nyuuzyou/wikis huggingface dataset:
Law of Sines/Also presented as

== Law of Sines: Also presented as ==
In their presentation of the law of sines, some sources do not include the relation with the circumradius, but instead merely present:
:$\dfrac a {\sin A} = \dfrac b {\sin B} = \dfrac c {\sin C}$
== Sources ==
* : $\S 5$: Trigonometric Functions: $5.92$
* 
* 
* 
* : Chapter $5$: Eternal Triangles: Logarithms
Category:Law of Sines
Law of Sines/Proof 1

== Theorem ==
== Proof ==
Construct the altitude from $B$.
:
From the definition of sine:
:$\sin A = \dfrac h c$ and $\sin C = \dfrac h a$
Thus:
:$h = c \sin A$
and:
:$h = a \sin C$
This gives:
:$c \sin A = a \sin C$
So:
:$\dfrac a {\sin A} = \dfrac c {\sin C}$
Similarly, constructing the altitude from $A$ gives:
:$\dfrac b {\sin B} = \dfrac c {\sin C}$
Category:Law of Sines
Law of Sines/Proof 2

== Theorem ==
== Proof ==
Construct the circumcircle of $\triangle ABC$, let $O$ be the circumcenter and $R$ be the circumradius.
Construct $\triangle AOB$ and let $E$ be the foot of the altitude of $\triangle AOB$ from $O$.
:
By the Inscribed Angle Theorem:
:$\angle ACB = \dfrac {\angle AOB} 2$
From the definition of the circumcenter:
:$AO = BO$
From the definition of altitude and the fact that all right angles are congruent:
:$\angle AEO = \angle BEO$
Therefore from Pythagoras's Theorem:
:$AE = BE$
and then from Triangle Side-Side-Side Congruence:
:$\angle AOE = \angle BOE$
Thus:
:$\angle AOE = \dfrac {\angle AOB} 2$
and so:
:$\angle ACB = \angle AOE$
Then by the definition of sine:
:$\sin C = \map \sin {\angle AOE} = \dfrac {c / 2} R$
and so:
:$\dfrac c {\sin C} = 2 R$
The same argument holds for all three angles in the triangle, and so:
:$\dfrac c {\sin C} = \dfrac b {\sin B} = \dfrac a {\sin A} = 2 R$
Category:Law of Sines
Law of Sines/Proof 3

== Theorem ==
== Proof ==
=== Acute Case ===
Let $\triangle ABC$ be acute.
:
Construct the circumcircle of $\triangle ABC$.
Let its radius be $R$.
Construct its diameter $BX$ through $B$.
By Thales' Theorem, $\angle BAX$ is a right angle.
From Angles in Same Segment of Circle are Equal:
:$\angle AXB = \angle ACB$
Then:
The same construction can be applied to each of the remaining vertices of $\triangle ABC$.
Hence the result.
Let $\triangle ABC$ be obtuse.
:
As for the acute case, construct the circumcircle of $\triangle ABC$.
Let its radius be $R$.
Construct its diameter $BX$ through $B$.
By Thales' Theorem, $\angle BCX$ is a right angle.
We note that $\Box ABXC$ is a cyclic quadrilateral.
From Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles:
:$\angle BXC = 180 \degrees - A$
Hence using a similar argument to the acute case:
and the result follows.
== Sources ==
* : $\text V$. Trigonometry: The sine formula
Category:Law of Sines
Law of Species

== Theorem ==
Let $T$ be a right spherical triangle whose angles are $A$, $B$ and $C$ and whose respective sides opposite those angles are $a$, $b$ and $c$.
Let $C$ be the right angle of $T$.
Then:
:$(1): \quad$ $A$ and $a$ are of the same species, and $B$ and $b$ are of the same species
:$(2): \quad$ if $c < 90 \degrees$, then $a$ and $b$ are of the same species, as are $A$ and $B$
:$(3): \quad$ if $c > 90 \degrees$, then $a$ and $b$ are of the opposite species, as are $A$ and $B$.
== Proof ==
== Examples ==
== Also known as ==
== Sources ==
* 
* 
* 
* 
Category:Law of Species
Category:Species
Category:Spherical Triangles
Law of Species/Examples/Arbitrary Example 1

