nyuuzyou/wikis huggingface dataset:
Convergent Real Sequence/Examples/n^3 + 5 n^2 + 2 over 2 n^3 + 9

== Example of Convergent Real Sequence ==
:$\ds \lim_{n \mathop \to \infty} \paren {\dfrac {n^3 + 5 n^2 + 2} {2 n^3 + 9} } = \dfrac 1 2$
== Proof ==
== Sources ==
* : $\S 4$: Convergent Sequences: Exercise $\S 4.20 \ (1)$
Category:Examples of Convergent Real Sequences
Convergent Real Sequence/Examples/x (n+1) = k over 1 + x n

== Example of Convergent Real Sequence ==
Let $h, k \in \R_{>0}$.
Let $\sequence {x_n}$ be the real sequence defined as:
:$x_n = \begin {cases} h & : n = 1 \\ \dfrac k {1 + x_{n - 1} } & : n > 1 \end {cases}$
Then $\sequence {x_n}$ is convergent to the positive root of the quadratic equation:
:$x^2 + x = k$
== Proof ==
First some lemmata:
=== Lemma 1 ===
=== Lemma 2 ===
From Lemma 2, We have that both $\sequence {x_{2 n} }$ and $\sequence {x_{2 n - 1} }$ is strictly monotone (one strictly increasing and the other strictly decreasing).
From Lemma 1, they are both bounded above by $k$ and bounded below by $0$.
Hence from the Monotone Convergence Theorem (Real Analysis), they both converge.
Let:
:$x_{2 n} \to l$ as $n \to \infty$
:$x_{2 n - 1} \to m$ as $n \to \infty$
Then:
Hence the result.
== Sources ==
* : $\S 5$: Subsequences: Exercise $\S 5.7 \ (3)$
Category:Examples of Convergent Real Sequences
Category:Convergent Real Sequence/Examples/x (n+1) = k over 1 + x n
Convergent Real Sequence/Examples/x (n+1) = k over 1 + x n/Lemma 1

== Example of Convergent Real Sequence ==
Let $h, k \in \R_{>0}$.
Let $\sequence {x_n}$ be the real sequence defined as:
:$x_n = \begin {cases} h & : n = 1 \\ \dfrac k {1 + x_{n - 1} } & : n > 1 \end {cases}$
Then:
:$\forall n \in \N_{>1}: k > x_n > 0$
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{>1}$, let $\map P n$ be the proposition:
:$k > x_n > 0$
=== Basis for the Induction ===
$\map P 2$ is the case:
:$k > x_2 > 0$
We have:
Also, as $k > 0$ and $x_1 > 0$ we have that:
:$\dfrac k {1 + x_1} > 0$
Thus $\map P 2$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P r$ is true, where $r \ge 2$, then it logically follows that $\map P {r + 1}$ is true.
So this is the induction hypothesis:
:$k > x_r > 0$
from which it is to be shown that:
:$k > x_{r + 1} > 0$
=== Induction Step ===
This is the induction step:
Also, as $k > 0$ and $x_r > 0$ we have that:
:$\dfrac k {1 + x_r} > 0$
So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \N_{>1}: k > x_n > 0$
== Sources ==
* : $\S 5$: Subsequences: Exercise $\S 5.7 \ (3)$
Category:Convergent Real Sequence/Examples/x (n+1) = k over 1 + x n
Convergent Real Sequence/Examples/x (n+1) = k over 1 + x n/Lemma 2

== Example of Convergent Real Sequence ==
Let $h, k \in \R_{>0}$.
Let $\sequence {x_n}$ be the real sequence defined as:
:$x_n = \begin {cases} h & : n = 1 \\ \dfrac k {1 + x_{n - 1} } & : n > 1 \end {cases}$
Consider the subsequences $\sequence {x_{2 n} }$ and $\sequence {x_{2 n - 1} }$.
One of them is strictly increasing and the other is strictly decreasing.
== Proof ==
We have that:
and so $x_{n + 1} - x_{n - 1}$ has the opposite sign to $x_{n - 2} - x_n$.
It can be proved by induction that one of the sequences $\sequence {x_{2 n} }$ and $\sequence {x_{2 n - 1} }$ increases and one decreases.
In fact:
:$\sequence {x_{2 n - 1} }$ is strictly increasing $x_3 > x_1$ and is strictly decreasing $x_3 < x_1$.
== Sources ==
* : $\S 5$: Subsequences: Exercise $\S 5.7 \ (3)$
Category:Convergent Real Sequence/Examples/x (n+1) = k over 1 + x n
Convergent Real Sequence/Examples/x (n+1) = k over 1 + x n/Mistake

