nyuuzyou/wikis huggingface dataset:
Cauchy-Kovalevsky Theorem

== Theorem ==
Let $\KK$ denote the field of either the real or complex numbers.
Let $V = \KK^m$.
Let $W = \KK^n$.
Let $A_1, A_2, \ldots, A_{n − 1}$ be analytic functions defined on some neighborhood of $\tuple {0, 0}$ in $W \times V$, taking values in the $m \times m$ matrices.
Let $b$ be an analytic function with values in $V$ defined on the same neighborhood.
Then there exists a neighborhood of $0$ in $W$ on which the quasilinear Cauchy problem:
:$\partial_{x_n} f = \map {A_1} {x, f} \partial_{x_1} f + \cdots + \map {A_{n - 1} } {x, f} \partial_{x_{n - 1} } f + \map b {x, f}$
with initial condition:
:$\map f x = 0$
on the hypersurface:
:$x_n = 0$
has a unique analytic solution:
:$f: W \to V$
near $0$.
== Proof ==
== Also known as ==
== Historical Note ==
== Sources ==
* 
* 
Category:Cauchy-Kovalevsky Theorem
Category:Partial Differential Equations
Cauchy-Riemann Equations

== Theorem ==
Let $D \subseteq \C$ be an open subset of the set of complex numbers $\C$.
Let $f: D \to \C$ be a complex function on $D$.
Let $u, v: \set {\tuple {x, y} \in \R^2: x + i y = z \in D} \to \R$ be the two real-valued functions defined as:
where:
:$\map \Re {\map f z}$ denotes the real part of $\map f z$
:$\map \Im {\map f z}$ denotes the imaginary part of $\map f z$.
Then $f$ is complex-differentiable in $D$ :
:$u$ and $v$ are differentiable in their entire domain
and:
:The following two equations, known as the Cauchy-Riemann equations, hold for the continuous partial derivatives of $u$ and $v$:
If the conditions are true, then for all $z \in D$:
:$\map {f'} z = \map {\dfrac {\partial f} {\partial x} } z = -i \map {\dfrac {\partial f} {\partial y} } z$
=== Polar Form ===
== Proof ==
=== Necessary Condition ===
=== Sufficient Condition ===
=== Expression of Derivative ===
== Sources ==
* : $3$: Elementary Analytic Methods: $3.7$ Complex Numbers and Functions: Complex Functions, Cauchy-Riemann Equations: $3.7.30$
* 
* : $\S 1.3$
* 
* 
Category:Complex Analysis
Category:Cauchy-Riemann Equations
Cauchy-Riemann Equations/Expression of Derivative

== Theorem ==
Let $D \subseteq \C$ be an open subset of the set of complex numbers $\C$.
Let $f: D \to \C$ be a complex function on $D$.
Let $u, v: \set {\tuple {x, y} \in \R^2: x + i y = z \in D} \to \R$ be two real-valued functions defined as:
where:
:$\map \Re {\map f z}$ denotes the real part of $\map f z$
:$\map \Im {\map f z}$ denotes the imaginary part of $\map f z$.
Then $f$ is complex-differentiable in $D$ :
:$u$ and $v$ are differentiable in their entire domain
and:
:The following two equations, known as the Cauchy-Riemann equations, hold for the partial derivatives of $u$ and $v$:
If the conditions are true, then for all $z \in D$:
:$\map {f'} z = \map {\dfrac {\partial f} {\partial x} } z = -i \map {\dfrac {\partial f} {\partial y} } z$
== Proof ==
Let $z = x + i y$.
Then:
 
Similarly:
== Sources ==
* : $\S 1.3$
Category:Cauchy-Riemann Equations
Cauchy-Riemann Equations/Polar Form

== Theorem ==
Let $D \subseteq \C$ be an open subset of the set of complex numbers $\C$.
Let $f: D \to \C$ be a complex function on $D$.
Let $u, v: \set {\tuple {x, y} \in \R^2: x + i y = z \in D} \to \R$ be the two real-valued functions defined as:
where:
:$\map \Re {\map f z}$ denotes the real part of $\map f z$
:$\map \Im {\map f z}$ denotes the imaginary part of $\map f z$.
The Cauchy-Riemann equations can be expressed in polar form as:
where $z$ is expressed in exponential form as:
:$z = r e^{i \theta}$
== Proof ==
== Sources ==
* : $3$: Elementary Analytic Methods: $3.7$ Complex Numbers and Functions: Complex Functions, Cauchy-Riemann Equations: $3.7.31$
Category:Cauchy-Riemann Equations
Cauchy-Riemann Equations/Sufficient Condition

