nyuuzyou/wikis huggingface dataset:
St. Ives Problem/Historical Note

== Historical Note on St. Ives Problem ==
The similarities between the St. Ives Problem and the version found in the are too close to be coincidental.
's version in his of $1202$ and $1228$ is also close.
 suggested a link between this problem and the nursery rhyme , and went on to note that uses the same numbers as and performs the calculation in the same way.
It may of course be suggested that the St. Ives Problem and the Rhind papyrus problem are in fact the same problem, having been passed down virtually unchanged for over $36$ centuries.
Hence it may be considered one of the oldest mathematical problems ever.
Note that , who reports on this problem in more than one publication, does not commit himself one way or another as to whether there is a direct intellectual link or not.
It is not completely certain that the St. Ives in question is , as opposed to the better-known .
However, the former is the more likely location, as it is a popular market town whose Great Fair may have been the venue for the travellers who thus met.
== Sources ==
* 
* 
* : $7$
* : The World's Oldest Puzzle: $2$
* : $7$
::::although he cites as Pierce.
Category:Historical Notes
St. Ives Problem/Historical Note/Mistake

== Source Work ==
:
:The Dictionary
::$7$
:
:The Dictionary
::$7$
== Mistake ==
:Pierce comments that it seems to be of the same origin as the House that Jack Built, and that Leonardo uses the same numbers as Ahmes and makes his calculations in the same way.
== Correction ==
The Pierce being referred to here is a misspelling of .
== Sources ==
* 
* : $7$
* : $7$
Category:Mistakes/Curious and Interesting Numbers
St. Ives Problem/Linguistic Note

== Linguistic Note on St. Ives Problem ==
Prime.mover remembers this riddle as the first in his book of nursery rhymes which he had when he was somewhat younger.
In that edition, the cats, as documented here, each had $7$ kits.
The word is an archaic or poetic one for the more usual kittens, which can also be seen in some renditions of this riddle.
Some sources use the word kitts in order to reduce the possibility of the word being confused with the word kit meaning pack of useful tools and materials.
Category:Linguistic Notes
St. Ives Problem/Rhind Papyrus Variant

== Classic Problem ==
Problem $79$ of the , written by some time around $1650$ BCE, concerns:
:$7$ houses, each with:
:$7$ cats, each with:
:$7$ mice, each with:
:$7$ spelt, each with:
:$7$ hekat
== Solution ==
Thus we have:
As with the St. Ives Problem, the total can be calculated using the Sum of Geometric Sequence:
== Sources ==
* : Problem $79$
* : $7$
* : The World's Oldest Puzzle: $1$
* : $7$
* 
Category:St. Ives Problem
St. Ives Problem/Solution 1

== Classic Problem ==
== Solution ==
This is a trick puzzle, to which the trick answer is:
:$1$.
The narrator is travelling to .
The man he meets, with the colossal entourage, is, by assumption of the definition of the word meet, travelling in the opposite direction, from St. Ives.
Technically, as it is not specified whether the narrator is travelling alone or not, the answer should actually be:
:at least $1$.
== Sources ==
* : The World's Oldest Puzzle: $3$
* 
Category:St. Ives Problem
St. Ives Problem/Solution 2

== Classic Problem ==
== Solution ==
Trick question aside, one can suppose that in order for the narrator to find out all about the unusual nature of this man's ménage, he would have had to spend some considerable time in his company, more than just a tip of the hat as they passed each other.
Hence it can be assumed that they met at a meeting of the ways, for example, when the narrator's path merged with that of the man and his family, and they travelled to together.
So the number of entities mentioned in the riddle are counted as follows:
This is a geometric sequence:
... from which we subtract the number of sacks (that is, if it is tacitly understood that the count is to be of the sentient beings):
:$2 \, 801 - 49 = 2 \, 752$
... and don't forget the narrator (a surprising number of the treatments of this subject do just that):
:$2 \, 753$
Quite a crowd.
== Sources ==
* 
Category:St. Ives Problem
St. Petersburg Paradox

