nyuuzyou/wikis huggingface dataset:
Total Force on Body in Rotating Frame of Reference

== Theorem ==
Let $B$ be a body in a rotating frame of reference $\FF$.
The total force $\mathbf {F'}$ on $B$ is given by:
where:
:$\mathbf F$ is the vector sum of the forces acting on $B$
:$\bsomega$ is the angular velocity of $\FF$ relative to an inertial frame
:$\mathbf {r'}$ is the position vector of $B$ relative to $\FF$
:$\mathbf {v'}$ is the velocity of $B$ relative to $\FF$
:$\mathbf {a'}$ is the acceleration of $B$ relative to $\FF$.
The inertial forces as they are perceived in the rotating frame of reference $\FF$ are:
== Proof ==
== Sources ==
* 
* 
Category:Force
Category:Coriolis Forces
Category:Euler Forces
Category:Centrifugal Force
Total Force on Point Charge from 2 Point Charges

== Theorem ==
Let $q_1$, $q_2$ and $q_3$ be point charges.
Let $\mathbf F_{ij}$ denote the electrostatic force exerted on $q_j$ by $q_i$.
Let $\mathbf F_i$ denote the electrostatic force exerted on $q_i$ by the combined action of the other two point charges.
Then the electrostatic force $\mathbf F_1$ exerted on $q_1$ by the combined action of $q_2$ and $q_3$ is given by:
where:
:$\mathbf F_{21} + \mathbf F_{31}$ denotes the vector sum of $\mathbf F_{21}$ and $\mathbf F_{31}$
:$\mathbf r_{ij}$ denotes the displacement from $q_i$ to $q_j$
:$r_{ij}$ denotes the distance between $q_i$ and $q_j$
:$\varepsilon_0$ denotes the vacuum permittivity.
== Proof ==
:
By definition, the electrostatic force $\mathbf F_{21}$ and $\mathbf F_{31}$ are vector quantities.
Hence their resultant can be found by using the Parallelogram Law.
The result follows from Coulomb's Law of Electrostatics.
== Also presented as ==
== Also see ==
* Total Force on Point Charge from Multiple Point Charges
== Sources ==
* : Chapter $1$: Force and energy in electrostatics: $1.1$ Electric Charge
Category:Total Force on Point Charge from 2 Point Charges
Category:Electrostatic Force
Category:Point Charges
Total Force on Point Charge from 2 Point Charges/Also presented as

== Total Force on Point Charge from 2 Point Charges: Also presented as ==
This can also be presented as:
where:
:$\mathbf F_{21} + \mathbf F_{31}$ denotes the vector sum of $\mathbf F_{21}$ and $\mathbf F_{31}$
:$\mathbf r_{ij}$ denotes the displacement from $q_i$ to $q_j$
:$r_{ij}$ denotes the distance between $p_i$ and $p_j$
:$\varepsilon_0$ denotes the vacuum permittivity.
Category:Total Force on Point Charge from 2 Point Charges
Total Force on Point Charge from Multiple Point Charges