== Example of Use of Law of Species ==
Consider the right spherical triangle with:
:side $c$ equal to $30 \degrees$
:angle $B$ equal to $30 \degrees$.
The other sides are found by using Napier's Rules of Circular Parts.
Hence:
:$\sin b = \sin c \sin B$
and so:
:$\sin b = \dfrac 1 2 \cdot \dfrac 1 2 = \dfrac 1 4$
But from the Law of Species, $b$ and $B$ are of the same species.
We have that $B$ is an acute angle.
So $b$ must also be an acute angle.
Hence $b = \inv \sin {\dfrac 1 4}$ and less than $90 \degrees$.
So:
:$b = 14 \degrees \, 29 \minutes$
== Sources ==
* 
* 
Category:Examples of Use of Law of Species
Law of Subtraction

== Theorem ==
On the following number systems:
: integers $\Z$
: rational numbers $\Q$
: real numbers $\R$
: complex numbers $\C$
there exists a unique $x$ such that:
:$a + x = b$
for every given $a$ and $b$.
$x$ is then defined and denoted:
:$x := b - a$
== Proof ==
== Sources ==
* : $\S 1.1$. Number Systems: $\text{III}.$
Category:Numbers
Category:Subtraction
Law of Tangents

== Theorem ==
Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.
Then:
:$\dfrac {a + b} {a - b} = \dfrac {\tan \frac 1 2 \paren {A + B} } {\tan \frac 1 2 \paren {A - B} }$
=== Corollary ===
== Proof ==
Let $d = \dfrac a {\sin A}$.
From the Law of Sines, let:
:$d = \dfrac a {\sin A} = \dfrac b {\sin B}$
so that:
== Also see ==
* Law of Sines
* Law of Cosines
== Sources ==
* : $\S 5$: Trigonometric Functions: $5.94$
Category:Law of Tangents
Category:Triangles
Category:Tangent Function
Category:Named Theorems
Law of Tangents/Corollary

== Corollary to Law of Tangents ==
Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.
Then:
:$\tan \dfrac {A - B} 2 = \dfrac {a - b} {a + b} \cot \dfrac C 2$
== Proof 1 ==
== Proof 2 ==
== Also presented as ==
== Also known as ==
Some sources present this as the Law of Tangents itself.
== Examples ==
== Sources ==
* : $\text V$. Trigonometry: Formulae $(37)$
* : $\text V$. Trigonometry: Exercises $\text {XXXII}$: $13$.
* : The tangent rule
* : The tangent rule
Category:Law of Tangents
Law of Tangents/Corollary/Examples/Arbitrary Example 1

== Example of Use of Corollary to Law of Tangents ==
Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.
Let $\triangle ABC$ be such that:
where the units of measurement of $a$ and $b$ are arbitrary.
Then:
== Proof ==
First: 
Then:
Then we have:
== Sources ==
* : $\text V$. Trigonometry: Solution of triangles
Category:Law of Tangents
Laws of Logarithms

== Theorem ==
Let $x, y, b \in \R_{>0}$ be (strictly) positive real numbers.
Let $a \in \R$ be any real number such that $a > 0$ and $a \ne 1$.
Let $\log_a$ denote the logarithm to base $a$.
Then:
=== Change of Base of Logarithm ===
=== Sum of Logarithms ===
=== Logarithm of Power/Natural Logarithm ===
=== Logarithm of Power/General Logarithm ===
=== Difference of Logarithms ===
=== Logarithm of Reciprocal ===
== Examples ==
Category:Logarithms
Laws of Logarithms/Examples/Arbitrary Example 1

== Example of Use of Laws of Logarithms ==
:$\ln \dfrac {A^p B^q C^r} {D^s E^t} = p \ln A + q \ln B + r \ln C - s \ln D - t \ln E$
== Sources ==
* : Chapter $1$: Computations Using Logarithms
Category:Examples of Use of Laws of Logarithms
Leading Coefficient of Matrix/Examples