== Paradox ==
Let $A$ and $B$ be a players in a gambling game whose mechanics are as follows:
$A$ tosses a coin until the first appearance of a head.
Let this occur at the $k$th toss.
$A$ then pays $2^k$ monetary units to $B$.
For the game to be fair, $B$ must pay $A$ in advance the amount equal to the mean payment $A$ can expect to make to $B$.
This mean payment in monetary units is:
Hence it is not possible for $B$ to pay that amount, and a fair game is impossible.
However, while his expectation is infinite, $B$ can receive only a finite amount in a finite time.
Indeed, in the long run this sum will be $8$ monetary units or less in $87 \cdotp 5 \%$ of all such games played, and will exceed $64$ in less than $1 \cdotp 6 \%$ of such games.
== Resolution ==
== Also known as ==
== Historical Note ==
== Sources ==
* 
Category:St. Petersburg Paradox
Category:Veridical Paradoxes
Stabilizer in Group of Transformations

== Theorem ==
Let $X$ be any set with $n$ elements (where $n \in \Z_{>0}$).
Consider the symmetric group on $n$ letters $S_n$ as a group of transformations on $X$.
Let $x \in X$.
Then the stabilizer of $x$ is isomorphic to $S_{n - 1}$.
== Proof ==
Consider the initial segment of natural numbers $\N_n = \set {1, 2, \ldots, n}$.
By the definition of cardinality, $H$ is equivalent to $\N_n$.
 we can consider $S_n$ acting directly on $\N_n$.
The stabilizer of $n$ in $\N_n$ is all the permutations of $S_n$ which fix $n$, which is clearly $S_{n - 1}$.
A permutation can be applied to $\N_n$ so that $i \to n$ for any $i$.
Thus one can build an isomorphism to show the result for a general $i$.
== Sources ==
* : $\S 5.6$. Stabilizers: Example $107$
* : Chapter $10$: The Orbit-Stabiliser Theorem: Example $10.10$
Category:Stabilizers
Category:Symmetric Groups
Stabilizer is Normal iff Stabilizer of Each Element of Orbit

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $S$ be a set.
Let $*: G \times S \to S$ be a group action.
Let $x \in S$.
Let $\Stab x$ denote the stabilizer of $x$ under $*$.
Let $\Orb x$ denote the orbit of $x$ under $*$.
Then $\Stab x$ is normal in $G$ $\Stab x$ is also the stabilizer of every element in $\Orb x$.
== Proof ==
=== Necessary Condition ===
Let $\Stab x$ be normal in $G$.
Let $t \in \Stab x$.
Let $y \in \Orb x$.
By :
:$(1): \quad y = g * x$
for a $g \in G$.
By :
:${\Stab x} \circ g = g \circ \Stab x$
In particular:
:$(2): \quad t \circ g = g \circ t'$
for a $t' \in \Stab x$.
Therefore:
That is:
:$t \in \Stab y$
=== Sufficient Condition ===
Let $\Stab x$ be the stabilizer of every element in $\Orb x$.
We shall show , i.e.
:$\forall g \in G : g^{-1} \circ \Stab x \circ g \subseteq \Stab x$
:$\forall g \in G : g \circ \Stab x \circ g^{-1} \subseteq \Stab x$
It suffices to verify the first inclusion, from which the second one follows by choosing $g^{-1}$ instead of $g$.
To this end, let $g \in G$.
Let $t \in \Stab x$.
By :
:$g * x \in \Orb x$
Thus :
:$(3): \quad t \in \Stab {g * x}$
Therefore:
That is:
:$g^{-1} \circ t \circ g \in \Stab x$
== Sources ==
* : Chapter $6$: Cosets: Exercise $12$
Category:Stabilizers
Category:Normal Subgroups
Stabilizer is Subgroup

