nyuuzyou/wikis huggingface dataset:
Subring Module is Module

== Theorem ==
Let $\struct {R, +, \times}$ be a ring.
Let $\struct {S, +_S, \times_S}$ be a subring of $R$.
Let $\struct {G, +_G, \circ}_R$ be an $R$-module.
Let $\circ_S$ be the restriction of $\circ$ to $S \times G$.
Let $\struct {G, +_G, \circ_S}_S$ be subring module induced by $S$.
Then $\struct {G, +_G, \circ_S}_S$ is an $S$-module.
=== Unitary Subring ===
=== Special Case ===
== Proof ==
We have that:
:$\forall a, b \in S: a +_S b = a + b$
:$\forall a, b \in S: a \times_S b = a \times b$
:$\forall a \in S: \forall x \in G = a \circ_S x = a \circ x$
as $+_S$, $\times_S$ and $\circ_S$ are restrictions.
Let us verify the module axioms.
=== ===
We need to show that:
:$\forall a \in S: \forall x, y \in G: a \circ_S \paren {x +_G y} = a \circ_S x +_G a \circ_S y$
We have:
=== ===
We need to show that:
:$\forall a, b \in S: \forall x \in G: \paren {a +_S b} \circ_S x = a \circ_S x +_G b \circ_S y$
We have:
=== ===
We need to show that:
:$\forall a, b \in S: \forall x \in G: \paren {a \times_S b} \circ_S x = a \circ_S \paren {b \circ_S x}$
We have:
Thus $\struct {G, +_G, \circ_S}_S$ is an $S$-module.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 26$. Vector Spaces and Modules: Example $26.3$
Category:Subrings
Category:Module Theory
Category:Subring Module is Module
Subring Module is Module/Special Case

== Theorem ==
Let $S$ be a subring of the ring $\struct {R, +, \circ}$.
Let $\circ_S$ be the restriction of $\circ$ to $S \times R$.
Then $\struct {R, +, \circ_S}_S$ is an $S$-module.
=== Unitary Module ===
== Proof ==
From Ring is Module over Itself, it follows that:
:$\struct {R, +, \circ}_R$ is an $R$-module.
The result follows directly from Subring Module is Module.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 26$. Vector Spaces and Modules: Example $26.2$
Category:Subring Module is Module
Subring Module is Module/Special Case/Unitary Module

== Theorem ==
Let $S$ be a subring of the ring $\struct {R, +, \circ}$.
Let $\circ_S$ be the restriction of $\circ$ to $S \times R$.
Let $\struct {R, +, \circ}$ be a ring with unity such that $1_R$ is that unity.
Let $1_R \in S$.
Then $\struct {R, +, \circ_S}_S$ is a unitary $S$-module.
== Proof ==
From Subring Module is Module: Special Case, we have that $\struct {R, +, \circ_S}_S$ is an $S$-module.
Then $1_R$ is the unity of $\struct {S, +, \circ_S}$.
Thus $\struct {S, +, \circ_S}$ is also a ring with unity.
It follows from Ring with Unity is Module over Itself that $\struct {R, +, \circ_S}_S$ is a unitary module.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 26$. Vector Spaces and Modules: Example $26.2$
Category:Subring Module is Module
Subring Module is Module/Unitary

== Theorem ==
Let $\struct {R, +, \times}$ be a ring.
Let $\struct {S, +_S, \times_S}$ be a subring of $R$.
Let $\struct {G, +_G, \circ}_R$ be an $R$-module.
Let $\circ_S$ be the restriction of $\circ$ to $S \times G$.
Let $\struct {R, +, \times}$ be a ring with unity.
Let $\struct {G, +_G, \circ}_R$ be a unitary $R$-module.
Let $1_R \in S$.
Then $\struct{G, +_G, \circ_S}_S$ is also unitary.
== Proof ==
From Subring Module is Module, we have that $\struct {G, +_G, \circ_S}_S$ is an $S$-module.
It remains to be demonstrated that $\struct{G, +_G, \circ_S}_S$ is unitary.
To show this, we must prove that:
:$\forall x \in G: 1_R \circ_S x = x$
Since $1_R \in S$ by assumption, the product $1_R \circ_S x$ is defined. 
We now have:
and the proof is complete.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 26$. Vector Spaces and Modules: Example $26.3$
Category:Subring Module is Module
Subring Test