== Source Work ==
:
: Chapter $\S 5$: Subsequences
:: $\S 5.7$: Solution to Exercise $(3)$
This mistake can be seen in the $1977$ edition as published by Cambridge University Press: ISBN 0 521 29167 4
== Mistake ==
:It is easily shown by induction that $k > x_n > 0$ ($n = 1, 2, \ldots$).
== Correction ==
The condition does not necessarily hold for $n = 1$.
$x_1$ is specified as being (strictly) positive, but it is not stated (and indeed, it is not necessary) that $x_1 < k$.
Hence the line should read:
:It is easily shown by induction that $k > x_n > 0$ ($n = 2, 3, \ldots$).
== Sources ==
* : $\S 5$: Subsequences: Exercise $\S 5.7 \ (3)$
Category:Mistakes/Mathematical Analysis: A Straightforward Approach
Convergent Real Sequence/Examples/x (n+1) = x n^2 + k

== Example of Convergent Real Sequence ==
Let $\sequence {x_n}$ be the real sequence defined as:
:$x_n = \begin {cases} h & : n = 1 \\ {x_{n - 1} }^2 + k & : n > 1 \end {cases}$
where:
:$0 < k < \dfrac 1 4$
:$a < h < b$, where $a$ and $b$ are the roots of the quadratic equation: $x^2 - x + k = 0$.
Then $\sequence {x_n}$ is convergent such that:
:$\ds \lim_{n \mathop \to \infty} x_n = a$
== Proof ==
First some lemmata:
=== Lemma 1 ===
=== Lemma 2 ===
From Lemma 1 and Lemma 2 we have that:
:$0 < a < x_n < b$
for all $n \in \N_{>0}$.
Then:
Hence:
:$0 < a < x_{n + 1} < x_n < b$
Thus $\sequence {x_n}$ is decreasing and bounded below by $a$.
Hence by the Monotone Convergence Theorem (Real Analysis), $\sequence {x_n}$ converges to its infimum.
It remains to be shown that $\map \inf {x_n} = a$.
Suppose that:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
Thus:
But $b$ cannot be the infimum of $\sequence {x_n}$ because it is not a lower bound.
Hence:
:$\ds \lim_{n \mathop \to \infty} x_n = a$
and the result follows.
== Sources ==
* : $\S 5$: Subsequences: Exercise $\S 5.7 \ (2)$
Category:Examples of Convergent Real Sequences
Category:Convergent Real Sequence/Examples/x (n+1) = x n^2 + k
Convergent Real Sequence/Examples/x (n+1) = x n^2 + k/Lemma 1

== Example of Convergent Real Sequence ==
Let $\sequence {x_n}$ be the real sequence defined as:
:$x_n = \begin {cases} h & : n = 1 \\ {x_{n - 1} }^2 + k & : n > 1 \end {cases}$
where:
:$0 < k < \dfrac 1 4$
:$a < h < b$, where $a$ and $b$ are the roots of the quadratic equation $x^2 - x + k = 0$.
Then:
:$\forall n \in \N_{>0}: a < x_n < b$
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$a < x_n < b$
=== Basis for the Induction ===
$\map P 1$ is the case:
:$a < x_1 < b$
By assertion:
:$a < h < b$
and:
:$x_1 = h$
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$a < x_k < b$
from which it is to be shown that:
:$a < x_{k + 1} < b$
=== Induction Step ===
This is the induction step:
Similarly:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \N_{>0}: a < x_n < b$
== Sources ==
* : $\S 5$: Subsequences: Exercise $\S 5.7 \ (2)$
Category:Convergent Real Sequence/Examples/x (n+1) = x n^2 + k
Convergent Real Sequence/Examples/x (n+1) = x n^2 + k/Lemma 2

