nyuuzyou/wikis huggingface dataset:
Set of Rational Numbers is not G-Delta Set in Reals

== Theorem ==
Let $\Q$ be the set of rational numbers.
Let $\struct {\R, \tau}$ denote the real number line with the usual (Euclidean) topology.
Then $\Q$ is not a $G_\delta$ set in $\R$.
== Proof ==
 $\Q$ is a $G_\delta$ set in $\R$.
Let $\Q = \ds \bigcap_{i \mathop \in \N} V_i$.
Since Rational Numbers are Countably Infinite, there exists an enumeration of $\Q$.
Write $\Q = \set {q_i}_{i \mathop \in \N}$.
Define $U_i = V_i \setminus \set {q_i}$.
We show that $U_i$ is dense in $\R$.
:Let $A \subseteq \R$ be an open set of $\struct {\R, \tau}$.
:From Basis for Euclidean Topology on Real Number Line, the set of all open real intervals of $\R$ form a basis for $\struct {\R, \tau}$.
:So there exists $\openint a b \subseteq A$ for some $a < b$.
:One of $q_i \le a$, $q_i > a$ must hold.
:Suppose $q_i \le a$.
:By Between two Real Numbers exists Rational Number we have:
::$\exists r \in Q: a < r < b$.
:Since $r > a \ge q_i$:
::$r \in \Q \setminus \set {q_i} \subseteq V_i \setminus \set {q_i} = U_i$.
:Suppose $q_i > a$.
:By Between two Real Numbers exists Rational Number we have:
::$\exists r \in Q: a < r < \min \set {q_i, b}$.
:Since $r < q_i$:
::$r \in \Q \setminus \set {q_i} \subseteq V_i \setminus \set {q_i} = U_i$.
:In both cases we see that $r \in \openint a b \cap U_i \subseteq A \cap U_i$.
:Therefore $A \cap U_i \ne \O$.
:Thus $U_i$ is dense in $\R$.
Also we have:
By Real Number Line is Complete Metric Space and Baire Category Theorem, $\struct {\R, \tau}$ is a Baire space.
By definition of Baire Space, $\ds \bigcap_{i \mathop \in \N} U_i$ is dense in $\R$.
But Empty Set is Nowhere Dense.
This is a contradiction.
Hence $\Q$ is not a $G_\delta$ set in $\R$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $30$. The Rational Numbers: $2$
Category:Rational Numbers
Category:Real Number Line with Euclidean Topology
Category:G-Delta Sets
Set of Rational Numbers whose Numerator Divisible by p is Closed under Addition

== Theorem ==
Let $p$ be a prime number.
Let $A_p$ be the set of all rational numbers which, when expressed in canonical form has a numerator which is divisible by $p$.
Then $A_p$ is closed under rational addition.
== Proof ==
Let $a, b \in A_p$.
Then $a = \dfrac {p n_1} {d_1}, b = \dfrac {p n_1} {d_1}$ where:
:$n_1, n_2 \in \Z$
:$d_1, d_2 \in \Z_{>0}$
:$p n_1 \perp d_1, p n_2 \perp d_2$
Then:
From Euclid's Lemma for Prime Divisors, if $p \divides d_1 d_2$ then either $p \divides d_1$ or $p \divides d_2$.
But neither of these is the case, so $p \nmid d_1 d_2$.
Hence by Prime not Divisor implies Coprime:
:$p \perp d_1 d_2$
where $\perp$ denotes coprimeness.
So when $\dfrac {p \paren {n_1 d_2 + n_2 d_1} } {d_1 d_2}$ is expressed in canonical form, $p$ will still be a divisor of the numerator.
Hence the result.
== Sources ==
* : $\S 2.4$: The rational numbers and some finite fields: Exercise $6$
Category:Prime Numbers
Category:Rational Addition
Category:Algebraic Closure
Set of Rational Numbers whose Numerator Divisible by p is Closed under Multiplication

