nyuuzyou/wikis huggingface dataset:
Hermitian Operator has Real Eigenvalues

== Theorem ==
The eigenvalues of a Hermitian operator are real.
== Proof ==
Let $\hat H$ be a Hermitian operator on an inner product space $V$ over the field of complex numbers, $\C$. That is, $\hat H = \hat H^\dagger$.
Then for an eigenvector $\left\vert{x}\right\rangle \in V, \left\vert{x}\right\rangle \ne \left\vert{0}\right\rangle$ and eigenvalue $\lambda \in \C$:
:$\left.{\hat H}\middle\vert{x}\right\rangle = \left.{\lambda}\middle\vert{x}\right\rangle$ 
We know for a general operator $\hat A$ on $V$, the following holds:
:$\forall \left\vert{x}\right\rangle, \left\vert{y}\right\rangle \in V: \left\langle{x}\middle\vert{\hat A}\middle\vert{y}\right\rangle = \left\langle{y}\middle\vert{\hat A^\dagger}\middle\vert{x }\right\rangle^*$
where $^*$ denotes the complex conjugate.
Noting $\hat H = \hat H^\dagger$ gives:
:$\left\langle{x}\middle\vert{\hat H}\middle\vert{y}\right\rangle = \left\langle{y}\middle\vert{\hat H}\middle\vert{x}\right\rangle^*$
Now we compute:
Using our previous result:
Equating the two previous equations gives:
:$\lambda \left\langle{x}\middle\vert{x}\right\rangle = \left({\lambda \left\langle{x}\middle\vert{x}\right\rangle}\right)^*$
Recalling the conjugate symmetry property of the inner product, we can see that:
:$\left\langle{x}\middle\vert{x}\right\rangle = \left\langle{x}\middle\vert{x}\right\rangle^*$ 
which is true $\left\langle{x}\middle\vert{x}\right\rangle \in \R$.
So
:$\lambda \left\langle{x}\middle\vert{x}\right\rangle = \lambda^* \left\langle{x}\middle\vert{x}\right\rangle$
and so:
:$\lambda = \lambda^*$
Therefore $\lambda \in \R$.
== Also see ==
* Hermitian Operator has Orthogonal Eigenvectors
Category:Hermitian Operators
Hero's Method

== Theorem ==
Let $a \in \R$ be a real number such that $a > 0$.
Let $x_1 \in \R$ be a real number such that $x_1 > 0$.
Let $\sequence {x_n}$ be the sequence in $\R$ defined recursively by:
:$\forall n \in \N_{>0}: x_{n + 1} = \dfrac {x_n + \dfrac a {x_n} } 2$
Then $x_n \to \sqrt a$ as $n \to \infty$.
== Lemmata ==
First we have the following lemmata:
=== Lemma 1 ===
=== Lemma 2 ===
== Proof 1 ==
== Proof 2 ==
== Examples ==
== Also known as ==
== Sources ==
* : $1$: Review of some real analysis: $\S 1.2$: Real Sequences: Example $1.2.1 \, \text {(g)}$
* : $\S 5$: Subsequences: $\S 5.5, \ \S 5.6$: Example
* 
* 
* 
* 
Category:Real Analysis
Category:Square Roots
Category:Hero's Method
Hero's Method/Lemma 1

== Lemma for Hero's Method ==
Let $a \in \R$ be a real number such that $a > 0$.
Let $x_1 \in \R$ be a real number such that $x_1 > 0$.
Let $\sequence {x_n}$ be the sequence in $\R$ defined recursively by:
:$\forall n \in \N_{>0}: x_{n + 1} = \dfrac {x_n + \dfrac a {x_n} } 2$
Then:
:$\forall n \in \N_{>0}: x_n > 0$
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:$x_n > 0$
=== Basis for the Induction ===
$\map P 1$ is the case:
:$x_1 > 0$
which is assumed.
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$x_k > 0$
from which it is to be shown that:
:$x_{k + 1} > 0$
=== Induction Step ===
This is the induction step:
We have that:
:$x_{k + 1} = \dfrac {x_k + \dfrac a {x_k} } 2$
But as $x_k > 0$ and $a > 0$, it follows that:
:$\dfrac a {x_k} > 0$
Then as $x_k > 0$ and $\dfrac a {x_k} > 0$, it follows that:
:$\dfrac 1 2 \paren {x_k + \dfrac a {x_k} } > 0$
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \N_{>0}: x_n > 0$
== Sources ==
* : $\S 5$: Subsequences: $\S 5.5$: Example
Category:Hero's Method
Hero's Method/Lemma 2