== Example of Convergent Real Sequence ==
Let $a$ and $b$ be the roots of the quadratic equation:
:$(1): \quad x^2 - x + k = 0$
Let:
:$0 < k < \dfrac 1 4$
Then $a$ and $b$ are both strictly positive real numbers.
== Proof ==
First we investigate the consequences of the condition $k < \dfrac 1 4$.
By Solution to Quadratic Equation with Real Coefficients:
:In order for the quadratic equation $a x^2 + b x + c$ to have real roots, its discriminant $b^2 - 4 a c$ needs to be strictly positive.
The discriminant $D$ of $(1)$ is:
Thus:
and so when $k < \dfrac 1 4$, $(1)$ has real roots.
Next we investigate the consequences of the condition $0 < k$.
By Solution to Quadratic Equation:
We have that:
That is, when $0 < k$ both roots of $(1)$ are strictly positive.
Hence when $0 < k < \dfrac 1 4$, both roots of $(1)$ are strictly positive real numbers.
== Sources ==
* : $\S 5$: Subsequences: Exercise $\S 5.7 \ (2)$
Category:Convergent Real Sequence/Examples/x (n+1) = x n^2 + k
Convergent Real Sequence/Examples/x + x^n over 1 + x^n

== Example of Convergent Real Sequence ==
The sequence $\sequence {a_n}$ defined as:
:$a_n = \dfrac {x + x^n} {1 + x^n}$
is convergent for $x \ne -1$.
Then:
:$\ds \lim_{n \mathop \to \infty} \dfrac {x + x^n} {1 + x^n} = \begin {cases} 1 & : x \ge 1 \\ x & : -1 < x < 1 \\ 1 & : x < -1 \\ \text {undefined} & : x = -1 \end {cases}$
== Proof ==
Let $\size x < 1$.
We have:
Now let $\size x > 1$.
We have:
Then when $x = 1$:
When $x = -1$:
Hence the result.
== Sources ==
* : $\S 4$: Convergent Sequences: Exercise $\S 4.20 \ (2)$
Category:Examples of Convergent Real Sequences
Convergent Real Sequence/Examples/x n = root x n-1 y n-1, 1 over y n = half (1 over x n + 1 over y n-1)

== Example of Convergent Real Sequence ==
Let $\sequence {x_n}$ and $\sequence {y_n}$ be the real sequences defined as:
:$x_n = \begin {cases} \dfrac 1 2 & : n = 1 \\ \sqrt {x_{n - 1} y_{n - 1} } & : n > 1 \end {cases}$
:$\dfrac 1 {y_n} = \begin {cases} 1 & : n = 1 \\ \dfrac 1 2 \paren {\dfrac 1 {x_n} + \dfrac 1 {y_{n - 1} } } & : n > 1 \end {cases}$
Then both $\sequence {x_n}$ and $\sequence {y_n}$ converge to the limit $\dfrac \pi 4$.
== Proof ==
By definition, we have that:
:$\sqrt {x_{n - 1} y_{n - 1} }$ is the geometric mean of $x_{n - 1}$ and $y_{n - 1}$
:$\dfrac 1 2 \paren {\dfrac 1 {x_n} + \dfrac 1 {y_{n - 1} } }$ is the reciprocal of the harmonic mean of $x_n$ and $y_{n - 1}$.
We are given that:
:$\dfrac 1 2 = x_1 < y_1 = 1$
Assuming $x_{n - 1} < y_{n - 1}$, it follows from Geometric Mean of two Positive Real Numbers is Between them that:
:$x_{n - 1} < x_n < y_{n - 1}$
Given that $x_n < y_{n - 1}$, it follows from Harmonic Mean of two Positive Real Numbers is Between them that:
:$x_n < y_n < y_{n - 1}$
It follows from the Principle of Mathematical Induction that:
:$x_{n - 1} < x_n < y_n < y_{n - 1}$ for $n = 2, 3, \ldots$}
Hence we have:
:$\sequence {x_n}$ is strictly increasing and bounded above by $y_1 = 1$.
:$\sequence {y_n}$ is strictly decreasing and bounded below by $x_1 = \dfrac 1 2$.
So by the Monotone Convergence Theorem (Real Analysis), both $\sequence {x_n}$ and $\sequence {y_n}$ converge.
Let:
:$x_n \to l$ as $n \to \infty$
:$y_n \to m$ as $n \to \infty$
Then:
:$l^2 = l m$
and:
:$\dfrac 1 m = \dfrac 1 2 \paren {\dfrac 1 l + \dfrac 1 m}$
from which it follows that:
:$l = m$
It remains to be shown that $l = m = \dfrac \pi 4$.
== Sources ==
* : $\S 5$: Subsequences: Exercise $\S 5.7 \ (4)$
Category:Examples of Convergent Real Sequences
Convergent Real Sequence has Unique Limit/Proof 1