== Theorem ==
Let $\struct {G, \circ}$ be a group which acts on a set $X$.
Let $\Stab x$ be the stabilizer of $x$ by $G$.
Then for each $x \in X$, $\Stab x$ is a subgroup of $G$.
=== Corollary ===
== Proof ==
From the :
:$e * x = x \implies e \in \Stab x$
and so $\Stab x$ cannot be empty.
Let $g, h \in \Stab x$.
Let $g \in \Stab x$.
Then:
:$x = \paren {g^{-1} \circ g} * x = g^{-1} * \paren {g * x} = g^{-1} * x$
Hence $g^{-1} \in \Stab x$.
Thus the conditions for the Two-Step Subgroup Test are fulfilled, and $\Stab x \le G$.
== Sources ==
* : $\S 5.6$. Stabilizers
* : Chapter $2$: The Sylow Theorems: $\S 54$
* : Chapter $6$: An Introduction to Groups: Exercise $5$
* : $\S 3.3$: Group actions and coset decompositions
* : Chapter $10$: The Orbit-Stabiliser Theorem: Proposition $10.9$
Category:Group Actions
Category:Stabilizers
Category:Stabilizer is Subgroup
Stabilizer is Subgroup/Corollary

== Corollary to Stabilizer is Subgroup ==
Let $G$ be a group whose identity is $e$.
Let $G$ act on a set $X$.
Let $x \in X$.
Then:
:$\forall g, h \in G: g * x = h * x \iff g^{-1} h \in \Stab x$
== Proof ==
== Sources ==
* : $\S 6.5$. Orbits: Lemma $\text {(ii)}$
Category:Stabilizer is Subgroup
Stabilizer of Cartesian Product of Group Actions

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $S$ and $T$ be sets.
Let $*_S: G \times S \to S$ and $*_T: G \times T \to T$ be group actions.
Let the group action $*: G \times \paren {S \times T} \to S \times T$ be defined as:
:$\forall \tuple {g, \tuple {s, t} } \in G \times \paren {S \times T}: g * \tuple {s, t} = \tuple {g *_S s, g *_T t}$
Then the stabilizer of $\tuple {s, t} \in S \times T$ is given by:
:$\Stab {s, t} = \Stab s \cap \Stab t$
where $\Stab s$ and $\Stab t$ are the stabilizers of $s$ and $t$ under $*_S$ and $*_T$ respectively.
== Proof ==
By definition, the stabilizer of an element $x$ of $S$ is defined as:
:$\Stab x := \set {g \in G: g * x = x}$
where $*$ denotes the group action.
So:
== Sources ==
* : $\S 3.3$: Group actions and coset decompositions: Exercise $3$
Category:Group Actions
Category:Cartesian Product
Category:Stabilizers
Stabilizer of Conjugacy Action on Subgroup is Normalizer

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $X$ be the set of all subgroups of $G$.
Let $*$ be the conjugacy action on $H$:
:$\forall g \in G, H \in X: g * H = g \circ H \circ g^{-1}$
Then the stabilizer of $H$ in $\powerset G$ is given by:
:$\Stab H = \map {N_G} H$
where $\map {N_G} H$ is the normalizer of $H$ in $G$.
== Proof 1 ==
From Stabilizer of Subset Product Action on Power Set, we have that:
:$\Stab H = \set {g \in G: g * H = H}$
Therefore:
:$\Stab H = \set {g \in G: g \circ H \circ g^{-1} = H}$
Hence:
:$\Stab H = \set {g \in G: H^g = H}$
where $H^g$ is the $G$-conjugate of $H$ by $g$.
The result follows from the definition of normalizer.
== Proof 2 ==
From Stabilizer of Coset under Group Action on Coset Space we have:
:$\Stab H = \set {g \in G: g \circ H \circ g^{-1} = H}$
Hence:
:$\Stab H = \set {g \in G: H^g = H}$
where $H^g$ is the $G$-conjugate of $H$ by $g$.
The result follows from the definition of normalizer.
== Also see ==
* Conjugacy Action on Subgroups is Group Action
* Orbit of Conjugacy Action on Subgroup is Set of Conjugate Subgroups
== Sources ==
* : $\S 5.6$. Stabilizers: Example $110$
* : Chapter $10$: The Orbit-Stabiliser Theorem: Example $10.10$
Category:Conjugacy Action
Category:Normalizers
Stabilizer of Coset Action on Set of Subgroups

