nyuuzyou/wikis huggingface dataset:
Convergent Sequence in Normed Vector Space is Weakly Convergent/Proof 2

== Theorem ==
== Proof ==
Let $\struct {X^\ast, \norm \cdot_{X^\ast} }$ be the normed dual space of $\struct {X, \norm \cdot}$.
Then, for each $f \in X^\ast$:
Category:Convergent Sequence in Normed Vector Space is Weakly Convergent
Convergent Sequence in P-adic Numbers has Unique Limit

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $\sequence {x_n} $ be a sequence in $\Q_p$.
Then $\sequence {x_n}$ can have at most one limit.
== Proof ==
From P-adic Metric on P-adic Numbers is Non-Archimedean Metric the $p$-adic metric is a metric on $\Q_p$.
By definition, the sequence $\sequence {x_n}$ converges in $\Q_p$ :
:$\sequence {x_n}$ converges in the $p$-adic metric. 
The result then follows from Convergent Sequence in Metric Space has Unique Limit.
Category:Sequences
Category:Convergence
Category:P-adic Number Theory
Convergent Sequence in Particular Point Space

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a particular point space.
Let $\sequence {a_i}$ be a convergent sequence in $T$.
Except for a finite number of indices, the terms of $\sequence {a_i}$ for which $a_i \ne p$ are all equal.
== Proof ==
Let $\sequence {a_i}$ be a convergent sequence in $T$ whose limit is $\alpha$.
Then by definition every open set in $T$ containing $\alpha$ contains all but a finite number of terms of $\sequence {a_i}$.
This includes the open set $\set {\alpha, p}$.
Thus $\sequence {a_i}$ is such that, except for a finite number of terms, all are equal either to $\alpha$ or $p$.
Thus $\sequence {a_i}$ is such that, except for a finite number of terms, all such that $a_i \ne p$, are equal to $\alpha$.
The result follows.
== Sources ==
* : Part $\text {II}$: Counterexamples: $8 \text { - } 10$. Particular Point Topology: $1$
Category:Particular Point Topology
Category:Convergent Sequences
Convergent Sequence in Set of Integers

== Theorem ==
Let $\sequence{x_n}_{n \in \N}$ be a sequence in the set $\Z$ of integers considered as a subspace of the real number line $\R$ under the Euclidean metric.
Then $\sequence{x_n}_{n \in \N}$ converges in $\R$ to a limit :
:$\exists k \in \N: \forall m \in \N: m > k: x_m = x_k$
That is, the sequence reaches some value of $\Z$ and "stays there".
=== Corollary ===
== Proof ==
Suppose $\sequence{x_n}_{n \in \N}$ converges to a limit $l$.
Consider the open set in $\R$:
:$U := \openint{l - \dfrac 1 2}{l + \dfrac 1 2}$
Then $\forall x \in \Z: x \in U \implies x = l$
It follows by definition of convergence that:
:$\exists k \in \N: \forall m \in \N: m > k: x_m = x_k$
where $x_k = l$.
Now suppose that:
:$\exists k \in \N: \forall m \in \N: m > k: x_m = x_k$
Then trivially $\sequence{x_n}_{n \in \N}$ converges to the limit $x_k$.
Category:Sequences
Category:Convergence
Convergent Sequence in Set of Integers/Corollary

== Theorem ==
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence of distinct terms in the set $\Z$.
Then $\sequence {x_n}_{n \mathop \in \N}$ is not convergent.
== Proof ==
By Convergent Sequence in Set of Integers, $\sequence {x_n}_{n \mathop \in \N}$ is convergent :
:$\exists k \in \N: \forall m \in \N: m > k: x_m = x_k$
But since $\sequence {x_n}_{n \mathop \in \N}$ is a sequence of distinct terms, for all $k \ne m$, one has $x_m \ne x_k$.
The result follows.
Category:Sequences
Category:Convergence
Convergent Sequence in Test Function Space multiplied by Smooth Function

