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-10400 | Surjection iff Right Cancellable/Sufficient Condition | Let $f$ be a mapping which is right cancellable.
Then $f$ is a surjection. | Let $f: X \to Y$ be a right cancellable mapping.
Let $Y$ contain exactly one element.
Then by definition $Y$ is a singleton.
From Mapping to Singleton is Surjection it follows that $f$ is a surjection.
So let $Y$ contain at least two elements.
Call those two elements $a$ and $b$, and we note that $a \ne b$.
We define t... | Let $f$ be a [[Definition:Mapping|mapping]] which is [[Definition:Right Cancellable Mapping|right cancellable]].
Then $f$ is a [[Definition:Surjection|surjection]]. | Let $f: X \to Y$ be a [[Definition:Right Cancellable Mapping|right cancellable mapping]].
Let $Y$ contain exactly one element.
Then by definition $Y$ is a [[Definition:Singleton|singleton]].
From [[Mapping to Singleton is Surjection]] it follows that $f$ is a [[Definition:Surjection|surjection]].
So let $Y$ contai... | Surjection iff Right Cancellable/Sufficient Condition/Proof 2 | https://proofwiki.org/wiki/Surjection_iff_Right_Cancellable/Sufficient_Condition | https://proofwiki.org/wiki/Surjection_iff_Right_Cancellable/Sufficient_Condition/Proof_2 | [
"Surjection iff Right Cancellable"
] | [
"Definition:Mapping",
"Definition:Right Cancellable Mapping",
"Definition:Surjection"
] | [
"Definition:Right Cancellable Mapping",
"Definition:Singleton",
"Mapping to Singleton is Surjection",
"Definition:Surjection",
"Definition:Right Cancellable Mapping",
"Definition:Surjection"
] |
proofwiki-10401 | Absolute Value Function on Integers induces Equivalence Relation | Let $\Z$ be the set of integers.
Let $\RR$ be the relation on $\Z$ defined as:
:$\forall x, y \in \Z: \tuple {x, y} \in \RR \iff \size x = \size y$
where $\size x$ denotes the absolute value of $x$.
Then $\RR$ is an equivalence relation. | $\RR$ is shown to be an equivalence relation thus: | Let $\Z$ be the [[Definition:Integer|set of integers]].
Let $\RR$ be the [[Definition:Relation|relation]] on $\Z$ defined as:
:$\forall x, y \in \Z: \tuple {x, y} \in \RR \iff \size x = \size y$
where $\size x$ denotes the [[Definition:Absolute Value|absolute value]] of $x$.
Then $\RR$ is an [[Definition:Equivalence... | $\RR$ is shown to be an [[Definition:Equivalence Relation|equivalence relation]] thus: | Absolute Value Function on Integers induces Equivalence Relation | https://proofwiki.org/wiki/Absolute_Value_Function_on_Integers_induces_Equivalence_Relation | https://proofwiki.org/wiki/Absolute_Value_Function_on_Integers_induces_Equivalence_Relation | [
"Examples of Equivalence Relations",
"Integers",
"Absolute Value Function"
] | [
"Definition:Integer",
"Definition:Relation",
"Definition:Absolute Value",
"Definition:Equivalence Relation"
] | [
"Definition:Equivalence Relation",
"Definition:Equivalence Relation"
] |
proofwiki-10402 | Absolute Value induces Equivalence not Compatible with Integer Addition | Let $\Z$ be the set of integers.
Let $\RR$ be the relation on $\Z$ defined as:
:$\forall x, y \in \Z: \tuple {x, y} \in \RR \iff \size x = \size y$
where $\size x$ denotes the absolute value of $x$.
Then $\RR$ is not a congruence relation for integer addition. | From Absolute Value Function on Integers induces Equivalence Relation, $\RR$ is an equivalence relation.
However, consider that:
{{begin-eqn}}
{{eqn | l = \size {-1} = \size 1
| o = \leadsto
| r = -1 \mathop \RR 1
| c =
}}
{{eqn | l = \size 2 = \size 2
| o = \leadsto
| r = 2 \mathop \RR 2... | Let $\Z$ be the [[Definition:Integer|set of integers]].
Let $\RR$ be the [[Definition:Relation|relation]] on $\Z$ defined as:
:$\forall x, y \in \Z: \tuple {x, y} \in \RR \iff \size x = \size y$
where $\size x$ denotes the [[Definition:Absolute Value|absolute value]] of $x$.
Then $\RR$ is not a [[Definition:Congruen... | From [[Absolute Value Function on Integers induces Equivalence Relation]], $\RR$ is an [[Definition:Equivalence Relation|equivalence relation]].
However, consider that:
{{begin-eqn}}
{{eqn | l = \size {-1} = \size 1
| o = \leadsto
| r = -1 \mathop \RR 1
| c =
}}
{{eqn | l = \size 2 = \size 2
... | Absolute Value induces Equivalence not Compatible with Integer Addition | https://proofwiki.org/wiki/Absolute_Value_induces_Equivalence_not_Compatible_with_Integer_Addition | https://proofwiki.org/wiki/Absolute_Value_induces_Equivalence_not_Compatible_with_Integer_Addition | [
"Examples of Congruence Relations",
"Integer Addition",
"Absolute Value Function"
] | [
"Definition:Integer",
"Definition:Relation",
"Definition:Absolute Value",
"Definition:Congruence Relation",
"Definition:Addition/Integers"
] | [
"Absolute Value Function on Integers induces Equivalence Relation",
"Definition:Equivalence Relation",
"Definition:Addition/Integers",
"Definition:Congruence Relation",
"Definition:Addition/Integers",
"Proof by Counterexample"
] |
proofwiki-10403 | Equivalence Relation is Congruence for Right Operation | Every equivalence relation is a congruence for the right operation $\rightarrow$. | Let $\RR$ be an equivalence relation on the structure $\struct {S, \rightarrow}$.
Then:
:$x_1 \rightarrow y_1 = y_1$
:$x_2 \rightarrow y_2 = y_2$
Suppose $x_1 \mathrel \RR x_2 \land y_1 \mathrel \RR y_2$.
It follows directly that:
:$\paren {x_1 \rightarrow y_1} \mathrel \RR \paren {x_2 \rightarrow y_2}$
{{Qed}} | Every [[Definition:Equivalence Relation|equivalence relation]] is a [[Definition:Congruence Relation|congruence]] for the [[Definition:Right Operation|right operation]] $\rightarrow$. | Let $\RR$ be an [[Definition:Equivalence Relation|equivalence relation]] on the structure $\struct {S, \rightarrow}$.
Then:
:$x_1 \rightarrow y_1 = y_1$
:$x_2 \rightarrow y_2 = y_2$
Suppose $x_1 \mathrel \RR x_2 \land y_1 \mathrel \RR y_2$.
It follows directly that:
:$\paren {x_1 \rightarrow y_1} \mathrel \RR \paren... | Equivalence Relation is Congruence for Right Operation | https://proofwiki.org/wiki/Equivalence_Relation_is_Congruence_for_Right_Operation | https://proofwiki.org/wiki/Equivalence_Relation_is_Congruence_for_Right_Operation | [
"Right Operation",
"Examples of Equivalence Relations",
"Examples of Congruence Relations"
] | [
"Definition:Equivalence Relation",
"Definition:Congruence Relation",
"Definition:Right Operation"
] | [
"Definition:Equivalence Relation"
] |
proofwiki-10404 | Equivalence Relation is Congruence for Left Operation | Every equivalence relation is a congruence for the left operation $\leftarrow$. | Let $\RR$ be an equivalence relation on the structure $\struct {S, \leftarrow}$.
Then:
:$x_1 \leftarrow y_1 = x_1$
:$x_2 \leftarrow y_2 = x_2$
Suppose $x_1 \mathrel \RR x_2 \land y_1 \mathrel \RR y_2$.
It follows directly that:
:$\paren {x_1 \leftarrow y_1} \mathrel \RR \paren {x_2 \leftarrow y_2}$
{{Qed}} | Every [[Definition:Equivalence Relation|equivalence relation]] is a [[Definition:Congruence Relation|congruence]] for the [[Definition:Left Operation|left operation]] $\leftarrow$. | Let $\RR$ be an [[Definition:Equivalence Relation|equivalence relation]] on the structure $\struct {S, \leftarrow}$.
Then:
:$x_1 \leftarrow y_1 = x_1$
:$x_2 \leftarrow y_2 = x_2$
Suppose $x_1 \mathrel \RR x_2 \land y_1 \mathrel \RR y_2$.
It follows directly that:
:$\paren {x_1 \leftarrow y_1} \mathrel \RR \paren {x_... | Equivalence Relation is Congruence for Left Operation | https://proofwiki.org/wiki/Equivalence_Relation_is_Congruence_for_Left_Operation | https://proofwiki.org/wiki/Equivalence_Relation_is_Congruence_for_Left_Operation | [
"Left Operation",
"Examples of Equivalence Relations",
"Examples of Congruence Relations"
] | [
"Definition:Equivalence Relation",
"Definition:Congruence Relation",
"Definition:Left Operation"
] | [
"Definition:Equivalence Relation"
] |
proofwiki-10405 | Left Coset Space forms Partition | The left coset space of $H$ forms a partition of its group $G$, and hence:
{{begin-eqn}}
{{eqn | l = x \equiv^l y \pmod H
| o = \iff
| r = x H = y H
}}
{{eqn | l = \neg \paren {x \equiv^l y} \pmod H
| o = \iff
| r = x H \cap y H = \O
}}
{{end-eqn}} | Follows directly from:
:Left Congruence Modulo Subgroup is Equivalence Relation
:Relation Partitions Set iff Equivalence
{{qed}} | The [[Definition:Left Coset Space|left coset space]] of $H$ forms a [[Definition:Set Partition|partition]] of its [[Definition:Group|group]] $G$, and hence:
{{begin-eqn}}
{{eqn | l = x \equiv^l y \pmod H
| o = \iff
| r = x H = y H
}}
{{eqn | l = \neg \paren {x \equiv^l y} \pmod H
| o = \iff
| r... | Follows directly from:
:[[Left Congruence Modulo Subgroup is Equivalence Relation]]
:[[Relation Partitions Set iff Equivalence]]
{{qed}} | Left Coset Space forms Partition | https://proofwiki.org/wiki/Left_Coset_Space_forms_Partition | https://proofwiki.org/wiki/Left_Coset_Space_forms_Partition | [
"Coset Space forms Partition"
] | [
"Definition:Coset Space/Left Coset Space",
"Definition:Set Partition",
"Definition:Group"
] | [
"Left Congruence Modulo Subgroup is Equivalence Relation",
"Relation Partitions Set iff Equivalence"
] |
proofwiki-10406 | Right Coset Space forms Partition | The right coset space of $H$ forms a partition of its group $G$:
{{begin-eqn}}
{{eqn | l = x \equiv^r y \pmod H
| o = \iff
| r = H x = H y
}}
{{eqn | l = \neg \paren {x \equiv^r y} \pmod H
| o = \iff
| r = H x \cap H y = \O
}}
{{end-eqn}} | Follows directly from:
:Right Congruence Modulo Subgroup is Equivalence Relation
:Relation Partitions Set iff Equivalence.
{{qed}} | The [[Definition:Right Coset Space|right coset space]] of $H$ forms a [[Definition:Set Partition|partition]] of its [[Definition:Group|group]] $G$:
{{begin-eqn}}
{{eqn | l = x \equiv^r y \pmod H
| o = \iff
| r = H x = H y
}}
{{eqn | l = \neg \paren {x \equiv^r y} \pmod H
| o = \iff
| r = H x \c... | Follows directly from:
:[[Right Congruence Modulo Subgroup is Equivalence Relation]]
:[[Relation Partitions Set iff Equivalence]].
{{qed}} | Right Coset Space forms Partition | https://proofwiki.org/wiki/Right_Coset_Space_forms_Partition | https://proofwiki.org/wiki/Right_Coset_Space_forms_Partition | [
"Coset Space forms Partition"
] | [
"Definition:Coset Space/Right Coset Space",
"Definition:Set Partition",
"Definition:Group"
] | [
"Right Congruence Modulo Subgroup is Equivalence Relation",
"Relation Partitions Set iff Equivalence"
] |
proofwiki-10407 | Right Cosets are Equal iff Product with Inverse in Subgroup | Let $H x$ denote the right coset of $H$ by $x$.
Then:
:$H x = H y \iff x y^{-1} \in H$ | {{begin-eqn}}
{{eqn | l = H x
| r = H y
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = x
| o = \equiv^r
| r = y \bmod H
| c = Right Coset Space forms Partition
}}
{{eqn | ll= \leadstoandfrom
| l = x y^{-1}
| o = \in
| r = H
| c = Equivalent Statements for Congrue... | Let $H x$ denote the [[Definition:Right Coset|right coset]] of $H$ by $x$.
Then:
:$H x = H y \iff x y^{-1} \in H$ | {{begin-eqn}}
{{eqn | l = H x
| r = H y
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = x
| o = \equiv^r
| r = y \bmod H
| c = [[Right Coset Space forms Partition]]
}}
{{eqn | ll= \leadstoandfrom
| l = x y^{-1}
| o = \in
| r = H
| c = [[Equivalent Statements for C... | Right Cosets are Equal iff Product with Inverse in Subgroup | https://proofwiki.org/wiki/Right_Cosets_are_Equal_iff_Product_with_Inverse_in_Subgroup | https://proofwiki.org/wiki/Right_Cosets_are_Equal_iff_Product_with_Inverse_in_Subgroup | [
"Cosets"
] | [
"Definition:Coset/Right Coset"
] | [
"Right Coset Space forms Partition",
"Equivalent Statements for Congruence Modulo Subgroup"
] |
proofwiki-10408 | Left Cosets are Equal iff Product with Inverse in Subgroup | Let $x H$ denote the left coset of $H$ by $x$.
Then:
:$x H = y H \iff x^{-1} y \in H$ | {{begin-eqn}}
{{eqn | l = x H
| r = y H
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = x
| o = \equiv^l
| r = y \pmod H
| c = Left Coset Space forms Partition
}}
{{eqn | ll= \leadstoandfrom
| l = x^{-1} y
| o = \in
| r = H
| c = Equivalent Statements for Congruen... | Let $x H$ denote the [[Definition:Left Coset|left coset]] of $H$ by $x$.
Then:
:$x H = y H \iff x^{-1} y \in H$ | {{begin-eqn}}
{{eqn | l = x H
| r = y H
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = x
| o = \equiv^l
| r = y \pmod H
| c = [[Left Coset Space forms Partition]]
}}
{{eqn | ll= \leadstoandfrom
| l = x^{-1} y
| o = \in
| r = H
| c = [[Equivalent Statements for Co... | Left Cosets are Equal iff Product with Inverse in Subgroup | https://proofwiki.org/wiki/Left_Cosets_are_Equal_iff_Product_with_Inverse_in_Subgroup | https://proofwiki.org/wiki/Left_Cosets_are_Equal_iff_Product_with_Inverse_in_Subgroup | [
"Cosets"
] | [
"Definition:Coset/Left Coset"
] | [
"Left Coset Space forms Partition",
"Equivalent Statements for Congruence Modulo Subgroup"
] |
proofwiki-10409 | Left Congruence Class Modulo Subgroup is Left Coset | Let $\RR^l_H$ be the equivalence defined as left congruence modulo $H$.
The equivalence class $\eqclass g {\RR^l_H}$ of an element $g \in G$ is the left coset $g H$.
This is known as the '''left congruence class of $g \bmod H$'''. | Let $x \in \eqclass g {\RR^l_H}$.
Then:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \eqclass g {\RR^l_H}
| c =
}}
{{eqn | ll= \leadsto
| q = \exists h \in H
| l = g^{-1} x
| r = h
| c = {{Defof|Left Congruence Modulo Subgroup|Left Congruence Modulo $H$}}
}}
{{eqn | ll= \leadst... | Let $\RR^l_H$ be the [[Definition:Equivalence Relation|equivalence]] defined as [[Definition:Left Congruence Modulo Subgroup|left congruence modulo $H$]].
The [[Definition:Equivalence Class|equivalence class]] $\eqclass g {\RR^l_H}$ of an element $g \in G$ is the [[Definition:Left Coset|left coset]] $g H$.
This is k... | Let $x \in \eqclass g {\RR^l_H}$.
Then:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \eqclass g {\RR^l_H}
| c =
}}
{{eqn | ll= \leadsto
| q = \exists h \in H
| l = g^{-1} x
| r = h
| c = {{Defof|Left Congruence Modulo Subgroup|Left Congruence Modulo $H$}}
}}
{{eqn | ll= \lead... | Left Congruence Class Modulo Subgroup is Left Coset | https://proofwiki.org/wiki/Left_Congruence_Class_Modulo_Subgroup_is_Left_Coset | https://proofwiki.org/wiki/Left_Congruence_Class_Modulo_Subgroup_is_Left_Coset | [
"Cosets"
] | [
"Definition:Equivalence Relation",
"Definition:Congruence Modulo Subgroup/Left Congruence",
"Definition:Equivalence Class",
"Definition:Coset/Left Coset"
] | [
"Definition:Group",
"Definition:Group",
"Definition:Equivalence Class",
"Definition:Coset/Left Coset"
] |
proofwiki-10410 | Right Congruence Class Modulo Subgroup is Right Coset | Let $\RR^r_H$ be the equivalence defined as right congruence modulo $H$.
The equivalence class $\eqclass g {\RR^r_H}$ of an element $g \in G$ is the right coset $H g$.
This is known as the '''right congruence class of $g \bmod H$'''. | Let $x \in \eqclass g {\RR^r_H}$.
Then:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \eqclass g {\RR^r_H}
| c =
}}
{{eqn | ll= \leadsto
| q = \exists h \in H
| l = x g^{-1}
| r = h
| c = {{Defof|Right Congruence Modulo Subgroup|Right Congruence Modulo $H$}}
}}
{{eqn | ll= \lead... | Let $\RR^r_H$ be the [[Definition:Equivalence Relation|equivalence]] defined as [[Definition:Right Congruence Modulo Subgroup|right congruence modulo $H$]].
The [[Definition:Equivalence Class|equivalence class]] $\eqclass g {\RR^r_H}$ of an element $g \in G$ is the [[Definition:Right Coset|right coset]] $H g$.
This ... | Let $x \in \eqclass g {\RR^r_H}$.
Then:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \eqclass g {\RR^r_H}
| c =
}}
{{eqn | ll= \leadsto
| q = \exists h \in H
| l = x g^{-1}
| r = h
| c = {{Defof|Right Congruence Modulo Subgroup|Right Congruence Modulo $H$}}
}}
{{eqn | ll= \le... | Right Congruence Class Modulo Subgroup is Right Coset | https://proofwiki.org/wiki/Right_Congruence_Class_Modulo_Subgroup_is_Right_Coset | https://proofwiki.org/wiki/Right_Congruence_Class_Modulo_Subgroup_is_Right_Coset | [
"Cosets"
] | [
"Definition:Equivalence Relation",
"Definition:Congruence Modulo Subgroup/Right Congruence",
"Definition:Equivalence Class",
"Definition:Coset/Right Coset"
] | [
"Definition:Group",
"Definition:Group",
"Definition:Equivalence Class",
"Definition:Coset/Right Coset"
] |
proofwiki-10411 | Right Congruence Modulo Subgroup is Equivalence Relation | Let $x \equiv^r y \pmod H$ denote the relation that $x$ is right congruent modulo $H$ to $y$
Then the relation $\equiv^r$ is an equivalence relation. | Let $G$ be a group whose identity is $e$.
Let $H$ be a subgroup of $G$.
For clarity of expression, we will use the notation:
:$\tuple {x, y} \in \RR^r_H$
for:
:$x \equiv^r y \pmod H$
From the definition of right congruence modulo a subgroup, we have:
:$\RR^r_H = \set {\tuple {x, y} \in G \times G: x y^{-1} \in H}$
We s... | Let $x \equiv^r y \pmod H$ denote the [[Definition:Relation|relation]] that $x$ is [[Definition:Right Congruence Modulo Subgroup|right congruent modulo $H$]] to $y$
Then the [[Definition:Relation|relation]] $\equiv^r$ is an [[Definition:Equivalence Relation|equivalence relation]]. | Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
For clarity of expression, we will use the notation:
:$\tuple {x, y} \in \RR^r_H$
for:
:$x \equiv^r y \pmod H$
From the definition of [[Definition:Right Congruence Mo... | Right Congruence Modulo Subgroup is Equivalence Relation | https://proofwiki.org/wiki/Right_Congruence_Modulo_Subgroup_is_Equivalence_Relation | https://proofwiki.org/wiki/Right_Congruence_Modulo_Subgroup_is_Equivalence_Relation | [
"Congruence Modulo Subgroup",
"Examples of Equivalence Relations"
] | [
"Definition:Relation",
"Definition:Congruence Modulo Subgroup/Right Congruence",
"Definition:Relation",
"Definition:Equivalence Relation"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Subgroup",
"Definition:Congruence Modulo Subgroup/Right Congruence",
"Definition:Equivalence Relation",
"Definition:Subgroup",
"Definition:Group",
"Definition:Equivalence Relation"
] |
proofwiki-10412 | Group Epimorphism is Isomorphism iff Kernel is Trivial | Let $\struct {G, \oplus}$ and $\struct {H, \odot}$ be groups.
Let $\phi: \struct {G, \oplus} \to \struct {H, \odot}$ be a group epimorphism.
Let $e_G$ and $e_H$ be the identities of $G$ and $H$ respectively.
Let $K = \map \ker \phi$ be the kernel of $\phi$.
Then:
:the epimorphism $\phi$ is an isomorphism
{{iff}}
:$K = ... | === Necessary Condition ===
Let $\phi$ be an isomorphism.
Then by definition $\phi$ is a bijective homomorphism.
Thus by definition of bijection, $\phi$ is an injection.
By definition of injection, there exists exactly one element $x$ of $G$ such that $\map \phi x = e_H$.
From Epimorphism Preserves Identity, that eleme... | Let $\struct {G, \oplus}$ and $\struct {H, \odot}$ be [[Definition:Group|groups]].
Let $\phi: \struct {G, \oplus} \to \struct {H, \odot}$ be a [[Definition:Group Epimorphism|group epimorphism]].
Let $e_G$ and $e_H$ be the [[Definition:Identity Element|identities]] of $G$ and $H$ respectively.
Let $K = \map \ker \phi... | === Necessary Condition ===
Let $\phi$ be an [[Definition:Group Isomorphism|isomorphism]].