== Theorem ==
Let $S$ be a subset of a ring $\struct {R, +, \circ}$.
Then $\struct {S, +, \circ}$ is a subring of $\struct {R, +, \circ}$ these all hold:
:$(1): \quad S \ne \O$
:$(2): \quad \forall x, y \in S: x + \paren {-y} \in S$
:$(3): \quad \forall x, y \in S: x \circ y \in S$
== Proof ==
=== Necessary Condition ===
If $S$ is a subring of $\struct {R, +, \circ}$, the conditions hold by virtue of the ring axioms as applied to $S$.
=== Sufficient Condition ===
Conversely, suppose the conditions hold. We check that the ring axioms hold for $S$.
:$(1): \quad$ : By $1$ and $2$ above, it follows from:
:::: the One-Step Subgroup Test
::::Subgroup of Abelian Group is Abelian
:::that $\struct {S, +}$ is an abelian subgroup of $\struct {R, +}$, and therefore an abelian group.
:$(2): \quad$ : From $3$, $\struct {S, \circ}$ is closed.
:$(3): \quad$ : From Restriction of Associative Operation is Associative, $\circ$ is associative on $R$, therefore also associative on $S$.
:$(4): \quad$ : From Restriction of Operation Distributivity, $\circ$ distributes over $+$ for the whole of $R$, therefore for $S$ also.
So $\struct {S, +, \circ}$ is a ring, and therefore a subring of $\struct {R, +, \circ}$.
== Also defined as ==
Some sources insist that for $\struct {S, +, \circ}$ to be a subring of $\struct {R, +, \circ}$, the unity (if there is one) must be the same for both, but this is an extra condition as this is not necessarily the general case.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $21$. Rings and Integral Domains: Theorem $21.4$
* : Chapter $5$: Rings: $\S 19$. Subrings: Theorems $31, \ 32$
* : $\S 2.1$: Subrings: $2.2$ Lemma
* : $\S 56.2$ Subrings and subfields
Category:Subrings
Subring is not necessarily Ideal

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring.
Let $\struct {S, +_S, \circ_S}$ be a subring of $R$.
Then it is not necessarily the case that $S$ is also an ideal of $R$.
== Proof ==
Consider the field of real numbers $\struct {\R, +, \times}$.
We have that a field is by definition a ring, hence so is $\struct {\R, +, \times}$.
From Rational Numbers form Subfield of Real Numbers and Integers form Subdomain of Rationals, it follows that the integers $\struct {\Z, +, \times}$ are a subring of $\struct {\R, +, \times}$.
Consider $1 \in \Z$, and consider $\dfrac 1 2 \in \R$.
We have that $1 \times \dfrac 1 2 = \dfrac 1 2 \notin \Z$.
From this counterexample it is seen that $\Z$ is not an ideal of $R$.
Hence the result, again by Proof by Counterexample.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old
Category:Ideal Theory
Subring of Commutative Ring is Commutative

== Definition ==
Let $\struct{R, +, *}$ be a commutatve ring.
Let $\struct{S, +_S, *_S}$ be a subring of $\struct{R, +, *}$.
Then:
:$\struct{S, +_S, *_S}$ is a commutatve ring
== Proof ==
We have:
Category:Commutative Rings
Subring of Integers is Ideal

== Theorem ==
Let $\struct {\Z, +}$ be the additive group of integers.
Every subring of $\struct {\Z, +, \times}$ is an ideal of the ring $\struct {\Z, +, \times}$.
== Proof ==
Follows directly from:
:Subrings of Integers are Sets of Integer Multiples
and:
:Subgroup of Integers is Ideal.
== Sources ==
* : $\S 2.3$: Some properties of subrings and ideals
Category:Subrings
Category:Integers
Category:Ideal Theory
Subring of Non-Archimedean Division Ring