== Theorem ==
Let $p$ be a prime number.
Let $A_p$ be the set of all rational numbers which, when expressed in canonical form has a numerator which is divisible by $p$.
Then $A_p$ is closed under rational multiplication.
== Proof ==
Let $a, b \in A_p$.
Then $a = \dfrac {p n_1} {d_1}, b = \dfrac {p n_1} {d_1}$ where:
:$n_1, n_2 \in \Z$
:$d_1, d_2 \in \Z_{>0}$
:$p n_1 \perp d_1, p n_2 \perp d_2$
Then:
From Euclid's Lemma for Prime Divisors, if $p \divides d_1 d_2$ then either $p \divides d_1$ or $p \divides d_2$.
But neither of these is the case, so $p \nmid d_1 d_2$.
Hence by Prime not Divisor implies Coprime:
:$p \perp d_1 d_2$
where $\perp$ denotes coprimeness.
So when $\dfrac {p^2 \paren {n_1 n_2} } {d_1 d_2}$ is expressed in canonical form, $p$ will still be a divisor of the numerator.
Hence the result.
== Sources ==
* : $\S 2.4$: The rational numbers and some finite fields: Exercise $6$
Category:Prime Numbers
Category:Rational Multiplication
Category:Algebraic Closure
Set of Rationals Greater than Root 2 has no Smallest Element

== Theorem ==
Let $B$ be the set of all positive rational numbers $p$ such that $p^2 > 2$.
Then $B$ has no smallest element.
== Proof ==
 $p \in B$ is the smallest element of $B$.
Then by definition of $B$:
:$p^2 > 2$
Let $q = p - \dfrac {p_2 - 2} {2 p}$.
Then $q > p$, and:
Hence:
:$0 < q < p$
and so:
That means $q \in B$.
This contradicts our assertion that $p$ is the smallest element of $B$.
It follows that $B$ can have no smallest element.
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Introduction: $1.1$ Example
Category:Rational Numbers
Set of Rationals Less than Root 2 has no Greatest Element

== Theorem ==
Let $A$ be the set of all positive rational numbers $p$ such that $p^2 < 2$.
Then $A$ has no greatest element.
== Proof ==
 $p \in A$ is the greatest element of $A$.
Then by definition of $A$:
:$p^2 < 2$
Let $h \in \Q$ be a rational number such that $0 < h < 1$ such that:
:$h < \dfrac {2 - p^2} {2 p + 1}$
This is always possible, because by definition $2 - p^2 > 0$ and $2 p + 1 > 0$.
Let $q = p + h$.
Then $q > p$, and:
That means $q \in A$.
This contradicts our assertion that $p$ is the greatest element of $A$.
It follows that $A$ can have no greatest element.
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Introduction: $1.1$ Example
Category:Rational Numbers
Set of Real Numbers is Bounded Below iff Set of Negatives is Bounded Above

== Theorem ==
Let $S$ be a subset of the real numbers $\R$.
Let $T = \set {x \in \R: -x \in S}$ be the set of negatives of the elements of $S$.
Then:
:$S$ is bounded below
:
:$T$ is bounded above.
=== Corollary ===
== Proof ==
=== Sufficient Condition ===
Let $S$ be bounded below.
Then $S$ has a lower bound.
Let $B$ be a lower bound for $S$.
From Negative of Lower Bound of Set of Real Numbers is Upper Bound of Negatives:
:$B$ is a lower bound for $S$
:$-B$ is an upper bound for $T$.
It follows that $T$ is bounded above.
=== Necessary Condition ===
Let $T$ be bounded above.
Then $T$ has an upper bound.
From Negative of Upper Bound of Set of Real Numbers is Lower Bound of Negatives:
:$U$ is an upper bound for $T$
:
:$-U$ is a lower bound for $S$.
It follows that $S$ is bounded below.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 5$: Limits
Category:Real Analysis
Set of Real Numbers is Bounded Below iff Set of Negatives is Bounded Above/Corollary

== Corollary to Set of Real Numbers is Bounded Below iff Set of Negatives is Bounded Above ==
Let $S$ be a subset of the real numbers $\R$.
Let $T = \set {x \in \R: -x \in S}$ be the set of negatives of the elements of $S$.
Then:
:$S$ is bounded above
:
:$T$ is bounded below.
== Proof ==
Let $V$ be the set defined as:
:$V = \set {x \in \R: -x \in T}$
From Set of Real Numbers is Bounded Below iff Set of Negatives is Bounded Above:
:$T$ is bounded below
:
:$V$ is bounded above.
Then we have:
The result follows.
Category:Real Analysis
Set of Reciprocals of Positive Integers is Nowhere Dense in Reals