== Theorem ==
Let $\alpha \in \map {C^\infty} {\R^d}$ be a smooth real multivariable function.
Let $\sequence {\phi_n}_{n \mathop \in \N}$ be a sequence in the test function space $\map \DD {\R^d}$.
Suppose $\sequence {\phi_n}_{n \mathop \in \N}$ converges in the test function space:
:$\phi_n \stackrel \DD {\longrightarrow} \phi$
Then $\sequence {\alpha \phi_n}_{n \mathop \in \N}$ converges in the test function space.
:$\alpha \phi_n \stackrel \DD {\longrightarrow} \alpha \phi$
== Proof ==
== Sources ==
* : Chapter $\S 6.4$: A glimpse of distribution theory. Multiplication by $C^\infty$ functions
Category:Leibniz's Rule
Convergent Sequence in Topological Vector Space is Cauchy

== Theorem ==
Let $\struct {X, \tau}$ be a topological vector space.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a convergent sequence with $x_n \to x$.
Then $\sequence {x_n}_{n \mathop \in \N}$ is Cauchy. 
== Proof ==
Let $V$ be an open neighborhood of ${\mathbf 0}_X$. 
From , there exists a symmetric open neighborhood $U$ of ${\mathbf 0}_X$ such that: 
:$U + U \subseteq V$
Since $x_n \to x$, there exists $N \in \N$ such that: 
:$x_n \in x + U$ 
for $n \ge N$. 
Since $U$ is symmetric, we have $U = -U$ and so: 
:$-x_m \in -x + U$ 
for $m \ge N$.
Then we have: 
:$x_n - x_m \in U + U \subseteq V$ 
for $n, m \ge N$.
So $\sequence {x_n}_{n \mathop \in \N}$ is Cauchy.
Category:Topological Vector Spaces
Category:Cauchy Sequences in Topological Vector Spaces
Convergent Sequence is Cauchy Sequence/Metric Space

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Every convergent sequence in $A$ is a Cauchy sequence.
== Proof ==
Let $\sequence {x_n}$ be a sequence in $A$ that converges to the limit $l \in A$.
Let $\epsilon > 0$. 
Then also $\dfrac \epsilon 2 > 0$.
Because $\sequence {x_n}$ converges to $l$ in $A$, we have:
:$\exists N_1 \in \R_{>0}: \forall n > N_1: \map d {x_n, l} < \dfrac \epsilon 2$
Because $\sequence {x_n}$ converges to $l$, there must also exist at least one sequence $\sequence {x_m}$ in $A$ that converges to the limit $l \in A$, for some $m \in \R_{>0}$.
We have:
:$\exists N_2 \in \R_{>0}: \forall m > N_2: \map d {x_m, l} < \dfrac \epsilon 2$
So if $N = \map \max {N_1, N_2}$, such that $m > N$ and $n > N$, then:
Thus $\sequence {x_n}$ is a Cauchy sequence.
== Also see ==
* Definition:Complete Metric Space, where the converse is true.
* Cauchy's Convergence Criterion on Real Numbers
* Cauchy's Convergence Criterion on Complex Numbers
== Sources ==
* : $3.11a$
* : $7.2$: Sequential compactness: Proposition $7.2.4$
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces: Complete Metric Spaces
* $4$: Complete Normed Spaces
Category:Convergent Sequence is Cauchy Sequence
Category:Convergent Sequences (Metric Space)
Convergent Sequence is Cauchy Sequence/Normed Division Ring

== Theorem ==
Let $\struct {R, \norm {\,\cdot\,}} $ be a normed division ring.
Every convergent sequence in $R$ is a Cauchy sequence.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Cauchy's Convergence Criterion on Real Numbers
* Cauchy's Convergence Criterion on Complex Numbers
== Sources ==
* : $\S 1.2$: Normed Fields: Definition $1.7$
Category:Convergent Sequence in Normed Division Ring is Cauchy Sequence
Category:Convergent Sequence is Cauchy Sequence
Category:Convergent Sequences in Normed Division Rings
Convergent Sequence is Cauchy Sequence/Normed Division Ring/Proof 1

== Theorem ==
== Proof ==
Let $\sequence {x_n}$ be a sequence in $R$ that converges to the limit $l \in R$.
Let $\epsilon > 0$. 
Then also $\dfrac \epsilon 2 > 0$.
Because $\sequence {x_n}$ converges to $l$, we have:
:$\exists N: \forall n > N: \norm {x_n - l} < \dfrac \epsilon 2$
So if $m > N$ and $n > N$, then:
Thus $\sequence {x_n}$ is a Cauchy sequence.
== Also see ==
* Cauchy's Convergence Criterion on Real Numbers
* Cauchy's Convergence Criterion on Complex Numbers
== Sources ==
* : $\S 1.2$: Normed Fields, Exercise $11 \ (1)$
Category:Convergent Sequence in Normed Division Ring is Cauchy Sequence
Convergent Sequence is Cauchy Sequence/Normed Division Ring/Proof 2