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $\powerset G$ denote the power set of $G$.
Let $\HH \subseteq \powerset G$ denote the set of subgroups of $G$.
Let $*$ be the subset product action on $\HH \subseteq \powerset G$ defined as:
:$\forall g \in G: \forall H \in \HH: g * H = g \circ H$
where $g \circ H$ is the (left) coset of $g$ by $H$.
Then the stabilizer of $H$ in $\powerset G$ is $H$ itself:
:$\Stab H = H$
== Proof ==
From the definition of Stabilizer of Subset Product Action on Power Set:
:$\Stab H = H = \set {g \in G: g * H = H}$
The result follows from Left Coset Equals Subgroup iff Element in Subgroup.
== Also see ==
* Subset Product Action is Group Action
* Stabilizer of Subset Product Action on Power Set
* Orbit of Subgroup under Coset Action is Coset Space
== Sources ==
* : Chapter $5$: Subgroups: Exercise $19$
* : $\S 3.3$: Group actions and coset decompositions
* : Chapter $10$: The Orbit-Stabiliser Theorem: Example $10.10$
Category:Subset Product Action
Category:Stabilizers
Stabilizer of Coset under Group Action on Coset Space

== Theorem ==
Let $G$ be a group whose identity is $e$.
Let $H$ be a subgroup of $G$.
Let $*: G \times G / H \to G / H$ be the action on the (left) coset space:
:$\forall g \in G, \forall g' H \in G / H: g * \paren {g' H} := \paren {g g'} H$
Then the stabilizer of $a H$ under $*$ is given by:
:$\Stab {a H} = a H a^{-1}$
== Proof ==
It is established in Action of Group on Coset Space is Group Action that $*$ is a group action.
Then:
== Sources ==
* : Chapter $6$: Cosets: Exercise $9$
Category:Group Action on Coset Space
Category:Stabilizers
Stabilizer of Element after Group Action

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $S$ be a set.
Let $*_S: G \times S \to S$ be a group action.
Let $x \in S, a \in G$.
Then:
:$\Stab {a * x} = a^{-1} \circ \Stab x \circ a$
== Proof ==
== Sources ==
* : Chapter $10$: The Orbit-Stabiliser Theorem: Exercise $2$
Category:Stabilizers
Stabilizer of Element of Group Acting on Itself is Trivial

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $*$ be the group action of $\struct {G, \circ}$ on itself by the rule:
:$\forall g, h \in G: g * h = g \circ h$
Then the stabilizer of an element $x \in G$ is given by:
:$\Stab x = \set e$
== Proof ==
Let $g \in \Stab x$.
Then:
Hence the result, by definition of trivial subgroup.
== Also see ==
* Group Acts on Itself
* Orbit of Element of Group Acting on Itself is Group
== Sources ==
* : Chapter $10$: The Orbit-Stabiliser Theorem: Example $10.10$
Category:Stabilizers
Stabilizer of Element under Conjugacy Action is Centralizer

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $*$ be the conjugacy action on $G$ defined by the rule:
:$\forall g, h \in G: g * h = g \circ h \circ g^{-1}$
Let $x \in G$.
Then the stabilizer of $x$ under this conjugacy action is:
:$\Stab x = \map {C_G} x$
where $\map {C_G} x$ is the centralizer of $x$ in $G$.
== Proof ==
From the definition of centralizer:
:$\map {C_G} x = \set {g \in G: g \circ x = x \circ g}$
Then:
== Also see ==
* Conjugacy Action on Group Elements is Group Action
* Orbit of Element under Conjugacy Action is Conjugacy Class
== Sources ==
* : $\S 5.6$. Stabilizers: Example $108$
* : Chapter $10$: The Orbit-Stabiliser Theorem: Example $10.10$
Category:Conjugacy Action
Category:Centralizers
Category:Stabilizers
Stabilizer of Polynomial

