proofwiki.org/wiki/Closed_Elements_Uniquely_Determine_Closure_Operator | https://proofwiki.org/wiki/Closed_Elements_Uniquely_Determine_Closure_Operator | [
"Closure Operators"
] | [
"Definition:Ordered Set",
"Definition:Closure Operator/Ordering",
"Definition:Closed Element"
] | [
"Definition:Set",
"Definition:Closed Element",
"Definition:Succeed",
"Closure is Smallest Closed Successor",
"Definition:Smallest Element",
"Definition:Closed Element",
"Definition:Successor",
"Definition:Smallest Element",
"Definition:Upper Closure/Element",
"Smallest Element is Unique",
"Equal... |
proofwiki-7492 | Equivalence of Definitions of Closed Element | Let $\struct {S, \preceq}$ be an ordered set.
Let $\cl$ be a closure operator on $S$.
Let $x \in S$.
{{TFAE|def = Closed Element}} | Let $\struct {S, \preceq}$ be an ordered set.
Let $\cl: S \to S$ be a closure operator on $S$.
Let $x \in S$.
By the definition of closure operator, $\cl$ is idempotent.
Thus by Fixed Point of Idempotent Mapping:
:An element of $S$ is a fixed point of $\cl$ {{iff}} it is in the image of $\cl$.
Thus the above definition... | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $\cl$ be a [[Definition:Closure Operator (Order Theory)|closure operator]] on $S$.
Let $x \in S$.
{{TFAE|def = Closed Element}} | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $\cl: S \to S$ be a [[Definition:Closure Operator|closure operator]] on $S$.
Let $x \in S$.
By the definition of [[Definition:Closure Operator|closure operator]], $\cl$ is [[Definition:Idempotent Mapping|idempotent]].
Thus by [[Fixed Point... | Equivalence of Definitions of Closed Element | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Closed_Element | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Closed_Element | [
"Closed Elements"
] | [
"Definition:Ordered Set",
"Definition:Closure Operator/Ordering"
] | [
"Definition:Ordered Set",
"Definition:Closure Operator",
"Definition:Closure Operator",
"Definition:Idempotence/Mapping",
"Fixed Point of Idempotent Mapping",
"Definition:Fixed Point",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Category:Closed Elements"
] |
proofwiki-7493 | Square of Number Always Exists | Let $x$ be a number.
Then its square $x^2$ is guaranteed to exist. | Whatever flavour of number under discussion, the algebraic structure $\struct {\mathbb K, +, \times}$ in which this number sits is at least a semiring.
The binary operation that is multiplication is therefore closed on that algebraic structure.
Therefore:
: $\forall x \in \mathbb K: x \times x \in \mathbb K$
{{qed}}
Ca... | Let $x$ be a [[Definition:Number|number]].
Then its [[Definition:Square (Algebra)|square]] $x^2$ is guaranteed to exist. | Whatever flavour of [[Definition:Number|number]] under discussion, the [[Definition:Algebraic Structure|algebraic structure]] $\struct {\mathbb K, +, \times}$ in which this number sits is at least a [[Definition:Semiring|semiring]].
The [[Definition:Binary Operation|binary operation]] that is [[Definition:Multiplicati... | Square of Number Always Exists | https://proofwiki.org/wiki/Square_of_Number_Always_Exists | https://proofwiki.org/wiki/Square_of_Number_Always_Exists | [
"Numbers"
] | [
"Definition:Number",
"Definition:Square/Function"
] | [
"Definition:Number",
"Definition:Algebraic Structure",
"Definition:Semiring",
"Definition:Operation/Binary Operation",
"Definition:Multiplication",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Algebraic Structure",
"Category:Numbers"
] |
proofwiki-7494 | Schröder Rule | Let $A$, $B$ and $C$ be relations on a set $S$.
{{TFAE}}
:$(1): \quad A \circ B \subseteq C$
:$(2): \quad A^{-1} \circ \overline C \subseteq \overline B$
:$(3): \quad \overline C \circ B^{-1} \subseteq \overline A$
where:
:$\circ$ denotes relation composition
:$A^{-1}$ denotes the inverse of $A$
:$\overline A$ denotes ... | === $(1)$ iff $(2)$ ===
By the definition of relation composition and subset we have that statement $(1)$ may be written as:
:$(1'): \quad \forall x, y, z \in S: \paren {\tuple {y, z} \in A \land \tuple {x, y} \in B \implies \tuple {x, z} \in C}$
Using a different arrangement of variable names, statement $(2)$ can be ... | Let $A$, $B$ and $C$ be [[Definition:Endorelation|relations]] on a [[Definition:Set|set]] $S$.