== Examples of Leading Coefficient of Matrix ==
Consider:
:$\mathbf A = \begin {bmatrix} 1 & 5 & 4 & 2 \\ 0 & 0 & 5 & 7 \\ 0 & 6 & 0 & 9 \\ 0 & 0 & 0 & 0 \\ \end {bmatrix}$
The leading coefficient of row $1$ is $1$.
The leading coefficient of row $2$ is $5$.
The leading coefficient of row $3$ is $6$.
Row $4$ has no leading coefficient.
Category:Leading Coefficients of Matrices
Leading Coefficient of Product of Polynomials over Integral Domain

== Theorem ==
Let $R$ be an integral domain.
Let $f, g \in R \sqbrk x$ be polynomials.
Let $c$ and $d$ be their leading coefficients.
Then $f g$ has leading coefficient $c d$.
== Proof ==
Let $p = \deg f$ and $q = \deg g$ be their degrees.
By Degree of Product of Polynomials over Integral Domain, $\map \deg {f g} = p + q$.
For a natural number $n \ge 0$, let:
:$a_n$ be the coefficient of the monomial $x^n$ in $f$.
:$b_n$ be the coefficient of the monomial $x^n$ in $g$.
By Coefficients of Product of Two Polynomials, the coefficient of $x^{p + q}$ in $f g$ is the sum:
:$\ds \sum_{k \mathop = p}^p a_k b_{p + q - k}$
By Indexed Summation over Interval of Length One:
:$\ds \sum_{k \mathop = p}^p a_k b_{p + q - k} = a_p b_q = c d$
Category:Polynomial Theory
Leaf of Rooted Tree is on One Branch

== Theorem ==
Let $T$ be a rooted tree with root node $r_T$.
Let $t$ be a leaf node of $T$.
Then there exists a unique branch $\Gamma$ of $T$ such that $t \in \Gamma$.
== Proof ==
Let $t$ be on the branch $\Gamma$.
By definition of branch, $\Gamma$ is a path from the root node $r_T$ to $t$.
By Path in Tree is Unique, such $\Gamma$ is unique.
== Sources ==
* : $\S 1.7$: Tableaus
Category:Rooted Trees
Least Fixed Point of Enumeration Operator/Lemma

== Lemma ==
Let $\psi : \powerset \N \to \powerset \N$ be an enumeration operator.
Let $A_i$ be defined recursively as:
:$A_0 = \empty$
:$A_{n + 1} = \map \psi {A_n}$
Then $A_i$ is an increasing sequence with respect to the subset ordering.
== Proof ==
By definition of enumeration operator, let $\phi \subseteq \N$ be some recursively enumerable set that yields $\psi$.
Define:
:$\map P i \iff \forall j < i: A_j \subseteq A_i$
Proceed by induction on $i$.
=== Basis for the Induction ===
As there are no natural numbers $j < 0$, $\map P 0$ is satisfied vacuously.
=== Induction Hypothesis ===
Suppose the induction hypothesis holds.
That is, suppose that, for $k \in \N$:
:$\map P k$
or, equivalently:
:$\forall j < k: A_j \subseteq A_k$
=== Induction Step ===
Let $j \le k$ be arbitrary.
By the induction hypothesis, as well as Set is Subset of Itself:
:$A_j \subseteq A_k$
Therefore, by Subset Relation is Transitive:
:$A_k \subseteq A_{k + 1} \implies A_j \subseteq A_{k + 1}$
Thus, it suffices to show that $A_k \subseteq A_{k + 1}$.
Suppose $k = 0$.
Then, by definition, $A_0 = \empty$.
Then, $A_0 \subseteq A_1$ follows from Empty Set is Subset of All Sets.
Otherwise, let $k > 0$.
Let $x \in A_k$ be arbitrary.
By definition of enumeration operator, there is a finite subset $B \subseteq A_{k - 1}$ such that:
:$\map \pi {x, b} \in \phi$
where $b$ is the finite set coding of $B$.
But, by the induction hypothesis, $A_{k - 1} \subseteq A_k$.
Therefore, by Subset Relation is Transitive:
:$B \subseteq A_k$
Additionally, as $B$ is finite by definition, it follows that $B$ is a finite subset of $A_k$.
As $\map \pi {x, b} \in \phi$, it follows that $x \in A_{k + 1}$ by definition.
As $x$ was arbitrary, it follows from the definition that:
:$A_k \subseteq A_{k + 1}$
As the Basis for the Induction holds, and the Induction Step follows from the Induction Hypothesis, we conclude by the Principle of Mathematical Induction that:
:$\forall i \in \N, j < i: A_j \subseteq A_i$
Therefore, the sequence $A_i$ is increasing with respect to the subset ordering by definition.
Least Fixed Point of Enumeration Operator is Recursively Enumerable