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring with non-archimedean norm $\norm {\, \cdot \,}$.
Let $\struct {S, \norm {\, \cdot \,}_S }$ be a normed division subring of $R$.
Then:
:$\norm {\, \cdot \,}_S$ is a non-archimedean norm.
== Proof ==
$\forall x, y \in S$:
Category:Normed Division Rings
Subring of Polynomials over Integral Domain Contains that Domain

== Theorem ==
Let $\struct {R, +, \circ}$ be a commutative ring.
Let $\struct {D, +, \circ}$ be an integral subdomain of $R$.
Let $x \in R$.
Let $D \sqbrk x$ denote the ring of polynomials in $x$ over $D$.
Then $D \sqbrk x$ contains $D$ as a subring and $x$ as an element.
== Proof ==
We have that $\ds \sum_{k \mathop = 0}^m a_k \circ x^k$ is a polynomial for all $m \in \Z_{\ge 0}$.
Set $m = 0$:
:$\ds \sum_{k \mathop = 0}^0 a_k \circ x^k = a_k \circ x^0 = a_k \circ 1_D = a_k$
Thus:
:$\ds \forall a_k \in D: \sum_{k \mathop = 0}^0 a_k \circ x^k \in D$
It follows directly that $D$ is a subring of $D \sqbrk x$ by applying the Subring Test on elements of $D$.
== Sources ==
* : $\S 64.1$ Polynomial rings over an integral domain
Category:Polynomial Theory
Category:Subrings
Subring of Polynomials over Integral Domain is Smallest Subring containing Element and Domain

== Theorem ==
Let $\struct {R, +, \circ}$ be a commutative ring.
Let $\struct {D, +, \circ}$ be an integral subdomain of $R$.
Let $x \in R$.
Let $D \sqbrk x$ denote the ring of polynomials in $x$ over $D$.
Then $D \sqbrk x$ is the smallest subring of $R$ which contains $D$ as a subring and $x$ as an element.
== Proof ==
== Sources ==
* : $\S 64.1$ Polynomial rings over an integral domain
Category:Polynomial Theory
Category:Subrings
Subrings of Integers are Sets of Integer Multiples

== Theorem ==
Let $\struct {\Z, +, \times}$ be the integral domain of integers.
The subrings of $\struct {\Z, +, \times}$ are the rings of integer multiples:
:$\struct {n \Z, +, \times}$
where $n \in \Z: n \ge 0$.
There are no other subrings of $\struct {\Z, +, \times}$ but these.
== Proof ==
From Integer Multiples form Commutative Ring, it is clear that $\struct {n \Z, +, \times}$ is a subring of $\struct {\Z, +, \times}$ when $n \ge 1$.
We also note that when $n = 0$, we have:
:$\struct {n \Z, +, \times} = \struct {0, +, \times}$
which is the null ring.
When $n = 1$, we have:
:$\struct {n \Z, +, \times} = \struct {\Z, +, \times}$
From Null Ring and Ring Itself are Subrings, these extreme cases are also subrings of $\struct {\Z, +, \times}$.
From Subgroups of Additive Group of Integers, the only additive subgroups of $\struct {\Z, +, \times}$ are $\struct {n \Z, +}$.
So there can be no subrings of $\struct {\Z, +, \times}$ which do not have $\struct {n \Z, +}$ as their additive group.
Hence the result.
== Examples ==
== Sources ==
* : Chapter $5$: Rings: $\S 19$. Subrings: Example $32$
* : $\S 2.3$: Some properties of subrings and ideals: Lemma $2.17$
Category:Subrings of Integers are Sets of Integer Multiples
Category:Subrings
Category:Integers
Subrings of Integers are Sets of Integer Multiples/Examples/Even Integers

== Theorem ==
Let $2 \Z$ be the set of even integers.
Then $\struct {2 \Z, +, \times}$ is a subring of $\struct {\Z, +, \times}$.
== Proof ==
From Subrings of Integers are Sets of Integer Multiples, a ring of the form $\struct {n \Z, +, \times}$ is a subring of $\struct {\Z, +, \times}$ when $n \ge 1$.
$\struct {2 \Z, +, \times}$ is such an example.
== Sources ==
* : Chapter $5$: Rings: $\S 19$. Subrings: Example $31$
* 
Category:Subrings of Integers are Sets of Integer Multiples
Subsemigroup/Examples/2x2 Matrices with One Non-Zero Entry