== Theorem ==
Let $q_1, q_2, \ldots, q_n$ be point charges.
For all $i$ in $\set {1, 2, \ldots, n}$ where $i \ne j$, let $\mathbf F_{i j}$ denote the force exerted on $q_j$ by $q_i$.
For all $i$ in $\set {1, 2, \ldots, n}$, let $\mathbf F_i$ denote the force exerted on $q_i$ by the combined action of all the other point charges.
Then the electrostatic force $\mathbf F_i$ exerted on $q_i$ by the combined action of all the other point charges is given by:
where:
:the summation denotes the vector sum of $\mathbf F_{ji}$
:$\mathbf r_{ij}$ denotes the displacement from $q_i$ to $q_j$
:$r_{ij}$ denotes the distance between $q_i$ and $q_j$
:$\varepsilon_0$ denotes the vacuum permittivity.
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition:
:$\ds \mathbf F_i = \dfrac 1 {4 \pi \varepsilon_0} \sum_{\substack {1 \mathop \le j \mathop \le n \\ i \mathop \ne j} } \dfrac {q_i q_j} {r_{j i}^3} \mathbf r_{j i}$
$\map P 2$ is the case:
$\mathbf F_1 = \dfrac 1 {4 \pi \varepsilon_0} \dfrac {q_1 q_2} {r_{2 1}^3} \mathbf r_{2 1}$
which is Coulomb's Law of Electrostatics.
Thus $\map P 0$ is seen to hold.
=== Basis for the Induction ===
$\map P 3$ is the case:
:$\mathbf F_1 = \dfrac {q_2 q_1} {4 \pi \varepsilon_0 r_{2 1}^3} \mathbf r_{2 1} + \dfrac {q_3 q_1} {4 \pi \varepsilon_0 r_{3 1}^3} \mathbf r_{3 1}$
which is proved in Total Force on Point Charge from 2 Point Charges.
Thus $\map P 3$ 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:
:$\ds \mathbf F_i = \dfrac 1 {4 \pi \varepsilon_0} \sum_{\substack {1 \mathop \le j \mathop \le k \\ i \mathop \ne j} } \dfrac {q_i q_j} {r_{j i}^3} \mathbf r_{j i}$
from which it is to be shown that:
:$\ds \mathbf F_i = \dfrac 1 {4 \pi \varepsilon_0} \sum_{\substack {1 \mathop \le j \mathop \le {k + 1} \\ i \mathop \ne j} } \dfrac {q_i q_j} {r_{j i}^3} \mathbf r_{j i}$
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 2}: \ds \mathbf F_i = \dfrac 1 {4 \pi \varepsilon_0} \sum_{\substack {1 \mathop \le j \mathop \le n \\ i \mathop \ne j} } \dfrac {q_i q_j} {r_{j i}^3} \mathbf r_{j i}$
</onlyinclude>
== Also presented as ==
This can also be presented as:
:$\ds \mathbf F_i = \dfrac 1 {4 \pi \varepsilon_0} \sum_{\substack {1 \mathop \le j \mathop \le n \\ i \mathop \ne j} } \dfrac {q_i q_j} {\size {\mathbf r_j - \mathbf r_i}^3} \paren {\mathbf r_j - \mathbf r_i}$
where $\mathbf r_i$ denotes the position vector of $q_i$ with respect to the origin of the coordinate system used as a frame of reference.
== Sources ==
* : Chapter $1$: Force and energy in electrostatics: $1.1$ Electric Charge: $(1.3)$
Category:Electrostatic Force
Category:Point Charges
Total Number of Set Partitions

== Theorem ==
Let $S$ be a finite set of cardinality $n$.
Then the number of different partitions of $S$ is $B_n$, where $B_n$ is the $n$th Bell number.
== Proof ==
The number of different partitions of $S$ is defined as $B_n$.
From Bell Number as Summation over Lower Index of Stirling Numbers of the Second Kind:
:$B_n = \ds \sum_{k \mathop = 0}^n {n \brace k}$
where $\ds {n \brace k}$ denotes a Stirling number of the second kind.
From Number of Set Partitions by Number of Components, $\ds {n \brace k}$ is the number of partitions of $S$ into $k$ components.
Hence the result.
=== Examples ===
Category:Set Partitions
Total Number of Set Partitions/Examples/2

== Example of Total Number of Set Partitions ==
Let $S$ be a set whose cardinality is $2$.
Then the number of partitions of $S$ is $2$.
== Proof ==
Let $p \paren n$ denote the cardinality of the set of partitions of a set whose cardinality is $n$.
From Total Number of Set Partitions, $p \paren n$ is the $n$th Bell number $B_n$.
Thus:
=== Illustration ===
== Sources ==
* : Chapter $1$: Sets and Logic: Exercise $16 \ \text{(i)}$
Category:Examples of Set Partitions
Total Number of Set Partitions/Examples/3