== Theorem ==
Let $n \in \Z: n > 0$.
Let $\map f {x_1, x_2, \ldots, x_n}$ be a polynomial in $n$ variables $x_1, x_2, \ldots, x_n$.
Let $S_n$ denote the symmetric group on $n$ letters.
Let $\pi, \rho \in S_n$.
Let the group action $\pi * f$ be defined as the permutation on the polynomial $f$ by $\pi$.
Then the stabilizer of $f$ is the set of permutations on $n$ letters which fix $f$.
== Proof ==
Follows directly from the definition of the stabilizer of $f$.
== Sources ==
* : Chapter $6$: Cosets: Exercise $10$
* : Chapter $10$: The Orbit-Stabiliser Theorem: Example $10.10$
Category:Polynomial Theory
Category:Stabilizers
Stabilizer of Subgroup Action is Identity

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $\struct {H, \circ}$ be a subgroup of $G$.
Let $*: H \times G \to G$ be the subgroup action defined for all $h \in H, g \in G$ as:
:$\forall h \in H, g \in G: h * g := h \circ g$
The stabilizer of $x \in G$ is $\set e$:
:$\Stab x = \set e$
== Proof ==
From Subgroup Action is Group Action we have that $*$ is a group action.
Let $x \in G$.
Then:
Hence the result, by definition of right coset.
== Sources ==
* : $\S 5.6$. Stabilizers: Example $111$
* : Chapter $2$: The Sylow Theorems: $\S 54 \alpha$
Category:Subgroup Action
Category:Stabilizers
Stabilizer of Subgroup Action on Left Coset Space

== Theorem ==
Let $G$ be a group.
Let $H$ and $K$ be subgroups of $G$.
Let $K$ act on the left coset space $G / H^l$ by:
:$\forall \tuple {k, g H} \in K \times G / H^l: k * \paren {g H} := \paren {k g} H$
The stabilizer of $g H$ is $K \cap H^g$, where $H^g$ denotes the $G$-conjugate of $H$ by $g$.
== Proof ==
Let $\RR^l_H$ be the equivalence defined as left congruence modulo $H$.
By Left Congruence Class Modulo Subgroup is Left Coset, $\RR^l_H$ is an equivalence relation. 
We check that $*$ is a well-defined mapping over $\RR^l_H$.
Suppose:
:$\eqclass g {\RR^l_H} = \eqclass {g'} {\RR^l_H} $
for some $g, g' \in G$, where $\eqclass g {\RR^l_H}$ and $\eqclass {g'} {\RR^l_H}$ are the equivalence classes of $g \in G$ and $g' \in G$ respectively.
From Left Congruence Class Modulo Subgroup is Left Coset:
:$\eqclass g {\RR^l_H} = g H$
And:
:$\eqclass {g'} {\RR^l_H} = g' H$
Then:
:$\forall g, g' \in G: \paren {g H} = \paren {g' H}$
We have:
Thus $*$ is a well-defined mapping over $\RR^l_H$.
Hence:
== Sources ==
* : Chapter $2$: The Sylow Theorems: $\S 54 \gamma$
Category:Stabilizers
Category:Group Action on Coset Space
Stabilizer of Subset Product Action on Power Set

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $\powerset G$ be the power set of $\struct {G, \circ}$.
Let $*: G \times \powerset G \to \powerset G$ be the subset product action on $\powerset G$ defined as:
:$\forall g \in G: \forall S \in \powerset G: g * S = g \circ S$
where $g \circ S$ is the subset product $\set g \circ S$.
Then the stabilizer of $S$ in $\powerset G$ is the set:
:$\Stab S = S$
== Proof ==
From the definition of stabilizer:
:$\Stab S = \set {g \in G: g * S = S}$
The result follows from the definition of the group action $*$ given.
== Also see ==
* Subset Product Action is Group Action
* Stabilizer of Coset Action on Set of Subgroups
* Orbit of Subgroup under Coset Action is Coset Space
== Sources ==
* : $\S 5.6$. Stabilizers: Example $109$
* : Chapter $10$: The Orbit-Stabiliser Theorem: Example $10.10$
Category:Subset Product Action
Category:Stabilizers
Stabilizer of Subspace stabilizes Orthogonal Complement