== Lemma for Hero's Method ==
Let $a \in \R$ be a real number such that $a > 0$.
Let $x_1 \in \R$ be a real number such that $x_1 > 0$.
Let $\sequence {x_n}$ be the sequence in $\R$ defined recursively by:
:$\forall n \in \N_{>0}: x_{n + 1} = \dfrac {x_n + \dfrac a {x_n} } 2$
Then:
:$\forall n \ge 2: x_n \ge \sqrt a$
== Proof ==
=== Lemma 1 ===
We have:
This is a quadratic equation in $x_n$.
We know that this equation must have a real solution with respect to $x_n$, because $x_n$ has been explicitly constructed by the iterative process.
Thus its discriminant is $b^2 - 4 a c \ge 0$, where:
:$a = 1$
:$b = -2 x_{n + 1}$
:$c = a$
Thus $x_{n + 1}^2 \ge a$.
From Lemma 1:
:$x_{n + 1} > 0$ 
It follows that:
:$\forall n \ge 1: x_{n + 1} \ge \sqrt a$ for $n \ge 1$
Thus:
:$\forall n \ge 2: x_n \ge \sqrt a$ for $n \ge 2$
== Sources ==
* : $\S 5$: Subsequences: $\S 5.5$: Example
Category:Hero's Method
Hero's Method/Proof 1

== Theorem ==
== Proof ==
First we have the following lemmata:
=== Lemma 1 ===
=== Lemma 2 ===
Consider $x_n - x_{n + 1}$.
So, providing we ignore the first term (about which we can state nothing), the sequence $\sequence {x_n}$ is decreasing and bounded below by $\sqrt a$.
Thus by the Monotone Convergence Theorem (Real Analysis), $x_n \to l$ as $n \to \infty$, where $l \ge \sqrt a$.
Now we want to find exactly what that value of $l$ actually is.
By Limit of Subsequence equals Limit of Real Sequence we also have $x_{n + 1} \to l$ as $n \to \infty$.
But $x_{n + 1} = \dfrac {x_n + \dfrac a {x_n} } 2$.
Because $l \ge \sqrt a$ it follows that $l \ne 0$.
So by the Combination Theorem for Sequences:
:$x_{n + 1} = \dfrac {x_n + \dfrac a {x_n} } 2 \to \dfrac {l + \dfrac a l} 2$ as $n \to \infty$
Since a Convergent Real Sequence has Unique Limit, that means:
:$l = \dfrac {l + \dfrac a l} 2$
and so (after some straightforward algebra):
:$l^2 = a$
Thus:
:$l = \pm \sqrt a$
and as $l \ge +\sqrt a$ it follows that:
:$l = +\sqrt a$
Hence the result.
== Sources ==
* : $\S 5$: Subsequences: $\S 5.5$: Example
Category:Hero's Method
Hero's Method/Proof 2