== Theorem ==
Let $N$ be the set defined as:
:$N := \set {\dfrac 1 n: n \in \Z_{>0} }$
where $\Z_{>0}$ is the set of (strictly) positive integers.
Let $\R$ denote the real number line with the usual (Euclidean) metric.
Then $N$ is nowhere dense in $\R$.
== Proof ==
From Zero is Limit Point of Integer Reciprocal Space, the only limit point of $N$ is $0$.
Hence:
:$\map \cl N = \set {\dfrac 1 n: n \in \Z_{>0} } \cup \set 0$
where $\map \cl N$ denotes the closure of $N$ in $\R$.
Trivially, $\map \cl N$ contains no open real intervals.
Hence no subset of $\map \cl N$ is open in $\R$.
Hence the union of all the subset of $\map \cl N$ which are open in $\R$ is empty.
That is, by definition, the interior of $N$ is empty.
That is:
:$\paren {\map \cl N}^\circ = \O$
and the result follows by definition of nowhere dense.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Example $3.7.27$
Category:Nowhere Denseness
Category:Real Number Line with Euclidean Metric
Category:Integer Reciprocal Space
Set of Relations can be Ordered by Subset Relation

== Theorem ==
Let $S \times T$ be the product of two sets.
Let $\RR$ be a set of relations on $S \times T$.
Then $\RR$ can be ordered by the subset relation.
== Proof ==
Let $R$ be a relation on $S \times T$.
By the definition of relation, $R$ is associated with a subset $R \subseteq S \times T$.
Thus $\RR$ is a subset of the power set $\powerset {S \times T}$.
The result follows from Subset Relation is Ordering.
== Also see ==
* Set of Mappings can be Ordered by Subset Relation
Category:Order Theory
Category:Subset Relation
Set of Residue Classes/Examples/2

== Example of Set of Residue Classes ==
The elements of $\Z_2$, the set of residue classes modulo $2$, are:
== Sources ==
* : Appendix $\text{A}.3$: Equivalence Relations: Problem Set $\text{A}.3$: $17$
* : Chapter $2$: Maps and relations on sets: Example $2.23$
Category:Examples of Sets of Residue Classes
Set of Residue Classes/Examples/3

== Example of Set of Residue Classes ==
The elements of $\Z_3$, the set of residue classes modulo $3$, are:
== Sources ==
* : Chapter $2$. Equivalence Relations: Exercise $5 \ \text{(ii)}$
* : Appendix $\text{A}.3$: Equivalence Relations: Problem Set $\text{A}.3$: $17$
* : Chapter $2$: 'And you do addition?': $\S 2.3$: Relations: Example $2.3.5$
Category:Examples of Sets of Residue Classes
Set of Residue Classes/Examples/4

== Example of Set of Residue Classes ==
The elements of $\Z_4$, the set of residue classes modulo $4$, are:
== Sources ==
* : $\S 2.5$. Congruence of integers: Example $40$
* : Chapter $1$: Integral Domains: $\S 6$. The Residue Classes: Example $7$
Category:Examples of Sets of Residue Classes
Set of Ring Elements forming Zero Product with given Element is Ideal

== Theorem ==
Let $\struct {R, +, \circ}$ be a commutative ring whose zero is $0_R$.
Let $a \in R$ be an arbitrary element of $R$.
Let $A$ be the subset of $R$ defined as:
:$A = \set {x \in R: x \circ a = 0_R}$
Then $A$ is an ideal of $A$.
== Proof ==
By definition of ring zero:
:$\forall x \in R: x \circ 0_R = 0_R$
Hence $0_R \in A$ and so $A \ne \O$.
Let $a, b \in A$.
Thus:
Then:
Hence the result, from Test for Ideal:
== Sources ==
* : Chapter $9$: Rings: Exercise $4$
Category:Ideal Theory
Set of Rotations in Space about Fixed Point forms Infinite Group