== Example of Total Number of Set Partitions ==
Let $S$ be a set whose cardinality is $3$.
Then the number of partitions of $S$ is $5$.
== Proof ==
Let $\map p n$ denote the cardinality of the set of partitions of a set whose cardinality is $n$.
From Total Number of Set Partitions, $\map p n$ is the $n$th Bell number $B_n$.
Thus:
=== Illustration ===
== Sources ==
* : Chapter $1$: Sets and Logic: Exercise $16 \ \text{(ii)}$
Category:Examples of Set Partitions
Total Number of Set Partitions/Examples/3/Illustration

== Example of Total Number of Set Partitions ==
Let a set $S$ of cardinality $3$ be exemplified by $S = \set {a, b, c}$.
Then the partitions of $S$ are:
:$\set {a, b, c}$
:$\set {a, b \mid c}$
:$\set {a, c \mid b}$
:$\set {b, c \mid a}$
:$\set {a \mid b \mid c}$
Category:Examples of Set Partitions
Total Number of Set Partitions/Examples/4

== Example of Total Number of Set Partitions ==
Let $S$ be a set whose cardinality is $4$.
Then the number of partitions of $S$ is $15$.
== Proof ==
Let $\map p n$ denote the cardinality of the set of partitions of a set whose cardinality is $n$.
From Total Number of Set Partitions, $\map p n$ is the $n$th Bell number $B_n$.
Thus:
=== Illustration ===
== Sources ==
* : Chapter $1$: Sets and Logic: Exercise $16 \ \text{(iii)}$
Category:Examples of Set Partitions
Total Number of Set Partitions/Examples/4/Illustration

== Example of Total Number of Set Partitions ==
Let a set $S$ of cardinality $4$ be exemplified by $S = \set {a, b, c, d}$.
Then the partitions of $S$ are:
:$\set {a, b, c, d}$
:$\set {\set a, \set {b, c, d} }$
:$\set {\set b, \set {a, c, d} }$
:$\set {\set c, \set {a, b, d} }$
:$\set {\set d, \set {a, b, c} }$
:$\set {\set {a, b}, \set {c, d} }$
:$\set {\set {a, c}, \set {b, d} }$
:$\set {\set {a, d}, \set {b, c} }$
:$\set {\set a, \set b, \set {c, d} }$
:$\set {\set a, \set c, \set {b, d} }$
:$\set {\set a, \set d, \set {b, c} }$
:$\set {\set b, \set c, \set {a, d} }$
:$\set {\set b, \set d, \set {a, c} }$
:$\set {\set c, \set d, \set {a, b} }$
:$\set {\set a, \set b, \set c, \set d}$
== Sources ==
* : $\S 6$. Indexed families; partitions; equivalence relations: Example $6.2$
Category:Examples of Set Partitions
Total Ordering/Examples/Monarchy

== Example of Total Ordering ==
Let $K$ denote the set of .
Let $\MM$ denote the relation on $K$ defined as:
:$a \mathrel \MM b$ $a$ was after or at the same time as $b$.
Its dual $\MM^{-1}$ is defined as:
:$a \mathrel {\MM^{-1} } b$ $a$ was before or at the same time as $b$.
Then $\MM$ and $\MM^{-1}$ are total orderings on $K$. 
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings
Category:Examples of Total Orderings
Total Ordering/Examples/Usual Ordering on Real Numbers

== Example of Total Ordering ==
Let $\R$ denote the set of real numbers.
The usual ordering $\le$ on $\R$ and its dual $\ge$ are total orderings on $\R$.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings
Category:Examples of Total Orderings
Total Ordering is Total Relation

== Theorem ==
Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a total ordering.
Then $\RR$ is a total relation.
== Proof ==
By definition of total ordering:
:$\RR$ is a reflexive relation on the strength of being an ordering
:$\RR$ is a connected relation on the strength of being a total ordering.
Category:Total Orderings
Category:Total Relations
Total Probability Theorem