== Theorem ==
Let $D \subseteq \C$ be an open subset of the set of complex numbers $\C$.
Let $f: D \to \C$ be a complex function on $D$.
Let $u, v: \set {\paren {x, y} \in \R^2: x + i y = z \in D} \to \R$ be two real-valued functions defined as:
:$\map u {x, y} = \map \Re {\map f z}$
:$\map v {x, y} = \map \Im {\map f z}$
where:
:$\map \Re {\map f z}$ denotes the real part of $\map f z$
:$\map \Im {\map f z}$ denotes the imaginary part of $\map f z$.
Let:
:$u$ and $v$ be differentiable in their entire domain
and:
:The following two equations, known as the Cauchy-Riemann equations, hold for the continuous partial derivatives of $u$ and $v$:
::$(1): \quad \dfrac {\partial u} {\partial x} = \dfrac {\partial v} {\partial y}$
::$(2): \quad \dfrac {\partial u} {\partial y} = -\dfrac {\partial v} {\partial x}$
Then:
:$f$ is complex-differentiable in $D$
and:
:for all $z \in D$:
::$\map {f'} z = \map {\dfrac {\partial f} {\partial x} } z = -i \map {\dfrac {\partial f} {\partial y} } z$
== Proof ==
Suppose that the Cauchy-Riemann equations hold for $u$ and $v$ in their entire domain.
Let $h, k \in \R \setminus \set 0$, and put $t = h + i k \in \C$.
Let $\tuple {x, y} \in \R^2$ be a point in the domain of $u$ and $v$.
Put:
:$a = \map {\dfrac {\partial u} {\partial x} } {x, y} = \map {\dfrac {\partial v} {\partial y} } {x, y}$
and:
:$b = -\map {\dfrac {\partial u} {\partial y} } {x, y} = \map {\dfrac {\partial v} {\partial x} } {x, y}$
From the Epsilon-Function Differentiability Condition, it follows that:
:$\map u {x + h, y} = \map u {x, y} + h \paren {a + \map {\epsilon_{u x} } h}$
:$\map u {x, y + k} = \map u {x, y} + k \paren {-b + \map {\epsilon_{u y} } k}$
:$\map v {x + h, y} = \map v {x, y} + h \paren {b + \map {\epsilon_{v x} } k}$
:$\map v {x, y + k} = \map v {x, y} + k \paren {a + \map {\epsilon_{v y} } h}$
where $\epsilon_{u x}, \epsilon_{u y}, \epsilon_{v x}, \epsilon_{v y}$ are real functions that converge to $0$ as $h$ and $k$ tend to $0$.
We have that the partial derivatives are continuous.
Then Epsilon-Function Differentiability Condition shows that $\epsilon_{u x}, \epsilon_{u y}, \epsilon_{v x}, \epsilon_{v y}$ are continuous real functions.
With $z = x + i y$, it follows that:
Now, define $\map \epsilon t$ as:
By Modulus Larger than Real Part and Imaginary Part:
:$\cmod k \le \cmod t$
and:
:$\cmod h \le \cmod t$
Hence dividing by $\cmod t$ we have:
:$\cmod {\dfrac h t} \le 1$
and:
:$\cmod {\dfrac k t} \le 1$
From Composite of Continuous Mappings is Continuous, it follows that $\map \epsilon t$ is continuous.
Thus:
This shows that:
:$\ds \lim_{t \mathop \to 0} \map \epsilon t = 0$
Then the Epsilon-Function Complex Differentiability Condition shows that:
:$\map {f'} x = a + i b$
== Sources ==
* $\S 1.3$
Category:Cauchy-Riemann Equations
Cauchy-Schwarz Inequality/Complex Numbers