== Theorem ==
== Proof ==
 that $\sequence {s_n}$ converges to $l$ and also to $m$.
That is, suppose that:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
and:
:$\ds \lim_{n \mathop \to \infty} x_n = m$
, assume that $l \ne m$.
Let:
:$\epsilon = \dfrac {\size {l - m} } 2$
As $l \ne m$, it follows that $\epsilon > 0$.
As $\sequence {s_n} \to l$:
:$\exists N_1 \in \N: \forall n \in \N: n > N_1: \size {s_n - l} < \epsilon$
Similarly, since $\sequence {s_n} \to m$:
:$\exists N_2 \in \N: \forall n \in \N: n > N_2: \size {s_n - m} < \epsilon$
Now set $N = \max \set {N_1, N_2}$.
We have:
This constitutes a contradiction.
It follows from Proof by Contradiction that $l = m$.
== Sources ==
* : $1$: Review of some real analysis: $\S 1.2$: Real Sequences: Proposition $1.2.3$
* : $\S 4$: Convergent Sequences: Some simple properties of convergent sequences: $\S 4.22$
Category:Convergent Real Sequence has Unique Limit
Convergent Real Sequence is Bounded

== Theorem ==
Let $\sequence {x_n}$ be a sequence in $\R$.
Let $l \in A$ such that $\ds \lim_{n \mathop \to \infty} x_n = l$.
Then $\sequence {x_n}$ is bounded.
That is, all convergent real sequences are bounded.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 5$: Limits: Exercise $4$
* : $3.2 \, \text{c}$
Category:Real Sequences
Category:Limits of Sequences
Category:Convergent Real Sequence is Bounded
Convergent Real Sequence is Bounded/Proof 1

== Theorem ==
== Proof ==
From Real Number Line is Metric Space, the set $\R$ under the usual metric is a metric space.
By Convergent Sequence in Metric Space is Bounded it follows that:
:$\exists M > 0: \forall n, m \in \N: \size {x_n - x_m} \le M$
Then for $n \in \N$, by the Triangle Inequality for Real Numbers:
Hence $\sequence {x_n}$ is bounded.
== Sources ==
Category:Convergent Real Sequence is Bounded
Convergent Real Sequence is Bounded/Proof 2

== Theorem ==
== Proof ==
Let $\sequence {x_n}$ be a sequence in $\R$.
Let $x_n \to l$ as $n \to \infty$.
To show that $\sequence {x_n}$ is bounded sequence, we need to find $K$ such that:
:$\forall n \in \N: \size {x_n} \le K$
Because $\sequence {x_n}$ converges:
:$\forall \epsilon > 0: \exists N: n > N \implies \size {x_n - l} < \epsilon$
In particular, this is true when $\epsilon = 1$.
That is:
:$\exists N_1: \forall n > N_1: \size {x_n - l} < 1$
By Backwards Form of Triangle Inequality:
:$\forall n > N_1: \size {x_n} - \size l \le \size {x_n - l} < 1$
That is:
:$\size {x_n} < \size l + 1$
So we set:
:$K = \max \set {\size {x_1}, \size {x_2}, \ldots, \size {x_{N_1} }, \size l + 1}$
and the result follows.
== Sources ==
* : $\S 4$: Convergent Sequences: Some simple properties of convergent sequences: $\S 4.25$
Category:Convergent Real Sequence is Bounded
Convergent Sequence Minus Limit

== Theorem ==
Let $X$ be one of the standard number fields $\Q, \R, \C$.
Let $\sequence {x_n}$ be a sequence in $X$ which converges to $l$.
That is:
: $\ds \lim_{n \mathop \to \infty} x_n = l$
Then:
: $\ds \lim_{n \mathop \to \infty} \cmod {x_n - l} = 0$
== Proof 1 ==
== Proof 2 ==
Category:Limits of Sequences
Category:Convergent Sequence Minus Limit
Convergent Sequence Minus Limit/Proof 1