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\set {B_1, B_2, \ldots}$ be a partition of $\Omega$ such that $\forall i: \map \Pr {B_i} > 0$.
Then:
:$\ds \forall A \in \Sigma: \map \Pr A = \sum_i \condprob A {B_i} \map \Pr {B_i}$
=== Theorem for Conditional Probabilities ===
== Proof ==
== Also known as ==
This theorem is also called the Partition Theorem, but as there are already quite a few theorems with such a name, prefers to use this somewhat more distinctive name.
Other names include:
:Law of Alternatives
:Law of Total Probability
:Total Probability Law
:Theorem of Total Probability
 and appear to be dismissive of them:
:::This theorem has several other fancy names, such as 'the theorem of total probability'; ...
:::::from 
== Also see ==
* Bayes' Theorem
== Sources ==
* : $\S 1.8$: The partition theorem: Theorem $1 \text{B}$
* 
Category:Probability Theory
Category:Named Theorems
Category:Total Probability Theorem
Total Probability Theorem/Conditional Probabilities

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\set {B_1, B_2, \ldots}$ be a partition of $\Omega$ such that $\forall i: \map \Pr {B_i} > 0$.
Let $C \in \Sigma$ be an event independent to any of the $B_i$.
Let $\map \Pr C > 0$.
Then:
:$\ds \forall A \in \Sigma: \condprob A C = \sum_i \condprob A {C \cap B_i} \, \map \Pr {B_i}$
== Proof ==
First define $Q_C := \condprob {\, \cdot} C$.
Then, from Conditional Probability Defines Probability Space, $\struct {\Omega, \Sigma, Q_C}$ is a probability space.
Moreover:
Therefore the Total Probability Theorem also holds true.
Hence we have:
Category:Total Probability Theorem
Total Semilattice has Unique Total Ordering

== Theorem ==
Let $\struct {S, \vee}$ be a total semilattice.
There exists a unique total ordering $\preccurlyeq$ on $S$ such that:
:$x \vee y = \max \set {x, y}$
where:
:$\max \set {x, y} := \begin {cases} b & : a \preccurlyeq b \\ a & : b \preccurlyeq a \end {cases}$
== Proof ==
Recall the definition of total semilattice:
A total semilattice $\struct {S, \odot}$ is a semilattice which has the property that every subset of $\struct {S, \odot}$ is a subsemilattice.
That is, such that every subset of $\struct {S, \odot}$ is closed under $\odot$.
Let us define a relation $\RR$ on $S$ as:
:$\forall x, y \in S: x \mathrel \RR y \iff x \vee y = y$
It is to be shown that $\RR$ is a total ordering.
From Semilattice Induces Ordering, we have that $\RR$ is an ordering.
=== Connectedness ===
Let $a, b \in S$ be arbitrary such that $a \ne b$.
We have that:
:$\forall x, y \in \set {a, b}: x \vee y \in \set {a, b}$
That is, either:
:$a \vee b = a$
or:
:$a \vee b = b$
Hence, by definition of $\RR$, either:
:$a \mathrel \RR b$
or:
:$b \mathrel \RR a$
That is, $\RR$ is a connected relation.
Summarising:
:$\RR$ is an ordering
and
:$\RR$ is connected.
Hence by definition $\RR$ is a total ordering.
Let us denote this total ordering by $\preccurlyeq$.
We have:
and:
Hence by definition:
:$x \vee y = \max \set {x, y}$
=== Uniqueness ===
It remains to be proved that the total ordering $\preccurlyeq$, with the property that:
:$x \vee y = \max \set {x, y}_\preccurlyeq$
is unique.
Indeed, suppose there exists another total ordering $\preccurlyeq'$ such that:
:$x \vee y = \max \set {x, y}_{\preccurlyeq'}$
We have:
The proof is complete.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.22 \ \text {(c)}$
Category:Total Semilattices
Category:Total Orderings
Total Solid Angle Subtended by Spherical Surface