== Theorem ==
:$\ds \paren {\sum \cmod {w_i}^2} \paren {\sum \cmod {z_i}^2} \ge \cmod {\sum w_i z_i}^2$
where all of $w_i, z_i \in \C$.
== Proof ==
Let $w_1, w_2, \ldots, w_n$ and $z_1, z_2, \ldots, z_n$ be arbitrary complex numbers.
Take the Binet-Cauchy Identity:
:$\ds \paren {\sum_{i \mathop = 1}^n a_i c_i} \paren {\sum_{j \mathop = 1}^n b_j d_j} = \paren {\sum_{i \mathop = 1}^n a_i d_i} \paren {\sum_{j \mathop = 1}^n b_j c_j} + \sum_{1 \mathop \le i \mathop < j \mathop \le n} \paren {a_i b_j - a_j b_i} \paren {c_i d_j - c_j d_i}$
and set $a_i = w_i, b_j = \overline {z_j}, c_i = \overline {w_i}, d_j = z_j $.
This gives us:
Hence the result.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $168$
* : $\S 1.2.3$: Sums and Products: Exercise $30$
Category:Complex Analysis
Category:Cauchy-Bunyakovsky-Schwarz Inequality
Cauchy Condensation Test

== Theorem ==
Let $\sequence {a_n}: n \mapsto \map a n$ be a decreasing sequence of strictly positive terms in $\R$ which converges with a limit of zero.
That is, for every $n$ in the domain of $\sequence {a_n}$: $a_n > 0$, $a_n \ge a_{n + 1}$, and $a_n \to 0$ as $n \to +\infty$.
Then the series:
:$\ds \sum_{n \mathop = 1}^\infty a_n$ 
converges the condensed series:
:$\ds \sum_{n \mathop = 1}^\infty 2^n \map a {2^n}$
converges.
== Proof ==
=== Necessary Condition ===
We will first show that:
:if the condensed series $\ds \sum_{n \mathop = 1}^\infty 2^n \map a {2^n}$ converges
then:
:$\ds \sum_{n \mathop = 1}^\infty a_n$ converges as well.
Assume $\ds \sum_{n \mathop = 1}^\infty 2^n \map a {2^n}$ converges.
Consider the graph of $\sequence {a_n}$ and the partial sums of $\sum 2^n \map a {2^n}$:
:
The dotted black line represents the sequence $\sequence {a_n}$.
The $n$th rectangle has:
:Base equal to $2^n$
:Altitude equal to $\map a {2^n}$
The series of partial sums of the areas of the rectangles are:
:$\ds \sum_{n \mathop = 1}^N 2^n \map a {2^n}$
which is precisely the defined condensed series.
The series $\sum a_n$ can be viewed as the sum of the areas of rectangles with width $1$ and height $a_n$. 
The diagram suggests that the partial sums of $\sum a_n$ are not greater than the condensed partial sums:
:$\ds \sum_{n \mathop = 1}^N a_n \le \sum_{n \mathop = 1}^N 2^n \map a {2^n}$
To formalize this claim, observe that because $\sequence {a_n}$ is decreasing:
:$a_1 + \underbrace {a_2 + a_3}_{\le 2 a_2} + \underbrace {a_4 + a_5 + a_6 + a_7}_{\le 4 a_4} + \cdots + a_N \le a_1 + 2 a_2 + 4 a_4 + \cdots + 2^N \map a {2^N}$
As $n \to +\infty$, the converges by hypothesis.
Hence $\ds \sum_{n \mathop = 1}^\infty a_n$ also converges, by the Comparison Test.
=== Sufficient Condition ===
We will show that if $\ds \sum_{n \mathop = 1}^\infty a_n$ converges, the condensed series $\ds \sum_{n \mathop = 1}^\infty 2^n \map a {2^n}$ converges as well.
Assume $\ds \sum_{n \mathop = 1}^\infty a_n$ converges.
From the Combination Theorem for Sequences, $2\ds \sum_{n \mathop = 1}^\infty a_n = \ds \sum_{n \mathop = 1}^\infty 2 a_n$ converges as well.
Consider, then, the graph of $\sequence {a_n}$ and of the partial sums of what will be shown to equal $\ds \frac 1 2 \sum 2^n \map a {2^n}$:
:
The dotted black line represents the sequence $\sequence {a_n}$.
The $n$th rectangle has:
:Base equal to $2^n$
:Altitude equal to $\map a {2 \cdot 2^n}$
The series of partial sums of the areas of the rectangles are:
From Deletion of Terms from a Sequence, it is justified to examine only the behavior of $n$ for $n \ge 2$.
The series $\sum a_n$ can be viewed as the sum of the area of rectangles with height $a_n$ and width $1$.
The diagram suggests that the partial sums of $\sum a_n$ satisfy the following inequality:
To formalize this claim, observe that because $\sequence {a_n}$ is decreasing: 
By hypothesis, the converges as $n \to +\infty$.
Hence the condensed series converges as well, by the Comparison Test.
== Also see ==
* Cauchy Condescension Test
== Sources ==
* 
* : $\S 13.10$
Category:Series
Category:Convergence Tests
Category:Analytic Geometry
Cauchy Distribution is Symmetric about Median