== Theorem ==
== Proof ==
Let $\epsilon > 0$.
We need to show that there exists $N$ such that:
:$\forall n > N: \size {\paren {\size {x_n - l} - 0} } < \epsilon$
But:
:$\size {\paren {\size {x_n - l} - 0} } = \size {x_n - l}$
So what needs to be shown is just:
:$x_n \to l$ as $n \to \infty$
which is the definition of $\ds \lim_{n \mathop \to \infty} x_n = l$.
== Sources ==
* : $\S 4$: Convergent Sequences: Exercise $\S 4.29 \ (1) \ \text {(i)}$
Category:Convergent Sequence Minus Limit
Convergent Sequence Minus Limit/Proof 2

== Theorem ==
== Proof ==
We note that all of $\Q, \R, \C$ can be considered as metric spaces.
Then under the usual metric:
: $\map d {x_n, l} = \cmod {x_n - l}$.
The result follows from the definition of metric:
:$\map d {x_n, l} = 0 \iff x_n = l$.
Category:Convergent Sequence Minus Limit
Convergent Sequence in Hausdorff Space has Unique Limit

== Theorem ==
Let $T = \struct {S, \tau}$ be a Hausdorff space.
Let $\sequence {x_n}$ be a convergent sequence in $T$.
Then $\sequence {x_n}$ has exactly one limit.
== Proof ==
From the definition of convergent sequence, we have that $\sequence {x_n}$ converges to at least one limit.
 $l$ and $m$ are both limits of $\sequence{x_n}$.
As $T$ is Hausdorff, there exist $U, V \in \tau$ such that:
:$l \in U, m \in V$ and $U \cap V = \O$
Then, from the definition of convergent sequence:
Taking $N = \max \set {N_U, N_V}$ we then have:
:$\exists N \in \R: n > N \implies x_n \in U, x_n \in V$
But $U \cap V = \O$.
From that contradiction we can see that there can be no such two distinct $l$ and $m$.
Hence the result.
== Sources ==
* : $4$: The Hausdorff condition: $4.2$: Separation axioms: Proposition $4.2.2$
Category:Limits of Sequences
Category:Hausdorff Spaces
Convergent Sequence in Metric Space has Unique Limit

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $\sequence {x_n}$ be a sequence in $M$.
Then $\sequence {x_n}$ can have at most one limit in $M$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 5$: Limits: Exercise $2$
Category:Limits of Sequences
Category:Metric Spaces
Category:Convergent Sequence in Metric Space has Unique Limit
Convergent Sequence in Metric Space has Unique Limit/Proof 1

== Theorem ==
== Proof ==
Suppose $\ds \lim_{n \mathop \to \infty} x_n = l$ and $\ds \lim_{n \mathop \to \infty} x_n = m$.
Let $\epsilon > 0$.
Then, provided $n$ is sufficiently large:
So $0 \le \dfrac {\map d {l, m} } 2 < \epsilon$.
This holds for any value of $\epsilon > 0$.
Thus from Real Plus Epsilon it follows that $\dfrac {\map d {l, m} } 2 = 0$, that is, that $l = m$.
Category:Convergent Sequence in Metric Space has Unique Limit
Convergent Sequence in Metric Space is Bounded

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $\sequence {x_n}$ be a sequence in $M$ which is convergent, and so $x_n \to l$ as $n \to \infty$.
Then $\sequence {x_n}$ is bounded.
=== Convergent Sequence in Normed Division Ring is Bounded ===
=== Convergent Real Sequence is Bounded ===
== Proof ==
Let $M = \struct {A, d}$ be a metric space.
Let $\sequence {x_n}$ be a sequence in $M$ which is convergent, and so $x_n \to l$ as $n \to \infty$.
From the definition, in order to prove boundedness, all we need to do is find $K \in \R$ such that $\forall n \in \N: \map d {x_n, l} \le K$.
Since $\sequence {x_n}$ converges, it is true that:
:$\forall \epsilon > 0: \exists N: n > N \implies \map d {x_n, l} < \epsilon$
In particular, this is true when $\epsilon = 1$, for example.
That is:
:$\exists N_1: \forall n > N_1: \map d {x_n, l} < 1$
So now we set:
:$K = \max \set {\map d {x_1, l}, \map d {x_2, l}, \ldots, \map d {x_{N_1}, l}, 1}$
The result follows.
== Sources ==
* : $3.2 \, \text{c}$
Category:Limits of Sequences
Category:Metric Spaces
Convergent Sequence in Normed Division Ring is Bounded