Then by definition $\phi$ is a [[Definition:Bijection|bijective]] [[Definition:Homomorphism|homomorphism]].
Thus by definition of [[Definition:Bijection|bijection]], $\phi$ is an [[Definition:Injection|injection]].
By definit... | Group Epimorphism is Isomorphism iff Kernel is Trivial/Proof 1 | https://proofwiki.org/wiki/Group_Epimorphism_is_Isomorphism_iff_Kernel_is_Trivial | https://proofwiki.org/wiki/Group_Epimorphism_is_Isomorphism_iff_Kernel_is_Trivial/Proof_1 | [
"Quotient Groups",
"Group Epimorphisms",
"Group Isomorphisms",
"Group Epimorphism is Isomorphism iff Kernel is Trivial"
] | [
"Definition:Group",
"Definition:Group Epimorphism",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Kernel of Group Homomorphism",
"Definition:Group Epimorphism",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism"
] | [
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Bijection",
"Definition:Homomorphism",
"Definition:Bijection",
"Definition:Injection",
"Definition:Injection",
"Definition:Element",
"Epimorphism Preserves Identity",
"Definition:Kernel of Group Homomorphism",
"Quotient The... |
proofwiki-10413 | Group Epimorphism is Isomorphism iff Kernel is Trivial | Let $\struct {G, \oplus}$ and $\struct {H, \odot}$ be groups.
Let $\phi: \struct {G, \oplus} \to \struct {H, \odot}$ be a group epimorphism.
Let $e_G$ and $e_H$ be the identities of $G$ and $H$ respectively.
Let $K = \map \ker \phi$ be the kernel of $\phi$.
Then:
:the epimorphism $\phi$ is an isomorphism
{{iff}}
:$K = ... | From Kernel is Trivial iff Group Monomorphism, $\phi$ is a monomorphism {{iff}} $K = \set {e_G}$.
By definition, a group $G$ is an epimorphism is an isomorphism {{iff}} $G$ is also a monomorphism.
Hence the result.
{{qed}} | Let $\struct {G, \oplus}$ and $\struct {H, \odot}$ be [[Definition:Group|groups]].
Let $\phi: \struct {G, \oplus} \to \struct {H, \odot}$ be a [[Definition:Group Epimorphism|group epimorphism]].
Let $e_G$ and $e_H$ be the [[Definition:Identity Element|identities]] of $G$ and $H$ respectively.
Let $K = \map \ker \phi... | From [[Kernel is Trivial iff Group Monomorphism]], $\phi$ is a [[Definition:Group Monomorphism|monomorphism]] {{iff}} $K = \set {e_G}$.
By definition, a [[Definition:Group|group]] $G$ is an [[Definition:Group Epimorphism|epimorphism]] is an [[Definition:Group Isomorphism|isomorphism]] {{iff}} $G$ is also a [[Definitio... | Group Epimorphism is Isomorphism iff Kernel is Trivial/Proof 2 | https://proofwiki.org/wiki/Group_Epimorphism_is_Isomorphism_iff_Kernel_is_Trivial | https://proofwiki.org/wiki/Group_Epimorphism_is_Isomorphism_iff_Kernel_is_Trivial/Proof_2 | [
"Quotient Groups",
"Group Epimorphisms",
"Group Isomorphisms",
"Group Epimorphism is Isomorphism iff Kernel is Trivial"
] | [
"Definition:Group",
"Definition:Group Epimorphism",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Kernel of Group Homomorphism",
"Definition:Group Epimorphism",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism"
] | [
"Kernel is Trivial iff Monomorphism/Group",
"Definition:Group Monomorphism",
"Definition:Group",
"Definition:Group Epimorphism",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Group Monomorphism"
] |
proofwiki-10414 | Real Numbers are not Well-Ordered under Conventional Ordering | Let $\struct {\R, \leqslant}$ be the ordered structure consisting of the real numbers under the usual ordering.
Then $\struct {\R, \leqslant}$ is not a well-ordered set. | ;Proof by Counterexample
Consider the set:
:$A := \set {x \in \R: x > 1}$
Suppose $a$ were the smallest element of $A$.
Then $a > 1$.
But then:
:$1 < \dfrac {a + 1} 2 < a$
and so $a$ has been shown not to be the smallest element of $A$ after all.
Hence $A$ has no smallest element.
Thus there exists a subset of $\R$ whi... | Let $\struct {\R, \leqslant}$ be the [[Definition:Ordered Structure|ordered structure]] consisting of the [[Definition:Real Number|real numbers]] under the [[Definition:Usual Ordering|usual ordering]].
Then $\struct {\R, \leqslant}$ is not a [[Definition:Well-Ordered Set|well-ordered set]]. | ;[[Proof by Counterexample]]
Consider the [[Definition:Set|set]]:
:$A := \set {x \in \R: x > 1}$
Suppose $a$ were the [[Definition:Smallest Element|smallest element]] of $A$.
Then $a > 1$.
But then:
:$1 < \dfrac {a + 1} 2 < a$
and so $a$ has been shown not to be the [[Definition:Smallest Element|smallest element]]... | Real Numbers are not Well-Ordered under Conventional Ordering | https://proofwiki.org/wiki/Real_Numbers_are_not_Well-Ordered_under_Conventional_Ordering | https://proofwiki.org/wiki/Real_Numbers_are_not_Well-Ordered_under_Conventional_Ordering | [
"Real Numbers",
"Well-Orderings"
] | [
"Definition:Ordered Structure",
"Definition:Real Number",
"Definition:Usual Ordering",
"Definition:Well-Ordered Set"
] | [
"Proof by Counterexample",
"Definition:Set",
"Definition:Smallest Element",
"Definition:Smallest Element",
"Definition:Smallest Element",
"Definition:Subset",
"Definition:Smallest Element",
"Definition:Well-Ordered Set"
] |
proofwiki-10415 | Supremum is Unique | Let $\struct {S, \preceq}$ be an ordered set.
Let $T$ be a non-empty subset of $S$.
Then $T$ has at most one supremum in $S$. | Let $c$ and $c'$ both be suprema of $T$ in $S$.
From the definition of supremum, $c$ and $c'$ are upper bounds of $T$ in $S$.
By that definition:
:$c$ is an upper bound of $T$ in $S$ and $c'$ is a supremum of $T$ in $S$ implies that $c' \preceq c$
:$c'$ is an upper bound of $T$ in $S$ and $c$ is a supremum of $T$ in $S... | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $T$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $S$.
Then $T$ has at most one [[Definition:Supremum of Set|supremum]] in $S$. | Let $c$ and $c'$ both be [[Definition:Supremum of Set|suprema]] of $T$ in $S$.
From the definition of [[Definition:Supremum of Set|supremum]], $c$ and $c'$ are [[Definition:Upper Bound of Set|upper bounds]] of $T$ in $S$.
By that definition:
:$c$ is an [[Definition:Upper Bound of Set|upper bound]] of $T$ in $S$ and... | Supremum is Unique | https://proofwiki.org/wiki/Supremum_is_Unique | https://proofwiki.org/wiki/Supremum_is_Unique | [
"Order Theory"
] | [
"Definition:Ordered Set",
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Supremum of Set"
] | [
"Definition:Supremum of Set",
"Definition:Supremum of Set",
"Definition:Upper Bound of Set",
"Definition:Upper Bound of Set",
"Definition:Supremum of Set",
"Definition:Upper Bound of Set",
"Definition:Supremum of Set",
"Definition:Antisymmetric Relation",
"Definition:Ordering"
] |
proofwiki-10416 | Infimum is Unique | Let $\struct {S, \preceq}$ be an ordered set.
Let $T$ be a non-empty subset of $S$.
Then $T$ has at most one infimum in $S$. | Let $c$ and $c'$ both be infima of $T$ in $S$.
From the definition of infimum, $c$ and $c'$ are lower bounds of $T$ in $S$.
By that definition:
:$c$ is a lower bound of $T$ in $S$ and $c'$ is an infimum of $T$ in $S$ implies that $c \preceq c'$
:$c'$ is a lower bound of $T$ in $S$ and $c$ is an infimum of $T$ in $S$ im... | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $T$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $S$.
Then $T$ has at most one [[Definition:Infimum of Set|infimum]] in $S$. | Let $c$ and $c'$ both be [[Definition:Infimum of Set|infima]] of $T$ in $S$.
From the definition of [[Definition:Infimum of Set|infimum]], $c$ and $c'$ are [[Definition:Lower Bound of Set|lower bounds]] of $T$ in $S$.
By that definition:
:$c$ is a [[Definition:Lower Bound of Set|lower bound]] of $T$ in $S$ and $c'$... | Infimum is Unique | https://proofwiki.org/wiki/Infimum_is_Unique | https://proofwiki.org/wiki/Infimum_is_Unique | [
"Infima"
] | [
"Definition:Ordered Set",
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Infimum of Set"
] | [
"Definition:Infimum of Set",
"Definition:Infimum of Set",
"Definition:Lower Bound of Set",
"Definition:Lower Bound of Set",
"Definition:Infimum of Set",
"Definition:Lower Bound of Set",
"Definition:Infimum of Set",
"Definition:Antisymmetric Relation",
"Definition:Ordering"
] |
proofwiki-10417 | Infimum of Subgroups in Lattice | Then:
:$\inf \set {H, K} = H \cap K$ | Let $H, K \in \mathbb G$.
From Set of Subgroups forms Complete Lattice:
:$\inf \set {H, K} = H \cap K$
{{qed}} | Then:
:$\inf \set {H, K} = H \cap K$ | Let $H, K \in \mathbb G$.
From [[Set of Subgroups forms Complete Lattice]]:
:$\inf \set {H, K} = H \cap K$
{{qed}} | Infimum of Subgroups in Lattice | https://proofwiki.org/wiki/Infimum_of_Subgroups_in_Lattice | https://proofwiki.org/wiki/Infimum_of_Subgroups_in_Lattice | [
"Complete Lattices",
"Infima",
"Subgroups"
] | [] | [
"Set of Subgroups forms Complete Lattice"
] |
proofwiki-10418 | Supremum of Subgroups in Lattice | Let either $H$ or $K$ be normal in $G$.
Then:
:$\sup \set {H, K} = H \circ K$
where $H \circ K$ denotes subset product. | Recall that Set of Subgroups forms Complete Lattice.
Let $L = \sup \set {H, K}$.
Let either $H$ or $K$ be normal in $G$.
Since $L$ contains $H$ and $K$, then $L$ contains $H \circ K$.
The smallest subgroup of $G$ containing $H$ and $K$ is:
:$\gen {H, K}$
the subgroup generated by $H$ and $K$.
From Subset Product with N... | Let either $H$ or $K$ be [[Definition:Normal Subgroup|normal]] in $G$.
Then:
:$\sup \set {H, K} = H \circ K$
where $H \circ K$ denotes [[Definition:Subset Product|subset product]]. | Recall that [[Set of Subgroups forms Complete Lattice]].
Let $L = \sup \set {H, K}$.
Let either $H$ or $K$ be [[Definition:Normal Subgroup|normal]] in $G$.
Since $L$ contains $H$ and $K$, then $L$ contains $H \circ K$.
The smallest [[Definition:Subgroup|subgroup]] of $G$ containing $H$ and $K$ is:
:$\gen {H, K}$
t... | Supremum of Subgroups in Lattice | https://proofwiki.org/wiki/Supremum_of_Subgroups_in_Lattice | https://proofwiki.org/wiki/Supremum_of_Subgroups_in_Lattice | [
"Complete Lattices",
"Suprema",
"Subgroups"
] | [
"Definition:Normal Subgroup",
"Definition:Subset Product"
] | [
"Set of Subgroups forms Complete Lattice",
"Definition:Normal Subgroup",
"Definition:Subgroup",
"Definition:Generator of Subgroup",
"Subset Product with Normal Subgroup as Generator",
"Definition:Normal Subgroup"
] |
proofwiki-10419 | Lexicographic Order on Pair of Totally Ordered Sets is Total Ordering | Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be ordered sets.
Let $\preccurlyeq_l$ be the lexicographic order on $S_1 \times S_2$''':
:$\tuple {x_1, x_2} \preccurlyeq_l \tuple {y_1, y_2} \iff \paren {x_1 \prec_1 y_1} \lor \paren {x_1 = y_1 \land x_2 \preccurlyeq_2 y_2}$
Then:
:$\preccurlyeq_l... | Recall that from Lexicographic Order is Ordering we have that $\struct {S_1 \times S_2, \preccurlyeq_l}$ is an ordered set. | Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be [[Definition:Ordered Set|ordered sets]].
Let $\preccurlyeq_l$ be the [[Definition:Lexicographic Order|lexicographic order on $S_1 \times S_2$]]''':
:$\tuple {x_1, x_2} \preccurlyeq_l \tuple {y_1, y_2} \iff \paren {x_1 \prec_1 y_1} \lor \paren {... | Recall that from [[Lexicographic Order is Ordering]] we have that $\struct {S_1 \times S_2, \preccurlyeq_l}$ is an [[Definition:Ordered Set|ordered set]]. | Lexicographic Order on Pair of Totally Ordered Sets is Total Ordering | https://proofwiki.org/wiki/Lexicographic_Order_on_Pair_of_Totally_Ordered_Sets_is_Total_Ordering | https://proofwiki.org/wiki/Lexicographic_Order_on_Pair_of_Totally_Ordered_Sets_is_Total_Ordering | [
"Lexicographic Order",
"Total Orderings"
] | [
"Definition:Ordered Set",
"Definition:Lexicographic Order",
"Definition:Total Ordering",
"Definition:Total Ordering"
] | [
"Lexicographic Order is Ordering",
"Definition:Ordered Set"
] |
proofwiki-10420 | Lexicographic Order is Ordering | Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be ordered sets.
Let $\preccurlyeq_l$ be the lexicographic order on $S_1 \times S_2$''':
:$\tuple {x_1, x_2} \preccurlyeq_l \tuple {y_1, y_2} \iff \paren {x_1 \prec_1 y_1} \lor \paren {x_1 = y_1 \land x_2 \preccurlyeq_2 y_2}$
Then $\preccurlyeq_l$ ... | In the following, $\tuple {x_1, x_2}, \tuple {y_1, y_2}, \tuple {z_1, z_2} \in S_1 \times S_2$.
Checking in turn each of the criteria for an ordering: | Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be [[Definition:Ordered Set|ordered sets]].
Let $\preccurlyeq_l$ be the [[Definition:Lexicographic Order|lexicographic order on $S_1 \times S_2$]]''':
:$\tuple {x_1, x_2} \preccurlyeq_l \tuple {y_1, y_2} \iff \paren {x_1 \prec_1 y_1} \lor \paren {... | In the following, $\tuple {x_1, x_2}, \tuple {y_1, y_2}, \tuple {z_1, z_2} \in S_1 \times S_2$.
Checking in turn each of the criteria for an [[Definition:Ordering|ordering]]: | Lexicographic Order is Ordering | https://proofwiki.org/wiki/Lexicographic_Order_is_Ordering | https://proofwiki.org/wiki/Lexicographic_Order_is_Ordering | [
"Lexicographic Order"
] | [
"Definition:Ordered Set",
"Definition:Lexicographic Order",
"Definition:Ordering"
] | [
"Definition:Ordering",
"Definition:Ordering",
"Definition:Ordering",
"Definition:Ordering",
"Definition:Ordering",
"Definition:Ordering"
] |
proofwiki-10421 | Lexicographic Order on Pair of Well-Ordered Sets is Well-Ordering | Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be ordered sets.
Let $\preccurlyeq_l$ denote the lexicographic order on $S_1 \times S_2$''':
:$\tuple {x_1, x_2} \preccurlyeq_l \tuple {y_1, y_2} \iff \tuple {x_1 \prec_1 y_1} \lor \tuple {x_1 = y_1 \land x_2 \preccurlyeq_2 y_2}$
Then:
:$\preccurly... | From Lexicographic Order is Ordering we have that $\struct {S_1 \times S_2, \preccurlyeq_l}$ is an ordered set. | Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be [[Definition:Ordered Set|ordered sets]].
Let $\preccurlyeq_l$ denote the [[Definition:Lexicographic Order|lexicographic order on $S_1 \times S_2$]]''':
:$\tuple {x_1, x_2} \preccurlyeq_l \tuple {y_1, y_2} \iff \tuple {x_1 \prec_1 y_1} \lor \tup... | From [[Lexicographic Order is Ordering]] we have that $\struct {S_1 \times S_2, \preccurlyeq_l}$ is an [[Definition:Ordered Set|ordered set]]. | Lexicographic Order on Pair of Well-Ordered Sets is Well-Ordering | https://proofwiki.org/wiki/Lexicographic_Order_on_Pair_of_Well-Ordered_Sets_is_Well-Ordering | https://proofwiki.org/wiki/Lexicographic_Order_on_Pair_of_Well-Ordered_Sets_is_Well-Ordering | [
"Lexicographic Order",
"Well-Orderings"
] | [
"Definition:Ordered Set",
"Definition:Lexicographic Order",
"Definition:Well-Ordering",
"Definition:Well-Ordering"
] | [
"Lexicographic Order is Ordering",
"Definition:Ordered Set"
] |
proofwiki-10422 | Powers of Commuting Elements of Monoid Commute | :$\forall m, n \in \N: \paren {\circ^m a} \circ \paren {\circ^n b} = \paren {\circ^n b} \circ \paren {\circ^m a}$ | Because $\struct {S, \circ}$ is a monoid, it is a fortiori also a semigroup.
From Powers of Commuting Elements of Semigroup Commute:
:$\forall m, n \in \N_{>0}: \paren {\circ^m a} \circ \paren {\circ^n b} = \paren {\circ^n b} \circ \paren {\circ^m a}$
That is:
:$\forall m, n \in \N_{>0}: a^m \circ b^n = b^n \circ a^m$
... | :$\forall m, n \in \N: \paren {\circ^m a} \circ \paren {\circ^n b} = \paren {\circ^n b} \circ \paren {\circ^m a}$ | Because $\struct {S, \circ}$ is a [[Definition:Monoid|monoid]], it is [[Definition:A Fortiori|a fortiori]] also a [[Definition:Semigroup|semigroup]].
From [[Powers of Commuting Elements of Semigroup Commute]]:
:$\forall m, n \in \N_{>0}: \paren {\circ^m a} \circ \paren {\circ^n b} = \paren {\circ^n b} \circ \paren {\c... | Powers of Commuting Elements of Monoid Commute | https://proofwiki.org/wiki/Powers_of_Commuting_Elements_of_Monoid_Commute | https://proofwiki.org/wiki/Powers_of_Commuting_Elements_of_Monoid_Commute | [
"Monoids",
"Commutativity",
"Powers (Abstract Algebra)"
] | [] | [
"Definition:Monoid",
"Definition:A Fortiori",
"Definition:Semigroup",
"Powers of Commuting Elements of Semigroup Commute"
] |
proofwiki-10423 | Power of Product of Commuting Elements in Monoid equals Product of Powers | :$\forall n \in \N: \circ^n \paren {a \circ b} = \paren {\circ^n a} \circ \paren {\circ^n b}$ | Because $\struct {S, \circ}$ is a monoid, it is a fortiori also a semigroup.
From Power of Product of Commuting Elements in Semigroup equals Product of Powers:
:$\forall n \in \N_{>0}: \circ^n \paren {a \circ b} = \paren {\circ^n a} \circ \paren {\circ^n b}$
That is:
:$\forall n \in \N_{>0}: \paren {a \circ b}^n = a^n ... | :$\forall n \in \N: \circ^n \paren {a \circ b} = \paren {\circ^n a} \circ \paren {\circ^n b}$ | Because $\struct {S, \circ}$ is a [[Definition:Monoid|monoid]], it is [[Definition:A Fortiori|a fortiori]] also a [[Definition:Semigroup|semigroup]].
From [[Power of Product of Commuting Elements in Semigroup equals Product of Powers]]:
:$\forall n \in \N_{>0}: \circ^n \paren {a \circ b} = \paren {\circ^n a} \circ \pa... | Power of Product of Commuting Elements in Monoid equals Product of Powers | https://proofwiki.org/wiki/Power_of_Product_of_Commuting_Elements_in_Monoid_equals_Product_of_Powers | https://proofwiki.org/wiki/Power_of_Product_of_Commuting_Elements_in_Monoid_equals_Product_of_Powers | [
"Monoids",
"Commutativity",
"Powers (Abstract Algebra)"
] | [] | [
"Definition:Monoid",
"Definition:A Fortiori",
"Definition:Semigroup",
"Power of Product of Commuting Elements in Semigroup equals Product of Powers"
] |
proofwiki-10424 | Index Laws/Product of Indices/Monoid | Let $\struct {S, \circ}$ be a monoid whose identity element is $e$.
For $a \in S$, let $\circ^n a = a^n$ be the $n$th power of $a$.
Then:
:$\forall m, n \in \N: a^{n m} = \paren {a^n}^m = \paren {a^m}^n$ | Because $\struct {S, \circ}$ is a monoid, it is a fortiori a semigroup.
Hence, from Index Laws for Semigroup: Product of Indices:
:$\forall m, n \in \N_{>0}: \circ^{n m} a = \circ^m \paren {\circ^n a} = \circ^n \paren {\circ^m a}$
That is:
:$\forall m, n \in \N_{>0}: a^{n m} = \paren {a^n}^m = \paren {a^m}^n$
It remain... | Let $\struct {S, \circ}$ be a [[Definition:Monoid|monoid]] whose [[Definition:Identity Element|identity element]] is $e$.
For $a \in S$, let $\circ^n a = a^n$ be the [[Definition:Power of Element of Magma with Identity|$n$th power of $a$]].
Then:
:$\forall m, n \in \N: a^{n m} = \paren {a^n}^m = \paren {a^m}^n$ | Because $\struct {S, \circ}$ is a [[Definition:Monoid|monoid]], it is [[Definition:A Fortiori|a fortiori]] a [[Definition:Semigroup|semigroup]].
Hence, from [[Index Laws for Semigroup/Product of Indices|Index Laws for Semigroup: Product of Indices]]:
:$\forall m, n \in \N_{>0}: \circ^{n m} a = \circ^m \paren {\circ^n... | Index Laws/Product of Indices/Monoid | https://proofwiki.org/wiki/Index_Laws/Product_of_Indices/Monoid | https://proofwiki.org/wiki/Index_Laws/Product_of_Indices/Monoid | [
"Index Laws",
"Monoids"
] | [
"Definition:Monoid",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Power of Element/Magma with Identity"
] | [
"Definition:Monoid",
"Definition:A Fortiori",
"Definition:Semigroup",
"Index Laws/Product of Indices/Semigroup",
"Zero Element of Multiplication on Numbers",
"Zero Element of Multiplication on Numbers"
] |
proofwiki-10425 | Cardinality of Proper Subset of Finite Set | Let $A$ and $B$ be finite sets such that $A \subsetneqq B$.