== Example of Subsemigroup ==
Let $\struct {S, \times}$ be the semigroup formed by the set of order $2$ square matrices over the real numbers $R$ under (conventional) matrix multiplication.
Let $T$ be the subset of $S$ consisting of the matrices of the form $\begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix}$ for $x \in \R$.
Then $\struct {T, \times}$ is a subsemigroup of $\struct {S, \times}$.
== Proof ==
From the Subsemigroup Closure Test it is sufficient to demonstrate that $\struct {T, \times}$ is closed.
Let $A = \begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} y & 0 \\ 0 & 0 \end{bmatrix}$.
Then:
Hence the result.
== Sources ==
* : $\S 32$ Identity element and inverses
Category:Examples of Subsemigroups
Subsemigroup/Examples/x+y-xy on Integers

== Example of Subsemigroup ==
Let $\struct {\Z, \circ}$ be the semigroup where $\circ: \Z \times \Z$ is the operation defined on the integers $\Z$ as:
:$\forall x, y \in \Z: x \circ y := x + y - x y$
Let $T$ be the set $\set {x \in \Z: x \le 1}$.
Then $\struct {T, \circ}$ is a subsemigroup of $\struct {\Z, \circ}$.
== Proof ==
It is established in Operation Defined as $x + y - x y$ on Integers that $\struct {\Z, \circ}$ is a semigroup.
From the Subsemigroup Closure Test it is sufficient to demonstrate that $\struct {T, \circ}$ is closed.
Let $x, y \in \Z_{\le 1}$.
:$\paren {x - 1}\le 0$ and $\paren {y - 1} \le 0$.
So:
So:
: $x \le 1, y \le 1 \implies x \circ y \le 1$
and so $\struct {T, \circ}$ is closed.
The result follows by definition of subsemigroup.
== Sources ==
* : Chapter $5$: Semigroups: Exercise $1$
Category:Examples of Subsemigroups
Subsemigroup Closure Test

== Theorem ==
To show that an algebraic structure $\struct {T, \circ_T}$ is a subsemigroup of a semigroup $\struct {S, \circ}$, we need to show only that:
:$(1): \quad T \subseteq S$
:$(2): \quad \circ$ is a closed operation in $T$.
where $\circ_T$ denotes the restriction of $\circ$ to $T$.
== Proof ==
From Restriction of Associative Operation is Associative, if $\circ$ is associative on $\struct {S, \circ}$, then it will also be associative on $\struct {T, \circ_T}$.
Thus we do not need to check for associativity in $\struct {T, \circ_T}$, as that has been inherited from its extension $\struct {S, \circ}$.
So, once we have established that $T \subseteq S$, all we need to do is to check for $\circ$ to be a closed operation in $T$.
== Sources ==
* : $\S 5.1$. Subsets closed to an operation
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets
* : $\S 32.1$ Identity element and inverses
Category:Subsemigroups
Subsemigroup of Cancellable Mappings is Subgroup of Invertible Mappings

== Theorem ==
Let $S$ be a set.
Let $S^S$ denote the set of mappings from $S$ to itself.
Let $\CC \subseteq S^S$ denote the set of cancellable mappings on $S$.
Let $\MM \subseteq S^S$ denote the set of invertible mappings on $S$.
Then:
:the subsemigroup $\struct {\CC, \circ}$ of $\struct {S^S, \circ}$ coincides with the subgroup $\struct {\MM, \circ}$ of $\struct {S^S, \circ}$
where $\circ$ denotes composition of mappings.
== Proof ==
From Set of Invertible Mappings forms Symmetric Group, we have that $\struct {\MM, \circ}$ is a group.
Hence, by definition, $\struct {\MM, \circ}$ is a subgroup of $\struct {S^S, \circ}$.
Recall from Bijection iff Left and Right Inverse that a mapping is invertible it is a bijection.
By definition, a cancellable mapping is a mapping both left cancellable and right cancellable.
From Injection iff Left Cancellable and Surjection iff Right Cancellable, a cancellable mapping is both an injection and a surjection.
That is, a mapping is cancellable mapping it is a bijection.
That is, $\struct {\CC, \circ}$ is exactly the same as $\struct {\MM, \circ}$.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.10 \ \text{(a)}$
Category:Cancellability
Category:Inverse Mappings
Category:Examples of Subgroups
Category:Examples of Subsemigroups
Subsemigroup of Monoid is not necessarily Monoid