== Theorem ==
== Proof ==
Let $d$ be the metric induced on $R$ be the norm $\norm {\,\cdot\,}$.
Let $\sequence {x_n}$ be a sequence in $R$ that converges to the limit $l$ in $\struct {R, \norm {\,\cdot\,}}$.
Thus, by definition, $\sequence {x_n} $ converges to the limit $l$ in $\struct {R, d}$.
By Convergent Sequence is Cauchy Sequence in metric space, $\sequence {x_n} $ is a Cauchy sequence in $\struct {R, d}$.
Thus, by definition, $\sequence {x_n} $ is a Cauchy sequence in $\struct {R, \norm {\,\cdot\,}}$.
== Also see ==
* Cauchy's Convergence Criterion on Real Numbers
* Cauchy's Convergence Criterion on Complex Numbers
== Sources ==
* : $\S 1.2$: Normed Fields, Exercise $11 \ (1)$
Category:Convergent Sequence in Normed Division Ring is Cauchy Sequence
Convergent Sequence is Cauchy Sequence/Normed Vector Space

== Theorem ==
Let $\struct{X, \norm{\,\cdot\,} }$ be a normed vector space.
Every convergent sequence in $X$ is a Cauchy sequence.
== Proof ==
Let $\sequence {x_n}$ be a sequence in $X$ that converges to the limit $L \in X$.
Let $\epsilon > 0$. 
Then also $\dfrac \epsilon 2 > 0$.
Because $\sequence {x_n}$ converges to $L$, we have:
:$\exists N: \forall n > N: \norm {x_n - L} < \dfrac \epsilon 2$
So if $m > N$ and $n > N$, then:
Thus $\sequence {x_n}$ is a Cauchy sequence.
== Sources ==
* : $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces
Category:Convergent Sequences (Normed Vector Spaces)
Category:Convergent Sequence is Cauchy Sequence
Convergent Sequence of Continuous Real Functions is Integrable Termwise

== Theorem ==
Let $I = \closedint a b \subseteq \R$ be a closed real interval.
Let $\struct {\map CI, \norm {\, \cdot \,}_\infty}$ be the normed vector space of real-valued functions continuous on $I$ equipped with the supremum norm.
Let $\sequence {f_k}_{k \mathop \in \N_{>0} } \in C \closedint a b$ be a sequence.
Suppose $\ds \sum_{k \mathop = 1}^\infty f_k$ converges to $f$ in $\struct {\map C I, \norm {\, \cdot \,}_\infty}$.
Then:
:$\ds \sum_{k \mathop = 1}^\infty \int_a^b f_k \rd t = \int_a^b \map f t \rd t$
== Proof ==
== Sources ==
* : Chapter $\S 2.3$: The normed space $\map {CL} {X,Y}$
Category:Operator Theory
Category:Definite Integrals
Category:Linear Transformations
Category:Continuous Transformations
Convergent Sequence with Finite Elements Prepended is Convergent Sequence

== Theorem ==
Let $\struct {R, \norm { \, \cdot \, } }$ be a normed division ring.
Let $\sequence {x_n}$ be a sequence in $R$.
Let $N \in \N$.
Let $\sequence {y_n}$ be the sequence defined by:
:$\forall n \in \N: y_n = x_{N + n}$
Let $\sequence {y_n}$ be a convergent sequence in $R$ with limit $l$.
Then:
:$\sequence {x_n}$ is a convergent sequence in $R$ with limit $l$.
== Proof ==
Let $\epsilon \in \R_{>0}$ be given.
By the definition of a convergent sequence in $R$ with limit $l$:
:$\exists N' \in \R_{>0}: \forall n \in \N: n > N' \implies \norm {y_n - l} < \epsilon$
Hence:
The result follows.
== Sources ==
Category:Cauchy Sequences
Category:Normed Division Rings
Category:Convergent Sequences in Normed Division Rings
Convergent Sequence with Finite Number of Terms Deleted is Convergent