Let $\card B = n$, where $\card {\, \cdot \,}$ denotes cardinality.
Then $\card A < n$. | The proof proceeds by the Principle of Mathematical Induction on $n$, the cardinality of $B$.
Let $S$ be the set of $n \in \N$ such that every proper subset of any set with $m$ elements is finite and has (strictly) fewer than $n$ elements. | Let $A$ and $B$ be [[Definition:Finite Set|finite sets]] such that $A \subsetneqq B$.
Let $\card B = n$, where $\card {\, \cdot \,}$ denotes [[Definition:Cardinality|cardinality]].
Then $\card A < n$. | The proof proceeds by the [[Principle of Mathematical Induction]] on $n$, the [[Definition:Cardinality|cardinality]] of $B$.
Let $S$ be the set of $n \in \N$ such that every [[Definition:Proper Subset|proper subset]] of any [[Definition:Set|set]] with $m$ [[Definition:Element|elements]] is [[Definition:Finite Set|fin... | Cardinality of Proper Subset of Finite Set | https://proofwiki.org/wiki/Cardinality_of_Proper_Subset_of_Finite_Set | https://proofwiki.org/wiki/Cardinality_of_Proper_Subset_of_Finite_Set | [
"Proper Subsets",
"Cardinality",
"Proofs by Induction"
] | [
"Definition:Finite Set",
"Definition:Cardinality"
] | [
"Principle of Mathematical Induction",
"Definition:Cardinality",
"Definition:Proper Subset",
"Definition:Set",
"Definition:Element",
"Definition:Finite Set",
"Definition:Element",
"Definition:Proper Subset",
"Definition:Set",
"Definition:Element",
"Definition:Finite Set",
"Definition:Element",... |
proofwiki-10426 | First-Countable Space is Hausdorff iff All Convergent Sequences have Unique Limit | Let $T = \struct {S, \tau}$ be a first-countable topological space.
Then $T$ is Hausdorff {{iff}} all convergent sequences on $T$ have a unique limit. | === Sufficient Condition ===
This is shown in Convergent Sequence in Hausdorff Space has Unique Limit.
Note that it does not require first-countability.
{{qed|lemma}} | Let $T = \struct {S, \tau}$ be a [[Definition:First-Countable Space|first-countable]] [[Definition:Topological Space|topological space]].
Then $T$ is [[Definition:Hausdorff Space|Hausdorff]] {{iff}} all [[Definition:Convergent Sequence|convergent sequences]] on $T$ have a [[Definition:Unique|unique]] [[Definition:Limi... | === Sufficient Condition ===
This is shown in [[Convergent Sequence in Hausdorff Space has Unique Limit]].
Note that it does not require [[Definition:First-Countable Space|first-countability]].
{{qed|lemma}} | First-Countable Space is Hausdorff iff All Convergent Sequences have Unique Limit | https://proofwiki.org/wiki/First-Countable_Space_is_Hausdorff_iff_All_Convergent_Sequences_have_Unique_Limit | https://proofwiki.org/wiki/First-Countable_Space_is_Hausdorff_iff_All_Convergent_Sequences_have_Unique_Limit | [
"First-Countable Spaces",
"Hausdorff Spaces"
] | [
"Definition:First-Countable Space",
"Definition:Topological Space",
"Definition:T2 Space",
"Definition:Convergent Sequence",
"Definition:Unique",
"Definition:Limit of Sequence/Topological Space"
] | [
"Convergent Sequence in T2 Space has Unique Limit",
"Definition:First-Countable Space",
"Definition:First-Countable Space"
] |
proofwiki-10427 | Topologically Distinguishable Points are Distinct | Let $T = \struct {X, \tau}$ be a topological space.
Let $x, y \in X$ be topologically distinguishable.
Then the singleton sets $\set x$ and $\set y$ are disjoint and so:
:$x \ne y$ | Let $x$ and $y$ be '''topologically distinguishable'''.
Then either:
:$\exists U \in \tau: x \in U \subseteq N_x \subseteq X: y \notin N_x$
or:
:$\exists V \in \tau: y \in V \subseteq N_y \subseteq X: x \notin N_y$
{{AimForCont}} $x = y$.
Then:
:$\map \neg {\exists U \in \tau: x \in U \subseteq N_x \subseteq X: y \noti... | Let $T = \struct {X, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $x, y \in X$ be [[Definition:Topologically Distinguishable|topologically distinguishable]].
Then the [[Definition:Singleton|singleton sets]] $\set x$ and $\set y$ are [[Definition:Disjoint Sets|disjoint]] and so:
:$x \ne y$ | Let $x$ and $y$ be '''topologically distinguishable'''.
Then either:
:$\exists U \in \tau: x \in U \subseteq N_x \subseteq X: y \notin N_x$
or:
:$\exists V \in \tau: y \in V \subseteq N_y \subseteq X: x \notin N_y$
{{AimForCont}} $x = y$.
Then:
:$\map \neg {\exists U \in \tau: x \in U \subseteq N_x \subseteq X: y \... | Topologically Distinguishable Points are Distinct | https://proofwiki.org/wiki/Topologically_Distinguishable_Points_are_Distinct | https://proofwiki.org/wiki/Topologically_Distinguishable_Points_are_Distinct | [
"Topological Distinguishability"
] | [
"Definition:Topological Space",
"Definition:Topologically Distinguishable",
"Definition:Singleton",
"Definition:Disjoint Sets"
] | [
"Proof by Contradiction",
"Singleton Equality"
] |
proofwiki-10428 | Restriction of Strict Total Ordering is Strict Total Ordering | Let $\struct {S, \prec}$ be a strict total ordering.
Let $T \subseteq S$.
Let $\prec \restriction_T$ be the restriction of $\prec$ to $T$.
Then $\prec \restriction_T$ is a strict total ordering of $T$. | By definition of strict total ordering, $\prec$ is:
:$(1): \quad$ a relation which is transitive and antireflexive
:$(2): \quad$ a relation which is connected.
By Reflexive Reduction of Transitive Antisymmetric Relation is Strict Ordering:
:$\prec \restriction_T$ is a strict ordering.
It follows from Restriction of Co... | Let $\struct {S, \prec}$ be a [[Definition:Strict Total Ordering|strict total ordering]].
Let $T \subseteq S$.
Let $\prec \restriction_T$ be the [[Definition:Restriction of Ordering|restriction]] of $\prec$ to $T$.
Then $\prec \restriction_T$ is a [[Definition:Strict Total Ordering|strict total ordering]] of $T$. | By definition of [[Definition:Strict Total Ordering|strict total ordering]], $\prec$ is:
:$(1): \quad$ a [[Definition:Relation|relation]] which is [[Definition:Transitive|transitive]] and [[Definition:Antireflexive Relation|antireflexive]]
:$(2): \quad$ a [[Definition:Relation|relation]] which is [[Definition:Connecte... | Restriction of Strict Total Ordering is Strict Total Ordering | https://proofwiki.org/wiki/Restriction_of_Strict_Total_Ordering_is_Strict_Total_Ordering | https://proofwiki.org/wiki/Restriction_of_Strict_Total_Ordering_is_Strict_Total_Ordering | [
"Total Orderings",
"Strict Orderings"
] | [
"Definition:Strict Total Ordering",
"Definition:Restriction of Ordering",
"Definition:Strict Total Ordering"
] | [
"Definition:Strict Total Ordering",
"Definition:Relation",
"Definition:Transitive",
"Definition:Antireflexive Relation",
"Definition:Relation",
"Definition:Connected Relation",
"Reflexive Reduction of Transitive Antisymmetric Relation is Strict Ordering",
"Definition:Strict Ordering",
"Restriction o... |
proofwiki-10429 | Principle of Recursive Definition | Let $\N$ be the natural numbers.
Let $T$ be a class (which may be a set).
Let $a \in T$.
Let $g: T \to T$ be a mapping.
Then there exists exactly one mapping $f: \N \to T$ such that:
:<nowiki>$\forall x \in \N: \map f x = \begin{cases}
a & : x = 0 \\
\map g {\map f n} & : x = n + 1
\end{cases}$</nowiki> | Consider $\N$, defined as a naturally ordered semigroup $\struct {S, \circ, \preceq}$.
Let the mapping $f$ be defined as:
:<nowiki>$\map f x = \begin{cases}
a & : x = 0 \\
\map s {\map f n} & : x = n \circ 1 \end{cases}$</nowiki>
if $\map f n$ is defined.
Let $S' = \set {n \in S: \map f n \text{ is defined} }$.
Then:
:... | Let $\N$ be the [[Definition:Natural Numbers|natural numbers]].
Let $T$ be a [[Definition:Class (Class Theory)|class]] (which may be a [[Definition:Set|set]]).
Let $a \in T$.
Let $g: T \to T$ be a [[Definition:Mapping|mapping]].
Then there exists [[Definition:Exactly One|exactly one]] [[Definition:Mapping|mapping]... | Consider $\N$, defined as a [[Definition:Naturally Ordered Semigroup|naturally ordered semigroup]] $\struct {S, \circ, \preceq}$.
Let the [[Definition:Mapping|mapping]] $f$ be defined as:
:<nowiki>$\map f x = \begin{cases}
a & : x = 0 \\
\map s {\map f n} & : x = n \circ 1 \end{cases}$</nowiki>
if $\map f n$ is defi... | Principle of Recursive Definition/Fallacious Proof | https://proofwiki.org/wiki/Principle_of_Recursive_Definition | https://proofwiki.org/wiki/Principle_of_Recursive_Definition/Fallacious_Proof | [
"Named Theorems",
"Mapping Theory",
"Natural Numbers",
"Recursive Definitions",
"Principle of Recursive Definition"
] | [
"Definition:Natural Numbers",
"Definition:Class (Class Theory)",
"Definition:Set",
"Definition:Mapping",
"Definition:Unique",
"Definition:Mapping"
] | [
"Definition:Naturally Ordered Semigroup",
"Definition:Mapping",
"Principle of Mathematical Induction/Naturally Ordered Semigroup",
"Definition:Domain",
"Definition:Mapping",
"Definition:Set",
"Definition:Logic",
"Definition:Set Theory",
"Definition:Mapping",
"Definition:Mapping",
"Definition:Set... |
proofwiki-10430 | Principle of Recursive Definition | Let $\N$ be the natural numbers.
Let $T$ be a class (which may be a set).
Let $a \in T$.
Let $g: T \to T$ be a mapping.
Then there exists exactly one mapping $f: \N \to T$ such that:
:<nowiki>$\forall x \in \N: \map f x = \begin{cases}
a & : x = 0 \\
\map g {\map f n} & : x = n + 1
\end{cases}$</nowiki> | Consider $\N$ defined as a Peano structure $\struct {P, 0, s}$.
The result follows from Principle of Recursive Definition for Peano Structure.
{{qed}} | Let $\N$ be the [[Definition:Natural Numbers|natural numbers]].
Let $T$ be a [[Definition:Class (Class Theory)|class]] (which may be a [[Definition:Set|set]]).
Let $a \in T$.
Let $g: T \to T$ be a [[Definition:Mapping|mapping]].
Then there exists [[Definition:Exactly One|exactly one]] [[Definition:Mapping|mapping]... | Consider $\N$ defined as a [[Definition:Peano Structure|Peano structure]] $\struct {P, 0, s}$.
The result follows from [[Principle of Recursive Definition for Peano Structure]].
{{qed}} | Principle of Recursive Definition/Proof 1 | https://proofwiki.org/wiki/Principle_of_Recursive_Definition | https://proofwiki.org/wiki/Principle_of_Recursive_Definition/Proof_1 | [
"Named Theorems",
"Mapping Theory",
"Natural Numbers",
"Recursive Definitions",
"Principle of Recursive Definition"
] | [
"Definition:Natural Numbers",
"Definition:Class (Class Theory)",
"Definition:Set",
"Definition:Mapping",
"Definition:Unique",
"Definition:Mapping"
] | [
"Definition:Peano Structure",
"Principle of Recursive Definition for Peano Structure"
] |
proofwiki-10431 | Principle of Recursive Definition | Let $\N$ be the natural numbers.
Let $T$ be a class (which may be a set).
Let $a \in T$.
Let $g: T \to T$ be a mapping.
Then there exists exactly one mapping $f: \N \to T$ such that:
:<nowiki>$\forall x \in \N: \map f x = \begin{cases}
a & : x = 0 \\
\map g {\map f n} & : x = n + 1
\end{cases}$</nowiki> | Consider $\N$ defined as the von Neumann construction of the natural numbers.
The result follows from Principle of Recursive Definition for Minimally Inductive Set.
{{qed}} | Let $\N$ be the [[Definition:Natural Numbers|natural numbers]].
Let $T$ be a [[Definition:Class (Class Theory)|class]] (which may be a [[Definition:Set|set]]).
Let $a \in T$.
Let $g: T \to T$ be a [[Definition:Mapping|mapping]].
Then there exists [[Definition:Exactly One|exactly one]] [[Definition:Mapping|mapping]... | Consider $\N$ defined as the [[Definition:Von Neumann Construction of Natural Numbers|von Neumann construction of the natural numbers]].
The result follows from [[Principle of Recursive Definition for Minimally Inductive Set]].
{{qed}} | Principle of Recursive Definition/Proof 2 | https://proofwiki.org/wiki/Principle_of_Recursive_Definition | https://proofwiki.org/wiki/Principle_of_Recursive_Definition/Proof_2 | [
"Named Theorems",
"Mapping Theory",
"Natural Numbers",
"Recursive Definitions",
"Principle of Recursive Definition"
] | [
"Definition:Natural Numbers",
"Definition:Class (Class Theory)",
"Definition:Set",
"Definition:Mapping",
"Definition:Unique",
"Definition:Mapping"
] | [
"Definition:Natural Numbers/Von Neumann Construction",
"Principle of Recursive Definition for Minimally Inductive Set"
] |
proofwiki-10432 | Principle of Recursive Definition | Let $\N$ be the natural numbers.
Let $T$ be a class (which may be a set).
Let $a \in T$.
Let $g: T \to T$ be a mapping.
Then there exists exactly one mapping $f: \N \to T$ such that:
:<nowiki>$\forall x \in \N: \map f x = \begin{cases}
a & : x = 0 \\
\map g {\map f n} & : x = n + 1
\end{cases}$</nowiki> | Recall the general result:
{{:Principle of Recursive Definition/General Result}}
The result follows from setting $p = 0$.
{{qed}} | Let $\N$ be the [[Definition:Natural Numbers|natural numbers]].
Let $T$ be a [[Definition:Class (Class Theory)|class]] (which may be a [[Definition:Set|set]]).
Let $a \in T$.
Let $g: T \to T$ be a [[Definition:Mapping|mapping]].
Then there exists [[Definition:Exactly One|exactly one]] [[Definition:Mapping|mapping]... | Recall the [[Principle of Recursive Definition/General Result|general result]]:
{{:Principle of Recursive Definition/General Result}}
The result follows from setting $p = 0$.
{{qed}} | Principle of Recursive Definition/Proof 3 | https://proofwiki.org/wiki/Principle_of_Recursive_Definition | https://proofwiki.org/wiki/Principle_of_Recursive_Definition/Proof_3 | [
"Named Theorems",
"Mapping Theory",
"Natural Numbers",
"Recursive Definitions",
"Principle of Recursive Definition"
] | [
"Definition:Natural Numbers",
"Definition:Class (Class Theory)",
"Definition:Set",
"Definition:Mapping",
"Definition:Unique",
"Definition:Mapping"
] | [
"Principle of Recursive Definition/General Result"
] |
proofwiki-10433 | Principle of Recursive Definition | Let $\N$ be the natural numbers.
Let $T$ be a class (which may be a set).
Let $a \in T$.
Let $g: T \to T$ be a mapping.
Then there exists exactly one mapping $f: \N \to T$ such that:
:<nowiki>$\forall x \in \N: \map f x = \begin{cases}
a & : x = 0 \\
\map g {\map f n} & : x = n + 1
\end{cases}$</nowiki> | From the Principle of Recursive Definition: Strong Version:
{{:Principle of Recursive Definition/Strong Version}}
Let $h: A \to A$ be defined as:
:$\forall x \in A: \map h x := \map g {a, x}$ for arbitrary $a \in \omega$
That is:
:$\forall y \in \omega: \map g {y, x} = \map h x$
Then {{apriori}} there exists exactly on... | Let $\N$ be the [[Definition:Natural Numbers|natural numbers]].
Let $T$ be a [[Definition:Class (Class Theory)|class]] (which may be a [[Definition:Set|set]]).
Let $a \in T$.
Let $g: T \to T$ be a [[Definition:Mapping|mapping]].
Then there exists [[Definition:Exactly One|exactly one]] [[Definition:Mapping|mapping]... | From the [[Principle of Recursive Definition/Strong Version|Principle of Recursive Definition: Strong Version]]:
{{:Principle of Recursive Definition/Strong Version}}
Let $h: A \to A$ be defined as:
:$\forall x \in A: \map h x := \map g {a, x}$ for arbitrary $a \in \omega$
That is:
:$\forall y \in \omega: \map g {y, ... | Principle of Recursive Definition/Proof 4 | https://proofwiki.org/wiki/Principle_of_Recursive_Definition | https://proofwiki.org/wiki/Principle_of_Recursive_Definition/Proof_4 | [
"Named Theorems",
"Mapping Theory",
"Natural Numbers",
"Recursive Definitions",
"Principle of Recursive Definition"
] | [
"Definition:Natural Numbers",
"Definition:Class (Class Theory)",
"Definition:Set",
"Definition:Mapping",
"Definition:Unique",
"Definition:Mapping"
] | [
"Principle of Recursive Definition/Strong Version",
"Definition:Unique",
"Definition:Mapping"
] |
proofwiki-10434 | Principle of Recursive Definition/General Result | Let $p \in \N$.
Let $p^\ge$ be the upper closure of $p$ in $\N$:
:$p^\ge := \set {x \in \N: x \ge p} = \set {p, p + 1, p + 2, \ldots}$
Then there exists exactly one mapping $f: p^\ge \to T$ such that:
:<nowiki>$\forall x \in p^\ge: \map f x = \begin{cases}
a & : x = p \\
\map g {\map f n} & : x = n + 1
\end{cases}$</no... | Consider $\N$, defined as a naturally ordered semigroup $\struct {S, \circ, \preceq}$.
For simplicity, let $S' = p^\ge$.
First an '''admissible mapping''' is defined.
Let $n \in S'$.
The mapping $h: \closedint p n \to T$ is defined as an '''admissible mapping''' for $n$ {{iff}}:
:<nowiki>$\forall r \in \closedint p n: ... | Let $p \in \N$.
Let $p^\ge$ be the [[Definition:Upper Closure of Element|upper closure]] of $p$ in $\N$:
:$p^\ge := \set {x \in \N: x \ge p} = \set {p, p + 1, p + 2, \ldots}$
Then there exists [[Definition:Exactly One|exactly one]] [[Definition:Mapping|mapping]] $f: p^\ge \to T$ such that:
:<nowiki>$\forall x \in p... | Consider $\N$, defined as a [[Definition:Naturally Ordered Semigroup|naturally ordered semigroup]] $\struct {S, \circ, \preceq}$.
For simplicity, let $S' = p^\ge$.
First an '''admissible mapping''' is defined.
Let $n \in S'$.
The [[Definition:Mapping|mapping]] $h: \closedint p n \to T$ is defined as an '''admissib... | Principle of Recursive Definition/General Result | https://proofwiki.org/wiki/Principle_of_Recursive_Definition/General_Result | https://proofwiki.org/wiki/Principle_of_Recursive_Definition/General_Result | [
"Mapping Theory",
"Natural Numbers",
"Principle of Recursive Definition"
] | [
"Definition:Upper Closure/Element",
"Definition:Unique",
"Definition:Mapping"
] | [
"Definition:Naturally Ordered Semigroup",
"Definition:Mapping",
"Definition:Closed Interval/Integer Interval",
"Principle of Mathematical Induction/Naturally Ordered Semigroup",
"Definition:Closed Interval/Integer Interval",
"Definition:Mapping",
"Definition:Mapping",
"Principle of Mathematical Induct... |
proofwiki-10435 | Integers are not Densely Ordered | The integers $\Z$ are not densely ordered.
That is:
:$\forall n \in \Z: \not \exists m \in \Z: n < m < n + 1$ | By definition of immediate successor element, this is equivalent to the statement:
:$\forall n \in \Z: n + 1$ is the immediate successor to $n$
We have that Integers form Ordered Integral Domain.
From One Succeeds Zero in Well-Ordered Integral Domain:
:$\not \exists r \in \Z: 0 < r < 1$
From Properties of Ordered Ring:... | The [[Definition:Integer|integers]] $\Z$ are not [[Definition:Densely Ordered|densely ordered]].
That is:
:$\forall n \in \Z: \not \exists m \in \Z: n < m < n + 1$ | By definition of [[Definition:Immediate Successor Element|immediate successor element]], this is equivalent to the statement:
:$\forall n \in \Z: n + 1$ is the [[Definition:Immediate Successor Element|immediate successor]] to $n$
We have that [[Integers form Ordered Integral Domain]].
From [[One Succeeds Zero in Well... | Integers are not Densely Ordered | https://proofwiki.org/wiki/Integers_are_not_Densely_Ordered | https://proofwiki.org/wiki/Integers_are_not_Densely_Ordered | [
"Integers",
"Densely Ordered"
] | [
"Definition:Integer",
"Definition:Densely Ordered"
] | [
"Definition:Immediate Successor Element",
"Definition:Immediate Successor Element",
"Integers form Ordered Integral Domain",
"One Succeeds Zero in Well-Ordered Integral Domain",
"Properties of Ordered Ring"
] |
proofwiki-10436 | Real Zero is Zero Element | :$\forall x \in \R: 0 \times x = 0$ | {{begin-eqn}}
{{eqn | l = 0 \times x
| r = \paren {0 + 0} \times x
| c = {{Real-number-axiom|A3}}
}}
{{eqn | r = 0 \times x + 0 \times x
| c = {{Real-number-axiom|D}}
}}
{{eqn | ll= \leadsto
| l = 0 \times x
| r = 0
| c = Real Addition Identity is Zero: Corollary
}}
{{end-eqn}}
{{qed... | :$\forall x \in \R: 0 \times x = 0$ | {{begin-eqn}}
{{eqn | l = 0 \times x
| r = \paren {0 + 0} \times x
| c = {{Real-number-axiom|A3}}
}}
{{eqn | r = 0 \times x + 0 \times x
| c = {{Real-number-axiom|D}}
}}
{{eqn | ll= \leadsto
| l = 0 \times x
| r = 0
| c = [[Real Addition Identity is Zero/Corollary|Real Addition Ident... | Real Zero is Zero Element | https://proofwiki.org/wiki/Real_Zero_is_Zero_Element | https://proofwiki.org/wiki/Real_Zero_is_Zero_Element | [
"Real Numbers",
"Zero"
] | [] | [
"Real Addition Identity is Zero/Corollary"
] |
proofwiki-10437 | Negative of Real Zero equals Zero | Let $0$ denote the identity for addition in the real numbers $\R$.