== Theorem ==
Let $\SS$ be a rigid body in space.
Let $O$ be a fixed point in space.
The set of all rotations of $\SS$ through some straight line through $O$ forms an infinite group.
== Proof ==
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 3$: Examples of Infinite Groups: $\text{(iii)}$
Category:Examples of Groups
Set of Rotations is Subgroup of Symmetry Group

== Theorem ==
Let $G$ be a symmetry group.
Let $H$ be the subset of $G$ consisting of the rotations in $G$ about a given axis.
Then $H$ is a subgroup of $G$.
== Proof ==
== Sources ==
* : Chapter $4$: Subgroups: Example $4.5$
Category:Symmetry Groups
Set of Row Matrices under Entrywise Addition forms Vector Space

== Corollary to Set of Matrices under Entrywise Addition forms Vector Space ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $\map {\MM_\GF} {1, n}$ be set of all row matrices over a field $\GF$.
Let $\struct {\map {\MM_\GF} {1, n}, +}$ denote the abelian group formed from $\map {\MM_\GF} {1, n}$ under matrix entrywise addition.
Let $\struct {\map {\MM_\GF} {1, n}, +, \times}_\GF$ denote the unitary module over $\GF$ where $\times$ denotes the matrix scalar product.
Then $\struct {\map {\MM_\GF} {1, n}, +, \times}_\GF$ forms a vector space.
== Proof ==
This is an instance of Set of Matrices under Entrywise Addition forms Vector Space with $m = 1$.
== Sources ==
* : Chapter $1$: Introductory Concepts: $1$. Basic Building Blocks: $4$. LINEAR VECTOR SPACE: Example $5$
Category:Set of Matrices under Entrywise Addition forms Vector Space
Category:Vector Spaces of Matrices
Category:Row Matrices
Set of Sequence Codes is Primitive Recursive

== Theorem ==
Let $\operatorname{Seq}$ be the set of all code numbers of finite sequences in $\N$.
Then $\operatorname{Seq}$ is primitive recursive.
== Proof ==
By the definition of a primitive recursive set, it is sufficient to show that the characteristic function $\chi_{\operatorname{Seq}}$ of $\operatorname{Seq}$ is primitive recursive.
Let $p: \N \to \N$ be the prime enumeration function.
Let $\map {\operatorname{len} } n$ be the length of $n$.
We note that $\map {\chi_{\operatorname{Seq}} } n = 1$ $\map p y$ divides $n$ for $1 \le y \le \map {\operatorname{len} } n$.
That is, $\map {\operatorname{div} } {n, \map p y} = 1$ for $1 \le y \le \map {\operatorname{len} } n$, where $\operatorname{div}$ is the divisor relation.
We then see that $\map {\operatorname{div} } {n, \map p y} = 1$ for $1 \le y \le \map {\operatorname{len} } n$ their product equals $1$.
So we can define $\chi_{\operatorname{Seq}}$ by:
:$\map {\chi_{\operatorname{Seq}} } n = \begin{cases}
\ds \prod_{y \mathop = 1}^{\map {\operatorname{len} } n} \map {\operatorname{div} } {n, \map p y} & : n > 1 \\
0 & : \text{otherwise}
\end{cases}$
Then we define $g: \N^2 \to \N$ as:
:$\map g {n, z} = \begin{cases}
1 & : z = 0 \\
\ds \prod_{y \mathop = 1}^z \map {\operatorname{div} } {n, \map p y} & : z \ne 0
\end{cases}$
We then apply Bounded Product is Primitive Recursive to the primitive recursive function $\map {\operatorname{div} } {n, \map p y}$, and see that $g$ is primitive recursive.
Finally, we have that:
:$\map {\chi_{\operatorname{Seq}} } n = \begin{cases}
\map g {n, \map {\operatorname{len} } n} & : n > 1 \\
0 & : \text{otherwise}
\end{cases}$
is obtained by substitution from:
* the primitive recursive function $\operatorname{len}$
* the primitive recursive function $g$
* the primitive recursive relation $>$
* the constant $1$
So $\chi_{\operatorname{Seq}}$ is primitive recursive.
Hence the result.
Category:Primitive Recursive Functions
Set of Sets/Examples/Set of Arbitrary Sets 1