== Theorem ==
Let $X$ be a continuous random variable with a Cauchy distribution:
:$\map {f_X} x = \dfrac 1 {\pi \lambda \paren {1 + \paren {\frac {x - \gamma} \lambda }^2} }$
for:
:$\lambda \in \R_{>0}$
:$\gamma \in \R$
$X$ is (reflectionally) symmetric about the vertical line through the median $\gamma$.
== Proof ==
Recall from Median of Cauchy Distribution‎ that $\gamma$ is indeed the median of $X$.
So $X$ is symmetric about the line $x = \gamma$.
== Sources == 
* 
Category:Cauchy Distribution
Category:Medians
Cauchy Integral Test

== Theorem ==
Let $f$ be a real function which is continuous, positive and decreasing on the interval $\hointr 1 {+\infty}$.
Let the sequence $\sequence {\Delta_n}$ be defined as:
:$\ds \Delta_n = \sum_{k \mathop = 1}^n \map f k - \int_1^n \map f x \rd x$
Then $\sequence {\Delta_n} $ is decreasing and bounded below by zero.
Hence it converges.
== Proof ==
From Darboux's Theorem, we have that:
:$\ds m \paren {b - a} \le \int_a^b \map f x \rd x \le M \paren {b - a}$
where:
:$M$ is the maximum
and:
:$m$ is the minimum
of $\map f x$ on $\closedint a b$.
Since $f$ decreases, $M = \map f a$ and $m = \map f b$.
Thus it follows that:
:$\ds \forall k \in \N_{>0}: \map f {k + 1} \le \int_k^{k + 1} \map f x \rd x \le \map f k$
as $\paren {k + 1} - k = 1$.
Thus:
Thus $\sequence {\Delta_n}$ is decreasing.
Also:
Hence the result.
== Also known as ==
== Motivation ==
== Sources ==
* : $\S 13.32$
* : $\S 1.2.2$: Summary of convergence tests: Theorem $1.2$
* 
* 
Category:Cauchy Integral Test
Category:Integral Calculus
Category:Convergence Tests
Category:Series
Cauchy Integral Test/Also known as

== Cauchy Integral Test: Also known as ==
The Cauchy Integral Test is also known as the Euler-Maclaurin Summation Formula, but that result properly refer to a more precise theorem of which this is a simple corollary.
Some sources just use the name Integral Test.
== Sources ==
* : $\S 13.32$
* : $\S 1.2.2$: Summary of convergence tests: Theorem $1.2$
* 
* 
Category:Cauchy Integral Test
Cauchy Integral Test/Motivation

== Cauchy Integral Test: Motivation ==
It follows from the Cauchy Integral Test that if $f$ is continuous, positive and decreasing on $\hointr 1 \infty$, then the series $\ds \sum_{k \mathop = 1}^\infty \map f k$ and the improper integral $\ds \int_1^{\mathop \to +\infty} \map f x \rd x$ either both converge or both diverge.
So this theorem provides a test for the convergence of both a series and an improper integral.
== Sources ==
* : $\S 13.32$
Category:Cauchy Integral Test
Cauchy Mean Value Theorem

== Theorem ==
Let $f$ and $g$ be real functions which are continuous on the closed interval $\closedint a b$ and differentiable on the open interval $\openint a b$.
Suppose:
:$\forall x \in \openint a b: \map {g'} x \ne 0$
Then:
:$\exists \xi \in \openint a b: \dfrac {\map {f'} \xi} {\map {g'} \xi} = \dfrac {\map f b - \map f a} {\map g b - \map g a}$
== Proof ==
First we check $\map g a \ne \map g b$.
 $\map g a = \map g b$.
From Rolle's Theorem:
:$\exists \xi \in \openint a b: \map {g'} \xi = 0$.
This contradicts $\forall x \in \openint a b: \map {g'} x \ne 0$.
Thus by Proof by Contradiction $\map g a \ne \map g b$.
Let $h = \dfrac {\map f b - \map f a} {\map g b - \map g a}$.
Let $F$ be the real function defined on $\closedint a b$ by:
:$\map F x = \map f x - h \map g x$.
Then:
=== Also presented as ===
=== Geometrical Interpretation ===
== Also known as ==
== Example ==
== Also see ==
* Mean Value Theorem
== Sources ==
* 
Category:Cauchy Mean Value Theorem
Category:Differential Calculus
Cauchy Mean Value Theorem/Also presented as