Then:
:$-0 = 0$ | {{begin-eqn}}
{{eqn | l = -0 + 0
| r = 0
| c = {{Real-number-axiom|A4}}
}}
{{eqn | ll= \leadsto
| l = -0
| r = 0
| c = Real Addition Identity is Zero: Corollary
}}
{{end-eqn}}
{{qed}} | Let $0$ denote the [[Real Addition Identity is Zero|identity for addition]] in the [[Definition:Real Number|real numbers]] $\R$.
Then:
:$-0 = 0$ | {{begin-eqn}}
{{eqn | l = -0 + 0
| r = 0
| c = {{Real-number-axiom|A4}}
}}
{{eqn | ll= \leadsto
| l = -0
| r = 0
| c = [[Real Addition Identity is Zero/Corollary|Real Addition Identity is Zero: Corollary]]
}}
{{end-eqn}}
{{qed}} | Negative of Real Zero equals Zero | https://proofwiki.org/wiki/Negative_of_Real_Zero_equals_Zero | https://proofwiki.org/wiki/Negative_of_Real_Zero_equals_Zero | [
"Real Numbers"
] | [
"Real Addition Identity is Zero",
"Definition:Real Number"
] | [
"Real Addition Identity is Zero/Corollary"
] |
proofwiki-10438 | Negative of Negative Real Number | :$\forall x \in \R: -\paren {-x} = x$ | {{begin-eqn}}
{{eqn | l = 0
| r = \paren {-x} + x
| c = {{Real-number-axiom|A4}}
}}
{{eqn | ll= \leadsto
| l = -\paren {-x} + 0
| r = -\paren {-x} + \paren {-x} + x
| c = adding $-\paren {-x}$ to both sides
}}
{{eqn | ll= \leadsto
| l = -\paren {-x} + 0
| r = \paren {-\paren {-... | :$\forall x \in \R: -\paren {-x} = x$ | {{begin-eqn}}
{{eqn | l = 0
| r = \paren {-x} + x
| c = {{Real-number-axiom|A4}}
}}
{{eqn | ll= \leadsto
| l = -\paren {-x} + 0
| r = -\paren {-x} + \paren {-x} + x
| c = adding $-\paren {-x}$ to both sides
}}
{{eqn | ll= \leadsto
| l = -\paren {-x} + 0
| r = \paren {-\paren {-... | Negative of Negative Real Number | https://proofwiki.org/wiki/Negative_of_Negative_Real_Number | https://proofwiki.org/wiki/Negative_of_Negative_Real_Number | [
"Real Numbers"
] | [] | [] |
proofwiki-10439 | Multiplication by Negative Real Number | :$\forall x, y \in \R: x \times \paren {-y} = -\paren {x \times y} = \paren {-x} \times y$ | {{begin-eqn}}
{{eqn | l = x \times \paren {\paren {-y} + y}
| r = x \times 0
| c = {{Real-number-axiom|A4}}
}}
{{eqn | r = 0
| c = Real Zero is Zero Element
}}
{{eqn | ll= \leadsto
| l = \paren {x \times \paren {-y} } + \paren {x \times y}
| r = 0
| c = {{Real-number-axiom|D}}
}}
{{e... | :$\forall x, y \in \R: x \times \paren {-y} = -\paren {x \times y} = \paren {-x} \times y$ | {{begin-eqn}}
{{eqn | l = x \times \paren {\paren {-y} + y}
| r = x \times 0
| c = {{Real-number-axiom|A4}}
}}
{{eqn | r = 0
| c = [[Real Zero is Zero Element]]
}}
{{eqn | ll= \leadsto
| l = \paren {x \times \paren {-y} } + \paren {x \times y}
| r = 0
| c = {{Real-number-axiom|D}}
}}... | Multiplication by Negative Real Number | https://proofwiki.org/wiki/Multiplication_by_Negative_Real_Number | https://proofwiki.org/wiki/Multiplication_by_Negative_Real_Number | [
"Real Multiplication"
] | [] | [
"Real Zero is Zero Element",
"Real Zero is Zero Element"
] |
proofwiki-10440 | Negative of Sum of Real Numbers | :$\forall x, y \in \R: -\paren {x + y} = -x - y$ | {{begin-eqn}}
{{eqn | l = -\paren {x + y}
| r = \paren {-1} \times \paren {x + y}
| c = {{Corollary|Multiplication by Negative Real Number}}
}}
{{eqn | r = \paren {\paren {-1} \times x} + \paren {\paren {-1} \times y}
| c = {{Real-number-axiom|D}}
}}
{{eqn | r = \paren {-x} + \paren {-y}
| c = {... | :$\forall x, y \in \R: -\paren {x + y} = -x - y$ | {{begin-eqn}}
{{eqn | l = -\paren {x + y}
| r = \paren {-1} \times \paren {x + y}
| c = {{Corollary|Multiplication by Negative Real Number}}
}}
{{eqn | r = \paren {\paren {-1} \times x} + \paren {\paren {-1} \times y}
| c = {{Real-number-axiom|D}}
}}
{{eqn | r = \paren {-x} + \paren {-y}
| c = {... | Negative of Sum of Real Numbers | https://proofwiki.org/wiki/Negative_of_Sum_of_Real_Numbers | https://proofwiki.org/wiki/Negative_of_Sum_of_Real_Numbers | [
"Real Addition",
"Real Subtraction"
] | [] | [] |
proofwiki-10441 | Negative of Sum of Real Numbers/Corollary | :$\forall x, y \in \R: -\paren {x - y} = -x + y$ | {{begin-eqn}}
{{eqn | l = -\paren {x - y}
| r = -\paren {x + \paren {-y} }
| c = {{Defof|Real Subtraction}}
}}
{{eqn | r = -x - \paren {-y}
| c = Negative of Sum of Real Numbers
}}
{{eqn | r = -x + \paren {-\paren {-y} }
| c = {{Defof|Real Subtraction}}
}}
{{eqn | r = -x + y
| c = Negative... | :$\forall x, y \in \R: -\paren {x - y} = -x + y$ | {{begin-eqn}}
{{eqn | l = -\paren {x - y}
| r = -\paren {x + \paren {-y} }
| c = {{Defof|Real Subtraction}}
}}
{{eqn | r = -x - \paren {-y}
| c = [[Negative of Sum of Real Numbers]]
}}
{{eqn | r = -x + \paren {-\paren {-y} }
| c = {{Defof|Real Subtraction}}
}}
{{eqn | r = -x + y
| c = [[Ne... | Negative of Sum of Real Numbers/Corollary | https://proofwiki.org/wiki/Negative_of_Sum_of_Real_Numbers/Corollary | https://proofwiki.org/wiki/Negative_of_Sum_of_Real_Numbers/Corollary | [
"Real Addition",
"Real Subtraction"
] | [] | [
"Negative of Sum of Real Numbers",
"Negative of Negative Real Number"
] |
proofwiki-10442 | Real Multiplication Identity is One/Corollary | :$\forall x \in \R_{\ne 0}: x \times y = x \implies y = 1$ | {{begin-eqn}}
{{eqn | l = x \times y
| r = x
| c =
}}
{{eqn | ll= \leadsto
| l = \frac 1 x \times \paren {x \times y}
| r = \frac 1 x \times x
| c = as long as $x \ne 0$
}}
{{eqn | ll= \leadsto
| l = \paren {\frac 1 x \times x} \times y
| r = \frac 1 x \times x
| c = {{R... | :$\forall x \in \R_{\ne 0}: x \times y = x \implies y = 1$ | {{begin-eqn}}
{{eqn | l = x \times y
| r = x
| c =
}}
{{eqn | ll= \leadsto
| l = \frac 1 x \times \paren {x \times y}
| r = \frac 1 x \times x
| c = as long as $x \ne 0$
}}
{{eqn | ll= \leadsto
| l = \paren {\frac 1 x \times x} \times y
| r = \frac 1 x \times x
| c = {{R... | Real Multiplication Identity is One/Corollary | https://proofwiki.org/wiki/Real_Multiplication_Identity_is_One/Corollary | https://proofwiki.org/wiki/Real_Multiplication_Identity_is_One/Corollary | [
"Real Multiplication"
] | [] | [
"Real Multiplication Identity is One"
] |
proofwiki-10443 | Real Number Divided by Itself | :$\forall x \in \R_{\ne 0}: \dfrac x x = 1$ | {{begin-eqn}}
{{eqn | q = \forall x \ne 0
| l = \frac x x
| r = x \times \frac 1 x
| c = {{Defof|Real Division}}
}}
{{eqn | r = 1
| c = {{Real-number-axiom|M4}}
}}
{{end-eqn}}
{{qed}} | :$\forall x \in \R_{\ne 0}: \dfrac x x = 1$ | {{begin-eqn}}
{{eqn | q = \forall x \ne 0
| l = \frac x x
| r = x \times \frac 1 x
| c = {{Defof|Real Division}}
}}
{{eqn | r = 1
| c = {{Real-number-axiom|M4}}
}}
{{end-eqn}}
{{qed}} | Real Number Divided by Itself | https://proofwiki.org/wiki/Real_Number_Divided_by_Itself | https://proofwiki.org/wiki/Real_Number_Divided_by_Itself | [
"Real Division"
] | [] | [] |
proofwiki-10444 | Real Division by One | :$\forall x \in \R: \dfrac x 1 = x$ | {{begin-eqn}}
{{eqn | l = \frac x 1
| r = x \times \frac 1 1
| c = {{Defof|Real Division}}
}}
{{eqn | r = x \times 1
| c = Real Number Divided by Itself
}}
{{eqn | r = x
| c = {{Real-number-axiom|M4}}
}}
{{end-eqn}}
{{qed}} | :$\forall x \in \R: \dfrac x 1 = x$ | {{begin-eqn}}
{{eqn | l = \frac x 1
| r = x \times \frac 1 1
| c = {{Defof|Real Division}}
}}
{{eqn | r = x \times 1
| c = [[Real Number Divided by Itself]]
}}
{{eqn | r = x
| c = {{Real-number-axiom|M4}}
}}
{{end-eqn}}
{{qed}} | Real Division by One | https://proofwiki.org/wiki/Real_Division_by_One | https://proofwiki.org/wiki/Real_Division_by_One | [
"Real Division"
] | [] | [
"Real Number Divided by Itself"
] |
proofwiki-10445 | Product of Reciprocals of Real Numbers | :$\forall x, y \in \R_{\ne 0}: \dfrac 1 x \times \dfrac 1 y = \dfrac 1 {x \times y}$ | {{begin-eqn}}
{{eqn | l = \frac 1 {x \times y} \times \paren {x \times y}
| r = 1
| c = {{Real-number-axiom|M4}}
}}
{{eqn | ll= \leadsto
| l = \frac 1 {x \times y} \times \paren {x \times y} \times \frac 1 y
| r = 1 \times \frac 1 y
| c = as $y \ne 0$
}}
{{eqn | ll= \leadsto
| l = \p... | :$\forall x, y \in \R_{\ne 0}: \dfrac 1 x \times \dfrac 1 y = \dfrac 1 {x \times y}$ | {{begin-eqn}}
{{eqn | l = \frac 1 {x \times y} \times \paren {x \times y}
| r = 1
| c = {{Real-number-axiom|M4}}
}}
{{eqn | ll= \leadsto
| l = \frac 1 {x \times y} \times \paren {x \times y} \times \frac 1 y
| r = 1 \times \frac 1 y
| c = as $y \ne 0$
}}
{{eqn | ll= \leadsto
| l = \p... | Product of Reciprocals of Real Numbers | https://proofwiki.org/wiki/Product_of_Reciprocals_of_Real_Numbers | https://proofwiki.org/wiki/Product_of_Reciprocals_of_Real_Numbers | [
"Real Multiplication",
"Reciprocals"
] | [] | [] |
proofwiki-10446 | Product of Quotients of Real Numbers | :$\forall x, w \in \R, y, z \in \R_{\ne 0}: \dfrac x y \times \dfrac w z = \dfrac {x \times w} {y \times z}$ | {{begin-eqn}}
{{eqn | l = \frac x y \times \frac w z
| r = x \times \frac 1 y \times w \times \frac 1 z
| c = {{Defof|Real Division}}
}}
{{eqn | r = x \times w \times \frac 1 y \times \frac 1 z
| c = {{Real-number-axiom|M2}}
}}
{{eqn | r = \paren {x \times w} \times \paren {\frac 1 y \times \frac 1 z}... | :$\forall x, w \in \R, y, z \in \R_{\ne 0}: \dfrac x y \times \dfrac w z = \dfrac {x \times w} {y \times z}$ | {{begin-eqn}}
{{eqn | l = \frac x y \times \frac w z
| r = x \times \frac 1 y \times w \times \frac 1 z
| c = {{Defof|Real Division}}
}}
{{eqn | r = x \times w \times \frac 1 y \times \frac 1 z
| c = {{Real-number-axiom|M2}}
}}
{{eqn | r = \paren {x \times w} \times \paren {\frac 1 y \times \frac 1 z}... | Product of Quotients of Real Numbers | https://proofwiki.org/wiki/Product_of_Quotients_of_Real_Numbers | https://proofwiki.org/wiki/Product_of_Quotients_of_Real_Numbers | [
"Real Multiplication",
"Real Division"
] | [] | [
"Product of Reciprocals of Real Numbers"
] |
proofwiki-10447 | Sum of Quotients of Real Numbers | :$\forall x, w \in \R, y, z \in \R_{\ne 0}: \dfrac x y + \dfrac w z = \dfrac {\paren {x \times z} + \paren {y \times w} } {y \times z}$ | {{begin-eqn}}
{{eqn | l = \frac x y + \frac w z
| r = \paren {x \times \frac 1 y} + \paren {w \times \frac 1 z}
| c = {{Defof|Real Division}}
}}
{{eqn | r = \paren {x \times \frac 1 y \times 1} + \paren {1 \times w \times \frac 1 z}
| c = {{Real-number-axiom|M3}}
}}
{{eqn | r = \paren {x \times \frac ... | :$\forall x, w \in \R, y, z \in \R_{\ne 0}: \dfrac x y + \dfrac w z = \dfrac {\paren {x \times z} + \paren {y \times w} } {y \times z}$ | {{begin-eqn}}
{{eqn | l = \frac x y + \frac w z
| r = \paren {x \times \frac 1 y} + \paren {w \times \frac 1 z}
| c = {{Defof|Real Division}}
}}
{{eqn | r = \paren {x \times \frac 1 y \times 1} + \paren {1 \times w \times \frac 1 z}
| c = {{Real-number-axiom|M3}}
}}
{{eqn | r = \paren {x \times \frac ... | Sum of Quotients of Real Numbers | https://proofwiki.org/wiki/Sum_of_Quotients_of_Real_Numbers | https://proofwiki.org/wiki/Sum_of_Quotients_of_Real_Numbers | [
"Real Addition",
"Real Division"
] | [] | [
"Product of Reciprocals of Real Numbers"
] |
proofwiki-10448 | Reciprocal of Real Number is Non-Zero | :$\forall x \in \R: x \ne 0 \implies \dfrac 1 x \ne 0$ | {{AimForCont}} that:
:$\exists x \in \R_{\ne 0}: \dfrac 1 x = 0$
From Real Zero is Zero Element
:$\dfrac 1 x \times x = 0$
But from {{Real-number-axiom|M4}}:
:$\dfrac 1 x \times x = 1$
The result follows by Proof by Contradiction.
{{qed}} | :$\forall x \in \R: x \ne 0 \implies \dfrac 1 x \ne 0$ | {{AimForCont}} that:
:$\exists x \in \R_{\ne 0}: \dfrac 1 x = 0$
From [[Real Zero is Zero Element]]
:$\dfrac 1 x \times x = 0$
But from {{Real-number-axiom|M4}}:
:$\dfrac 1 x \times x = 1$
The result follows by [[Proof by Contradiction]].
{{qed}} | Reciprocal of Real Number is Non-Zero | https://proofwiki.org/wiki/Reciprocal_of_Real_Number_is_Non-Zero | https://proofwiki.org/wiki/Reciprocal_of_Real_Number_is_Non-Zero | [
"Real Numbers",
"Reciprocals"
] | [] | [
"Real Zero is Zero Element",
"Proof by Contradiction"
] |
proofwiki-10449 | Reciprocal of Quotient of Real Numbers | :$\forall x, y \in \R_{\ne 0}: \dfrac 1 {x / y} = \dfrac y x$ | {{begin-eqn}}
{{eqn | l = \dfrac 1 {x / y}
| r = \frac 1 {x \times \dfrac 1 y}
| c = {{Defof|Real Division}}
}}
{{eqn | r = 1 \times \frac 1 {x \times \dfrac 1 y}
| c = {{Real-number-axiom|M3}}
}}
{{eqn | r = \paren {y \times \frac 1 y} \times \frac 1 {x \times \dfrac 1 y}
| c = {{Real-number-ax... | :$\forall x, y \in \R_{\ne 0}: \dfrac 1 {x / y} = \dfrac y x$ | {{begin-eqn}}
{{eqn | l = \dfrac 1 {x / y}
| r = \frac 1 {x \times \dfrac 1 y}
| c = {{Defof|Real Division}}
}}
{{eqn | r = 1 \times \frac 1 {x \times \dfrac 1 y}
| c = {{Real-number-axiom|M3}}
}}
{{eqn | r = \paren {y \times \frac 1 y} \times \frac 1 {x \times \dfrac 1 y}
| c = {{Real-number-ax... | Reciprocal of Quotient of Real Numbers | https://proofwiki.org/wiki/Reciprocal_of_Quotient_of_Real_Numbers | https://proofwiki.org/wiki/Reciprocal_of_Quotient_of_Real_Numbers | [
"Real Division",
"Reciprocals"
] | [] | [
"Product of Reciprocals of Real Numbers"
] |
proofwiki-10450 | Quotient of Quotients of Real Numbers | :$\forall x \in \R, y, w, z \in \R_{\ne 0}: \dfrac {x / y} {w / z} = \dfrac {x \times z} {y \times w}$ | {{begin-eqn}}
{{eqn | l = \frac {x / y} {w / z}
| r = \frac x y \times \frac 1 {w / z}
| c = {{Defof|Real Division}}
}}
{{eqn | r = \frac x y \times \frac z w
| c = Reciprocal of Quotient of Real Numbers
}}
{{eqn | r = \dfrac {x \times z} {y \times w}
| c = Product of Quotients of Real Numbers
}... | :$\forall x \in \R, y, w, z \in \R_{\ne 0}: \dfrac {x / y} {w / z} = \dfrac {x \times z} {y \times w}$ | {{begin-eqn}}
{{eqn | l = \frac {x / y} {w / z}
| r = \frac x y \times \frac 1 {w / z}
| c = {{Defof|Real Division}}
}}
{{eqn | r = \frac x y \times \frac z w
| c = [[Reciprocal of Quotient of Real Numbers]]
}}
{{eqn | r = \dfrac {x \times z} {y \times w}
| c = [[Product of Quotients of Real Num... | Quotient of Quotients of Real Numbers | https://proofwiki.org/wiki/Quotient_of_Quotients_of_Real_Numbers | https://proofwiki.org/wiki/Quotient_of_Quotients_of_Real_Numbers | [
"Real Division"
] | [] | [
"Reciprocal of Quotient of Real Numbers",
"Product of Quotients of Real Numbers"
] |
proofwiki-10451 | Product of Real Number with Quotient | :$\forall a, x \in \R, y \in \R_{\ne 0}: \dfrac {a \times x} y = a \times \dfrac x y$ | {{begin-eqn}}
{{eqn | l = \frac {a \times x} y
| r = \paren {a \times x} \times \frac 1 y
| c = {{Defof|Real Division}}
}}
{{eqn | r = a \times \paren {x \times \frac 1 y}
| c = {{Real-number-axiom|M1}}
}}
{{eqn | r = a \times \frac x y
| c = {{Defof|Real Division}}
}}
{{end-eqn}}
{{qed}} | :$\forall a, x \in \R, y \in \R_{\ne 0}: \dfrac {a \times x} y = a \times \dfrac x y$ | {{begin-eqn}}
{{eqn | l = \frac {a \times x} y
| r = \paren {a \times x} \times \frac 1 y
| c = {{Defof|Real Division}}
}}
{{eqn | r = a \times \paren {x \times \frac 1 y}
| c = {{Real-number-axiom|M1}}
}}
{{eqn | r = a \times \frac x y
| c = {{Defof|Real Division}}
}}
{{end-eqn}}
{{qed}} | Product of Real Number with Quotient | https://proofwiki.org/wiki/Product_of_Real_Number_with_Quotient | https://proofwiki.org/wiki/Product_of_Real_Number_with_Quotient | [
"Real Multiplication",
"Real Division"
] | [] | [] |
proofwiki-10452 | Neighborhood in Topological Subspace | Let $\struct {X, \tau}$ be a topological space.
Let $S \subseteq X$ be a subset of $X$.
Let $\tau_S$ denote the subspace topology on $S$.
Let $x \in S$ be an arbitrary point of $S$.
Let $E \subseteq S$.
Then:
:$E$ is a neighborhood of $x$ in $\struct {S, \tau_S}$
{{iff}}:
:$\exists D \subseteq X$ such that:
::$D$ is a ... | === Necessary Condition ===
Let $E$ be a neighborhood of $x$ in $\struct {S, \tau_S}$.
By the definition of neighborhood:
:$\exists U \in \tau_S : x \in U \subseteq E$
Now, by the definition of the subspace topology:
:$\exists V \in \tau: U = V \cap S$
Take $D := V \cup E$.