== Theorem ==
Let $\struct {S, \circ}$ be a monoid whose identity is $e_S$.
Let $\struct {T, \circ}$ be a subsemigroup of $\struct {S, \circ}$
Then it is not necessarily the case that $\struct {T, \circ}$ has an identity.
== Proof ==
Consider the set of integers under multiplication $\struct {\Z, \times}$.
From Integers under Multiplication form Monoid, $\struct {\Z, \times}$ is a monoid.
Let $n \in \Z$ such that $n > 1$.
Let $n \Z$ be the set of integer multiples of $n$:
:$\set {x \in \Z: n \divides x}$
where $\divides$ denotes divisibility.
From Integer Multiples under Multiplication form Semigroup, $\struct {n \Z, \times}$ is a semigroup which has no identity.
By construction, $n \Z$ is a subset of $\Z$.
Hence $\struct {n \Z, \times}$ is a subsemigroup of $\struct {\Z, \times}$ which has no identity.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets
Category:Identity Elements
Category:Subsemigroups
Category:Monoids
Subsemigroup of Ordered Semigroup is Ordered

== Theorem ==
Let $\struct {S, \circ, \preceq}$ be an ordered semigroup.
Let $\struct {T, \circ_T}$ be a subsemigroup of $\struct {S, \circ}$.
Then the ordered structure $\struct {T, \circ_T, \preceq_T}$ is also an ordered semigroup.
In the above:
:$\circ_T$ denotes the operation induced on $T$ by $\circ$
:$\preceq_T$ denotes the restriction of $\preceq$ to $T \times T$.
== Proof ==
It is necessary to ascertain that $\struct {T, \circ {\restriction_T} }$ fulfils the ordered semigroup axioms:
In this context, we see that $\text {OS} 0$ and $\text {OS} 1$ are fulfilled a fortiori by dint of $\struct {T, \circ {\restriction_T} }$ being a subsemigroup of $\struct {S, \circ}$.
We have that $\struct {S, \circ, \preceq}$ is an ordered semigroup.
From Restriction of Ordering is Ordering, we have that $\preceq_T$ is an ordering.
Hence:
and:
Hence $\preceq_T$ fulfils Ordered Semigroup Axiom $\text {OS} 2$ on $\struct {T, \circ_T, \preceq_T}$.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups
Category:Ordered Semigroups
Subsequence/Examples/Example 1

== Examples of Subsequences ==
Let $\sequence {n_r}$ be the sequence in $\N$ defined such that $n_r = r + 1$.
Then:
:$\sequence {x_{n_r} } = \sequence {x_{r + 1} } = x_2, x_3, x_4, \ldots$
== Sources ==
* : $\S 5$: Subsequences: $\S 5.1$: Subsequences
Category:Examples of Subsequences
Subsequence/Examples/Example 2

== Examples of Subsequences ==
Let $\sequence {n_r}$ be defined such that $n_r = 2 r$.
Then:
:$\sequence {x_{n_r} } = \sequence {x_{2 r} } = x_2, x_4, x_6, \ldots$
== Sources ==
* : $\S 5$: Subsequences: $\S 5.1$: Subsequences
Category:Examples of Subsequences
Subsequence/Examples/Example 3

== Examples of Subsequences ==
Let $\sequence {n_r}$ be defined such that $n_r = 3 r + 5$.
Then:
:$\sequence {x_{n_r} } = \sequence {x_{3 r + 5} } = x_8, x_{11}, x_{14}, x_{17}, \ldots$
== Sources ==
* : $\S 5$: Subsequences: $\S 5.1$: Subsequences
Category:Examples of Subsequences
Subsequence/Examples/Example 4