== Theorem ==
Let $\struct {R, \norm {\,\cdot\,}}$ be a normed division ring.
Let $\sequence {x_n}$ be a sequence in $R$.
Let $\sequence {x_n}$ be convergent in the norm $\norm {\,\cdot\,}$ to the following limit:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
Then $\sequence {x_n}$ is bounded.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Sources ==
* : $\S 1.2$: Normed Fields, Exercise $11 \ (1)$
Category:Sequences
Category:Convergence
Category:Normed Division Rings
Category:Convergent Sequences in Normed Division Rings
Category:Convergent Sequence in Normed Division Ring is Bounded
Convergent Sequence in Normed Division Ring is Bounded/Proof 1

== Theorem ==
== Proof ==
Let $\sequence {x_n}$ be convergent in the norm $\norm {\,\cdot\,}$ to the limit $l$, then:
:$\forall \epsilon \in \R_{>0}: \exists N \in \N : \forall n \ge N: \norm {x_n - l} < \epsilon$
Let $n_1$ satisfy:
:$\forall n \ge n_1: \norm {x_n - l} < 1$
Then $\forall n \ge n_1$:
Let $K = \max \set {\norm {x_1}, \norm {x_2}, \dots, \norm {x_{n_1 - 1} }, 1 + \norm l}$.
Then:
:$\forall n < n_1: \norm {x_n} \le K$
:$\forall n \ge n_1: \norm {x_n} \le 1 + \norm l \le K$
Thus, by definition, $\sequence {x_n}$ is bounded.
== Sources ==
* : $\S 1.2$: Normed Fields, Exercise $11 \ (1)$
Category:Convergent Sequence in Normed Division Ring is Bounded
Convergent Sequence in Normed Division Ring is Bounded/Proof 2

== Theorem ==
== Proof ==
Let $d$ be the metric induced on $R$ be the norm $\norm {\,\cdot\,}$.
Let $\sequence {x_n}$ be convergent to the limit $l$ in $\struct {R, \norm {\,\cdot\,}}$.
By the definition of a convergent sequence in a normed division ring, $\sequence {x_n} $ is convergent to the limit $l$ in $\struct {R, d}$.
By Convergent Sequence in Metric Space is Bounded, $\sequence {x_n} $ is a bounded sequence in $\struct {R, d}$.
By Sequence is Bounded in Norm iff Bounded in Metric, $\sequence {x_n} $ is a bounded sequence in $\struct {R, \norm {\,\cdot\,} }$.
== Sources ==
* : $\S 1.2$: Normed Fields, Exercise $11 \ (1)$
Category:Convergent Sequence in Normed Division Ring is Bounded
Convergent Sequence in Normed Division Ring is Bounded/Proof 3

== Theorem ==
== Proof ==
Let $\sequence {x_n}$ be convergent to the limit $l$ in $\struct {R, \norm {\,\cdot\,} }$.
By modulus of limit in normed division ring, $\sequence {\norm {x_n} }$ is a convergent sequence in $\R$.
By Convergent Real Sequence is Bounded, $\sequence {\norm {x_n} }$ is bounded.
That is:
:$\exists M \in \R_{> 0}: \forall n, \norm {x_n} = \size {\norm {x_n} } \le M$
Thus, by definition, $\sequence {x_n}$ is bounded.
== Sources ==
* : $\S 1.2$: Normed Fields, Exercise $11$ $(1)$
Category:Convergent Sequence in Normed Division Ring is Bounded
Convergent Sequence in Normed Division Ring is Bounded/Proof 4

== Theorem ==
== Proof ==
Let $\sequence {x_n}$ be convergent to the limit $l$ in $\struct {R, \norm {\,\cdot\,}}$.
By Convergent Sequence is Cauchy Sequence in Normed Division Ring, $\sequence {x_n}$ is a Cauchy sequence in $\struct {R, \norm {\,\cdot\,}}$.
By Cauchy Sequence in Normed Division Ring is Bounded, $\sequence {x_n}$ is a bounded sequence in $\struct {R, \norm {\,\cdot\,}}$.
== Sources ==
* : $\S 1.2$: Normed Fields, Exercise $11$ $(1)$
Category:Convergent Sequence in Normed Division Ring is Bounded
Convergent Sequence in Normed Vector Space has Unique Limit