We have that:
:$V \subseteq D$
and:
:$V \in ... | Let $\struct {X, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $S \subseteq X$ be a [[Definition:Subset|subset]] of $X$.
Let $\tau_S$ denote the [[Definition:Topological Subspace|subspace topology]] on $S$.
Let $x \in S$ be an arbitrary point of $S$.
Let $E \subseteq S$.
Then:
:$E$ is a [[De... | === Necessary Condition ===
Let $E$ be a [[Definition:Neighborhood of Point|neighborhood]] of $x$ in $\struct {S, \tau_S}$.
By the definition of [[Definition:Neighborhood of Point|neighborhood]]:
:$\exists U \in \tau_S : x \in U \subseteq E$
Now, by the definition of the [[Definition:Topological Subspace|subspace t... | Neighborhood in Topological Subspace | https://proofwiki.org/wiki/Neighborhood_in_Topological_Subspace | https://proofwiki.org/wiki/Neighborhood_in_Topological_Subspace | [
"Topology",
"Neighborhoods"
] | [
"Definition:Topological Space",
"Definition:Subset",
"Definition:Topological Subspace",
"Definition:Neighborhood (Topology)/Point",
"Definition:Neighborhood (Topology)/Point"
] | [
"Definition:Neighborhood (Topology)/Point",
"Definition:Neighborhood (Topology)/Point",
"Definition:Topological Subspace",
"Definition:Neighborhood (Topology)/Point",
"Definition:Neighborhood (Topology)/Point",
"Definition:Neighborhood (Topology)/Point",
"Definition:Topological Subspace",
"Definition:... |
proofwiki-10453 | Negative of Quotient of Real Numbers | :$\forall x \in \R, y \in \R_{\ne 0}: \dfrac {-x} y = -\dfrac x y = \dfrac x {-y}$ | {{begin-eqn}}
{{eqn | l = \frac {-x} y
| r = \frac {\paren {-1} \times x} y
| c = {{Corollary|Multiplication by Negative Real Number}}
}}
{{eqn | r = \paren {-1} \times \frac x y
| c = Product of Real Number with Quotient
}}
{{eqn | r = -\frac x y
| c = {{Corollary|Multiplication by Negative Rea... | :$\forall x \in \R, y \in \R_{\ne 0}: \dfrac {-x} y = -\dfrac x y = \dfrac x {-y}$ | {{begin-eqn}}
{{eqn | l = \frac {-x} y
| r = \frac {\paren {-1} \times x} y
| c = {{Corollary|Multiplication by Negative Real Number}}
}}
{{eqn | r = \paren {-1} \times \frac x y
| c = [[Product of Real Number with Quotient]]
}}
{{eqn | r = -\frac x y
| c = {{Corollary|Multiplication by Negative... | Negative of Quotient of Real Numbers | https://proofwiki.org/wiki/Negative_of_Quotient_of_Real_Numbers | https://proofwiki.org/wiki/Negative_of_Quotient_of_Real_Numbers | [
"Real Division"
] | [] | [
"Product of Real Number with Quotient",
"Real Number Divided by Itself",
"Product of Real Number with Quotient",
"Negative of Negative Real Number",
"Product of Quotients of Real Numbers"
] |
proofwiki-10454 | Compact Hausdorff Space is Locally Compact | Let $T = \struct {S, \tau}$ be a Hausdorff space which is compact.
Then $T$ is locally compact. | By Compact Space is Weakly Locally Compact, $T$ is weakly locally compact.
Thus $T$ is a locally compact Hausdorff space.
{{qed}} | Let $T = \struct {S, \tau}$ be a [[Definition:Hausdorff Space|Hausdorff space]] which is [[Definition:Compact Topological Space|compact]].
Then $T$ is [[Definition:Locally Compact Hausdorff Space|locally compact]]. | By [[Compact Space is Weakly Locally Compact]], $T$ is [[Definition:Weakly Locally Compact Space|weakly locally compact]].
Thus $T$ is a [[Definition:Locally Compact Hausdorff Space|locally compact Hausdorff space]].
{{qed}} | Compact Hausdorff Space is Locally Compact/Proof 1 | https://proofwiki.org/wiki/Compact_Hausdorff_Space_is_Locally_Compact | https://proofwiki.org/wiki/Compact_Hausdorff_Space_is_Locally_Compact/Proof_1 | [
"Compact Hausdorff Space is Locally Compact",
"Compact Topological Spaces",
"Locally Compact Hausdorff Spaces",
"Hausdorff Spaces"
] | [
"Definition:T2 Space",
"Definition:Compact Topological Space",
"Definition:Locally Compact Hausdorff Space"
] | [
"Compact Space is Weakly Locally Compact",
"Definition:Weakly Locally Compact Space",
"Definition:Locally Compact Hausdorff Space"
] |
proofwiki-10455 | Sum of Strictly Positive Real Numbers is Strictly Positive | :$x, y \in \R_{>0} \implies x + y \in \R_{>0}$ | {{begin-eqn}}
{{eqn | l = x
| o = >
| r = 0
| c =
}}
{{eqn | lo= \land
| l = y
| o = >
| r = 0
| c = Real Number Ordering is Compatible with Addition
}}
{{eqn | ll= \leadsto
| l = x + y
| o = >
| r = 0 + 0
| c = Real Number Inequalities can be Added
}}
{... | :$x, y \in \R_{>0} \implies x + y \in \R_{>0}$ | {{begin-eqn}}
{{eqn | l = x
| o = >
| r = 0
| c =
}}
{{eqn | lo= \land
| l = y
| o = >
| r = 0
| c = [[Real Number Ordering is Compatible with Addition]]
}}
{{eqn | ll= \leadsto
| l = x + y
| o = >
| r = 0 + 0
| c = [[Real Number Inequalities can be Adde... | Sum of Strictly Positive Real Numbers is Strictly Positive | https://proofwiki.org/wiki/Sum_of_Strictly_Positive_Real_Numbers_is_Strictly_Positive | https://proofwiki.org/wiki/Sum_of_Strictly_Positive_Real_Numbers_is_Strictly_Positive | [
"Real Addition"
] | [] | [
"Real Number Ordering is Compatible with Addition",
"Real Number Inequalities can be Added"
] |
proofwiki-10456 | Real Number is Greater than Zero iff its Negative is Less than Zero | :$\forall x \in \R: x > 0 \iff \paren {-x} < 0$ | Let $x > 0$.
{{begin-eqn}}
{{eqn | l = x
| o = >
| r = 0
| c =
}}
{{eqn | ll= \leadsto
| l = x + \paren {-x}
| o = >
| r = 0 + \paren {-x}
| c = {{Real-number-axiom|O2}}
}}
{{eqn | ll= \leadsto
| l = 0
| o = >
| r = 0 + \paren {-x}
| c = {{Real-number-a... | :$\forall x \in \R: x > 0 \iff \paren {-x} < 0$ | Let $x > 0$.
{{begin-eqn}}
{{eqn | l = x
| o = >
| r = 0
| c =
}}
{{eqn | ll= \leadsto
| l = x + \paren {-x}
| o = >
| r = 0 + \paren {-x}
| c = {{Real-number-axiom|O2}}
}}
{{eqn | ll= \leadsto
| l = 0
| o = >
| r = 0 + \paren {-x}
| c = {{Real-number-... | Real Number is Greater than Zero iff its Negative is Less than Zero | https://proofwiki.org/wiki/Real_Number_is_Greater_than_Zero_iff_its_Negative_is_Less_than_Zero | https://proofwiki.org/wiki/Real_Number_is_Greater_than_Zero_iff_its_Negative_is_Less_than_Zero | [
"Real Numbers"
] | [] | [] |
proofwiki-10457 | Order of Real Numbers is Dual of Order of their Negatives | :$\forall x, y \in \R: x > y \iff \paren {-x}< \paren {-y}$ | Let $x > y$.
{{begin-eqn}}
{{eqn | l = x
| o = >
| r = y
| c =
}}
{{eqn | ll= \leadsto
| l = x + \paren {-x}
| o = >
| r = y + \paren {-x}
| c = {{Real-number-axiom|O1}}
}}
{{eqn | ll= \leadsto
| l = 0
| o = >
| r = y + \paren {-x}
| c = {{Real-number-a... | :$\forall x, y \in \R: x > y \iff \paren {-x}< \paren {-y}$ | Let $x > y$.
{{begin-eqn}}
{{eqn | l = x
| o = >
| r = y
| c =
}}
{{eqn | ll= \leadsto
| l = x + \paren {-x}
| o = >
| r = y + \paren {-x}
| c = {{Real-number-axiom|O1}}
}}
{{eqn | ll= \leadsto
| l = 0
| o = >
| r = y + \paren {-x}
| c = {{Real-number-... | Order of Real Numbers is Dual of Order of their Negatives/Proof 1 | https://proofwiki.org/wiki/Order_of_Real_Numbers_is_Dual_of_Order_of_their_Negatives | https://proofwiki.org/wiki/Order_of_Real_Numbers_is_Dual_of_Order_of_their_Negatives/Proof_1 | [
"Real Numbers",
"Inequalities",
"Order of Real Numbers is Dual of Order of their Negatives"
] | [] | [] |
proofwiki-10458 | Order of Real Numbers is Dual of Order of their Negatives | :$\forall x, y \in \R: x > y \iff \paren {-x}< \paren {-y}$ | {{begin-eqn}}
{{eqn | l = x
| o = >
| r = y
}}
{{eqn | ll= \leadstoandfrom
| l = y - x
| o = <
| r = 0
| c = Inequality iff Difference is Positive
}}
{{eqn | ll= \leadstoandfrom
| lr = -x + y
| o = <
| r = 0
| c = {{Real-number-axiom|A2}}
}}
{{eqn | ll= \leads... | :$\forall x, y \in \R: x > y \iff \paren {-x}< \paren {-y}$ | {{begin-eqn}}
{{eqn | l = x
| o = >
| r = y
}}
{{eqn | ll= \leadstoandfrom
| l = y - x
| o = <
| r = 0
| c = [[Inequality iff Difference is Positive]]
}}
{{eqn | ll= \leadstoandfrom
| lr = -x + y
| o = <
| r = 0
| c = {{Real-number-axiom|A2}}
}}
{{eqn | ll= \l... | Order of Real Numbers is Dual of Order of their Negatives/Proof 2 | https://proofwiki.org/wiki/Order_of_Real_Numbers_is_Dual_of_Order_of_their_Negatives | https://proofwiki.org/wiki/Order_of_Real_Numbers_is_Dual_of_Order_of_their_Negatives/Proof_2 | [
"Real Numbers",
"Inequalities",
"Order of Real Numbers is Dual of Order of their Negatives"
] | [] | [
"Inequality iff Difference is Positive",
"Negative of Negative Real Number",
"Inequality iff Difference is Positive"
] |
proofwiki-10459 | Square of Non-Zero Real Number is Strictly Positive | :$\forall x \in \R: x \ne 0 \implies x^2 > 0$ | There are two cases to consider:
:$(1): \quad x > 0$
:$(2): \quad x < 0$
Let $x > 0$.
Then:
{{begin-eqn}}
{{eqn | l = x \times x
| o = >
| r = 0
| c = Strictly Positive Real Numbers are Closed under Multiplication
}}
{{end-eqn}}
Let $x < 0$.
Then:
{{begin-eqn}}
{{eqn | l = x
| o = <
| r = ... | :$\forall x \in \R: x \ne 0 \implies x^2 > 0$ | There are two cases to consider:
:$(1): \quad x > 0$
:$(2): \quad x < 0$
Let $x > 0$.
Then:
{{begin-eqn}}
{{eqn | l = x \times x
| o = >
| r = 0
| c = [[Strictly Positive Real Numbers are Closed under Multiplication]]
}}
{{end-eqn}}
Let $x < 0$.
Then:
{{begin-eqn}}
{{eqn | l = x
| o = <
... | Square of Non-Zero Real Number is Strictly Positive | https://proofwiki.org/wiki/Square_of_Non-Zero_Real_Number_is_Strictly_Positive | https://proofwiki.org/wiki/Square_of_Non-Zero_Real_Number_is_Strictly_Positive | [
"Square Function",
"Real Numbers"
] | [] | [
"Strictly Positive Real Numbers are Closed under Multiplication",
"Real Number Ordering is Compatible with Multiplication/Negative Factor",
"Real Zero is Zero Element"
] |
proofwiki-10460 | Minus One is Less than Zero | :$-1 < 0$ | {{begin-eqn}}
{{eqn | l = 0
| o = <
| r = 1
| c = Real Zero is Less than Real One
}}
{{eqn | ll= \leadsto
| l = -0
| o = >
| r = -1
| c = Order of Real Numbers is Dual of Order of their Negatives
}}
{{eqn | ll= \leadsto
| l = 0
| o = >
| r = -1
| c = Neg... | :$-1 < 0$ | {{begin-eqn}}
{{eqn | l = 0
| o = <
| r = 1
| c = [[Real Zero is Less than Real One]]
}}
{{eqn | ll= \leadsto
| l = -0
| o = >
| r = -1
| c = [[Order of Real Numbers is Dual of Order of their Negatives]]
}}
{{eqn | ll= \leadsto
| l = 0
| o = >
| r = -1
|... | Minus One is Less than Zero | https://proofwiki.org/wiki/Minus_One_is_Less_than_Zero | https://proofwiki.org/wiki/Minus_One_is_Less_than_Zero | [
"Real Numbers"
] | [] | [
"Real Zero is Less than Real One",
"Order of Real Numbers is Dual of Order of their Negatives",
"Negative of Real Zero equals Zero"
] |
proofwiki-10461 | Real Zero is Less than Real One | The real number $0$ is less than the real number $1$:
:$0 < 1$ | {{begin-eqn}}
{{eqn | l = 1 \times 1
| o = >
| r = 0
| c = Square of Non-Zero Real Number is Strictly Positive
}}
{{eqn | ll= \leadsto
| l = 1
| o = >
| r = 0
| c = {{Real-number-axiom|M3}}
}}
{{eqn | ll= \leadsto
| l = 0
| o = <
| r = 1
| c = {{Defof|Du... | The [[Definition:Real Number|real number]] $0$ is less than the [[Definition:Real Number|real number]] $1$:
:$0 < 1$ | {{begin-eqn}}
{{eqn | l = 1 \times 1
| o = >
| r = 0
| c = [[Square of Non-Zero Real Number is Strictly Positive]]
}}
{{eqn | ll= \leadsto
| l = 1
| o = >
| r = 0
| c = {{Real-number-axiom|M3}}
}}
{{eqn | ll= \leadsto
| l = 0
| o = <
| r = 1
| c = {{Defo... | Real Zero is Less than Real One | https://proofwiki.org/wiki/Real_Zero_is_Less_than_Real_One | https://proofwiki.org/wiki/Real_Zero_is_Less_than_Real_One | [
"Real Numbers"
] | [
"Definition:Real Number",
"Definition:Real Number"
] | [
"Square of Non-Zero Real Number is Strictly Positive"
] |
proofwiki-10462 | Product of Real Numbers is Positive iff Numbers have Same Sign | The product of two real numbers is greater than $0$ {{iff}} either both are greater than $0$ or both are less than $0$.
:$\forall x, y \in \R: x \times y > 0 \iff \paren {x, y \in \R_{>0} } \lor \paren {x, y \in \R_{<0} }$ | === Sufficient Condition ===
Let $x \times y > 0$.
{{AimForCont}} either $x = 0$ or $y = 0$.
Then from Real Zero is Zero Element:
:$x \times y = 0$
Therefore by Proof by Contradiction:
:$y \ne 0$ and $x \ne 0$
{{qed|lemma}}
Let $x > 0$.
{{AimForCont}} $y < 0$.
{{begin-eqn}}
{{eqn | l = x
| o = >
| r = 0
... | The [[Definition:Real Multiplication|product]] of two [[Definition:Real Number|real numbers]] is greater than $0$ {{iff}} either both are greater than $0$ or both are less than $0$.
:$\forall x, y \in \R: x \times y > 0 \iff \paren {x, y \in \R_{>0} } \lor \paren {x, y \in \R_{<0} }$ | === Sufficient Condition ===
Let $x \times y > 0$.
{{AimForCont}} either $x = 0$ or $y = 0$.
Then from [[Real Zero is Zero Element]]:
:$x \times y = 0$
Therefore by [[Proof by Contradiction]]:
:$y \ne 0$ and $x \ne 0$
{{qed|lemma}}
Let $x > 0$.
{{AimForCont}} $y < 0$.
{{begin-eqn}}
{{eqn | l = x
| o = >
... | Product of Real Numbers is Positive iff Numbers have Same Sign | https://proofwiki.org/wiki/Product_of_Real_Numbers_is_Positive_iff_Numbers_have_Same_Sign | https://proofwiki.org/wiki/Product_of_Real_Numbers_is_Positive_iff_Numbers_have_Same_Sign | [
"Real Multiplication"
] | [
"Definition:Multiplication/Real Numbers",
"Definition:Real Number"
] | [
"Real Zero is Zero Element",
"Proof by Contradiction",
"Real Number Ordering is Compatible with Multiplication/Negative Factor",
"Proof by Contradiction",
"Proof by Contradiction",
"Real Number Ordering is Compatible with Multiplication/Negative Factor"
] |
proofwiki-10463 | Reciprocal of Strictly Positive Real Number is Strictly Positive | :$\forall x \in \R: x > 0 \implies \dfrac 1 x > 0$ | Let $x > 0$.
{{AimForCont}} $\dfrac 1 x < 0$.
Then:
{{begin-eqn}}
{{eqn | l = x
| o = >
| r = 0
| c =
}}
{{eqn | ll= \leadsto
| l = x \times \dfrac 1 x
| o = <
| r = 0 \times 0
| c = Real Number Ordering is Compatible with Multiplication: Negative Factor
}}
{{eqn | ll= \leadst... | :$\forall x \in \R: x > 0 \implies \dfrac 1 x > 0$ | Let $x > 0$.
{{AimForCont}} $\dfrac 1 x < 0$.
Then:
{{begin-eqn}}
{{eqn | l = x
| o = >
| r = 0
| c =
}}
{{eqn | ll= \leadsto
| l = x \times \dfrac 1 x
| o = <
| r = 0 \times 0
| c = [[Real Number Ordering is Compatible with Multiplication/Negative Factor|Real Number Orderi... | Reciprocal of Strictly Positive Real Number is Strictly Positive | https://proofwiki.org/wiki/Reciprocal_of_Strictly_Positive_Real_Number_is_Strictly_Positive | https://proofwiki.org/wiki/Reciprocal_of_Strictly_Positive_Real_Number_is_Strictly_Positive | [
"Reciprocals",
"Real Numbers",
"Inequalities"
] | [] | [
"Real Number Ordering is Compatible with Multiplication/Negative Factor",
"Real Zero is Less than Real One",
"Proof by Contradiction"
] |
proofwiki-10464 | Order of Strictly Positive Real Numbers is Dual of Order of their Reciprocals | :$\forall x, y \in \R: x > y > 0 \implies \dfrac 1 x < \dfrac 1 y$ | From Reciprocal of Strictly Positive Real Number is Strictly Positive:
:$(1): \quad x > 0 \implies \dfrac 1 x > 0$
:$(2): \quad y > 0 \implies \dfrac 1 y > 0$
Then:
{{begin-eqn}}
{{eqn | l = x
| o = >
| r = y
| c =
}}
{{eqn | ll= \leadsto
| l = x \times \frac 1 x
| o = >
| r = y \ti... | :$\forall x, y \in \R: x > y > 0 \implies \dfrac 1 x < \dfrac 1 y$ | From [[Reciprocal of Strictly Positive Real Number is Strictly Positive]]:
:$(1): \quad x > 0 \implies \dfrac 1 x > 0$
:$(2): \quad y > 0 \implies \dfrac 1 y > 0$
Then:
{{begin-eqn}}
{{eqn | l = x
| o = >
| r = y
| c =
}}
{{eqn | ll= \leadsto
| l = x \times \frac 1 x
| o = >
| r =... | Order of Strictly Positive Real Numbers is Dual of Order of their Reciprocals | https://proofwiki.org/wiki/Order_of_Strictly_Positive_Real_Numbers_is_Dual_of_Order_of_their_Reciprocals | https://proofwiki.org/wiki/Order_of_Strictly_Positive_Real_Numbers_is_Dual_of_Order_of_their_Reciprocals | [
"Real Numbers"
] | [] | [
"Reciprocal of Strictly Positive Real Number is Strictly Positive"
] |
proofwiki-10465 | Mean of Unequal Real Numbers is Between them | :$\forall x, y \in \R: x < y \implies x < \dfrac {x + y} 2 < y$ | First note that:
{{begin-eqn}}
{{eqn | l = 0
| o = <
| r = 1
| c = Real Zero is Less than Real One
}}
{{eqn | ll= \leadsto
| l = 0 + 0
| o = <
| r = 1 + 1
| c = Real Number Inequalities can be Added
}}
{{eqn | ll= \leadsto
| l = 0
| o = <
| r = \frac 1 {1 + 1}... | :$\forall x, y \in \R: x < y \implies x < \dfrac {x + y} 2 < y$ | First note that:
{{begin-eqn}}
{{eqn | l = 0
| o = <
| r = 1
| c = [[Real Zero is Less than Real One]]
}}
{{eqn | ll= \leadsto
| l = 0 + 0
| o = <
| r = 1 + 1
| c = [[Real Number Inequalities can be Added]]
}}
{{eqn | ll= \leadsto
| l = 0
| o = <
| r = \frac ... | Mean of Unequal Real Numbers is Between them | https://proofwiki.org/wiki/Mean_of_Unequal_Real_Numbers_is_Between_them | https://proofwiki.org/wiki/Mean_of_Unequal_Real_Numbers_is_Between_them | [
"Real Numbers"
] | [] | [
"Real Zero is Less than Real One",
"Real Number Inequalities can be Added",
"Reciprocal of Strictly Positive Real Number is Strictly Positive"
] |
proofwiki-10466 | Intersection of Inductive Set as Subset of Real Numbers is Inductive Set | Let $\AA$ be a set of inductive sets defined as subsets of real numbers.
Then their intersection is an inductive set. | By definition of inductive set:
:$\forall S \in \AA: 1 \in S$
Thus by definition of set intersection:
:$\ds 1 \in \bigcap_{S \mathop \in \AA} S$
Also by definition of inductive set:
:$\forall S \in \AA: x \in S \implies x + 1 \in S$
So:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \bigcap_{S \mathop \in \AA}... | Let $\AA$ be a [[Definition:Set of Sets|set]] of [[Definition:Inductive Set as Subset of Real Numbers|inductive sets defined as subsets of real numbers]].