== Cauchy Mean Value Theorem: Also presented as ==
The Cauchy Mean Value Theorem can also be found presented as:
:$\exists \xi \in \openint a b: \map {f'} \xi \paren {\map g b - \map g a} = \map {g'} \xi \paren {\map f b - \map f a}$
where:
:$f$ and $g$ are continuous real functions on $\closedint a b$ and differentiable on $\openint a b$
:$\forall x \in \openint a b: \map {g'} x \ne 0$.
== Sources ==
* 
Category:Cauchy Mean Value Theorem
Cauchy Mean Value Theorem/Example

== Example of Use of Cauchy Mean Value Theorem ==
In the 2012 Olympics Usain Bolt won the 100 metres gold medal with a time of $9.63$ seconds.
By definition, his average speed was the total distance travelled divided by the total time it took:
The Mean Value Theorem gives:
:$\map {f'} c = \dfrac {\map f b - \map f a} {b - a}$
Hence, at some point Bolt was actually running at the average speed of $37.38 \ \mathrm {km/h}$
Asafa Powell was participating in that same race.
He achieved a time of $11.99 \ \mathrm s = 1.245 \times 9.63 \ \mathrm s$.
So Bolt's average speed was $1.245$ times the average speed of Powell.
The Cauchy Mean Value Theorem gives:
Hence, at some point, Bolt was actually running at a speed exactly $1.245$ times that of Powell's.
== Sources ==
* 
Category:Cauchy Mean Value Theorem
Cauchy Mean Value Theorem/Geometrical Interpretation

== Geometrical Interpretation of Cauchy Mean Value Theorem ==
Consider two functions $\map f x$ and $\map g x$:
:continuous on the closed interval $\closedint a b$
:differentiable on $\openint a b$.
For every $x \in \closedint a b$, we consider the point $\tuple {\map f x, \map g x}$.
If we trace out the points $\tuple {\map f x, \map g x}$ over every $x \in \closedint a b$, we get a curve in two dimensions, as shown in the graph:
:
In the drawing, the slope of the red line is $\dfrac {\map g b - \map g a} {\map f b - \map f a}$.
This is because:
:$\dfrac {\Delta y} {\Delta x} = \dfrac {\map g b - \map g a} {\map f b - \map f a}$
assuming that the vertical axis, which contains the value of $\map f x$, is the $y$-axis.
The slope of the green line is $\dfrac {\map {g'} c} {\map {f'} c}$.
This is because:
:$\valueat {\dfrac {\d g} {\d f} } {x \mathop = c} = \valueat {\dfrac {\d g / \d x} {\d f / \d x} } {x \mathop = c} = \dfrac {\map {g'} c} {\map {f'} c}$
The graph illustrates that for the value of $c$ chosen in the graph, the slopes of the red line and green line are the same.
That is:
:$\dfrac {\map g b - \map g a} {\map f b - \map f a} = \dfrac {\map {g'} c} {\map {f'} c}$
== Sources ==
* 
Category:Cauchy Mean Value Theorem
Cauchy Product of Absolutely Convergent Series

== Theorem ==
Let $\ds \sum_{n \mathop = 0}^\infty a_n$ and $\ds \sum_{n \mathop = 0}^\infty b_n$ be two real series that are absolutely convergent.
Then the Cauchy product of $\ds \sum_{n \mathop = 0}^\infty a_n$ and $\ds \sum_{n \mathop = 0}^\infty b_n$ is absolutely convergent.
== Proof ==
== Sources ==
* : $\S 1.2.3$: Operations with series: Theorem $1.7$
Category:Convergence
Category:Series
Cauchy Sequence Converges Iff Equivalent to Constant Sequence