== Theorem ==
Let $\struct {X, \norm {\,\cdot\,} }$ be a normed vector space.
Let $\sequence {x_n}$ be a sequence in $\struct {X, \norm {\,\cdot\,} }$.
Then $\sequence {x_n}$ can have at most one limit.
== Proof ==
 $\ds \lim_{n \mathop \to \infty} x_n = L_1$ and $\ds \lim_{n \mathop \to \infty} x_n = L_2$ such that $L_1 \ne L_2$.
Let $\epsilon = \dfrac {\norm {L_1 - L_2} } 3$.
From the norm axioms it follows that $\epsilon > 0$.
By definition:
:$\exists N_1 \in \N : \forall n > N_1 : \norm {x_n - L_1} < \epsilon$
:$\exists N_2 \in \N : \forall n > N_2 : \norm {x_n - L_2} < \epsilon$
Choose $n > N_1 + N_2$.
Then:
which implies that:
:$1 < \dfrac 2 3$
This is a contradiction.
Hence $L_1 = L_2$.
== Also see ==
* Convergent Sequence has Unique Limit
== Sources ==
* : $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces
Category:Limits of Sequences
Category:Normed Vector Spaces
Category:Convergent Sequences (Normed Vector Spaces)
Convergent Sequence in Normed Vector Space is Bounded

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space. 
Let $\sequence {x_n}_{n \in \N}$ be a convergent sequence. 
Then there exists $M > 0$ such that: 
:$\norm {x_n} \le M$
for each $n \in \N$. 
== Proof ==
Suppose that $x_n \to x$. 
From Convergent Sequence is Cauchy Sequence, $\sequence {x_n}_{n \in \N}$ is a Cauchy sequence.
So there exists $N \in \N$ such that:
:$\norm {x_n - x_m} < 1$ 
for each $n, m \ge N$. 
That is: 
:$\norm {x_n - x_N} < 1$
for $n \ge N$. 
From Reverse Triangle Inequality: Normed Vector Space, we have: 
:$\norm {x_n} - \norm {x_N} < 1$
so that: 
:$\norm {x_n} < 1 + \norm {x_N}$
for $n \ge N$. 
Now set: 
:$M = \max \set {\norm {x_1}, \norm {x_2}, \ldots, \norm {x_{N - 1} }, 1 + \norm {x_N} }$ 
Then we have: 
:$\norm {x_n} \le M$
for each $n \in \N$. 
Category:Convergent Sequences (Normed Vector Spaces)
Convergent Sequence in Normed Vector Space is Weakly Convergent

== Theorem ==
Let $\struct {X, \norm \cdot}$ be a normed vector space.
Let $x \in X$. 
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $X$ converging to $x$. 
Then $\sequence {x_n}_{n \mathop \in \N}$ converges weakly to $x$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* $27.1$: Weak Convergence
Category:Weak Convergence (Normed Vector Spaces)
Category:Convergent Sequences (Normed Vector Spaces)
Category:Convergent Sequence in Normed Vector Space is Weakly Convergent
Convergent Sequence in Normed Vector Space is Weakly Convergent/Proof 1

== Theorem ==
== Proof ==
Let $\struct {X^\ast, \norm \cdot_{X^\ast} }$ be the normed dual space of $\struct {X, \norm \cdot}$.
Let $f \in X^\ast$. 
If $\norm f_{X^\ast} = 0$, then $f = 0$ and: 
:$\map f {x_n} \to \map f x$
Take $f \ne 0$, so that $\norm f_{X^\ast} \ne 0$.
Let $\epsilon > 0$. 
We then have: 
Since $\sequence {x_n}_{n \mathop \in \N}$ converges to $x$:
:there exists $N \in \N$ such that $\ds \norm {x_n - x} < \frac \epsilon {\norm f_{X^\ast} }$ for all $n \ge N$.
Then, for $n \ge N$, we have: 
Since $\epsilon > 0$ was arbitrary, we obtain: 
:$\map f {x_n} \to \map f x$
Since $f \in X^\ast$ was arbitrary, we have: 
:$\sequence {x_n}_{n \mathop \in \N}$ converges weakly to $x$.
Category:Convergent Sequence in Normed Vector Space is Weakly Convergent