== Theorem ==
Let $\psi : \powerset \N \to \powerset \N$ be an enumeration operator.
Then there exists a recursively enumerable set $A$ such that:
:$A$ is a fixed point of $\psi$
:Every fixed point of $\psi$ is a superset of $A$
== Proof ==
By Least Fixed Point of Enumeration Operator, such an $A$ can be defined as:
:$\ds \bigcup_{i \mathop \in \N} A_i$
where:
:$A_0 = \O$
:$A_{n + 1} = \map \psi {A_n}$
By definition of enumeration operator, there exists a recursively enumerable set $\phi \subseteq \N$ such that:
:$\map \psi A = \set {x \in \N : \exists \text { finite } B \subseteq A : \map \pi {x, b} \in \phi}$
where:
:$b \in \N$ is the code number for $B$
:$\pi : \N^2 \to \N$ is the Cantor pairing function
Suppose that there is no $x_0 \in \N$ such that:
:$\map \pi {x_0, 0} \in \phi$
Then, for every $x \in \N$:
:$\map \pi {x, 0} \notin \phi$
By induction, we will show that every $A_i = \O$.
$A_0 = \O$ by definition.
If $A_i = \O$, then by Subset of Empty Set iff Empty, for every $B \subseteq A_i$:
:$B = \O$
Therefore:
:$b = 0$
But then:
:$\map \pi {x, b} \notin \phi$
for every $x \in \N$.
Thus, by definition of $A_{i + 1}$:
:$A_{i + 1} = \O$
By Principle of Mathematical Induction:
:$A_i = \O$
for every $i \in \N$.
Thus:
:$\ds A = \bigcup_{i \mathop \in \N} A_i = \empty$
As $\phi$ is recursively enumerable and $A = \phi$:
:$A$ is recursively enumerable
Now, suppose there is some $x_0 \in \N$ such that:
:$\map \pi {x_0, 0} \in \phi$
Then:
:$x_0 \in A_1 \subseteq A$
as $0$ codes the empty set.
Thus, $A$ is non-empty.
By Corollary to Recursively Enumerable Set is Image of Primitive Recursive Function:
:There exists a primitive recursive function $f : \N \to \N$ such that:
::$\Img f = \phi$
Define:
:$\map g {t, b} = \begin{cases}
\map {\pi_1} {\map f t} & : \map {\RR_\subseteq} {\map {\pi_2} {\map f t}, b} \\
x_0 & : \text {otherwise}
\end{cases}$
where:
:$\pi_1$ and $\pi_2$ are the projections of the Cantor pairing function
:$\map {\RR_\subseteq} {x, y}$ $x$ and $y$ respectively code finite sets $X$ and $Y$ such that $X \subseteq Y$
Least Number with Three Given Fractions

== Theorem ==
== Proof ==
Let $a, b, c$ be the given aliquot parts.
Let $d, e, f$ be the numbers called by the same name as the aliquot parts $a, b, c$.
From , let:
:$g = \lcm \set {d, e, f}$
So $g$ has aliquot parts called by the same name as $d, e, f$.
Therefore $g$ has the aliquot parts $a, b, c$.
Suppose there exists $h \in \N: h < g$ which has the aliquot parts $a, b, c$.
By , $h$ will be measured by numbers called by the same name as the aliquot parts $a, b, c$.
Therefore $h$ is measured by $d, e, f$.
But $h < g$ which is impossible.
Therefore there is no number less than $g$ which has the aliquot parts $a, b, c$.
== Sources ==
* : Book $\text{VII}$. Propositions
Category:Divisors
Least Ratio of Numbers