== Theorem ==
Let $\struct {X, d}$ be a metric space.
Let $\sequence {x_k}$ be a sequence in $X$.
Let $\sequence {x_k}$ be convergent.
Let a finite number of terms be deleted from $\sequence {x_k}$.
Then the resulting subsequence is convergent.
== Proof ==
Suppose the sequence $\sequence {x_k}$ converges to $x \in X$.
That is for every $\epsilon > 0$ there is some index $N$ such that $\map d {x_n, x} < \epsilon$ for all $n \ge N$.
The same $N$ will work for the new sequence with finitely many terms removed, so the new sequence converges to the same point $x$ as the original sequence.
For the second part note that also adding finitely many terms to a convergent sequence will still result in a convergent sequence.
This implies that removing finitely many terms from a divergent sequence will still result in a divergent sequence (if it were convergent then the original sequence must already have been convergent).
== Lemma 1 ==
Let $\struct {X, d}$ be a metric space.
Let $\sequence {x_k}$ be a sequence in $X$.
Let $\sequence {x_{n_k} }$ be a subsequence of $\sequence {x_k}$.
Then:
:$\forall k \in \N: k \le n_k$
=== Proof of Lemma 1 ===
Note that $1 \le n_1 < n_2 < \ldots < n_k < \ldots$ by definition of a subsequence.
We will proceed with induction on $k$ to prove our claim.
By definition of a subsequence:
:$1 \le n_1$
This is the basis for the induction.
Suppose $k \le n_k$. 
Then:
:$k + 1 \le n_k + 1$
By definition of a subsequence:
:$n_k < n_{k + 1}$
Therefore:
:$n_k + 1 \le n_{k + 1}$
Thus:
:$k + 1 \le n_{k + 1}$
== Lemma 2 ==
Let $\struct {X, d}$ be a metric space.
Let $\sequence {x_k}$ be a convergent sequence in $X$.
Then every subsequence of $\sequence {x_k}$ is convergent.
=== Proof of Lemma 2 ===
Let $\sequence {x_k}$ be a convergent sequence in $X$. 
Then $\exists L \in X$ such that $x_k \to L$ as $k \to \infty$.
Let $\sequence {x_{n_k} }$ be a subsequence of $\sequence {x_k}$.
We will show that $x_{n_k} \to L$ as $n_k \to \infty$.
Let $\epsilon >0$.
Then $\exists N \in \N$ such that:
:$\forall k \ge N: \map d {x_k, L} < \epsilon$
From Lemma 1:
:$n_k \ge k$
Therefore:
:$\forall n_k \ge k \ge N: \map d {x_{n_k}, L} < \epsilon$
Hence $x_{n_k} \to L$ as $n_k \to \infty$.
== Also see == 
* Tail of Convergent Series tends to Zero
Category:Series
Category:Sequences
Category:Convergence Tests
Convergent Sequences Characterize Metrizable Topology

== Theorem ==
Let $X$ be a set. 
Let $\tau_1$ and $\tau_2$ be metrizable topologies induced by metrics $d_1$ and $d_2$ respectively.
Suppose that:
:a sequence $\sequence {x_n}_{n \in \N} \subseteq X$ converges to $x \in X$ in $\struct {X, d_1}$ it converges to $x$ in $\struct {X, d_2}$. 
Then $\tau_1 = \tau_2$. 
== Proof ==
Let $U$ be open in $\struct {X, d_1}$. 
Then $X \setminus U$ is closed in $\struct {X, d_1}$.
That is, from the definition of closedness in a metric space, for each $x \in X \setminus U$ there exists a sequence $\sequence {x_n}_{n \in \N}$ in $X \setminus U$ converging to $x$ in $\struct {X, d_1}$.
By hypothesis, each such $\sequence {x_n}_{n \in \N}$ also converges to $x$ in $\struct {X, d_2}$.
Hence $X \setminus U$ is also cloesd in $\struct {X, d_2}$.
Hence $U$ is open in $\struct {X, d_2}$.
So $\tau_1 \subseteq \tau_2$.
Swapping $d_1$ and $d_2$ in the above we get $\tau_2 \subseteq \tau_1$ and hence $\tau_1 = \tau_2$. 
Category:Metric Spaces
Category:Convergent Sequences (Metric Space)
Convergent Sequences form Invariant Subspace of Bounded Sequences wrt Cesàro Summation Operator