== Theorem ==
The total solid angle subtended by a spherical surface is $4 \pi$.
== Proof ==
:
Let $\d S$ be an element of a surface $S$.
Let $\mathbf n$ be a unit normal on $\d S$ positive outwards.
From a point $O$, let a conical pencil touch the boundary of $S$.
Let $\mathbf r_1$ be a unit vector in the direction of the position vector $\mathbf r = r \mathbf r_1$ with respect to $O$.
Let spheres be drawn with centers at $O$ of radii $1$ and $r$.
Let $\d \omega$ be the area cut from the sphere of radius $1$.
We have:
:$\dfrac {\d \omega} {1^2} = \dfrac {\d S \cos \theta} {r^2}$
where $\theta$ is the angle between $\mathbf n$ and $\mathbf r$.
Thus $\d \omega$ is the solid angle subtended by $\d S$ at $O$.
Hence by definition of solid angle subtended:
:$\ds \Omega = \iint_S \frac {\mathbf {\hat r} \cdot \mathbf {\hat n} \rd S} {r^2}$
== Sources ==
* : Chapter $\text {VI}$: The Theorems of Gauss and Stokes: $2$. Gauss's Theorem and the Inverse Square Law
Category:Spheres
Category:Solid Angles
Total Surface Area of Lateral Faces of Regular Pyramid

== Theorem ==
Let $P$ be a regular pyramid.
The total surface area $S_L$ of the lateral faces of $P$ is:
:$S_L = \dfrac {s p} 2$
where:
:$s$ is the slant height of $P$
:$p$ is the perimeter of the base of $P$.
== Proof ==
== Sources ==
* 
* 
Category:Regular Pyramids
Category:Area Formulas
Total Variation is Non-Negative

== Theorem ==
Let $a, b$ be real numbers with $a < b$.
Let $f : \closedint a b \to \R$ be a function of bounded variation.
Let $\map {V_f} {\closedint a b}$ be the total variation of $f$ on $\closedint a b$.
Then: 
:$\map {V_f} {\closedint a b} \ge 0$
with equality $f$ is constant.
== Proof ==
We use the notation from the definition of bounded variation.
Note that by the definition of absolute value, we have: 
:$\size x \ge 0$
for all $x \in \R$.
Let $P$ be a finite subdivision of $\closedint a b$.
Write: 
:$P = \set {x_0, x_1, \ldots, x_n}$
with:
:$a = x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n = b$
Then: 
:$\ds \map {V_f} {P ; \closedint a b} = \sum_{i \mathop = 1}^n \size {\map f {x_i} - \map f {x_{i - 1} } } \ge 0$
So, by the definition of supremum, we have: 
:$\ds \map {V_f} {\closedint a b} = \sup_P \paren {\map {V_f} {P ; \closedint a b} } \ge 0$
Now consider $f$ a constant function.
We show that $\map {V_f} {\closedint a b} = 0$.
Then: 
:$\size {\map f {x_i} - \map f {x_{i - 1} } } = 0$
for all $x_{i - 1}, x_i \in \closedint a b$.
So, for any finite subdivision $P$ we have: 
We then have: 
So if $f$ is constant, then $f$ is of bounded variation with:
:$\map {V_f} {\closedint a b} = 0$
It remains to show that if $\map {V_f} {\closedint a b} = 0$, then $f$ is constant.
It suffices to show that if $f$ is not constant then $\map {V_f} {\closedint a b} > 0$.
Since $f$ is non-constant, we can pick $x \in \openint a b$ such that either: 
:$\map f x \ne \map f a$
or:
:$\map f x \ne \map f b$
So that either:
:$\size {\map f x - \map f a} > 0$
or:
:$\size {\map f b - \map f x} > 0$
Note that we then have: 
We must then have: 
:$\map {V_f} {\closedint a b} \ge \map {V_f} {\set {a, x, b} ; \closedint a b} > 0$
Hence the claim.
== Sources ==
* : $\S 6.4$: Total Variation
Category:Total Variation of Real Function
Total Variation is Smallest Measure exceeding Modulus of Measure