== Examples of Subsequences ==
Let $\sequence {n_r}$ be defined such that $n_r = 2^r$.
Then:
:$\sequence {x_{n_r} } = \sequence {x_{2^r} } = x_2, x_4, x_8, x_{16}, \ldots$
== Sources ==
* : $\S 5$: Subsequences: $\S 5.1$: Subsequences
Category:Examples of Subsequences
Subsequence is Equivalent to Cauchy Sequence

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\sequence {x_n}$ be a Cauchy sequence in $R$.
Let $\sequence {x_{m_n} }$ be a subsequence of $\sequence {x_n}$.
Then:
:$\ds \lim_{n \mathop \to \infty} {x_n - x_{m_n} } = 0$
== Proof ==
From Subsequence of Cauchy Sequence in Normed Division Ring is Cauchy Sequence:
:$\sequence {x_{m_n} }$ is a Cauchy sequence
Let $\epsilon > 0$.
By definition of a Cauchy sequence:
:$\exists N: \forall n, m > N: \norm {x_n - x_m } < \epsilon$
From Index of Subsequence not Less than its Index:
$\forall n \in \N : m_n \ge n$
Thus:
:$\exists N: \forall n > N: \norm {x_n - x_{m_n} } < \epsilon$
By definition of convergence:
:$\ds \lim_{n \mathop \to \infty} {x_n - x_{m_n} } = 0$
Category:Cauchy Sequences
Category:Normed Division Rings
Subsequence of Cauchy Sequence in Normed Division Ring is Cauchy Sequence

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring with zero: $0$.
Let $\sequence {x_n}$ be a Cauchy sequence in $R$.
Let $\sequence {x_{n_r} }$ be a subsequence of $\sequence {x_n}$.
Then:
:$\sequence {x_{n_r} }$ is a Cauchy sequence in $R$.
== Proof ==
Let $\epsilon > 0$.
Since $\sequence {x_n}$ is a Cauchy sequence then:
:$\exists N: \forall n,m > N: \norm {x_n - x_m } < \epsilon$
Now let $R = N$.
Then from Strictly Increasing Sequence of Natural Numbers‎:
:$\forall r, s > R: n_r \ge r$ and $n_s \ge s$
Thus $n_r, n_s > N$ and so:
:$\norm {x_{n_r} - x_{n_s} } < \epsilon$
The result follows.
== Sources ==
Category:Cauchy Sequences
Category:Normed Division Rings
Subsequence of Real Sequence Diverging to Negative Infinity Diverges to Negative Infinity

== Theorem ==
Let $\sequence {x_n}_{n \mathop \in \N}$ be a real sequence with $x_n \to -\infty$.
Let $\sequence {x_{n_j} }_{j \mathop \in \N}$ be a subsequence of $\sequence {x_n}_{n \mathop \in \N}$.
Then: 
:$x_{n_j} \to -\infty$
== Proof ==
Let $M > 0$ be a real number.
From the definition of divergence to $-\infty$, there exists $N \in \N$ such that: 
:$x_n < -M$
for each $n \ge N$.
From Strictly Increasing Sequence of Natural Numbers, we have: 
:$n_j \ge j$ for each $j$.
So, we have: 
:$x_{n_j} < -M$
for each $j \ge N$.
Since $M$ was arbitrary, we have: 
:$x_{n_j} \to -\infty$
by the definition of divergence to $-\infty$.
Category:Divergent Sequences
Category:Limits of Sequences
Subsequence of Real Sequence Diverging to Positive Infinity Diverges to Positive Infinity

== Theorem ==
Let $\sequence {x_n}_{n \mathop \in \N}$ be a real sequence with $x_n \to +\infty$.
Let $\sequence {x_{n_j} }_{j \mathop \in \N}$ be a subsequence of $\sequence {x_n}_{n \mathop \in \N}$.
Then: 
:$x_{n_j} \to +\infty$
== Proof ==
Let $M > 0$ be a real number.
From the definition of divergence to $+\infty$, there exists $N \in \N$ such that: 
:$x_n > M$
for each $n \ge N$.
From Strictly Increasing Sequence of Natural Numbers, we have: 
:$n_j \ge j$ for each $j$.
So, we have: 
:$x_{n_j} > M$
for each $j \ge N$.
Since $M$ was arbitrary, we have: 
:$x_{n_j} \to \infty$
by the definition of divergence to $+\infty$.
Category:Divergent Sequences
Category:Limits of Sequences
Subsequence of Sequence in Metric Space with Limit