== Theorem ==
Let $\ell^\infty$ be the space of bounded sequences.
Let $c$ be the space of convergent sequences.
Let $A : \ell^\infty \to \ell^\infty$ be the Cesàro summation operator.
Then $c$ is an invariant subspace of $\ell^\infty$ $A$.
== Proof ==
Let $\sequence {x_n}_{n \mathop \in \N} \in c$ be a sequence.
By definition, $\sequence {x_n}_{n \mathop \in \N}$ converges.
Let $\ds L = \lim_{n \mathop \to \infty} x_n$ be the limit of $\sequence {x_n}_{n \mathop \in \N}$.
Then:
:$\forall \epsilon \in \R_{> 0} : \exists N_1 \in \N : \forall n \in \N : n > N_1 \implies \size {x_n - L} < \epsilon$
Furthermore:
:$\exists M \in \R_{> 0} : \forall n \in \N : \size {x_n} < M$
=== $A \mathbf x$ converges to $L$ ===
Let $\epsilon' \in \R_{> 0 }$.
Let $n \in \N$ be such that:
:$\ds \frac {N_1} n \paren{M + \size L} + \paren {1 - \frac {N_1} n} \epsilon < \epsilon'$
Suppose $n > N_1$.
Then:
Suppose also that $\ds n > N_1 \frac {M + \size L}{\epsilon' - \epsilon}$.
Then:
Altogether, $n$ has to satisfy $n > \tilde N$ where:
:$\ds \tilde N = \max \set {N_1, N_1 \frac {M + \size L}{\epsilon' - \epsilon} }$.
$\epsilon'$ was arbitrary.
Hence:
:$\ds \forall \epsilon' \in \R{>0} : \exists \tilde N : \forall n \in \N : n > \tilde N \implies \size {A \mathbf x - L} < \epsilon'$
Therefore, $A \mathbf x \in c$.
Thus, $A c \subseteq c$.
By definition, $c$ is an invariant subspace of $\ell^\infty$ $A$.
== Sources ==
* : Chapter $\S 2.3$: The normed space $\map {CL} {X,Y}$. Operator norm and the normed space $\map {CL} {X, Y}$
Category:Convergent Real Sequences
Category:Cesàro Summation Operator
Convergent Sequences in Vector Spaces with Equivalent Norms Coincide

== Theorem ==
Let $M_a = \struct {X, \norm {\, \cdot \, }_a}$ and $M_b = \struct {X, \norm {\, \cdot \,}_b}$ be normed vector spaces.
Let $\sequence {x_n}_{n \mathop \in \N}$ be an convergent sequence in $M_a$.
Suppose, $\norm {\, \cdot \, }_a$ and $\norm {\, \cdot \,}_b$ are equivalent norms, i.e. $\norm {\, \cdot \, }_a \sim \norm {\, \cdot \,}_b$.
Then $\sequence {x_n}_{n \mathop \in \N}$ is also convergent in $M_b$. 
== Proof ==
Let $L \in X$.
Then:
:$\forall \epsilon_a \in \R_{> 0} : \exists N \in \N : \forall n \in \N : n > N \implies \norm {x_n - L}_a < \epsilon_a$
By equivalence of norms:
:$\exists M \in \R_{> 0} : \norm {x_n - L}_b \le M \norm {x_n - L}_a < M \epsilon_a$
Let $\epsilon_b := M \epsilon_a$
Then:
:$\forall \epsilon_b \in \R_{> 0}: \exists N \in \N : \forall n \in \N : n > N \implies \norm {x_n - L}_b < \epsilon_b$
Therefore, $\sequence{x_n}_{n \mathop \in \N}$ converges to $L$ also in $M_b$.
== Sources ==
* : $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces
Category:Equivalence Relations
Category:Norm Theory
Category:Vector Spaces
Category:Convergent Sequences (Normed Vector Spaces)
Convergent Series can be Added Term by Term

== Theorem ==
Let $\ds \sum_{n \mathop = 1}^\infty a_n$ and $\ds \sum_{n \mathop = 1}^\infty b_n$ be two convergent series converging to $A$ and $B$ respectively.
Then:
:$\ds \sum_{n \mathop = 1}^\infty \paren {a_n + b_n} = A + B$
== Proof ==
Category:Series
Category:Convergence
Convergent Series of Natural Numbers