== Theorem ==
== Proof ==
Let $A, B, C$ be the given numbers, as many as we please.
$A, B, C$ are either coprime or not.
Let $A, B, C$ be coprime.
Then by they are the least of those which have the same ratio with them.
Let $A, B, C$ not be coprime.
Then by their GCD $D$ can be found.
As many times as $D$ measures each of $A, B, C$, let those be $E, F, G$ respectively.
That is:
:$D E = A, D F = B, D G = C$
By , $E, F, G$ are in the same ratio with $A, B, C$.
Suppose $E, F, G$ are not the least of those which have the same ratio with $A, B, C$.
Let those numbers be $H, K, L$.
Then $H$ measures $A$ the same number of times that $K, L$ measure $B, C$.
Let this number of times be $M$.
We have that $H$ measures $A$ according to the units of $M$.
It follows from that $M$ also measures $A$ according to the units of $H$.
For the same reason, $M$ also measures $B$ and $C$ according to the units of $K$ and $L$.
Therefore $M$ measures $A$.
That is:
:$A = H M$
For the same reason:
:$A = E D$
That is:
:$E D = H M$
Therefore by :
:$E : H = M : D$
But $E > H$ and so $M > D$.
Also, $M$ measures $A$, $B$ and $C$.
But by hypothesis $D$ is the greatest common measure of $A, B, C$.
Therefore there cannot be any numbers less than $E, F, G$ which are in the same ratio with $A, B, C$.
Therefore $E, F, G$ are the least of those which are in the same ratio with $A, B, C$.
== Sources ==
* : Book $\text{VII}$. Propositions
Category:Greatest Common Divisor
Least Time at which Discrete-Time Adapted Stochastic Process equals or exceeds Real Number is Stopping Time

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \ge 0}, \Pr}$ be a filtered probability space.
Let $\sequence {X_n}_{n \ge 0}$ be a discrete-time $\sequence {\FF_n}_{n \ge 0}$-adapted stochastic process.
Let $\lambda \in \R$. 
Let:
:$T = \inf \set {n \in \Z_{\ge 0} : X_n \ge \lambda}$
Then $T$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$. 
== Proof ==
Note that for $t \in \Z_{\ge 0}$ and $\omega \in \Omega$, we have: 
:$\inf \set {n \in \Z_{\ge 0} : \map {X_n} \omega \ge \lambda} \le t$
:
:$\map {X_s} \omega \ge \lambda$ for some $s \le t$.
That is, we have: 
:$\ds \set {\omega \in \Omega : \inf \set {n \in \Z_{\ge 0} : \map {X_n} \omega \ge \lambda} \le t} = \bigcup_{s \in \Z_{\ge 0}, \, 0 \le s \le t} \set {\omega \in \Omega : \map {X_s} \omega \ge \lambda}$
Since $\sequence {X_n}_{n \ge 0}$ is $\sequence {\FF_n}_{n \ge 0}$-adapted, we have:
:$\set {\omega \in \Omega : \map {X_s} \omega \ge \lambda} \in \FF_s$ for each $0 \le s \le t$.
Since $\sequence {\FF_n}_{n \ge 0}$ is a filtration, we have: 
:$\FF_s \subseteq \FF_t$
So:
:$\set {\omega \in \Omega : \map {X_s} \omega \ge \lambda} \in \FF_t$ for each $s \in \Z_{\ge 0}$ with $0 \le s \le t$.
So since $\FF_t$ is closed under finite union, we conclude:
:$\set {\omega \in \Omega : \inf \set {n \in \Z_{\ge 0} : \map {X_n} \omega \ge \lambda} \le t} \in \FF_t$
for each $t \in \Z_{\ge 0}$.
So $T$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.
Category:Adapted Stochastic Processes
Category:Stopping Times
Least Upper Bound Property

== Theorem ==
Let $S \subset \R$ be a non-empty subset of the set of real numbers such that $S$ is bounded above.
Then $S$ admits a supremum in $\R$.
This is known as the least upper bound property of the real numbers.
== Proof 1 ==
== Proof 2 ==
== Also known as ==
== Also see ==
* Greatest Lower Bound Property
* Continuum Property
== Sources ==
* : Chapter $1$: Fourier Series: $1.1$ Basic Concepts: $1.1.2$ Definitions
* : $1$: Review of some real analysis: $\S 1.1$: Real Numbers: Axiom $1.1.4$
* : $\S 2$: Continuum Property: $\S 2.4$: The Continuum Property
* : $(5)$: Completeness property
* : $(5)$: Completeness property
Category:Least Upper Bound Property
Category:Continuum Property
Category:Named Theorems
Least Upper Bound Property/Also known as