{{TFAE}}
:$(1): \quad A \circ B \subseteq C$
:$(2): \quad A^{-1} \circ \overline C \subseteq \overline B$
:$(3): \quad \overline C \circ B^{-1} \subseteq \overline A$
where:
:$\circ$ denotes [[Definition:Composition of ... | === $(1)$ iff $(2)$ ===
By the definition of [[Definition:Composition of Relations|relation composition]] and [[Definition:Subset|subset]] we have that statement $(1)$ may be written as:
:$(1'): \quad \forall x, y, z \in S: \paren {\tuple {y, z} \in A \land \tuple {x, y} \in B \implies \tuple {x, z} \in C}$
Using a... | Schröder Rule/Proof 1 | https://proofwiki.org/wiki/Schröder_Rule | https://proofwiki.org/wiki/Schröder_Rule/Proof_1 | [
"Schröder Rule",
"Relation Theory"
] | [
"Definition:Endorelation",
"Definition:Set",
"Definition:Composition of Relations",
"Definition:Inverse Relation",
"Definition:Complement of Relation"
] | [
"Definition:Composition of Relations",
"Definition:Subset",
"Definition:Inverse Relation",
"Definition:Complement of Relation",
"Method of Truth Tables/Proof of Interderivability",
"Definition:Logical Equivalence",
"Definition:Main Connective"
] |
proofwiki-7495 | Schröder Rule | Let $A$, $B$ and $C$ be relations on a set $S$.
{{TFAE}}
:$(1): \quad A \circ B \subseteq C$
:$(2): \quad A^{-1} \circ \overline C \subseteq \overline B$
:$(3): \quad \overline C \circ B^{-1} \subseteq \overline A$
where:
:$\circ$ denotes relation composition
:$A^{-1}$ denotes the inverse of $A$
:$\overline A$ denotes ... | By the definition of relation composition and subset we have that statement $(1)$ may be written as:
:$(1') \quad \forall x, y, z \in S: \paren {\tuple {y, z} \in A \land \tuple {x, y} \in B \implies \tuple {x, z} \in C}$
{{explain|Actually, that only gets us to $\forall x, z \in S: \paren {\paren {\exists y \in S: \tu... | Let $A$, $B$ and $C$ be [[Definition:Endorelation|relations]] on a [[Definition:Set|set]] $S$.
{{TFAE}}
:$(1): \quad A \circ B \subseteq C$
:$(2): \quad A^{-1} \circ \overline C \subseteq \overline B$
:$(3): \quad \overline C \circ B^{-1} \subseteq \overline A$
where:
:$\circ$ denotes [[Definition:Composition of ... | By the definition of [[Definition:Composition of Relations|relation composition]] and [[Definition:Subset|subset]] we have that statement $(1)$ may be written as:
:$(1') \quad \forall x, y, z \in S: \paren {\tuple {y, z} \in A \land \tuple {x, y} \in B \implies \tuple {x, z} \in C}$
{{explain|Actually, that only gets ... | Schröder Rule/Proof 2 | https://proofwiki.org/wiki/Schröder_Rule | https://proofwiki.org/wiki/Schröder_Rule/Proof_2 | [
"Schröder Rule",
"Relation Theory"
] | [
"Definition:Endorelation",
"Definition:Set",
"Definition:Composition of Relations",
"Definition:Inverse Relation",
"Definition:Complement of Relation"
] | [
"Definition:Composition of Relations",
"Definition:Subset",
"Definition:Inverse Relation",
"Definition:Complement of Relation",
"Definition:Inverse Relation",
"Definition:Complement of Relation"
] |
proofwiki-7496 | Equivalence of Definitions of Dual Relation | {{TFAE|def = Dual Relation}}
Let $\RR \subseteq S \times T$ be a relation. | Let $\tuple {x, y} \in \paren {\overline \RR}^{-1}$.
Then:
{{begin-eqn}}
{{eqn | l = \tuple {x, y}
| o = \in
| r = \overline \RR
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = \tuple {y, x}
| o = \in
| r = \overline \RR
| c = {{Defof|Inverse Relation}}
}}
{{eqn | ll= \leadstoandfr... | {{TFAE|def = Dual Relation}}
Let $\RR \subseteq S \times T$ be a [[Definition:Binary Relation|relation]]. | Let $\tuple {x, y} \in \paren {\overline \RR}^{-1}$.
Then:
{{begin-eqn}}
{{eqn | l = \tuple {x, y}
| o = \in
| r = \overline \RR
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = \tuple {y, x}
| o = \in
| r = \overline \RR
| c = {{Defof|Inverse Relation}}
}}
{{eqn | ll= \leadstoan... | Equivalence of Definitions of Dual Relation | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Dual_Relation | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Dual_Relation | [
"Relation Theory"
] | [
"Definition:Relation"
] | [
"Category:Relation Theory"
] |
proofwiki-7497 | Trivial Gradation is Gradation | Let $\struct {R, +, \circ}$ be a ring.
Let $\struct {M, e, \cdot}$ be a monoid.
Let
:$\ds R = \bigoplus_{m \mathop \in M} R_m$
be the trivial $M$-gradation on $R$.
This is a gradation on $R$. | We are required to show that:
:$\forall x \in R_m, y \in R_n: x \circ y \in R_{m \cdot n}$
First suppose that $m = n = e$ are both the identity.
In this case, $R_m = R_n = R$.
Since by definition, $R$ is closed under $\circ$, it follows that
:$\forall x \in R, y \in R: x \circ y \in R$
as required.
Now suppose that eit... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $\struct {M, e, \cdot}$ be a [[Definition:Monoid|monoid]].