== Theorem ==
Let $H$ be a finite-dimensional real or complex Hilbert space (that is, inner product space).
Let $t: H \to H$ be a normal operator on $H$.
Let $t$ stabilize a subspace $V$.
Then $t$ also stabilizes its orthogonal complement $V^\perp$.
== Proof ==
Let $p: H \to V$ be the orthogonal projection of $H$ onto $V$.
Then the orthogonal projection of $H$ onto $V^\perp$ is $\mathbf 1 - p$, where $\mathbf 1$ is the identity map of $H$.
The fact that $t$ stabilizes $V$ can be expressed as:
:$\paren {\mathbf 1 - p} t p = 0$
or:
:$p t p = t p$
The goal is to show that:
:$p t \paren {\mathbf 1 - p} = 0$
We have that $\tuple {a, b} \mapsto \map \tr {a b^*}$ is an inner product on the space of endomorphisms of $H$.
Here, $b^*$ denotes the adjoint operator of $b$.
Thus it will suffice to show that $\map \tr {x x^*} = 0$ for $x = p t \paren {\mathbf 1 - p}$.
This follows from a direct computation, using properties of the trace and orthogonal projections:
Category:Hilbert Spaces
Stabilizers of Elements in Same Orbit are Conjugate Subgroups

== Theorem ==
Let $G$ be a group acting on a set $X$.
Let:
:$y, z \in \Orb x$
where $\Orb x$ denotes the orbit of some $x \in X$.
Then their stabilizers $\Stab y$ and $\Stab z$ are conjugate subgroups.
== Proof ==
From Stabilizer is Subgroup we have that both $\Stab y$ and $\Stab z$ are subgroups of $G$.
From definition of orbits:
:$\exists h_1, h_2 \in G: y = h_1 * x, z = h_2 * x$
Then $y = h_1 * \paren {h_2^{-1} * z} = h_1 h_2^{-1} * z$.
Thus:
This shows that $\Stab y$ and $\Stab z$ are conjugate.
Hence the result.
== Sources ==
* : Chapter $2$: The Sylow Theorems: $\S 54 \beta$
Category:Group Actions
Category:Stabilizers
Stadium Paradox

== Paradox ==
Consider three rows of bodies:
:$\begin{array} {ccccc}
\text {(A)} & 0 & 0 & 0 & 0 \\
\text {(B)} & 0 & 0 & 0 & 0 \\
\text {(C)} & 0 & 0 & 0 & 0 \\
\end{array}$
Let row $\text {(A)}$ be at rest, while row $\text {(B)}$ and row $\text {(C)}$ are travelling at the same speed in opposite directions.
:$\begin{array} {ccccccc}
\text {(A)} & & 0 & 0 & 0 & 0 & \\
\text {(B)} & 0 & 0 & 0 & 0 & & \\
\text {(C)} & & & 0 & 0 & 0 & 0 \\
\end{array}$
By the time they are all in the same part of the course, $\text {(B)}$ will have passed twice as many of the bodies in $\text {(C)}$ as $\text {(A)}$ has.
Therefore the time it takes to pass $\text {(A)}$ is twice as long as it takes to pass $\text {(C)}$.
But the time which $\text {(B)}$ and $\text {(C)}$ take to reach the position of $\text {(A)}$ is the same.
Therefore double the time is equal to half the time.
== Resolution ==
== Historical Note ==
== Sources ==
* : Chapter $\text{II}$: Modern Minds in Ancient Bodies
* 
* 
* 
* 
Category:Zeno's Paradoxes
Category:Antinomies
Standard Bounded Metric is Metric