== Least Upper Bound Property: Also known as == 
The least upper bound property of $\R$ is also known as:
:the supremum principle
:the completeness property
:the continuum property (although this is also used to encompass the Greatest Lower Bound Property, a complementary result).
== Sources ==
* : $1$: Review of some real analysis: $\S 1.1$: Real Numbers: Axiom $1.1.4$
* : $\S 2$: Continuum Property: $\S 2.4$: The Continuum Property
* : $(5)$: Completeness property
* : $(5)$: Completeness property
Category:Least Upper Bound Property
Least Upper Bound Property/Proof 2

== Theorem ==
== Proof ==
Let $S$ be bounded above.
Let $L$ be the set of real numbers defined as:
:$\alpha \in L \iff \exists x \in S: \alpha < x$
Let $R := \relcomp \R L$, where $\complement_\R$ denotes complement in $\R$.
By construction of $L$, every element of $L$ is less than some element of $S$.
Hence no element of $L$ is an upper bound of $S$.
By construction of $R$, for every element $x$ of $R$, there exists no element of $S$ which is greater than $x$.
Hence every element of $R$ is an upper bound of $S$.
So, to prove the existence of $\sup S$, it is sufficient to demonstrate that $R$ contains a smallest number.
We verify that $L$ and $R$ fulfil the conditions for Dedekind's Theorem to hold.
We confirm that $\tuple {L, R}$ is a Dedekind cut of $\R$:
:$(1): \quad \set {L, R}$ is a partition of $\R$
:$(2): \quad L$ does not have a greatest element
:$(3): \quad \forall x \in L: \forall y \in R: x < y$
By Union with Relative Complement:
:$L \cup R = \R$
By Set with Relative Complement forms Partition, $\tuple {L, R}$ forms a partition of $\R$.
So $(1)$ holds immediately.
Let $\alpha \in L$.
Then there exists $x \in S$ such that $\alpha < x$.
Let $\alpha'$ be such that $\alpha < \alpha' < x$.
Then $\alpha' \in L$
So whatever $\alpha \in L$ is, it cannot be the greatest element of $L$.
Thus $(2)$ holds.
Let $\alpha \in L$.
Let $\beta \in R$.
Then there exists $x \in S$ such that $\alpha < x$.
By construction of $R$, $x \le \beta$.
Thus $\alpha < \beta$ for all $\alpha \in L, \beta \in R$.
Thus $(3)$ holds.
By the corollary to Dedekind's Theorem, either $L$ contains a greatest element or $R$ contains a smallest element.
We have shown that $L$ does not contain a greatest element.
Hence $R$ contains a smallest element.
Hence if $S$ is bounded above, it has a supremum.
Thus $\R$ is Dedekind complete by definition.
Now let $S$ be bounded below.
By Dedekind Completeness is Self-Dual, it follows that $S$ admits an infimum.
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Real Numbers: $1.36$. Theorem
Category:Continuum Property
Lebesgue's Number Lemma/Compact Space