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be either a signed measure or complex measure on $\struct {X, \Sigma}$.
Let $\nu$ be a measure such that:
:$\cmod {\map \mu A} \le \map \nu A$ for each $A \in \Sigma$. 
Let $\cmod \mu$ be the variation of $\mu$.
Then:
:$\map {\cmod \mu} A \le \map \nu A$ for each $A \in \Sigma$. 
== Proof ==
Let $\map P A$ be the set of finite partitions of $A$ into $\Sigma$-measurable sets.
Let $\set {A_1, A_2, \ldots, A_n} \in \map P A$.
Then we have:
Taking the supremum over $\set {A_1, A_2, \ldots, A_n}$ we obtain:
:$\map {\cmod \mu} A \le \map \nu A$ 
Category:Variation of Complex Measure
Category:Variation of Signed Measure
Total Vector Area of Closed Surface is Zero

== Theorem ==
Let $S$ be a closed surface.
Let $\d \mathbf S$ be an infinitesimal vector area around some point $P$ of $S$.
Then the total surface area of $S$ is given by:
:$\ds \iint_S \d \mathbf S = \mathbf 0$
== Proof ==
== Sources ==
* : Chapter $\text {II}$: The Products of Vectors: $5$. Vector Area
Category:Vector Area
Category:Polyhedra
Total Vector Area of Polyhedron is Zero

== Theorem ==
Let $P$ be a polyhedron.
Let the positive direction be defined as outward.
Let $\mathbf T$ be the total vector area of all the faces of $P$.
Then:
:$\mathbf T = \mathbf 0$
== Proof ==
$P$ can be geometrically divided into a finite number of tetrahedra.
Every face of these tetrahedra which are internal to $P$ appears twice: once with a positive vector area, and once with a negative normal.
Hence for any polyhedron, the total vector area is the sum of the vector areas of all the tetrahedra.
The result follows from Total Vector Area of Tetrahedron is Zero.
== Sources ==
* : Chapter $\text {II}$: The Products of Vectors: $5$. Vector Area
Category:Vector Area
Category:Polyhedra
Total Vector Area of Tetrahedron is Zero

== Theorem ==
Let $T$ be a tetrahedron whose faces have vector area $\mathbf S_1$, $\mathbf S_2$, $\mathbf S_3$ and $\mathbf S_4$.
Let the positive direction be defined as outward.
:
Then the total vector area is zero:
:$\mathbf S_1 + \mathbf S_2 + \mathbf S_3 + \mathbf S_4 = \mathbf 0$
== Proof ==
Let the vector areas be resolved upon the faces of $T$.
Some of the projections will be positive and some negative.
Let $T$ be imagined in a fluid which is in equilibrium under pressure.
Each face experiences a force which is normal to its plane and proportional to its area.
Because the fluid inside is in equilibrium with the fluid outside, the resultant of the forces on its faces will equal zero.
Hence the sum of all the vector areas is also zero, as the pressure is the same on all faces.
Hence the result.
== Sources ==
* : Chapter $\text {II}$: The Products of Vectors: $5$. Vector Area
Category:Vector Area
Category:Tetrahedra
Totally Bounded Metric Space is Bounded

== Theorem ==
Let $M = \struct {A, d}$ be a totally bounded metric space.
Then $M$ is bounded.
== Proof ==
Let $M = \struct {A, d}$ be totally bounded.
Then there exist $n \in \N$ and points $x_0, \dots, x_n \in A$ such that:
:$\ds \inf_{0 \mathop \le i \mathop \le n} \map d {x_i, x} \le 1$
for all $x \in A$.
Let us set:
:$a := x_0$
:$\ds D := \max_{0 \mathop \le i \mathop \le n} \map d {x_0, x_i}$
:$K := D + 1$
Now let $x \in A$ be arbitrary.
Then by assumption there exists $i$ such that $\map d {x_i, x} \le 1$.
Hence:
:$\map d {a, x} \le \map d {a, x_i} + \map d {x_i, x} \le 1 + D = K$
So $M$ is bounded, as claimed.
== Also see ==
* Bounded Metric Space is not necessarily Totally Bounded, showing that the converse does not necessarily hold.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces: Complete Metric Spaces
Category:Bounded Metric Spaces
Category:Totally Bounded Metric Spaces
Totally Bounded Metric Space is Second-Countable