== Example of Set of Sets ==
Let:
Let $\mathscr S = \set {A, B, C}$.
Then:
:$\mathscr S = \set {\set {1, 2, 3, 4}, \set {a, 3, 4}, \set {2, a} }$
Note that none of $a, 1, 2, 3, 4$ are elements of $S$.
== Sources ==
* : $\S 1.8$. Sets of sets
Category:Examples of Sets of Sets
Set of Sets/Examples/Set of Arbitrary Sets 2

== Example of Set of Sets ==
Let $A$ be the set of (strictly) positive odd integers less than $8$.
Let $B$ be the set of (strictly) positive even integers less than $8$.
Then:
Let $\mathscr S = \set {A, B}$.
Then:
:$\mathscr S = \set {\set {1, 3, 5, 7}, \set {2, 4, 6} }$
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 2$: Sets and Subsets
Category:Examples of Sets of Sets
Set of Sets/Examples/Set of Initial Segments

== Example of Set of Sets ==
Let $\Z$ denote the set of integers.
Let $\map \Z n$ denote the initial segment of $\Z_{> 0}$:
:$\map \Z n = \set {1, 2, \ldots, n}$
Let $\mathscr S := \set {\map \Z n: n \in \Z_{> 0} }$
That is, $\mathscr S$ is the set of all initial segments of $\Z_{> 0}$.
Then:
:$\mathscr S := \set {\set 1, \set {1, 2}, \set {1, 2, 3}, \ldots}$
and we have that:
:$\mathscr S \subsetneq \powerset \Z$
where $\powerset \Z$ denotes the power set of $\Z$.
== Sources ==
* : $\S 1.8$. Sets of sets: Example $28$
Category:Examples of Sets of Sets
Set of Sets can be Defined as Family

== Theorem ==
Let $\Bbb S$ be a set of sets.
Then $\Bbb S$ can be defined as an indexed family of sets.
== Proof ==
Let $S: \Bbb S \to \Bbb S$ denote the identity mapping on $\Bbb S$:
:$\forall i \in \Bbb S: S_i = i$
where we use $S_i$ to mean the image of $i$ under $S$:
:$S_i := \map S i$
Then we can consider $S$ as an indexing function from $\Bbb S$ to $\Bbb S$.
Hence in this case $\Bbb S$ is at the same time both:
:an indexing set
and:
:the set indexed by itself.
It follows that each of the sets $i \in \Bbb S$ is both:
:an index
and:
:a term $S_i$ of the family of elements of $\Bbb S$ indexed by $\Bbb S$.
Thus we would write $\Bbb S$ as:
:$\family {S_i}_{i \mathop \in \Bbb S}$
== Sources ==
* : $\S 9$: Families
Category:Set Systems
Category:Indexed Families
Set of Singletons is Smallest Basis of Discrete Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a discrete topological space.
Let $\BB = \set {\set x : x \in S}$.
Then $\BB$ is the smallest basis of $T$.
That is:
:$\BB$ is a basis of $T$
and:
:for every basis $\CC$ of $T$, $\BB \subseteq \CC$.
== Proof ==
By Basis for Discrete Topology $\BB$ is a basis of $T$.
It remains to be shown that $\BB$ is the smallest basis of $T$.
Let $\CC$ be a basis of $T$.
Let $A \in \BB$.
By definition of the set $\BB$:
:$\exists x \in S: A = \set x$
By definition of basis:
:$\exists B \in \CC: x \in B \subseteq A$
Then by Singleton of Element is Subset:
:$\set x \subseteq B$
Hence $B = A$ by definition of set equality.
Thus $A \in \CC$.
== Sources ==
* 
* 
Category:Discrete Topology
Set of Strictly Negative Integers is Primitive Recursive