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $\sequence {x_n}$ be a sequence in $M$.
Let $x$ be a limit point of $S = \set {x_n: n \in \N}$, the set of members of $\sequence {x_n}$.
Then $\sequence {x_n}$ has a subsequence which converges to $x$.
== Proof ==
By Finite Subset of Metric Space has no Limit Points‎, $S$ is infinite (or it has no limit points).
We may assume that $x_n$ never equals $x$.
Otherwise we delete all instances of $x$ from $\sequence {x_n}$ and create a new sequence $x_m$ which is a subsequence of $x_n$ which does never equal $x$.
Then $S \setminus \set x$ is still infinite, and it still has $x$ as a limit point.
So, since $x$ is a limit point of $S$, there is an integer $\map n 1$, say, such that $x_{\map n 1} \in \map {B_1} x$, where $\map {B_1} x$ is the open $1$-ball of $x$.
Suppose we choose the integers $\map n 1 < \map n 2 < \cdots < \map n k$ so that $x_{\map n i} \in \map {B_{1/i} } x$ for $i = 1, 2, \ldots, k$.
Now we put $\epsilon = \min \set {\dfrac 1 {k+1}, \map d {x_1, x}, \map d {x_2, x}, \map d {x_{\map n k}, x} }$.
Since $x_n \ne x$ for any $n$, it follows that $\epsilon > 0$.
Since $x$ is a limit point of $S$, there exists an integer $\map n {k+1}$ such that $x_{\map n {k+1} } \in \map {B_\epsilon} x$.
Now $\epsilon$ has been chosen so as to force $\map n {k+1} > \map n k$, since $\map d {x_i, x} \ge \epsilon$ for all $i \le \map n k$.
Also, note that $x_{\map n {k+1} } \in \map {B_{\frac 1 {k+1}} } x$.
This completes the inductive step in constructing a subsequence.
This converges to $x$, because $\forall k_0 \in \N_{>0}$ and $\forall k \ge k_0$ we have $\map d {x_{\map n k}, x} < \dfrac 1 k \ge \dfrac 1 {k_0}$.
Hence the result.
== Sources ==
* : $7.2$: Sequential compactness: Lemma $7.2.5$
Category:Metric Spaces
Category:Limits of Sequences
Subsequence of Subsequence

== Theorem ==
Let $s$ be a set.
Let $\sequence {s_n}$ be a sequence in $S$.
Let $\sequence {s_m}$ be a subsequence of $\sequence {s_n}$.
Let $\sequence {s_k}$ be a subsequence of $\sequence {s_m}$.
Then $\sequence {s_k}$ is a subsequence of $\sequence {s_n}$.
== Proof ==
By definition, there exists a strictly increasing sequence $\sequence {n_r}$ in $\N$ such that:
:$\forall m \in \N: s_m = s_{n_r}$
Similarly, there exists a strictly increasing sequence $\sequence {m_s}$ in $\N$ such that:
:$\forall k \in \N: s_k = s_{m_s}$
We have that:
:$\forall k \in \N: s_k \in \sequence {s_m}$
and that:
:$\forall m \in \N: s_m \in \sequence {s_n}$
hence:
:$\forall k \in \N: s_k \in \sequence {s_n}$
Because $\sequence {s_k}$ is a subsequence of $\sequence {s_m}$, we have that:
:$k = m_s \implies k + 1 = m_s + p$ for some $p \in \Z_{>0}$
and:
:$m_s = n_r \implies m_s + p = n_r + q$ for some $q \in \Z_{>0}$
Hence:
:$k = n_r \implies k + 1 = n_r + q$
and so $\sequence {s_k}$ is a subsequence of $\sequence {s_n}$.
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 10$: Arbitrary Products: Exercise $4$
Category:Subsequences