Then their [[Definition:Set Intersection|intersection]] is an [[Definition:Inductive Set as Subset of Real Numbers|inductive set]]. | By definition of [[Definition:Inductive Set as Subset of Real Numbers|inductive set]]:
:$\forall S \in \AA: 1 \in S$
Thus by definition of [[Definition:Intersection of Set of Sets|set intersection]]:
:$\ds 1 \in \bigcap_{S \mathop \in \AA} S$
Also by definition of [[Definition:Inductive Set as Subset of Real Number... | Intersection of Inductive Set as Subset of Real Numbers is Inductive Set | https://proofwiki.org/wiki/Intersection_of_Inductive_Set_as_Subset_of_Real_Numbers_is_Inductive_Set | https://proofwiki.org/wiki/Intersection_of_Inductive_Set_as_Subset_of_Real_Numbers_is_Inductive_Set | [
"Inductive Sets"
] | [
"Definition:Set of Sets",
"Definition:Inductive Set/Subset of Real Numbers",
"Definition:Set Intersection",
"Definition:Inductive Set/Subset of Real Numbers"
] | [
"Definition:Inductive Set/Subset of Real Numbers",
"Definition:Set Intersection/Set of Sets",
"Definition:Inductive Set/Subset of Real Numbers",
"Definition:Inductive Set/Subset of Real Numbers"
] |
proofwiki-10467 | Non-Empty Subset of Initial Segment of Natural Numbers has Greatest Element | Let $n \in \N_{>0}$ be a non-zero natural number.
Let $\N^*_n$ denote the Initial segment $\set {1, 2, \ldots, n}$ of the non-zero natural numbers.
Then every non-empty subset of $\N^*_n$ has a greatest element. | The proof will proceed by the Principle of Mathematical Induction on $\N_{>0}$.
Let $S$ be the set defined as:
:$S := \leftset {n \in \N:}$ Every non-empty subset of $\N^*_n$ has a greatest element$\rightset{}$ | Let $n \in \N_{>0}$ be a non-zero [[Definition:Natural Number|natural number]].
Let $\N^*_n$ denote the [[Definition:Initial Segment of One-Based Natural Numbers|Initial segment]] $\set {1, 2, \ldots, n}$ of the non-zero [[Definition:Natural Number|natural numbers]].
Then every [[Definition:Non-Empty Set|non-empty]]... | The proof will proceed by the [[Principle of Mathematical Induction]] on $\N_{>0}$.
Let $S$ be the [[Definition:Set|set]] defined as:
:$S := \leftset {n \in \N:}$ Every [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $\N^*_n$ has a [[Definition:Greatest Element|greatest element]]$\rightset{}$ | Non-Empty Subset of Initial Segment of Natural Numbers has Greatest Element | https://proofwiki.org/wiki/Non-Empty_Subset_of_Initial_Segment_of_Natural_Numbers_has_Greatest_Element | https://proofwiki.org/wiki/Non-Empty_Subset_of_Initial_Segment_of_Natural_Numbers_has_Greatest_Element | [
"Natural Numbers"
] | [
"Definition:Natural Numbers",
"Definition:Initial Segment of Natural Numbers/One-Based",
"Definition:Natural Numbers",
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Greatest Element"
] | [
"Principle of Mathematical Induction",
"Definition:Set",
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Greatest Element",
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Greatest Element",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction",
"... |
proofwiki-10468 | Not every Non-Empty Subset of Natural Numbers has Greatest Element | Let $S \subseteq \N_{>0}$.
Then, despite Non-Empty Subset of Initial Segment of Natural Numbers has Greatest Element, it is not necessarily the case that $S$ has a greatest element. | Let:
:$S = \set {x \in \N_{>0}: x > 1}$
Then $S \subseteq \N_{>0}$.
{{AimForCont}} $S$ has a greatest element.
Let the greatest element of $S$ be $k$.
But $\N_{>0}$ is an inductive set.
Therefore $k + 1 \in \N_{>0}$.
By definition of $S$:
:$k + 1 \in S$
Therefore $k$ cannot be the greatest element of $S$.
By Proof by C... | Let $S \subseteq \N_{>0}$.
Then, despite [[Non-Empty Subset of Initial Segment of Natural Numbers has Greatest Element]], it is not necessarily the case that $S$ has a [[Definition:Greatest Element|greatest element]]. | Let:
:$S = \set {x \in \N_{>0}: x > 1}$
Then $S \subseteq \N_{>0}$.
{{AimForCont}} $S$ has a [[Definition:Greatest Element|greatest element]].
Let the [[Definition:Greatest Element|greatest element]] of $S$ be $k$.
But $\N_{>0}$ is an [[Definition:Inductive Set as Subset of Real Numbers|inductive set]].
Therefore ... | Not every Non-Empty Subset of Natural Numbers has Greatest Element | https://proofwiki.org/wiki/Not_every_Non-Empty_Subset_of_Natural_Numbers_has_Greatest_Element | https://proofwiki.org/wiki/Not_every_Non-Empty_Subset_of_Natural_Numbers_has_Greatest_Element | [
"Natural Numbers"
] | [
"Non-Empty Subset of Initial Segment of Natural Numbers has Greatest Element",
"Definition:Greatest Element"
] | [
"Definition:Greatest Element",
"Definition:Greatest Element",
"Definition:Inductive Set/Subset of Real Numbers",
"Definition:Greatest Element",
"Proof by Contradiction",
"Definition:Greatest Element",
"Non-Empty Subset of Initial Segment of Natural Numbers has Greatest Element",
"Definition:Initial Se... |
proofwiki-10469 | Existence of Integral on Union of Adjacent Intervals | Let $f$ be a real function defined on a closed interval $\closedint a b$ where $a < b$.
Let $c$ be a point in $\openint a b$.
Then:
:$f$ is Darboux integrable on $\closedint a c$ and $\closedint c b$
{{iff}}:
:$f$ is Darboux integrable on $\closedint a b$. | === Necessary Condition ===
We need to prove that $f$ is Darboux integrable on $\closedint a b$.
To do this it suffices to show that for all $\epsilon > 0$, there exists a subdivision $S$ of $\closedint a b$ such that $\map U S – \map L S < \epsilon$.
Here, $\map U S$ and $\map L S$ are, respectively, the upper Darboux... | Let $f$ be a [[Definition:Real Function|real function]] defined on a [[Definition:Closed Real Interval|closed interval]] $\closedint a b$ where $a < b$.
Let $c$ be a point in $\openint a b$.
Then:
:$f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint a c$ and $\closedint c b$
{{iff}}:... | === Necessary Condition ===
We need to prove that $f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint a b$.
To do this it suffices to show that for all $\epsilon > 0$, there exists a [[Definition:Subdivision of Interval|subdivision]] $S$ of $\closedint a b$ such that $\map U S – \map L... | Existence of Integral on Union of Adjacent Intervals | https://proofwiki.org/wiki/Existence_of_Integral_on_Union_of_Adjacent_Intervals | https://proofwiki.org/wiki/Existence_of_Integral_on_Union_of_Adjacent_Intervals | [
"Integral Calculus"
] | [
"Definition:Real Function",
"Definition:Real Interval/Closed",
"Definition:Darboux Integrable Function",
"Definition:Darboux Integrable Function"
] | [
"Definition:Darboux Integrable Function",
"Definition:Subdivision of Interval",
"Definition:Upper Darboux Sum",
"Definition:Lower Darboux Sum",
"Definition:Subdivision of Interval",
"Definition:Strictly Positive/Real Number",
"Definition:Darboux Integrable Function",
"Definition:Subdivision of Interva... |
proofwiki-10470 | Cartesian Product of Bijections is Bijection | Let $S_1 \times S_2$ be the Cartesian product of two sets $S_1$ and $S_2$.
Let $T_1 \times T_2$ be the Cartesian product of two sets $T_1$ and $T_2$.
Let $f_1: S_1 \to T_1$ and $f_2: S_2 \to T_2$ be bijections.
Let $f_1 \times f_2: S_1 \times S_2 \to T_1 \times T_2$ be the Cartesian product of $f_1$ and $f_2$ defined a... | Because $f_1$ and $f_2$ are both bijections, it follows by definition that they are both surjections.
Let $\left({t_1, t_2}\right) \in T_1 \times T_2$.
Then:
:as $f_1$ is a surjection, $\exists s_1 \in S_1: f_1 \left({s_1}\right) = t_1$
:as $f_2$ is a surjection, $\exists s_2 \in S_2: f_2 \left({s_2}\right) = t_2$
Thus... | Let $S_1 \times S_2$ be the [[Definition:Cartesian Product|Cartesian product]] of two [[Definition:Set|sets]] $S_1$ and $S_2$.
Let $T_1 \times T_2$ be the [[Definition:Cartesian Product|Cartesian product]] of two [[Definition:Set|sets]] $T_1$ and $T_2$.
Let $f_1: S_1 \to T_1$ and $f_2: S_2 \to T_2$ be [[Definition:Bi... | Because $f_1$ and $f_2$ are both [[Definition:Bijection|bijections]], it follows by definition that they are both [[Definition:Surjection|surjections]].
Let $\left({t_1, t_2}\right) \in T_1 \times T_2$.
Then:
:as $f_1$ is a [[Definition:Surjection|surjection]], $\exists s_1 \in S_1: f_1 \left({s_1}\right) = t_1$
:as ... | Cartesian Product of Bijections is Bijection | https://proofwiki.org/wiki/Cartesian_Product_of_Bijections_is_Bijection | https://proofwiki.org/wiki/Cartesian_Product_of_Bijections_is_Bijection | [
"Bijections",
"Cartesian Product"
] | [
"Definition:Cartesian Product",
"Definition:Set",
"Definition:Cartesian Product",
"Definition:Set",
"Definition:Bijection",
"Definition:Cartesian Product of Relations",
"Definition:Bijection"
] | [
"Definition:Bijection",
"Definition:Surjection",
"Definition:Surjection",
"Definition:Surjection",
"Definition:Surjection",
"Definition:Bijection",
"Definition:Injection",
"Equality of Ordered Pairs",
"Definition:Injection",
"Definition:Injection",
"Definition:Injection",
"Definition:Surjectio... |
proofwiki-10471 | Cartesian Product of Bijections is Bijection/General Result | Let $I$ be an indexing set.
Let $\family {S_i}_{i \mathop \in I}$ and $\family {T_i}_{i \mathop \in I}$ be families of sets indexed by $I$.
Let $\ds \SS := \prod_{i \mathop \in I} S_i$ and $\ds \TT := \prod_{i \mathop \in I} T_i$ be the Cartesian products of $\family {S_i}_{i \mathop \in I}$ and $\family {T_i}_{i \math... | Because $f_i$ are bijections, it follows by definition that they are surjections.
Let $\mathbf t := \family {t_i}_{i \mathop \in I} \in \TT$.
Then as $f_i$ is a surjection:
:$\forall i \in I: \exists s_i \in S_i: \map {f_i} {s_i} = t_i$
Thus:
:$\exists \mathbf s \in \SS: \map \FF {\mathbf s} = \mathbf t$
So $\FF$ is a ... | Let $I$ be an [[Definition:Indexing Set|indexing set]].
Let $\family {S_i}_{i \mathop \in I}$ and $\family {T_i}_{i \mathop \in I}$ be [[Definition:Indexed Family of Sets|families of sets]] [[Definition:Indexing Set|indexed by $I$]].
Let $\ds \SS := \prod_{i \mathop \in I} S_i$ and $\ds \TT := \prod_{i \mathop \in I}... | Because $f_i$ are [[Definition:Bijection|bijections]], it follows by definition that they are [[Definition:Surjection|surjections]].
Let $\mathbf t := \family {t_i}_{i \mathop \in I} \in \TT$.
Then as $f_i$ is a [[Definition:Surjection|surjection]]:
:$\forall i \in I: \exists s_i \in S_i: \map {f_i} {s_i} = t_i$
Thu... | Cartesian Product of Bijections is Bijection/General Result | https://proofwiki.org/wiki/Cartesian_Product_of_Bijections_is_Bijection/General_Result | https://proofwiki.org/wiki/Cartesian_Product_of_Bijections_is_Bijection/General_Result | [
"Bijections",
"Cartesian Product",
"Indexed Families"
] | [
"Definition:Indexing Set",
"Definition:Indexing Set/Family of Sets",
"Definition:Indexing Set",
"Definition:Cartesian Product/Family of Sets",
"Definition:Bijection",
"Definition:Cartesian Product of Relations",
"Definition:Bijection"
] | [
"Definition:Bijection",
"Definition:Surjection",
"Definition:Surjection",
"Definition:Surjection",
"Definition:Bijection",
"Definition:Injection",
"Definition:Injection",
"Definition:Injection",
"Definition:Surjection",
"Definition:Injection",
"Definition:Bijection"
] |
proofwiki-10472 | First Order ODE/y' - f (y) phi' (x) over f' (y) = phi (x) phi' (x) over f' (y) | Let $\map f y$ and $\map \phi x$ be known real functions of $y$ and $x$ respectively.
The general solution of:
:$(1): \quad \dfrac {\d y} {\d x} - \dfrac {\map f y} {\map {f'} y} \map {\phi'} x = \dfrac {\map \phi x \map {\phi'} x} {\map {f'} y}$
is:
:$\map f y = C e^{\map \phi x} - \map \phi x - 1$ | Let $u = \map f y$
Then by the Chain Rule for Derivatives:
:$\dfrac {\d u} {\d x} = \map {f'} y \dfrac {\d y} {\d x}$
Then:
{{begin-eqn}}
{{eqn | l = \dfrac {\d y} {\d x} - \dfrac {\map f y} {\map {f'} y} \map {\phi'} x
| r = \dfrac {\map \phi x \map {\phi'} x} {\map {f'} y}
| c =
}}
{{eqn | ll= \leadsto
... | Let $\map f y$ and $\map \phi x$ be known [[Definition:Real Function|real functions]] of $y$ and $x$ respectively.
The [[Definition:General Solution to Differential Equation|general solution]] of:
:$(1): \quad \dfrac {\d y} {\d x} - \dfrac {\map f y} {\map {f'} y} \map {\phi'} x = \dfrac {\map \phi x \map {\phi'} x} {... | Let $u = \map f y$
Then by the [[Chain Rule for Derivatives]]:
:$\dfrac {\d u} {\d x} = \map {f'} y \dfrac {\d y} {\d x}$
Then:
{{begin-eqn}}
{{eqn | l = \dfrac {\d y} {\d x} - \dfrac {\map f y} {\map {f'} y} \map {\phi'} x
| r = \dfrac {\map \phi x \map {\phi'} x} {\map {f'} y}
| c =
}}
{{eqn | ll= \le... | First Order ODE/y' - f (y) phi' (x) over f' (y) = phi (x) phi' (x) over f' (y) | https://proofwiki.org/wiki/First_Order_ODE/y'_-_f_(y)_phi'_(x)_over_f'_(y)_=_phi_(x)_phi'_(x)_over_f'_(y) | https://proofwiki.org/wiki/First_Order_ODE/y'_-_f_(y)_phi'_(x)_over_f'_(y)_=_phi_(x)_phi'_(x)_over_f'_(y) | [
"Examples of First Order ODEs"
] | [
"Definition:Real Function",
"Definition:Differential Equation/Solution/General Solution"
] | [
"Derivative of Composite Function",
"Definition:Linear First Order Ordinary Differential Equation",
"Definition:Differential Equation/Solution/General Solution",
"Solution to Linear First Order Ordinary Differential Equation",
"Integration by Substitution",
"Integration by Substitution",
"Primitive of x... |
proofwiki-10473 | Cardinality of Set of All Mappings/Finite Sets | Let $S$ and $T$ be finite sets.
The cardinality of the set of all mappings from $S$ to $T$ (that is, the total number of mappings from $S$ to $T$) is:
:$\card {T^S} = \card T^{\card S}$ | Let $\card S = n$ and $\card T = m$.
First suppose that $n = 0$, that is, that $S = \O$.
From Cardinality of Set of All Mappings from Empty Set:
:$\card {T^\O} = 1 = m^0$
and the result is seen to hold for $n = 0$.
Next, suppose that $m = 0$, that is, that $T = \O$.
From Cardinality of Set of All Mappings to Empty Set:... | Let $S$ and $T$ be [[Definition:Finite Set|finite sets]].
The [[Definition:Cardinality|cardinality]] of the [[Definition:Set of All Mappings|set of all mappings]] from $S$ to $T$ (that is, the total number of [[Definition:Mapping|mappings]] from $S$ to $T$) is:
:$\card {T^S} = \card T^{\card S}$ | Let $\card S = n$ and $\card T = m$.
First suppose that $n = 0$, that is, that $S = \O$.
From [[Cardinality of Set of All Mappings from Empty Set]]:
:$\card {T^\O} = 1 = m^0$
and the result is seen to hold for $n = 0$.
Next, suppose that $m = 0$, that is, that $T = \O$.
From [[Cardinality of Set of All Mappings ... | Cardinality of Set of All Mappings/Finite Sets | https://proofwiki.org/wiki/Cardinality_of_Set_of_All_Mappings/Finite_Sets | https://proofwiki.org/wiki/Cardinality_of_Set_of_All_Mappings/Finite_Sets | [
"Cardinality of Set of All Mappings",
"Finite Sets"
] | [
"Definition:Finite Set",
"Definition:Cardinality",
"Definition:Set of All Mappings",
"Definition:Mapping"
] | [
"Cardinality of Set of All Mappings from Empty Set",
"Cardinality of Set of All Mappings to Empty Set",
"Definition:Bijection",
"Definition:Set of All Mappings",
"Definition:Bijection",
"Definition:Set of All Mappings",
"Definition:Restriction/Mapping",
"Definition:Element",
"Definition:Set Partitio... |
proofwiki-10474 | Cardinality of Set Union/General Case | Let $\sequence {S_n}_{n \mathop \in \N}$ be a sequence of sets.
Then:
{{begin-eqn}}
{{eqn | l = \card {\bigcup_{i \mathop = 1}^n S_i}
| r = \sum_{i \mathop = 1}^n \card {S_i}
| c =
}}
{{eqn | o =
| ro= -
| r = \sum_{1 \mathop \le i \mathop < j \mathop \le n} \card {S_i \cap S_j}
| c =... | By Cardinality is Additive Function, we can apply the Inclusion-Exclusion Principle:
If $f: \SS \to \R$ is an additive function, then:
{{Explain|What is $\SS$?}}
{{begin-eqn}}
{{eqn | l = \map f {\bigcup_{i \mathop = 1}^n S_i}
| r = \sum_{i \mathop = 1}^n \map f {S_i}
| c =
}}
{{eqn | o =
| ro= -... | Let $\sequence {S_n}_{n \mathop \in \N}$ be a [[Definition:Sequence|sequence]] of [[Definition:Set|sets]].
Then:
{{begin-eqn}}
{{eqn | l = \card {\bigcup_{i \mathop = 1}^n S_i}
| r = \sum_{i \mathop = 1}^n \card {S_i}
| c =
}}
{{eqn | o =
| ro= -
| r = \sum_{1 \mathop \le i \mathop < j \ma... | By [[Cardinality is Additive Function]], we can apply the [[Inclusion-Exclusion Principle]]:
If $f: \SS \to \R$ is an [[Definition:Additive Function (Measure Theory)|additive function]], then:
{{Explain|What is $\SS$?}}
{{begin-eqn}}
{{eqn | l = \map f {\bigcup_{i \mathop = 1}^n S_i}
| r = \sum_{i \mathop... | Cardinality of Set Union/General Case | https://proofwiki.org/wiki/Cardinality_of_Set_Union/General_Case | https://proofwiki.org/wiki/Cardinality_of_Set_Union/General_Case | [
"Cardinality of Set Union"
] | [
"Definition:Sequence",
"Definition:Set"
] | [
"Cardinality is Additive Function",
"Inclusion-Exclusion Principle",
"Definition:Additive Function (Measure Theory)"
] |
proofwiki-10475 | Structure of Inverse Completion of Commutative Semigroup | Let $\struct {S, \circ}$ be a commutative semigroup.
Let $\struct {C, \circ} \subseteq \struct {S, \circ}$ be the subsemigroup of cancellable elements of $\struct {S, \circ}$.
Let $\struct {T, \circ'}$ be an inverse completion of $\struct {S, \circ}$.
Then:
:$T = S \circ' C^{-1}$
where:
:$C^{-1}$ is the inverse of $C$ ... | Let $a \in C$.
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = S
| c =
}}
{{eqn | ll= \leadsto
| l = x
| r = x \circ \paren {a \circ' a^{-1} }
| c = {{Defof|Invertible Element}}
}}
{{eqn | ll= \leadsto
| l = x
| r = \paren {x \circ a} \circ' a^{-1}
| c = {{Defof|Associ... | Let $\struct {S, \circ}$ be a [[Definition:Commutative Semigroup|commutative semigroup]].
Let $\struct {C, \circ} \subseteq \struct {S, \circ}$ be the [[Definition:Subsemigroup|subsemigroup]] of [[Definition:Cancellable Element|cancellable elements]] of $\struct {S, \circ}$.
Let $\struct {T, \circ'}$ be an [[Definiti... | Let $a \in C$.
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = S
| c =
}}
{{eqn | ll= \leadsto
| l = x
| r = x \circ \paren {a \circ' a^{-1} }
| c = {{Defof|Invertible Element}}
}}
{{eqn | ll= \leadsto
| l = x
| r = \paren {x \circ a} \circ' a^{-1}
| c = {{Defof|Assoc... | Structure of Inverse Completion of Commutative Semigroup | https://proofwiki.org/wiki/Structure_of_Inverse_Completion_of_Commutative_Semigroup | https://proofwiki.org/wiki/Structure_of_Inverse_Completion_of_Commutative_Semigroup | [
"Inverse Completions",
"Commutative Semigroups"
] | [
"Definition:Commutative Semigroup",
"Definition:Subsemigroup",
"Definition:Cancellable Element",
"Definition:Inverse Completion",
"Definition:Inverse of Subset/Monoid",
"Definition:Subset Product"
] | [
"Inverse of Product",
"Union is Smallest Superset",
"Subset Product defining Inverse Completion of Commutative Semigroup is Commutative Semigroup",
"Definition:Commutative Semigroup",
"Definition:Semigroup",
"Definition:Generator of Subsemigroup",
"Definition:Generator of Subsemigroup",
"Definition:Se... |
proofwiki-10476 | Inverse in Monoid is Unique | Let $\struct {S, \circ}$ be a monoid.