== Theorem ==
Let $N \subseteq \N$ be the set of all $n \in \N$ such that:
:$n$ codes an integer $k$ such that $k < 0$.
Then $N$ is a primitive recursive set.
== Proof ==
By Set of Strictly Positive Integers is Primitive Recursive:
:$P = \set {n \in \N : k > 0}$
is primitive recursive.
By Complement of Primitive Recursive Set:
:$P^c = \set {n \in \N : k \le 0}$
is primitive recursive.
It is clear that:
:$N = P^c \setminus \set {n \in \N : k = 0}$
By Set Difference as Intersection with Relative Complement:
:$N = P^c \cap \relcomp \N {\set {n \in \N : k = 0}}$
By:
* Intersection of Primitive Recursive Sets
* Complement of Primitive Recursive Set
we only need to show that:
:$\set {n \in \N : k = 0}$
is primitive recursive.
As $k = 0 \le 0$:
:$n = - 2 k = 0$
Therefore:
:$\set {n \in \N : k = 0} = \set 0$
which is primitive recursive by:
:Set Containing Only Zero is Primitive Recursive
Category:Primitive Recursive Functions
Set of Strictly Positive Integers is Primitive Recursive

== Theorem ==
Let $P \subseteq \N$ be the set of all $n \in \N$ such that:
:$n$ codes an integer $k$ such that $k > 0$.
Then $P$ is a primitive recursive set.
== Theorem ==
Let $p : \N \to \N$ be defined as:
:$\map p n = \map {\operatorname{rem}} {n, 2}$
By:
* Constant Function is Primitive Recursive
* Remainder is Primitive Recursive
Suppose $n$ codes an integer $k$ such that $k > 0$.
Then:
:$n = 2 k - 1$
Therefore:
Suppose $n$ codes an integer $k$ such that $k \le 0$.
Then:
:$n = - 2 k$
Therefore:
By definition, $p$ is the characteristic function of $P$.
Thus, $P$ is primitive recursive by definition.
Category:Primitive Recursive Functions
Set of Strictly Positive Real Numbers has no Smallest Element

== Theorem ==
Let $\R_{>0}$ denote the set of strictly positive real numbers.
Then $\R_{>0}$ has no smallest element.
== Proof ==
 $\R_{>0}$ has a smallest element.
Let $m$ be that smallest element.
Then we have that:
:$0 < \dfrac m 2 < m$
But as $0 < \dfrac m 2$ it follows that $\dfrac m 2 \in \R_{>0}$.
This contradicts our assertion that $m$ is the smallest element of $\R_{>0}$.
Hence the result by Proof by Contradiction.
== Sources ==
* : $\S 2$: Continuum Property: Exercise $\S 2.10 \ (6)$
Category:Real Analysis
Set of Subfields forms Complete Lattice

== Theorem ==
Let $\struct {F, +, \circ}$ be a field.
Let $\mathbb F$ be the set of all subfields of $F$.
Then $\struct {\mathbb F, \subseteq}$ is a complete lattice.
== Proof ==
Let $\O \subset \mathbb S \subseteq \mathbb F$.
By Intersection of Subfields is Largest Subfield Contained in all Subfields:
:$\bigcap \mathbb S$ is the largest subfield of $F$ contained in each of the elements of $\mathbb S$.
By Intersection of Subfields Containing Subset is Smallest:
:The intersection of the set of all subfields of $F$ containing $\bigcup \mathbb S$ is the smallest subfield of $F$ containing $\bigcup \mathbb S$.
Thus:
:Not only is $\bigcap \mathbb S$ a lower bound of $\mathbb S$, but also the largest, and therefore an infimum.
:The supremum of $\mathbb S$ is the intersection of the set of all subfields of $R$.
Therefore $\struct {\mathbb F, \subseteq}$ is a complete lattice.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers: Theorem $23.1$
Category:Complete Lattices
Category:Subfields
Set of Subgroups forms Complete Lattice

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $\mathbb G$ be the set of all subgroups of $G$.
Then:
:$\struct {\mathbb G, \subseteq}$ is a complete lattice.
where for every set $\mathbb H$ of subgroups of $G$:
:the infimum of $\mathbb H$ necessarily admitted by $\mathbb H$ is $\ds \bigcap \mathbb H$.
== Proof 1 ==
== Proof 2 ==
Category:Examples of Complete Lattices
Category:Subgroups
Category:Set of Subgroups forms Complete Lattice
Set of Subgroups forms Complete Lattice/Proof 1