== Theorem ==
Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring.
Let $\CC$ be the ring of Cauchy sequences over $R$
Let $\NN$ be the set of null sequences.
Let $\\CC \,\big / \NN$ be the quotient ring of Cauchy sequences of $\CC$ by the maximal ideal $\NN$.
Let $\sequence {x_n} \in \CC$.
Then $\sequence {x_n}$ converges in $\struct {R, \norm {\,\cdot\,} }$ :
:$\exists a \in R: \sequence {x_n} \in \sequence {a, a, a, \dotsc} + \NN$
where $\sequence {a, a, a, \dotsc} + \NN$ is the left coset in $\CC \, \big / \NN$ that contains the constant sequence $\sequence {a, a, a, \dotsc}$.
== Proof ==
By definition, $\sequence {x_n}$ converges to $a \in R$ $\ds \lim_{n \mathop \to \infty} \norm {x_n - a} = 0$
Then:
Category:Normed Division Rings
Cauchy Sequence Is Eventually Bounded Away From Non-Limit

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\sequence {x_n}$ be a Cauchy sequence in $R$.
Suppose $\sequence {x_n}$ does not converge to $l \in R$.
Then:
:$\exists K \in \N$ and $C \in \R_{>0}: \forall n > K: C < \norm {x_n - l}$
== Proof ==
Because $\sequence {x_n}$ does not converge to $l$:
:$\exists \epsilon \in \R_{>0}: \forall n \in \N: \exists m \ge n: \norm {x_m - l} \ge \epsilon$
Because $\sequence {x_n}$ is a Cauchy sequence:
:$\exists K \in \N: \forall n, m \ge K: \norm {x_n - x_m} < \dfrac \epsilon 2$
Let $M \ge K: \norm {x_M - l} \ge \epsilon$.
Then $\forall n > K$:
Let $C = \dfrac \epsilon 2$ and the result follows.
Category:Cauchy Sequence in Normed Division Ring is Bounded
Cauchy Sequence in Metric Space is not necessarily Convergent

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $\sequence {x_n}$ be a Cauchy sequence in $M$.
Then it is not necessarily the case that $M$ is a convergent sequence in $M$.
== Proof ==
Let $A \subseteq \R$ be the set of all points on $\R$ defined as:
:$A := \set {\dfrac 1 n : n \in \Z_{>0} }$
Let $M = \struct {A, \tau_d}$ be the integer reciprocal space under the usual (Euclidean) topology.
Let $\sequence {x_n}$ be a sequence in $A$ that converges to the limit $l \in A$.
From Integer Reciprocal Space contains Cauchy Sequence with no Limit, $\sequence {x_n}$ is a Cauchy sequence in $M$ which does not converge in $M$.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces: Complete Metric Spaces
Category:Metric Spaces
Category:Cauchy Sequences
Cauchy Sequence in Positive Integers under Scaled Euclidean Metric

== Theorem ==
Let $\Z_{>0}$ be the set of (strictly) positive integers.
Let $\delta: \Z_{>0} \times \Z_{>0} \to \R$ be the scaled Euclidean metric on $\Z_{>0}$ defined as:
:$\forall x, y \in \Z_{>0}: \map \delta {x, y} = \dfrac {\size {x - y} } {x y}$
The sequence $\sequence {x_n}$ in $\Z_{>0}$ defined as:
:$\forall n \in \N: x_n = n$
is a Cauchy sequence in $\struct {\Z_{>0}, \delta}$.
== Proof ==
For a general $x_m, x_n \in \sequence {x_n}$ as defined:
Let $\epsilon \in \R_{>0}$.
Then by the Axiom of Archimedes:
:$\exists N \in \N: N > \dfrac 1 \epsilon$
from which it follows that:
:$\epsilon > \dfrac 1 N$
Thus:
Therefore $\sequence {x_n}$ is a Cauchy sequence in $\struct {\Z_{>0}, \delta}$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $32$. Special Subsets of the Real Line: $10$
Category:Scaled Euclidean Metric
Category:Cauchy Sequences
Cauchy Sequence in Positive Integers under Usual Metric is eventually Constant