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $\bar d: A^2 \to \R$ be the standard bounded metric of $d$:
:$\forall \tuple {x, y} \in A^2: \map {\bar d} {x, y} = \min \set {1, \map d {x, y} }$
Then $\bar d$ is a metric for $A$.
=== Topological Equivalence ===
== Proof ==
It is to be demonstrated that $\bar d$ satisfies all the metric space axioms.
=== Proof of ===
So holds for $\bar d$.
=== Proof of ===
Suppose that $\map d {x, y} < 1$ and $\map d {y, z} < 1$.
Then:
Now suppose that either $\map d {x, y} > 1$ or $\map d {x, y} > 1$.
, suppose $\map d {x, y} > 1$.
Then:
The same argument applies for $\map d {y, z} > 1$
So holds for $\bar d$.
=== Proof of ===
So holds for $\bar d$.
=== Proof of ===
So holds for $\bar d$.
Thus $\bar d$ satisfies all the metric space axioms and so is a metric.
== Sources ==
* : $2$: Continuity generalized: metric spaces: Exercise $2.6: 6$
Category:Examples of Metric Spaces
Category:Standard Bounded Metric is Metric
Standard Bounded Metric is Metric/Topological Equivalence

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $\bar d: A^2 \to \R$ be the standard bounded metric of $d$:
:$\forall \tuple {x, y} \in A^2: \map {\bar d} {x, y} = \min \set {1, \map d {x, y} }$
$\bar d$ is topologically equivalent to $d$.
== Proof ==
That $\bar d$ forms a metric on $M$ is demonstrated in Standard Bounded Metric is Metric.
We have that:
:$\forall x, y \in A^2: \map {\bar d} {x, y} \le \map d {x, y}$
Hence:
:$\map {B_\epsilon} {x; d} \subseteq \map {B_\epsilon} {x; \bar d}$
where $\map {B_\epsilon} {x; d}$ denotes the open $\epsilon$-ball of $x$ in $\struct {A, d}$.
Hence:
:if $U$ is $\bar d$-open, the $U$ is $d$-open
where $U$ is a subset of $A$.
Let $U$ be $d$-open.
Let $x \in U$.
Then $\map {B_\epsilon} {x; d} \subseteq U$ for some $\epsilon \in \R_{>0}$.
Let us take $\epsilon < 1$.
Then:
:$\map {B_\epsilon} {x; \bar d} = \map {B_\epsilon} {x; d} \subseteq U$
demonstrating that $U$ is $\bar d$-open.
The result follows.
== Sources ==
* : $2$: Continuity generalized: metric spaces: Exercise $2.6: 20$
Category:Standard Bounded Metric is Metric
Standard Continuous Uniform Distribution in terms of Exponential Distribution

== Theorem ==
Let $X$ and $Y$ be independent random variables.
Let $\beta$ be a strictly positive real number.
Let $X$ and $Y$ be random samples from the exponential distribution with parameter $\beta$. 
Then:
:$\dfrac X {X + Y} \sim \operatorname U \openint 0 1$
where $\operatorname U \openint 0 1$ is the uniform distribution on $\openint 0 1$.
== Proof ==
Note that the support of $\operatorname U \openint 0 1$ is $\openint 0 1$.
It is therefore sufficient to show that for $0 < z < 1$: 
:$\map \Pr {\dfrac X {X + Y} \le z} = z$
Note that if $x, y > 0$ then: 
:$0 < \dfrac x {x + y} < 1$
Note also that: 
:$\dfrac x {x + y} \le z$ 
with $0 < z < 1$ is equivalent to: 
:$x \le z x + z y$
which is in turn equivalent to: 
:$\paren {1 - z} x \le z y$
That is: 
:$x \le \dfrac z {1 - z} y$
We therefore have: 
:$\map \Pr {\dfrac X {X + Y} \le z} = \map \Pr {X \le \dfrac z {1 - z} Y}$
Let $f_{X, Y}$ be the joint probability density function of $X$ and $Y$. 
Let $f_X$ and $f_Y$ be the probability density function of $X$ and $Y$ respectively. 
From Condition for Independence from Joint Probability Density Function, we have for each $x, y \in \R_{> 0}$:
:$\map {f_{X, Y} } {x, y} = \map {f_X} x \map {f_Y} y$
We therefore have: 
Category:Continuous Uniform Distribution
Category:Exponential Distribution