== Theorem ==
== Proof ==
Let $\O \subset \mathbb H \subseteq \mathbb G$.
By Intersection of Subgroups: General Result, $\bigcap \mathbb H$ is the largest subgroup of $G$ contained in each of the elements of $\mathbb H$.
Thus, not only is $\ds \bigcap \mathbb H$ a lower bound of $\mathbb H$, but also the largest, and therefore an infimum.
The supremum of $\mathbb H$ is the smallest subgroup of $G$ containing $\bigcup \mathbb H$.
Therefore $\struct {\mathbb G, \subseteq}$ is a complete lattice.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Theorem $14.6$
Category:Set of Subgroups forms Complete Lattice
Set of Subgroups forms Complete Lattice/Proof 2

== Theorem ==
== Proof ==
From Group is Subgroup of Itself:
:$\struct {G, \circ} \in \mathbb G$
Let $\mathbb H$ be a non-empty subset of $\mathbb G$.
From Intersection of Subgroups is Subgroup: General Result:
:$\ds \bigcap \mathbb H \in \mathbb G$
Hence, from Set of Subsets which contains Set and Intersection of Subsets is Complete Lattice:
:$\struct {\mathbb G, \subseteq}$ is a complete lattice
where $\ds \bigcap \mathbb H$ is the infimum of $\mathbb H$.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.11 \ \text{(c) (2)}$
Category:Set of Subgroups forms Complete Lattice
Set of Subgroups of Abelian Group form Subsemigroup of Power Structure

== Theorem ==
Let $\struct {G, \circ}$ be an abelian group.
Let $\struct {\powerset G, \circ_\PP}$ denote the power structure of $\struct {G, \circ}$.
Let $\SS$ be the set of all subgroups of $G$.
Then $\struct {\SS, \circ_\PP}$ is a subsemigroup of $\struct {\powerset G, \circ_\PP}$.
== Proof ==
From Power Structure of Semigroup is Semigroup:
:$\struct {\powerset S, \circ_\PP}$ is a semigroup.
Let $A$ and $B$ be arbitrary subgroups of $G$.
We need to show that $A \circ_\PP B$ is also a subgroup of $G$.
Let $x$ and $y$ be arbitrary elements of $A \circ_\PP B$.
Then:
Hence the result from the One-Step Subgroup Test.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 9$: Compositions Induced on the Set of All Subsets: Exercise $9.9$
Category:Abelian Groups
Category:Power Structures
Category:Subsemigroups
Set of Submagmas of Magma under Subset Relation forms Complete Lattice

== Theorem ==
Let $\struct {A, \odot}$ be a magma.
Let $\SS$ be the set of submagmas of $A$.
Then:
:the ordered set $\struct {\SS, \subseteq}$ is a complete lattice
where for every subset $\TT$ of $\SS$:
:the infimum of $\TT$ necessarily admitted by $\TT$ is $\bigcap \TT$.
== Proof ==
From Magma is Submagma of Itself:
:$\struct {A, \odot} \in \SS$
Let $\TT$ be a non-empty subset of $\SS$.
From Intersection of Submagmas is Largest Submagma:
:$\bigcap \TT \in \SS$
Hence, from Set of Subsets which contains Set and Intersection of Subsets is Complete Lattice:
:$\struct {\SS, \subseteq}$ is a complete lattice
where $\bigcap \TT$ is the infimum of $\TT$.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.11 \ \text{(c) (1)}$
Category:Examples of Complete Lattices
Category:Submagmas
Set of Submodules of Module is Complete Lattice

== Theorem ==
Let $R$ be a ring.
Let $\struct {G, +_G}$ be an abelian group.
Let $M = \struct {G, +, \circ}_R$ be an $R$-module.
Ordered by $\subseteq$, the set of all submodules of $M$ is a complete lattice.
== Proof ==
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Theorem $27.2$: Corollary
Category:Submodules