== Theorem ==
Let $M = \struct {X, d}$ be a metric space.
Let $M$ be compact.
Then there exists a Lebesgue number for every open cover of $M$.
== Proof ==
Let $\UU$ be an open cover of $M$.
By definition of compact, there exists a finite subcover $\set {A_i}_{1 \le i \le n} \subseteq \UU$.
First, suppose some $A_i = X$.
Then, let $\epsilon = 1$.
For any $x \in X$:
:$\map {B_\epsilon} x \subseteq X = A_i \in \UU$
Therefore, $\epsilon$ is a Lebesgue number for $\UU$.
Now, suppose every $A_i \subsetneq X$.
Then, for every $1 \le i \le n$:
:$X \setminus A_i$
is non-empty.
By Distance in Pseudometric is Non-Negative:
:$\map d {x, y} \ge 0$
for every $x, y \in X$
Therefore, for every $x \in X$ and $1 \le i \le n$:
:$\set {\map d {x, y} : y \in X \setminus A_i}$
is bounded below.
Therefore, by the Greatest Lower Bound Property:
:$\ds \inf_{y \mathop \in X \setminus A_i} \map d {x, y}$
is well-defined.
For each $1 \le i \le n$, define $f_i : X \to \R_{\ge 0}$ as:
:$\ds \map {f_i} x = \inf_{y \mathop \in X \setminus A_i} \map d {x, y}$
We must show that every $f_i$ is continuous.
Let $\epsilon > 0$ and $a \in X$ be arbitrary.
Let $x \in X$ be arbitrary, satisfying:
:$\map d {x, a} < \dfrac \epsilon 2$
Combining the above, we have:
As $x \in X$ and $\epsilon > 0$ were arbitrary, it follows that $f_i$ is continuous at $a$ by definition.
But as $a \in X$ was also arbitrary, it follows that $f_i$ is continuous on $M$.
Define $f : X \to \R$ as:
:$\ds \map f x = \frac 1 n \sum_{i = 1}^n \map {f_i} x$
By the Sum Rule and Multiple Rule:
:$f$ is continuous on $M$.
By Continuous Image of Compact Space is Compact:
:$f \sqbrk X$ is compact
By Compact Subspace of Real Numbers is Closed and Bounded:
:$f \sqbrk X$ is closed and bounded.
By Infimum of Bounded Below Set of Reals is in Closure and Closed Set Equals its Closure:
:$\epsilon_0 = \ds \inf_{x \in X} \map f x \in f \sqbrk X$
from which it follows that there is some $x_0 \in X$ such that:
:$\map f {x_0} = \epsilon_0$
By definition of cover, there exists some $1 \le i \le n$ such that:
:$x_0 \in A_i$
By definition of open set, there exists some $\epsilon > 0$ such that:
:$\map {B_\epsilon} {x_0} \subseteq A_i$
Therefore:
Therefore:
:$\epsilon_0 > 0$
It remains to show that $\epsilon_0$ is a Lebesgue number for $\UU$.
Let $x \in X$ be arbitrary.
We want to show that:
:$\map {B_{\epsilon_0}} x \subseteq A_i$
for some $1 \le i \le n$.
By definition of infimum:
:$\map f x \ge \epsilon_0$
Therefore:
:$\ds \sum_{i \mathop = 1}^n \map {f_i} x \ge n \epsilon_0$
It follows that:
:$\map {f_i} x \ge \epsilon_0$
for some $1 \le i \le n$.
By definition of infimum again:
:$\forall y \in X \setminus A_i: \map d {x, y} \ge \epsilon_0$
That is:
:$\forall y \in X: \map d {x, y} < \epsilon_0 \implies y \in A_i$
which is precisely:
:$\map {B_{\epsilon_0}} x \subseteq A_i$
Category:Lebesgue's Number Lemma
Category:Compact Spaces
Lebesgue's Number Lemma/Sequentially Compact Space

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $M$ be sequentially compact.
Then there exists a Lebesgue number for every open cover of $M$.
== Proof ==
 $\UU$ is an open cover of $M$ for which no Lebesgue number exists.
Then for any $n \in \N$, there exists some $x_n \in M$ such that $\map {B_{1 / n} } {x_n} \subseteq U$ is false for each $U \in \UU$. (Otherwise $1/n$ would be a Lebesgue number for $\UU$.)
As $M$ is sequentially compact, $\sequence {x_n}$ has a subsequence, say $\sequence {x_{\map n r} }$ which converges to some $x \in M$.
Since $\UU$ covers $M$, $x \in U_0$ for some $U_0 \in \UU$.
Since $U_0$ is open, $\exists m \in \N: \map {B_{2 / m} } x \subseteq U_0$.
Now $\map {B_{1 / m} } x$ contains $x_{\map n r}$ for all $r \ge R$, say.
Choose $r \ge R$ such that $\map n r \ge m$ and write $s = \map n r$.
Then $\map {B_{1 / s} } {x_s} \subseteq \map {B_{2 / m} } x$ since:
So $\map {B_{1 / s} } {x_s} \subseteq U_0$ which contradicts the choice of $x_s$.
So, by Proof by Contradiction, there has to be a Lebesgue number for $\UU$.
== Sources ==
* : $7.2$: Sequential compactness: Lemma $7.2.12$
Category:Lebesgue's Number Lemma
Category:Sequentially Compact Spaces