== Theorem ==
Let $\sequence {a_n}_{n \mathop \in \N}$ be a sequence of natural numbers.
Then the following are equivalent:
$(1): \quad \ds \sum_{n \mathop = 1}^\infty a_n$ converges
$(2): \quad \exists N \in \N: \forall n \ge N: a_n = 0$
That is, $\ds \sum_{n \mathop = 1}^\infty a_n$ converges only finitely many of the $a_n$ are non-zero.
== Proof ==
$(1) \implies (2)$:
Suppose that there is an infinite subsequence $\sequence {a_{n_k} }_{k \mathop \in \N}$ such that for each $k$, $a_{n_k} \ne 0$.
For $N \in \N$ let 
:$\ds s_N = \sum_{n \mathop = 1}^N a_n$
To show that $s_N$ diverges it suffices to show that:
:$\forall M > 0: \exists N \in \N : \forall n > N : \size {s_n} > M$
Since for each $n$, $a_n \ge 0$, $s_N$ is a positive increasing sequence in $N$.
Therefore it suffices to show that:
:$\forall M > 0: \exists N \in \N: s_N > M$
Fix $M > 0$.
Let $k$ be any positive integer such that $n_k > M$.
Then we have:
Therefore the sequence $s_N$ diverges.
$(2) \implies (1)$:
Suppose there exists $N > 0$ such that $a_n = 0$ for all $n > N$.
Then we have, for all $L > N$:
:$\ds s_L = \sum_{n \mathop = 1}^L a_n = \sum_{n \mathop = 1}^N a_n = s_N$
In particular, for any $\epsilon > 0$ and all $L > N$:
:$\size {s_L - s_N} = 0 < \epsilon$
Therefore the sequence converges to $s_N$.
Category:Natural Numbers
Category:Series
Convergent Series of Positive Elements of C*-Algebra is Positive

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra. 
Let $\sequence {a_n}_{n \mathop \in \N}$ be a sequence of positive elements such that:
:$\ds \sum_{n \mathop = 1}^\infty a_n$ converges.
Then:
:$\ds \sum_{n \mathop = 1}^\infty a_n$ is positive.
== Proof ==
From Sum of Two Positive Elements of C*-Algebra is Positive: Corollary:
:$\ds \sum_{n \mathop = 1}^N a_j$ is positive.
From Set of Positive Elements of C*-Algebra is Closed, we then have:
:$\ds \sum_{n \mathop = 1}^\infty a_n = \lim_{N \mathop \to \infty} \sum_{n \mathop = 1}^N a_j$ is positive.
Category:C*-Algebras
Convergent Subsequence in Closed Interval

== Theorem ==
Let $\closedint a b$ be a closed real interval.
Then every sequence of points of $\closedint a b$ contains a subsequence which converges to a point in $\closedint a b$.
== Proof ==
Let $\sequence {x_n}$ be a sequence in $\closedint a b$.
Since $\closedint a b$ is bounded in $\R$, it follows that $\sequence {x_n}$ is a bounded sequence.
By the Bolzano-Weierstrass Theorem, $\sequence {x_n}$ has a subsequence $\sequence {x_{n_r} }$ which is convergent.
Suppose $x_{n_r} \to l$ as $n \to \infty$.
Since $a \le x_{n_r} \le b$, from Lower and Upper Bounds for Sequences it follows that $a \le l \le b$.
So $\sequence {x_{n_r} }$ converges to a point in $\closedint a b$.
== Sources ==
* : $\S 5$: Subsequences: Exercise $\S 5.21 \ (4)$
Category:Analysis
Category:Convergence
Category:Limits of Sequences
Convergent Subsequence of Cauchy Sequence/Metric Space