Then an element $x \in S$ can have at most one inverse for $\circ$. | Let $e$ be the identity element of $\struct {S, \circ}$.
Suppose $x \in S$ has two inverses: $y$ and $z$.
Then:
{{begin-eqn}}
{{eqn | l = y
| r = y \circ e
| c = {{Defof|Identity Element}}
}}
{{eqn | r = y \circ \paren {x \circ z}
| c = {{Defof|Inverse Element}}
}}
{{eqn | r = \paren {y \circ x} \circ... | Let $\struct {S, \circ}$ be a [[Definition:Monoid|monoid]].
Then an [[Definition:Element|element]] $x \in S$ can have at most one [[Definition:Inverse Element|inverse]] for $\circ$. | Let $e$ be the [[Definition:Identity Element|identity element]] of $\struct {S, \circ}$.
Suppose $x \in S$ has two [[Definition:Inverse Element|inverses]]: $y$ and $z$.
Then:
{{begin-eqn}}
{{eqn | l = y
| r = y \circ e
| c = {{Defof|Identity Element}}
}}
{{eqn | r = y \circ \paren {x \circ z}
| c ... | Inverse in Monoid is Unique | https://proofwiki.org/wiki/Inverse_in_Monoid_is_Unique | https://proofwiki.org/wiki/Inverse_in_Monoid_is_Unique | [
"Monoids",
"Inverse Elements"
] | [
"Definition:Monoid",
"Definition:Element",
"Definition:Inverse (Abstract Algebra)/Inverse"
] | [
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Unique"
] |
proofwiki-10477 | Inverse of Inverse/Monoid | Let $\struct {S, \circ}$ be a monoid.
Let $x \in S$ be invertible, and let its inverse be $x^{-1}$.
Then $x^{-1}$ is also invertible, and:
:$\paren {x^{-1} }^{-1} = x$ | By Inverse in Monoid is Unique, any inverse of $x$ is unique, and can be denoted $x^{-1}$.
From Inverse of Inverse in General Algebraic Structure:
:$x^{-1}$ is invertible and its inverse is $x$.
That is:
:$\paren {x^{-1} }^{-1} = x$
{{qed}} | Let $\struct {S, \circ}$ be a [[Definition:Monoid|monoid]].
Let $x \in S$ be [[Definition:Invertible Element|invertible]], and let its [[Definition:Inverse Element|inverse]] be $x^{-1}$.
Then $x^{-1}$ is also [[Definition:Invertible Element|invertible]], and:
:$\paren {x^{-1} }^{-1} = x$ | By [[Inverse in Monoid is Unique]], any [[Definition:Inverse Element|inverse]] of $x$ is [[Definition:Unique|unique]], and can be denoted $x^{-1}$.
From [[Inverse of Inverse in General Algebraic Structure]]:
:$x^{-1}$ is [[Definition:Invertible Element|invertible]] and its [[Definition:Inverse Element|inverse]] is $x$... | Inverse of Inverse/Monoid | https://proofwiki.org/wiki/Inverse_of_Inverse/Monoid | https://proofwiki.org/wiki/Inverse_of_Inverse/Monoid | [
"Monoids",
"Inverse Elements"
] | [
"Definition:Monoid",
"Definition:Invertible Element",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Invertible Element"
] | [
"Inverse in Monoid is Unique",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Unique",
"Inverse of Inverse/General Algebraic Structure",
"Definition:Invertible Element",
"Definition:Inverse (Abstract Algebra)/Inverse"
] |
proofwiki-10478 | Inverse of Product/Monoid | Let $\struct {S, \circ}$ be a monoid whose identity is $e$.
Let $a, b \in S$ be invertible for $\circ$, with inverses $a^{-1}, b^{-1}$.
Then $a \circ b$ is invertible for $\circ$, and:
:$\paren {a \circ b}^{-1} = b^{-1} \circ a^{-1}$ | {{begin-eqn}}
{{eqn | l = \paren {a \circ b} \circ \paren {b^{-1} \circ a^{-1} }
| r = \paren {\paren {a \circ b} \circ b^{-1} } \circ a^{-1}
| c = {{MonoidAxiom|1}}
}}
{{eqn | r = \paren {a \circ \paren {b \circ b^{-1} } } \circ a^{-1}
| c = {{MonoidAxiom|1}}
}}
{{eqn | r = \paren {a \circ e} \circ a... | Let $\struct {S, \circ}$ be a [[Definition:Monoid|monoid]] whose [[Definition:Identity Element|identity]] is $e$.
Let $a, b \in S$ be [[Definition:Invertible Element|invertible]] for $\circ$, with [[Definition:Inverse Element|inverses]] $a^{-1}, b^{-1}$.
Then $a \circ b$ is [[Definition:Invertible Element|invertible... | {{begin-eqn}}
{{eqn | l = \paren {a \circ b} \circ \paren {b^{-1} \circ a^{-1} }
| r = \paren {\paren {a \circ b} \circ b^{-1} } \circ a^{-1}
| c = {{MonoidAxiom|1}}
}}
{{eqn | r = \paren {a \circ \paren {b \circ b^{-1} } } \circ a^{-1}
| c = {{MonoidAxiom|1}}
}}
{{eqn | r = \paren {a \circ e} \circ a... | Inverse of Product/Monoid | https://proofwiki.org/wiki/Inverse_of_Product/Monoid | https://proofwiki.org/wiki/Inverse_of_Product/Monoid | [
"Monoids",
"Inverse Elements",
"Inverse of Product"
] | [
"Definition:Monoid",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Invertible Element",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Invertible Element"
] | [] |
proofwiki-10479 | Right Identity Element is Idempotent | Let $\struct {S, \circ}$ be an algebraic structure.
Let $e_R \in S$ be a right identity with respect to $\circ$.
Then $e_R$ is idempotent under $\circ$. | By the definition of a right identity:
:$\forall x \in S: x \circ e_R = x$
Thus in particular:
:$e_R \circ e_R = e_R$
Therefore $e_R$ is idempotent under $\circ$.
{{qed}} | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]].
Let $e_R \in S$ be a [[Definition:Right Identity|right identity]] with respect to $\circ$.
Then $e_R$ is [[Definition:Idempotent Element|idempotent]] under $\circ$. | By the definition of a [[Definition:Right Identity|right identity]]:
:$\forall x \in S: x \circ e_R = x$
Thus in particular:
:$e_R \circ e_R = e_R$
Therefore $e_R$ is [[Definition:Idempotent Element|idempotent]] under $\circ$.
{{qed}} | Right Identity Element is Idempotent | https://proofwiki.org/wiki/Right_Identity_Element_is_Idempotent | https://proofwiki.org/wiki/Right_Identity_Element_is_Idempotent | [
"Identities are Idempotent"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Identity (Abstract Algebra)/Right Identity",
"Definition:Idempotence/Element"
] | [
"Definition:Identity (Abstract Algebra)/Right Identity",
"Definition:Idempotence/Element"
] |
proofwiki-10480 | Left Identity Element is Idempotent | Let $\struct {S, \circ}$ be an algebraic structure.
Let $e_L \in S$ be a left identity with respect to $\circ$.
Then $e_L$ is idempotent under $\circ$. | By the definition of a left identity:
:$\forall x \in S: e_L \circ x = x$
Thus in particular:
:$e_L \circ e_L = e_L$
Therefore $e_L$ is idempotent under $\circ$.
{{qed}} | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]].
Let $e_L \in S$ be a [[Definition:Left Identity|left identity]] with respect to $\circ$.
Then $e_L$ is [[Definition:Idempotent Element|idempotent]] under $\circ$. | By the definition of a [[Definition:Left Identity|left identity]]:
:$\forall x \in S: e_L \circ x = x$
Thus in particular:
:$e_L \circ e_L = e_L$
Therefore $e_L$ is [[Definition:Idempotent Element|idempotent]] under $\circ$.
{{qed}} | Left Identity Element is Idempotent | https://proofwiki.org/wiki/Left_Identity_Element_is_Idempotent | https://proofwiki.org/wiki/Left_Identity_Element_is_Idempotent | [
"Identities are Idempotent"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Identity (Abstract Algebra)/Left Identity",
"Definition:Idempotence/Element"
] | [
"Definition:Identity (Abstract Algebra)/Left Identity",
"Definition:Idempotence/Element"
] |
proofwiki-10481 | More than one Right Identity then no Left Identity | Let $\struct {S, \circ}$ be an algebraic structure.
If $\struct {S, \circ}$ has more than one right identity, then it has no left identity. | Let $\struct {S, \circ}$ be an algebraic structure with more than one right identity.
Take any two of these, and call them $e_{R_1}$ and $e_{R_2}$, where $e_{R_1} \ne e_{R_2}$.
Suppose $\struct {S, \circ}$ has a left identity.
Call this left identity $e_L$.
Then, by the behaviour of $e_L$, $e_{R_1}$ and $e_{R_2}$:
:$e_... | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]].
If $\struct {S, \circ}$ has more than one [[Definition:Right Identity|right identity]], then it has no [[Definition:Left Identity|left identity]]. | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]] with more than one [[Definition:Right Identity|right identity]].
Take any two of these, and call them $e_{R_1}$ and $e_{R_2}$, where $e_{R_1} \ne e_{R_2}$.
Suppose $\struct {S, \circ}$ has a [[Definition:Left Iden... | More than one Right Identity then no Left Identity | https://proofwiki.org/wiki/More_than_one_Right_Identity_then_no_Left_Identity | https://proofwiki.org/wiki/More_than_one_Right_Identity_then_no_Left_Identity | [
"Identity Elements"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Identity (Abstract Algebra)/Right Identity",
"Definition:Identity (Abstract Algebra)/Left Identity"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Identity (Abstract Algebra)/Right Identity",
"Definition:Identity (Abstract Algebra)/Left Identity",
"Definition:Identity (Abstract Algebra)/Left Identity",
"Definition:Algebraic Structure/One Operation",
"Definition:Identity (Abstract Algebra)/R... |
proofwiki-10482 | Inverse Completion is Commutative Monoid | Let $\left({S, \circ}\right)$ be a commutative semigroup.
Let $\left ({C, \circ}\right) \subseteq \left({S, \circ}\right)$ be the subsemigroup of cancellable elements of $\left({S, \circ}\right)$.
Let $\left({T, \circ'}\right)$ be an inverse completion of $\left({S, \circ}\right)$.
Then $\left({T, \circ'}\right)$ is a ... | From Inverse Completion is Commutative Semigroup:
:$\left({T, \circ'}\right)$ is a commutative semigroup.
By definition of inverse completion:
:$\forall x \in C: \exists x^{-1} \in C: x \circ x^{-1} \in T$
Thus by definition of inverse element:
:$e := x \circ x^{-1}$ is an identity element of $T$.
By Identity is Unique... | Let $\left({S, \circ}\right)$ be a [[Definition:Commutative Semigroup|commutative semigroup]].
Let $\left ({C, \circ}\right) \subseteq \left({S, \circ}\right)$ be the [[Definition:Subsemigroup|subsemigroup]] of [[Definition:Cancellable Element|cancellable elements]] of $\left({S, \circ}\right)$.
Let $\left({T, \circ'... | From [[Inverse Completion is Commutative Semigroup]]:
:$\left({T, \circ'}\right)$ is a [[Definition:Commutative Semigroup|commutative semigroup]].
By definition of [[Definition:Inverse Completion|inverse completion]]:
:$\forall x \in C: \exists x^{-1} \in C: x \circ x^{-1} \in T$
Thus by definition of [[Definition:In... | Inverse Completion is Commutative Monoid | https://proofwiki.org/wiki/Inverse_Completion_is_Commutative_Monoid | https://proofwiki.org/wiki/Inverse_Completion_is_Commutative_Monoid | [
"Inverse Completions"
] | [
"Definition:Commutative Semigroup",
"Definition:Subsemigroup",
"Definition:Cancellable Element",
"Definition:Inverse Completion",
"Definition:Commutative Monoid"
] | [
"Inverse Completion is Commutative Semigroup",
"Definition:Commutative Semigroup",
"Definition:Inverse Completion",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Identity is Unique",
"Definition:Element",
"Definition:Semigroup",
"Defin... |
proofwiki-10483 | Inverse Completion is Commutative Semigroup | Let $\struct {S, \circ}$ be a commutative semigroup.
Let $\struct {C, \circ} \subseteq \struct {S, \circ}$ be the subsemigroup of cancellable elements of $\struct {S, \circ}$.
Let $\struct {T, \circ'}$ be an inverse completion of $\struct {S, \circ}$.
Then $T = S \circ' C^{-1}$, and is a commutative semigroup. | From Subset Product defining Inverse Completion of Commutative Semigroup is Commutative Semigroup:
:$S \circ' C^{-1}$ is a commutative semigroup.
From Structure of Inverse Completion of Commutative Semigroup:
:$T = S \circ' C^{-1}$
{{qed}} | Let $\struct {S, \circ}$ be a [[Definition:Commutative Semigroup|commutative semigroup]].
Let $\struct {C, \circ} \subseteq \struct {S, \circ}$ be the [[Definition:Subsemigroup|subsemigroup]] of [[Definition:Cancellable Element|cancellable elements]] of $\struct {S, \circ}$.
Let $\struct {T, \circ'}$ be an [[Definiti... | From [[Subset Product defining Inverse Completion of Commutative Semigroup is Commutative Semigroup]]:
:$S \circ' C^{-1}$ is a [[Definition:Commutative Semigroup|commutative semigroup]].
From [[Structure of Inverse Completion of Commutative Semigroup]]:
:$T = S \circ' C^{-1}$
{{qed}} | Inverse Completion is Commutative Semigroup | https://proofwiki.org/wiki/Inverse_Completion_is_Commutative_Semigroup | https://proofwiki.org/wiki/Inverse_Completion_is_Commutative_Semigroup | [
"Inverse Completions"
] | [
"Definition:Commutative Semigroup",
"Definition:Subsemigroup",
"Definition:Cancellable Element",
"Definition:Inverse Completion",
"Definition:Commutative Semigroup"
] | [
"Subset Product defining Inverse Completion of Commutative Semigroup is Commutative Semigroup",
"Definition:Commutative Semigroup",
"Structure of Inverse Completion of Commutative Semigroup"
] |
proofwiki-10484 | Complement of G-Delta Set is F-Sigma Set | Let $T = \struct {S, \tau}$ be a topological space.
Let $X$ be an $G_\delta$ set of $T$.
Then its complement $S \setminus X$ is an $F_\sigma$ set of $T$. | Let $X$ be a $G_\delta$ set of $T$.
Let $X = \bigcap \UU$ where $\UU$ is a countable set of open sets in $T$.
Then from De Morgan's Laws: Difference with Intersection we have:
:$\ds S \setminus X = S \setminus \bigcap \UU = \bigcup_{U \mathop \in \UU} \paren {S \setminus U}$
By definition of closed set, each of the $S ... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $X$ be an [[Definition:G-Delta Set|$G_\delta$ set]] of $T$.
Then its [[Definition:Relative Complement|complement]] $S \setminus X$ is an [[Definition:F-Sigma Set|$F_\sigma$ set]] of $T$. | Let $X$ be a [[Definition:G-Delta Set|$G_\delta$ set]] of $T$.
Let $X = \bigcap \UU$ where $\UU$ is a [[Definition:Countable Set|countable]] [[Definition:Set|set]] of [[Definition:Open Set (Topology)|open sets]] in $T$.
Then from [[De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Intersection... | Complement of G-Delta Set is F-Sigma Set | https://proofwiki.org/wiki/Complement_of_G-Delta_Set_is_F-Sigma_Set | https://proofwiki.org/wiki/Complement_of_G-Delta_Set_is_F-Sigma_Set | [
"F-Sigma Sets",
"G-Delta Sets"
] | [
"Definition:Topological Space",
"Definition:G-Delta Set",
"Definition:Relative Complement",
"Definition:F-Sigma Set"
] | [
"Definition:G-Delta Set",
"Definition:Countable Set",
"Definition:Set",
"Definition:Open Set/Topology",
"De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Intersection",
"Definition:Closed Set/Topology",
"Definition:Closed Set/Topology",
"Definition:Countable Set",
"Definitio... |
proofwiki-10485 | Bounded Piecewise Continuous Function has Improper Integrals | Let $f$ be a real function defined on a closed interval $\closedint a b$, $a < b$.
Let $f$ be piecewise continuous and bounded on $\closedint a b$.
Then $f$ is a piecewise continuous function with improper integrals. | Let $f$ be piecewise continuous and bounded on $\closedint a b$.
It is sufficient to prove that the improper integral $\ds \int_{ {x_{i - 1} }^+}^{ {x_i}^-} \map f x \rd x$ exists for every $i \in \set {1, 2, \ldots, n}$.
Let $i \in \set {1, 2, \ldots, n}$.
Let $c$ be a point in $\openint {x_{i − 1} } {x_i}$.
By defini... | Let $f$ be a [[Definition:Real Function|real function]] defined on a [[Definition:Closed Real Interval|closed interval]] $\closedint a b$, $a < b$.
Let $f$ be [[Definition:Bounded Piecewise Continuous Function|piecewise continuous and bounded]] on $\closedint a b$.
Then $f$ is a [[Definition:Piecewise Continuous Fu... | Let $f$ be [[Definition:Bounded Piecewise Continuous Function|piecewise continuous and bounded]] on $\closedint a b$.
It is sufficient to prove that the [[Definition:Improper Integral over Open Interval|improper integral]] $\ds \int_{ {x_{i - 1} }^+}^{ {x_i}^-} \map f x \rd x$ exists for every $i \in \set {1, 2, \ldot... | Bounded Piecewise Continuous Function has Improper Integrals | https://proofwiki.org/wiki/Bounded_Piecewise_Continuous_Function_has_Improper_Integrals | https://proofwiki.org/wiki/Bounded_Piecewise_Continuous_Function_has_Improper_Integrals | [
"Piecewise Continuous Functions"
] | [
"Definition:Real Function",
"Definition:Real Interval/Closed",
"Definition:Piecewise Continuous Function/Bounded",
"Definition:Piecewise Continuous Function/Improper Integrals"
] | [
"Definition:Piecewise Continuous Function/Bounded",
"Definition:Improper Integral/Open Interval",
"Definition:Improper Integral/Open Interval",
"Bounded Piecewise Continuous Function is Darboux Integrable",
"Definition:Darboux Integrable Function",
"Definition:Darboux Integrable Function",
"Definition:R... |
proofwiki-10486 | F-Sigma Set is not necessarily Closed Set | Let $T = \struct {S, \tau}$ be a topological space.
Let $X$ be an $F_\sigma$ set of $T$.
Then it is not necessarily the case that $X$ is a closed set of $T$. | Let $T = \struct {S, \tau}$ be a finite complement topology on an uncountable set $S$.
Let $X \subseteq S$ be an $F_\sigma$ set of $T$.
From $F_\sigma$ and $G_\delta$ Subsets of Uncountable Finite Complement Space:
:$X$ is neither open nor closed in $T$.
Hence the result.
{{qed}} | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $X$ be an [[Definition:F-Sigma Set|$F_\sigma$ set]] of $T$.
Then it is not necessarily the case that $X$ is a [[Definition:Closed Set (Topology)|closed set]] of $T$. | Let $T = \struct {S, \tau}$ be a [[Definition:Uncountable Finite Complement Topology|finite complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
Let $X \subseteq S$ be an [[Definition:F-Sigma Set|$F_\sigma$ set]] of $T$.
From [[F-Sigma and G-Delta Subsets of Uncountable Finite Complement Sp... | F-Sigma Set is not necessarily Closed Set | https://proofwiki.org/wiki/F-Sigma_Set_is_not_necessarily_Closed_Set | https://proofwiki.org/wiki/F-Sigma_Set_is_not_necessarily_Closed_Set | [
"F-Sigma Sets",
"Closed Sets"
] | [
"Definition:Topological Space",
"Definition:F-Sigma Set",
"Definition:Closed Set/Topology"
] | [
"Definition:Finite Complement Topology/Uncountable",
"Definition:Uncountable/Set",
"Definition:F-Sigma Set",
"F-Sigma and G-Delta Subsets of Uncountable Finite Complement Space",
"Definition:Open Set/Topology",
"Definition:Closed Set/Topology"
] |
proofwiki-10487 | Not every Closed Set is G-Delta Set | Let $T = \struct {S, \tau}$ be a topological space.
Let $V$ be a closed set of $T$.
Then it is not necessarily the case that $V$ is a $G_\delta$ set of $T$. | Let $T = \struct {S, \tau}$ be a finite complement topology on an uncountable set $S$.
Let $V$ be a closed set of $T$.
From Closed Set of Uncountable Finite Complement Topology is not $G_\delta$:
:$V$ is not a $G_\delta$ set of $T$.
Hence the result.
{{qed}} | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $V$ be a [[Definition:Closed Set (Topology)|closed set]] of $T$.
Then it is not necessarily the case that $V$ is a [[Definition:G-Delta Set|$G_\delta$ set]] of $T$. | Let $T = \struct {S, \tau}$ be a [[Definition:Uncountable Finite Complement Topology|finite complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
Let $V$ be a [[Definition:Closed Set (Topology)|closed set]] of $T$.
From [[Closed Set of Uncountable Finite Complement Topology is not G-Delta|Cl... | Not every Closed Set is G-Delta Set | https://proofwiki.org/wiki/Not_every_Closed_Set_is_G-Delta_Set | https://proofwiki.org/wiki/Not_every_Closed_Set_is_G-Delta_Set | [
"G-Delta Sets",
"Closed Sets"
] | [
"Definition:Topological Space",
"Definition:Closed Set/Topology",
"Definition:G-Delta Set"
] | [
"Definition:Finite Complement Topology/Uncountable",
"Definition:Uncountable/Set",
"Definition:Closed Set/Topology",
"Closed Set of Uncountable Finite Complement Topology is not G-Delta",
"Definition:G-Delta Set"
] |
proofwiki-10488 | Open Set of Uncountable Finite Complement Topology is not F-Sigma | Let $T = \struct {S, \tau}$ be a finite complement topology on an uncountable set $S$.
Let $U \in \tau$ be an open set of $T$.
Then $U$ is not an $F_\sigma$ set. | Let $U$ be an open set of $T$.
As $S$ is uncountable, then so is $U$.
By the definition of a finite complement topology, all closed sets of $T$ are finite.
From Countable Union of Countable Sets is Countable, $U$ can not be expressed as the union of a countable number of closed sets.