Let
:$\ds R = \bigoplus_{m \mathop \in M} R_m$
be the [[Definition:Trivial Gradation|trivial $M$-gradation]] on $R$.
This is a [[Definition:Gradation Compatible with Ring Struc... | We are required to show that:
:$\forall x \in R_m, y \in R_n: x \circ y \in R_{m \cdot n}$
First suppose that $m = n = e$ are both the [[Definition:Identity Element|identity]].
In this case, $R_m = R_n = R$.
Since by definition, $R$ is [[Definition:Closed Algebraic Structure|closed]] under $\circ$, it follows that
:... | Trivial Gradation is Gradation | https://proofwiki.org/wiki/Trivial_Gradation_is_Gradation | https://proofwiki.org/wiki/Trivial_Gradation_is_Gradation | [
"Ring Theory"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Monoid",
"Definition:Trivial Gradation",
"Definition:Gradation Compatible with Ring Structure"
] | [
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Null Ring",
"Ring Product with Zero",
"Definition:Abelian Group",
"Category:Ring Theory"
] |
proofwiki-7498 | Fixed Point of Mappings is Fixed Point of Composition | Let $S$ be a set.
Let $f, g: S \to S$ be mappings.
Let $x \in S$ be a fixed point of both $f$ and $g$.
Then $x$ is also a fixed point of $f \circ g$, the composition of $f$ and $g$. | Since $x$ is a fixed point of $g$:
: $\map g x = x$
Thus:
: $\map f {\map g x} = \map f x$
Since $x$ is a fixed point of $f$:
: $\map f x = x$
It follows that:
: $\map {\paren {f \circ g} } x = \map f {\map g x} = x$
Thus $x$ is a fixed point of $f \circ g$.
{{qed}}
Category:Mapping Theory
5vr2j2gkw62xlk2jb3eooda53pg2q... | Let $S$ be a [[Definition:Set|set]].
Let $f, g: S \to S$ be [[Definition:Mapping|mappings]].
Let $x \in S$ be a [[Definition:Fixed Point|fixed point]] of both $f$ and $g$.
Then $x$ is also a [[Definition:Fixed Point|fixed point]] of $f \circ g$, the [[Definition:Composition of Mappings|composition]] of $f$ and $g$. | Since $x$ is a [[Definition:Fixed Point|fixed point]] of $g$:
: $\map g x = x$
Thus:
: $\map f {\map g x} = \map f x$
Since $x$ is a [[Definition:Fixed Point|fixed point]] of $f$:
: $\map f x = x$
It follows that:
: $\map {\paren {f \circ g} } x = \map f {\map g x} = x$
Thus $x$ is a [[Definition:Fixed Point|fi... | Fixed Point of Mappings is Fixed Point of Composition | https://proofwiki.org/wiki/Fixed_Point_of_Mappings_is_Fixed_Point_of_Composition | https://proofwiki.org/wiki/Fixed_Point_of_Mappings_is_Fixed_Point_of_Composition | [
"Mapping Theory"
] | [
"Definition:Set",
"Definition:Mapping",
"Definition:Fixed Point",
"Definition:Fixed Point",
"Definition:Composition of Mappings"
] | [
"Definition:Fixed Point",
"Definition:Fixed Point",
"Definition:Fixed Point",
"Category:Mapping Theory"
] |
proofwiki-7499 | Fixed Point of Mappings is Fixed Point of Composition/General Result | Let $S$ be a set.
Let $n \in \N$ be a strictly positive integer.
Let $\N_n$ be the initial segment of $n$ in $\N$.
That is, let $\N_n = \set {0, 1, \dots, n-1}$.
For each $i \in \N_n$, let $f_i: S \to S$ be a mapping.
Let $x \in S$ be a fixed point of $f_i$ for each $i \in \N_n$.
Let $g = f_0 \circ f_1 \circ \dots \cir... | The proof proceeds by mathematical induction on $n$, the number of mappings. | Let $S$ be a [[Definition:Set|set]].
Let $n \in \N$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
Let $\N_n$ be the [[Definition:Initial Segment of Natural Numbers|initial segment]] of $n$ in $\N$.
That is, let $\N_n = \set {0, 1, \dots, n-1}$.
For each $i \in \N_n$, let $f_i: S \to S$ be... | The proof proceeds by [[Principle of Mathematical Induction|mathematical induction]] on $n$, the number of [[Definition:Mapping|mappings]]. | Fixed Point of Mappings is Fixed Point of Composition/General Result | https://proofwiki.org/wiki/Fixed_Point_of_Mappings_is_Fixed_Point_of_Composition/General_Result | https://proofwiki.org/wiki/Fixed_Point_of_Mappings_is_Fixed_Point_of_Composition/General_Result | [
"Mapping Theory"
] | [
"Definition:Set",
"Definition:Strictly Positive/Integer",
"Definition:Initial Segment of Natural Numbers",
"Definition:Mapping",
"Definition:Fixed Point",
"Definition:Composition of Mappings",
"Definition:Fixed Point"
] | [
"Principle of Mathematical Induction",
"Definition:Mapping"
] |
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