== Theorem ==
Let $\struct {A, d}$ be a metric space.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a Cauchy sequence in $A$. 
Let $x \in A$.
Then $\sequence {x_n}$ converges to $x$ it has a subsequence that converges to $x$.
== Proof ==
=== Necessary Condition ===
If $\sequence {x_n}$ converges to $x$, it trivially follows that $\sequence {x_n}$ is a subsequence of itself that converges to $x$.
=== Sufficient Condition ===
Suppose that $\sequence {x_{n_k} }$ is a subsequence of $\sequence {x_n}$ that converges to $x$.
Let $\epsilon$ be a strictly positive real number.
By the definition of a Cauchy sequence, there exists a real number $M$ such that:
:$\forall i, j \in \N: i, j > M \implies \map d {x_i, x_j} < \dfrac \epsilon 2$
By the definition of convergence, there exists a real number $N$ such that:
:$\forall k \in \N: k > N \implies \map d {x_{n_k}, x} < \dfrac \epsilon 2$
By the Axiom of Archimedes, there exists a natural number $K > \max \set {M, N}$.
Therefore, by :
:$\forall m \in \N: m > K \implies \map d {x_m, x} \le \map d {x_m, x_K} + \map d {x_K, x} < \dfrac \epsilon 2 + \dfrac \epsilon 2 = \epsilon$
That is, $\sequence {x_n}$ converges to $x$.
Category:Cauchy Sequences
Convergent Subsequence of Cauchy Sequence/Normed Division Ring

== Theorem ==
Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring.
Let $\sequence{x_n}_{n \mathop \in \N}$ be a Cauchy sequence in $\struct {R, \norm {\,\cdot\,} }$. 
Let $x \in R$.
Then $\sequence {x_n}$ converges to $x$ $\sequence {x_n}$ has a subsequence that converges to $x$.
== Proof ==
Let $d$ be the metric induced on $R$ be the norm $\norm {\,\cdot\,}$.
By the definition of a convergent sequence in a normed division ring then:
:$\sequence {x_n}$ converges to $x$ in $\struct {R, \norm {\,\cdot\,} }$ $\sequence {x_n}$ converges to $x$ in $\struct {R, d}$.
By Convergent Subsequence of Cauchy Sequence in Metric Space:
:$\sequence {x_n}$ converges to $x$ in $\struct {R, d}$ $\sequence {x_n}$ has a subsequence that converges to $x$ In $\struct {R, d}$.
By the definition of a convergent sequence in a normed division ring:
:$\sequence {x_n}$ has a subsequence that converges to $x$ In $\struct {R, d}$ $\sequence {x_n}$ has a subsequence that converges to $x$ in $\struct {R, \norm {\,\cdot\,} }$.
The result follows.
== Sources ==
* : $\S 1.2$: Normed Fields: Exercise $11$ $(2)$
Category:Convergent Sequences in Normed Division Rings
Convergent Subsequence of Cauchy Sequence/Normed Vector Space

== Theorem ==
Let $\struct {X, \norm {\,\cdot\,} }$ be a normed vector space.
Let $\sequence{x_n}_{n \mathop \in \N}$ be a Cauchy sequence in $\struct {X, \norm {\,\cdot\,} }$. 
Let $x \in X$.
Then $\sequence {x_n}$ converges to $x$ $\sequence {x_n}$ has a subsequence that converges to $x$.
== Proof ==
=== Necessary Condition ===
Suppose is a Cauchy sequence$\sequence {x_n}$ which converges to $x$.
By Limit of Subsequence equals Limit of Sequence, $\sequence{x_n}$ has a subsequence of itself that converges to $x$.
=== Sufficient Condition ===
Suppose that $\sequence {x_{n_k}}$ is a subsequence of $\sequence {x_n}$ that converges to $x$.
Let $\epsilon$ be a strictly positive real number.
By the definition of a Cauchy sequence, there exists a real number $M$ such that:
:$\ds \forall i, j \in \N: i, j > M \implies \norm {x_i - x_j} < \frac \epsilon 2$
By the definition of convergence, there exists a real number $N$ such that:
:$\ds \forall k \in \N: k > N \implies \norm {x_{n_k} - x} < \frac \epsilon 2$
By the Axiom of Archimedes, there exists a natural number $K > \max \set {M, N}$.
Therefore, by the triangle inequality:
:$\ds \forall m \in \N: m > K \implies \norm {x_m - x} \le \norm {x_m - x_K} + \norm {x_K - x} < \frac \epsilon 2 + \frac \epsilon 2 = \epsilon$
That is, $\sequence {x_n}$ converges to $x$.
== Sources ==
* : $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces
Category:Cauchy Sequences
Category:Convergent Sequences (Normed Vector Spaces)