== Theorem ==
Let $\Z_{>0}$ be the set of (strictly) positive integers.
Let $d: \Z_{>0} \times \Z_{>0} \to \R$ be the usual (Euclidean) metric on $\Z_{>0}$.
Let $\sequence {x_n}$ be a Cauchy sequence in $\struct {\Z_{>0}, d}$.
Then:
:$\exists m, n \in \Z_{>0}: \forall r > n: x_r = m$
That is, $\sequence {x_n}$ is eventually constant.
== Proof ==
Let $\sequence {x_n}$ be a Cauchy sequence in $\struct {\Z_{>0}, d}$.
By definition:
:$\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall m, n \in \N: m, n \ge N: \map d {x_n, x_m} < \epsilon$
Let $\epsilon < 1$, say: $\epsilon = \dfrac 1 2$.
By the definition of $d$:
:$\forall m, n \in \N: x_m \ne x_n \implies \map d {x_m, x_n} \ge 1$
So the only possible way for:
:$\forall m, n \in \N: m, n \ge N: \map d {x_n, x_m} < \epsilon$
is for $x_m = x_n$.
Hence the result.
== Sources ==
* : Part $\text {II}$: Counterexamples: $32$. Special Subsets of the Real Line: $10$
Category:Euclidean Metric
Category:Cauchy Sequences
Cauchy Sequence is Bounded/Metric Space

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Then every Cauchy sequence in $M$ is bounded.
== Proof ==
Let $\sequence {x_n}$ be a Cauchy sequence in $M$.
By definition:
:$\forall \epsilon > 0: \exists N \in \N: \forall m, n > N: \map d {x_n, x_m} < \epsilon$
Particularly, setting $\epsilon = 1$:
:$\exists N_1: \forall m, n > N_1: \map d {x_n, x_m} < 1$
Note that since $N_1 \ge N_1$, this means that:
:$\forall n \ge N_1: \map d {x_n, x_{N_1} } < 1$
To show $\sequence {x_n}$ is bounded, we need to show that there exists $a \in A$ and $K \in \R$ such that $\map d {x_n, a} \le K$ for all $x_n \in \sequence {x_n}$.
Let $K'$ be the maximum distance from $x_{N_1}$ to any of the earlier terms in the sequence.
That is, $K' = \max \set {\map d {x_{N_1}, x_1}, \map d {x_{N_1}, x_2}, \ldots, \map d {x_{N_1}, x_{N_1 - 1} } }$
Then:
:Each $x_n$ for $n \ge N_1$ satisfies $\map d {x_{N_1}, x_n} \le 1$ by choice of $N_1$ as mentioned above
:Each $x_n$ for $n < N_1$ satisfies $\map d {x_{N_1}, x_n} \le K'$ by choice of $K'$.
Thus, taking $a = x_{N_1}$ and $K = \max \set {K', 1}$, we have shown that each $x_n$ satisfies $\map d {a, x_n} \le K$.
So, $\sequence {x_n}$ is bounded.
== Sources ==
* : $\S 12$-$2$: Convergent and divergent sequences
Category:Cauchy Sequences
Cauchy Sequence is Bounded/Normed Division Ring

== Theorem ==
Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring.
Every Cauchy sequence in $R$ is bounded.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : $\S 1.2$: Normed Fields: Exercise $11 \ (1)$
Category:Cauchy Sequence is Bounded
Category:Cauchy Sequences in Normed Division Rings
Category:Cauchy Sequence in Normed Division Ring is Bounded
Cauchy Sequence is Bounded/Normed Division Ring/Proof 1

== Theorem ==
== Proof ==
Let $\sequence {x_n} $ be a Cauchy sequence in $R$.
Then by definition:
:$\forall \epsilon \in \R_{\gt 0}: \exists N \in \N : \forall n, m \ge N: \norm {x_n - x_m} < \epsilon$
Let $n_1$ satisfy:
:$\forall n, m \ge n_1: \norm {x_n - x_m} < 1$
Then $\forall n \ge n_1$:
Let $K = \max \set {\norm {x_1}, \norm {x_2}, \dots, \norm {x_{n_1 - 1} }, 1 + \norm {x_{n_1} } }$.
Then:
:$\forall n < n_1: \norm {x_n} \le K$
:$\forall n \ge n_1: \norm {x_n} \le 1 + \norm {x_{n_1} } \le K$
It follows by definition that $\sequence {x_n}$ is bounded.
== Sources ==
* : $\S 1.2$: Normed Fields, Exercise $11 \ (2)$
Category:Cauchy Sequence in Normed Division Ring is Bounded