== Theorem ==
Let $M = \struct {A, d}$ be a metric space which is totally bounded.
Then $M$ is second-countable.
== Proof 1 ==
== Proof 2 ==
Category:Metric Spaces
Category:Second-Countable Spaces
Category:Totally Bounded Metric Spaces
Category:Totally Bounded Metric Space is Second-Countable
Totally Bounded Metric Space is Second-Countable/Proof 1

== Theorem ==
== Proof ==
Let $M = \struct {A, d}$ be totally bounded.
Let $\epsilon = 1, \dfrac 1 2, \dfrac 1 3, \ldots$
As $M$ is totally bounded, for each $\epsilon$ there exists a finite $\epsilon$-net $\CC$ for $M$.
From Net forms Basis for Metric Space, $\CC$ is a countable basis for $M$.
That is, $M$ is second-countable.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces: Complete Metric Spaces
Category:Totally Bounded Metric Space is Second-Countable
Totally Bounded Metric Space is Separable

== Theorem ==
A totally bounded metric space is separable.
== Proof ==
Let $M = \struct {A, d}$ be a totally bounded metric space.
By the definition of total boundedness, we can use the axiom of countable choice to construct a sequence $\sequence {F_n}_{n \mathop \ge 1}$ such that:
:For all natural numbers $n \ge 1$, $F_n$ is a finite $\paren {1/n}$-net for $M$.
Let $\ds S = \bigcup_{n \mathop \ge 1} F_n$.
From Countable Union of Countable Sets is Countable, it follows that $S$ is countable.
It suffices to prove that $S$ is everywhere dense in $M$.
Let $S^-$ denote the closure of $S$.
Let $x \in X$.
Let $U \subseteq X$ be open in $M$ such that $x \in U$.
By definition, there exists a strictly positive real number $\epsilon$ such that $\map {B_\epsilon} x \subseteq U$.
That is, the open $\epsilon$-ball of $x$ in $M$ is contained in $U$.
By the Axiom of Archimedes, there exists a natural number $n > \dfrac 1 \epsilon$.
That is, $\dfrac 1 n < \epsilon$, and so $\map {B_{1 / n} } x \subseteq \map {B_\epsilon} x$.
Since $\subseteq$ is a transitive relation, we have $\map {B_{1/n} } x \subseteq U$.
By the definition of a net, there exists a $y \in F_n$ such that $x \in \map {B_{1/n} } y$.
By , it follows from the definition of an open ball that $y \in \map {B_{1/n} } x$.
Since $y \in S \cap U$, it follows that $x$ is an adherent point of $S$.
By definition of adherent point, we have $x \in S^-$.
That is, $X \subseteq S^-$, and so $S$ is everywhere dense in $M$.
Category:Separable Spaces
Category:Metric Spaces
Category:Totally Bounded Metric Spaces
Totally Disconnected Space is Punctiform

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which is totally disconnected.
Then $T$ is punctiform.
== Proof ==
Let $T = \struct {S, \tau}$ be totally disconnected.
Then by definition its components are singletons.
Thus by definition each of its connected sets are degenerate.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness: Biconnectedness and Continua
Category:Totally Disconnected Spaces
Category:Punctiform Spaces
Totally Disconnected Space is T1

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which is totally disconnected.
Then $T$ is a $T_1$ (Fréchet) space.
== Proof ==
Let $T = \struct {S, \tau}$ be totally disconnected.
Then as its components are singletons, it follows that each of its points is closed.
From Equivalence of Definitions of $T_1$ Space, it follows that $T$ is a $T_1$ (Fréchet) space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness: Disconnectedness
Category:T1 Spaces
Category:Totally Disconnected Spaces