So by definition $U$ is not an $F_\... | Let $T = \struct {S, \tau}$ be a [[Definition:Uncountable Finite Complement Topology|finite complement topology]] on an [[Definition:Uncountable Set|uncountable]] set $S$.
Let $U \in \tau$ be an [[Definition:Open Set (Topology)|open set]] of $T$.
Then $U$ is not an [[Definition:F-Sigma Set|$F_\sigma$ set]]. | Let $U$ be an [[Definition:Open Set (Topology)|open set]] of $T$.
As $S$ is [[Definition:Uncountable Set|uncountable]], then so is $U$.
By the definition of a [[Definition:Finite Complement Topology|finite complement topology]], all [[Definition:Closed Set (Topology)|closed sets]] of $T$ are [[Definition:Finite Set|f... | Open Set of Uncountable Finite Complement Topology is not F-Sigma | https://proofwiki.org/wiki/Open_Set_of_Uncountable_Finite_Complement_Topology_is_not_F-Sigma | https://proofwiki.org/wiki/Open_Set_of_Uncountable_Finite_Complement_Topology_is_not_F-Sigma | [
"Uncountable Finite Complement Topologies",
"Examples of F-Sigma Sets"
] | [
"Definition:Finite Complement Topology/Uncountable",
"Definition:Uncountable/Set",
"Definition:Open Set/Topology",
"Definition:F-Sigma Set"
] | [
"Definition:Open Set/Topology",
"Definition:Uncountable/Set",
"Definition:Finite Complement Topology",
"Definition:Closed Set/Topology",
"Definition:Finite Set",
"Countable Union of Countable Sets is Countable",
"Definition:Set Union",
"Definition:Countable Set",
"Definition:Closed Set/Topology",
... |
proofwiki-10489 | Closed Set of Uncountable Finite Complement Topology is not G-Delta | Let $T = \struct {S, \tau}$ be a finite complement topology on an uncountable set $S$.
Let $V \in \tau$ be a closed set of $T$.
Then $V$ is not a $G_\delta$ set. | Let $V$ be a closed set of $T$.
{{AimForCont}} $V$ is $G_\delta$ set.
Then by Complement of $G_\delta$ Set is $F_\sigma$ Set:
:$S \setminus V$ is an $F_\sigma$ set.
By definition of closed set, $S \setminus V$ is an open set of $T$.
But by Open Set of Uncountable Finite Complement Topology is not $F_\sigma$:
:$S \setmi... | Let $T = \struct {S, \tau}$ be a [[Definition:Uncountable Finite Complement Topology|finite complement topology]] on an [[Definition:Uncountable Set|uncountable]] set $S$.
Let $V \in \tau$ be a [[Definition:Closed Set (Topology)|closed set]] of $T$.
Then $V$ is not a [[Definition:G-Delta Set|$G_\delta$ set]]. | Let $V$ be a [[Definition:Closed Set (Topology)|closed set]] of $T$.
{{AimForCont}} $V$ is [[Definition:G-Delta Set|$G_\delta$ set]].
Then by [[Complement of G-Delta Set is F-Sigma Set|Complement of $G_\delta$ Set is $F_\sigma$ Set]]:
:$S \setminus V$ is an [[Definition:F-Sigma Set|$F_\sigma$ set]].
By definition of... | Closed Set of Uncountable Finite Complement Topology is not G-Delta | https://proofwiki.org/wiki/Closed_Set_of_Uncountable_Finite_Complement_Topology_is_not_G-Delta | https://proofwiki.org/wiki/Closed_Set_of_Uncountable_Finite_Complement_Topology_is_not_G-Delta | [
"Uncountable Finite Complement Topologies",
"Examples of G-Delta Sets"
] | [
"Definition:Finite Complement Topology/Uncountable",
"Definition:Uncountable/Set",
"Definition:Closed Set/Topology",
"Definition:G-Delta Set"
] | [
"Definition:Closed Set/Topology",
"Definition:G-Delta Set",
"Complement of G-Delta Set is F-Sigma Set",
"Definition:F-Sigma Set",
"Definition:Closed Set/Topology",
"Definition:Open Set/Topology",
"Open Set of Uncountable Finite Complement Topology is not F-Sigma",
"Definition:F-Sigma Set",
"Proof by... |
proofwiki-10490 | G-Delta Set is not necessarily Open Set | Let $T = \struct {S, \tau}$ be a topological space.
Let $X$ be a $G_\delta$ set of $T$.
Then it is not necessarily the case that $X$ is a open set of $T$. | Let $T = \struct {S, \tau}$ be a finite complement topology on an uncountable set $S$.
Let $X \subseteq S$ be a $G_\delta$ set of $T$.
From $F_\sigma$ and $G_\delta$ Subsets of Uncountable Finite Complement Space:
:$X$ is neither open nor closed in $T$.
Hence the result.
{{qed}} | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $X$ be a [[Definition:G-Delta Set|$G_\delta$ set]] of $T$.
Then it is not necessarily the case that $X$ is a [[Definition:Open Set (Topology)|open set]] of $T$. | Let $T = \struct {S, \tau}$ be a [[Definition:Uncountable Finite Complement Topology|finite complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
Let $X \subseteq S$ be a [[Definition:G-Delta Set|$G_\delta$ set]] of $T$.
From [[F-Sigma and G-Delta Subsets of Uncountable Finite Complement Spa... | G-Delta Set is not necessarily Open Set | https://proofwiki.org/wiki/G-Delta_Set_is_not_necessarily_Open_Set | https://proofwiki.org/wiki/G-Delta_Set_is_not_necessarily_Open_Set | [
"G-Delta Sets",
"Open Sets"
] | [
"Definition:Topological Space",
"Definition:G-Delta Set",
"Definition:Open Set/Topology"
] | [
"Definition:Finite Complement Topology/Uncountable",
"Definition:Uncountable/Set",
"Definition:G-Delta Set",
"F-Sigma and G-Delta Subsets of Uncountable Finite Complement Space",
"Definition:Open Set/Topology",
"Definition:Closed Set/Topology"
] |
proofwiki-10491 | Not every Open Set is F-Sigma Set | Let $T = \struct {S, \tau}$ be a topological space.
Let $V$ be an open set of $T$.
Then it is not necessarily the case that $V$ is an $F_\sigma$ set of $T$. | Let $T = \struct {S, \tau}$ be a finite complement topology on an uncountable set $S$.
Let $U$ be an open set of $T$.
From Closed Set of Uncountable Finite Complement Topology is not $F_\sigma$:
:$U$ is not an $F_\sigma$ set of $T$.
Hence the result.
{{qed}} | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $V$ be an [[Definition:Open Set (Topology)|open set]] of $T$.
Then it is not necessarily the case that $V$ is an [[Definition:F-Sigma Set|$F_\sigma$ set]] of $T$. | Let $T = \struct {S, \tau}$ be a [[Definition:Uncountable Finite Complement Topology|finite complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
Let $U$ be an [[Definition:Open Set (Topology)|open set]] of $T$.
From [[Open Set of Uncountable Finite Complement Topology is not F-Sigma|Closed ... | Not every Open Set is F-Sigma Set | https://proofwiki.org/wiki/Not_every_Open_Set_is_F-Sigma_Set | https://proofwiki.org/wiki/Not_every_Open_Set_is_F-Sigma_Set | [
"F-Sigma Sets",
"Open Sets"
] | [
"Definition:Topological Space",
"Definition:Open Set/Topology",
"Definition:F-Sigma Set"
] | [
"Definition:Finite Complement Topology/Uncountable",
"Definition:Uncountable/Set",
"Definition:Open Set/Topology",
"Open Set of Uncountable Finite Complement Topology is not F-Sigma",
"Definition:F-Sigma Set"
] |
proofwiki-10492 | Limit Point of Sequence may only be Adherent Point of Range | Let $T = \struct {S, \tau}$ be a topological space.
Let $A \subseteq S$.
Let $\sequence {x_n}$ be a sequence in $A$.
Let $\alpha$ be a limit of $\sequence {x_n}$.
Then $\alpha$ may be only an adherent point of $A$ and not a limit point of $A$. | Let $T = \struct {S, \tau}$ be the discrete space on $S$.
Let $x \in S$.
Then by definition of discrete space:
:$U = \set x$ is an open set of $T$.
Consider the sequence $\sequence {x_n}$ defined as:
:$\forall n \in \N: x_n = x$
That is:
:$\sequence {x_n} = \tuple {x, x, x, \ldots}$
From Limit Point of Sequence in Disc... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $A \subseteq S$.
Let $\sequence {x_n}$ be a [[Definition:Sequence|sequence]] in $A$.
Let $\alpha$ be a [[Definition:Limit of Sequence (Topology)|limit of $\sequence {x_n}$]].
Then $\alpha$ may be only an [[Definition:Adherent... | Let $T = \struct {S, \tau}$ be the [[Definition:Discrete Space|discrete space]] on $S$.
Let $x \in S$.
Then by definition of [[Definition:Discrete Space|discrete space]]:
:$U = \set x$ is an [[Definition:Open Set (Topology)|open set]] of $T$.
Consider the [[Definition:Sequence|sequence]] $\sequence {x_n}$ defined a... | Limit Point of Sequence may only be Adherent Point of Range | https://proofwiki.org/wiki/Limit_Point_of_Sequence_may_only_be_Adherent_Point_of_Range | https://proofwiki.org/wiki/Limit_Point_of_Sequence_may_only_be_Adherent_Point_of_Range | [
"Limit Points",
"Adherent Points of Sets"
] | [
"Definition:Topological Space",
"Definition:Sequence",
"Definition:Limit of Sequence/Topological Space",
"Definition:Adherent Point of Set",
"Definition:Limit Point/Topology/Set"
] | [
"Definition:Discrete Topology",
"Definition:Discrete Topology",
"Definition:Open Set/Topology",
"Definition:Sequence",
"Limit Point of Sequence in Discrete Space not always Limit Point of Open Set",
"Definition:Limit Point/Topology/Set",
"Limit Point of Sequence is Adherent Point of Range",
"Definitio... |
proofwiki-10493 | Limit Point of Sequence in Discrete Space not always Limit Point of Open Set | Let $T = \struct {S, \tau}$ be a discrete topological space.
Let $U \in \tau$ be an open set of $T$.
Let $\sequence {x_n}$ be a sequence in $U$.
Let $x$ be the limit of $\sequence {x_n}$.
Then $x$ is not always a limit point of $U$. | Let $x \in S$.
By definition of discrete space:
:$U = \set x$ is an open set of $T$.
Consider the sequence $\sequence {x_n}$ defined as:
:$\forall n \in \N: x_n = x$
That is:
:$\sequence {x_n} = \tuple {x, x, x, \ldots}$
Thus $x$ is the limit point of $\sequence {x_n}$.
But:
:$U \setminus \set x = \O$
and so $x$ is not... | Let $T = \struct {S, \tau}$ be a [[Definition:Discrete Topology|discrete topological space]].
Let $U \in \tau$ be an [[Definition:Open Set (Topology)|open set]] of $T$.
Let $\sequence {x_n}$ be a [[Definition:Sequence|sequence]] in $U$.
Let $x$ be the [[Definition:Limit of Sequence (Topology)|limit]] of $\sequence {... | Let $x \in S$.
By definition of [[Definition:Discrete Space|discrete space]]:
:$U = \set x$ is an [[Definition:Open Set (Topology)|open set]] of $T$.
Consider the [[Definition:Sequence|sequence]] $\sequence {x_n}$ defined as:
:$\forall n \in \N: x_n = x$
That is:
:$\sequence {x_n} = \tuple {x, x, x, \ldots}$
Thus $... | Limit Point of Sequence in Discrete Space not always Limit Point of Open Set | https://proofwiki.org/wiki/Limit_Point_of_Sequence_in_Discrete_Space_not_always_Limit_Point_of_Open_Set | https://proofwiki.org/wiki/Limit_Point_of_Sequence_in_Discrete_Space_not_always_Limit_Point_of_Open_Set | [
"Discrete Topologies",
"Limit Points",
"Limits"
] | [
"Definition:Discrete Topology",
"Definition:Open Set/Topology",
"Definition:Sequence",
"Definition:Limit of Sequence/Topological Space",
"Definition:Limit Point/Topology/Set"
] | [
"Definition:Discrete Topology",
"Definition:Open Set/Topology",
"Definition:Sequence",
"Definition:Limit of Sequence/Topological Space",
"Definition:Limit Point/Topology/Set"
] |
proofwiki-10494 | Accumulation Point of Sequence of Distinct Terms is Omega-Accumulation Point of Range | Let $T = \struct {S, \tau}$ be a topological space.
Let $\sequence {x_n}$ be a sequence of distinct terms of $S$.
Let $\alpha$ be an accumulation point of $\sequence {x_n}$.
Then $\alpha$ is also an $\omega$-accumulation point of $\set {x_n: n \in \N}$. | Let $U$ be an open set of $T$ containing $\alpha$.
By definition of accumulation point of $\sequence {x_n}$, $U$ contains infinitely many terms of $\sequence {x_n}$.
As $\sequence {x_n}$ is a sequence of distinct terms:
$U$ contains infinitely many elements of $\set {x_n: n \in \N}$.
Thus by definition $\alpha$ is an $... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $\sequence {x_n}$ be a [[Definition:Sequence of Distinct Terms|sequence of distinct terms]] of $S$.
Let $\alpha$ be an [[Definition:Accumulation Point of Sequence|accumulation point]] of $\sequence {x_n}$.
Then $\alpha$ is also... | Let $U$ be an [[Definition:Open Set (Topology)|open set]] of $T$ containing $\alpha$.
By definition of [[Definition:Accumulation Point of Sequence|accumulation point]] of $\sequence {x_n}$, $U$ contains [[Definition:Infinite Set|infinitely many]] [[Definition:Term of Sequence|terms]] of $\sequence {x_n}$.
As $\sequen... | Accumulation Point of Sequence of Distinct Terms is Omega-Accumulation Point of Range | https://proofwiki.org/wiki/Accumulation_Point_of_Sequence_of_Distinct_Terms_is_Omega-Accumulation_Point_of_Range | https://proofwiki.org/wiki/Accumulation_Point_of_Sequence_of_Distinct_Terms_is_Omega-Accumulation_Point_of_Range | [
"Accumulation Points",
"Omega-Accumulation Points"
] | [
"Definition:Topological Space",
"Definition:Sequence of Distinct Terms",
"Definition:Accumulation Point/Sequence",
"Definition:Omega-Accumulation Point"
] | [
"Definition:Open Set/Topology",
"Definition:Accumulation Point/Sequence",
"Definition:Infinite Set",
"Definition:Term of Sequence",
"Definition:Sequence of Distinct Terms",
"Definition:Infinite Set",
"Definition:Element",
"Definition:Omega-Accumulation Point"
] |
proofwiki-10495 | Limit Point of Sequence is Adherent Point of Range | Let $T = \struct{S, \tau}$ be a topological space.
Let $\sequence{x_n}$ be a sequence in $S$.
Let $\alpha$ be a limit of $\sequence {x_n}$.
Then $\alpha$ is an adherent point of $\set{x_n: n \in \N}$. | By definition of Limit of sequence:
:$\forall U \in \tau : \exists N \in \N : \forall n \ge N : x_n \in U$
Hence:
:$\forall U \in \tau : U \cap \set{x_n: n \in \N} \ne \O$.
By definition $\alpha$ is an adherent point of $\set{x_n: n \in \N}$.
{{qed}}
Category:Adherent Points of Sets
Category:Limits
20738664di2sm6nkc6li... | Let $T = \struct{S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $\sequence{x_n}$ be a [[Definition:Sequence|sequence]] in $S$.
Let $\alpha$ be a [[Definition:Limit of Sequence (Topology)|limit of $\sequence {x_n}$]].
Then $\alpha$ is an [[Definition:Adherent Point of Set|adherent point]] of ... | By definition of [[Definition:Limit of Sequence (Topology)|Limit of sequence]]:
:$\forall U \in \tau : \exists N \in \N : \forall n \ge N : x_n \in U$
Hence:
:$\forall U \in \tau : U \cap \set{x_n: n \in \N} \ne \O$.
By definition $\alpha$ is an [[Definition:Adherent Point of Set|adherent point]] of $\set{x_n: n \in ... | Limit Point of Sequence is Adherent Point of Range | https://proofwiki.org/wiki/Limit_Point_of_Sequence_is_Adherent_Point_of_Range | https://proofwiki.org/wiki/Limit_Point_of_Sequence_is_Adherent_Point_of_Range | [
"Adherent Points of Sets",
"Limits"
] | [
"Definition:Topological Space",
"Definition:Sequence",
"Definition:Limit of Sequence/Topological Space",
"Definition:Adherent Point of Set"
] | [
"Definition:Limit of Sequence/Topological Space",
"Definition:Adherent Point of Set",
"Category:Adherent Points of Sets",
"Category:Limits"
] |
proofwiki-10496 | Limit Point of Sequence of Distinct Terms is Omega-Accumulation Point of Range | Let $T = \struct {S, \tau}$ be a topological space.
Let $\sequence {x_n}$ be a sequence of distinct terms of $S$.
Let $\alpha$ be a limit point of $\sequence {x_n}$.
Then $\alpha$ is also an $\omega$-accumulation point of $\set {x_n: n \in \N}$. | Let $\alpha$ be an limit point of $\sequence {x_n}$.
From Limit of Sequence is Accumulation Point, $\alpha$ is an accumulation point of $\sequence {x_n}$.
From Accumulation Point of Sequence of Distinct Terms is Omega-Accumulation Point of Range, $\alpha$ is an $\omega$-accumulation point of $\set {x_n: n \in \N}$.
{{q... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $\sequence {x_n}$ be a [[Definition:Sequence of Distinct Terms|sequence of distinct terms]] of $S$.
Let $\alpha$ be a [[Definition:Limit Point of Set|limit point]] of $\sequence {x_n}$.
Then $\alpha$ is also an [[Definition:Ome... | Let $\alpha$ be an [[Definition:Limit Point of Set|limit point]] of $\sequence {x_n}$.
From [[Limit of Sequence is Accumulation Point]], $\alpha$ is an [[Definition:Accumulation Point of Sequence|accumulation point]] of $\sequence {x_n}$.
From [[Accumulation Point of Sequence of Distinct Terms is Omega-Accumulation P... | Limit Point of Sequence of Distinct Terms is Omega-Accumulation Point of Range | https://proofwiki.org/wiki/Limit_Point_of_Sequence_of_Distinct_Terms_is_Omega-Accumulation_Point_of_Range | https://proofwiki.org/wiki/Limit_Point_of_Sequence_of_Distinct_Terms_is_Omega-Accumulation_Point_of_Range | [
"Limit Points",
"Omega-Accumulation Points"
] | [
"Definition:Topological Space",
"Definition:Sequence of Distinct Terms",
"Definition:Limit Point/Topology/Set",
"Definition:Omega-Accumulation Point"
] | [
"Definition:Limit Point/Topology/Set",
"Limit of Sequence is Accumulation Point",
"Definition:Accumulation Point/Sequence",
"Accumulation Point of Sequence of Distinct Terms is Omega-Accumulation Point of Range",
"Definition:Omega-Accumulation Point"
] |
proofwiki-10497 | Equivalence of Definitions of Isolated Point | {{TFAE|def = Isolated Point of Subset|view = isolated point}}
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$ be a subset of $S$. | === Definition 1 implies Definition 2 ===
Let $x$ be an isolated point of $H$ by definition 1.
Then by definition:
:$\exists U \in \tau: U \cap H = \set x$
Thus we have an open set in $T$ such that $x \in U$ contains no other point of $H$ than $x$.
Thus, by definition, $x$ is not a limit point of $H$.
Thus $x$ is an is... | {{TFAE|def = Isolated Point of Subset|view = isolated point}}
Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $H \subseteq S$ be a [[Definition:Subset|subset]] of $S$. | === Definition 1 implies Definition 2 ===
Let $x$ be an [[Definition:Isolated Point of Subset/Definition 1|isolated point of $H$ by definition 1]].
Then by definition:
:$\exists U \in \tau: U \cap H = \set x$
Thus we have an [[Definition:Open Set (Topology)|open set]] in $T$ such that $x \in U$ contains no other poi... | Equivalence of Definitions of Isolated Point | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Isolated_Point | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Isolated_Point | [
"Isolated Points"
] | [
"Definition:Topological Space",
"Definition:Subset"
] | [
"Definition:Isolated Point of Subset/Definition 1",
"Definition:Open Set/Topology",
"Definition:Limit Point/Topology/Set",
"Definition:Isolated Point of Subset/Definition 2",
"Definition:Isolated Point of Subset/Definition 2",
"Definition:Limit Point/Topology/Set",
"Definition:Open Set/Topology",
"Def... |
proofwiki-10498 | Equivalence of Definitions of Closed Set | {{TFAE|def = Closed Set (Topology)|view = closed set|context = Topology (Mathematical Branch)|contextview = topology}}
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$. | Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$. | {{TFAE|def = Closed Set (Topology)|view = closed set|context = Topology (Mathematical Branch)|contextview = topology}}
Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $H \subseteq S$. | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $H \subseteq S$. | Equivalence of Definitions of Closed Set | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Closed_Set | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Closed_Set | [
"Closed Sets"
] | [
"Definition:Topological Space"
] | [
"Definition:Topological Space"
] |
proofwiki-10499 | Kuratowski's Closure-Complement Problem | Let $T = \struct {S, \tau}$ be a topological space.
Let $A \subseteq S$ be a subset of $T$.
By successive applications of the operations of complement relative to $S$ and the closure, there can be as many as $14$ distinct subsets of $S$ (including $A$ itself). | That there can be as many as $14$ will be demonstrated by example. | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $A \subseteq S$ be a [[Definition:Subset|subset]] of $T$.
By successive applications of the [[Definition:Unary Operation|operations]] of [[Definition:Relative Complement|complement relative to $S$]] and the [[Definition:Closure (... | That there can be as many as $14$ will be demonstrated by example. | Kuratowski's Closure-Complement Problem | https://proofwiki.org/wiki/Kuratowski's_Closure-Complement_Problem | https://proofwiki.org/wiki/Kuratowski's_Closure-Complement_Problem | [
"Kuratowski's Closure-Complement Problem",
"Set Closures",
"Relative Complement",
"14"
] | [
"Definition:Topological Space",
"Definition:Subset",
"Definition:Operation/Unary Operation",
"Definition:Relative Complement",
"Definition:Closure (Topology)",
"Definition:Distinct",
"Definition:Subset"
] | [] |
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