== Theorem ==
== Proof ==
First we have the following lemmata:
=== Lemma 1 ===
=== Lemma 2 ===
Let $a > 0$.
We make no statement about $x_1$.
We specify that:
:$x_{n + 1} = \dfrac {x_n + \dfrac a {x_n} } 2$
Now:
If we now assume that $x_1 \ge \sqrt a$, then it follows from Lemma 2 that $x_n \ge \sqrt a$.
So:
If $\size y < 1$, then $y^n \to 0$ as $n \to \infty$ from Sequence of Powers of Number less than One.
So, by Limit of Subsequence equals Limit of Real Sequence:
:$y^{2^n} \to 0$ as $n \to \infty$
Thus we see that:
:$x_n \to \sqrt a$ as $n \to \infty$
provided that:
:$\dfrac {x_1 - \sqrt a} {2 \sqrt a} < 1$
that is, that:
:$\sqrt a \le x_1 < 3 \sqrt a$
We assumed above that $x_1 \ge \sqrt a$.
Now we have shown that $x_n \to \sqrt a$ as $n \to \infty$ provided that $\sqrt a \le x_1 < 3 \sqrt a$.
However, we have already shown that $x_n \to \sqrt a$ as long as $x_1 \ge 0$.
The advantage to this analysis is that this gives us an opportunity to determine how close $x_n$ gets to $\sqrt a$.
== Sources ==
* : $\S 5$: Subsequences: $\S 5.6$: Example
Category:Hero's Method
Heron's Formula

== Theorem ==
Let $\triangle ABC$ be a triangle with sides $a$, $b$ and $c$ opposite vertices $A$, $B$ and $C$ respectively.
Then the area $\AA$ of $\triangle ABC$ is given by:
:$\AA = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$
where $s = \dfrac {a + b + c} 2$ is the semiperimeter of $\triangle ABC$.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Also known as ==
== Also see ==
* Brahmagupta's Formula
== Historical Note ==
== Sources ==
* : $\text V$. Trigonometry: Formulae $(45)$
* : $\S 4$: Geometric Formulas: Geometric Formulas: Triangle of Altitude $h$ and Base $b$: $4.5$
* 
* : Chapter $\text {A}.7$: Heron (first century A.D.)
* : Chapter $\text {B}.1$: The Pythagorean Theorem: Appendix: The Formulas of Heron and Brahmagupta
* : The Area Enclosed Against The Seashore: $31$
* 
* 
* : $\S 7$: Geometric Formulas: Triangle of Altitude $h$ and Base $b$: $7.5.$
* : Hero's formula
* : Hero's formula
Category:Heron's Formula
Category:Areas of Triangles
Heron's Formula/Historical Note

== Historical Note on Heron's Formula ==
Arabic sources from the Middle Ages inform us that Heron's Formula was actually due to .
However, 's is the earliest proof that survives.
== Sources ==
* : Chapter $\text {A}.7$: Heron (first century A.D.)
* : Chapter $\text {B}.1$: The Pythagorean Theorem: Appendix: The Formulas of Heron and Brahmagupta
Category:Historical Notes
Heron's Formula/Proof 1

== Theorem ==
== Proof ==
Construct the altitude from $A$.
Let the length of the altitude be $h$ and the foot of the altitude be $D$.
Let the distance from $D$ to $B$ be $z$.
:
From Pythagoras's Theorem:
:$\paren 1: \quad h^2 + \paren {a - z}^2 = b^2$
and:
:$\paren 2: \quad h^2 + z^2 = c^2$
By subtracting $\paren 1$ from $\paren 2$:
:$2 a z - a^2 = c^2 - b^2$
which can be expressed in terms of $z$ as:
:$z = \dfrac {a^2 + c^2 - b^2} {2 a}$
Substituting for $z$ in $\paren 2$ and simplifying yields:
:$h = \sqrt {c^2 - \paren {\dfrac {a^2 + c^2 - b^2} {2a} }^2}$
and so:
== Sources ==
* : Chapter $\text {B}.1$: The Pythagorean Theorem: Appendix: The Formulas of Heron and Brahmagupta
Category:Heron's Formula
Heron's Formula/Proof 2

== Theorem ==
== Proof ==
A triangle can be considered as a cyclic quadrilateral one of whose sides has degenerated to zero.
From Brahmagupta's Formula, the area of a cyclic quadrilateral is given by:
:$\sqrt {\paren {s - a} \paren {s - b} \paren {s - c} \paren {s - d}}$
where $s$ is the semiperimeter:
:$s = \dfrac {a + b + c + d} 2$
The result follows by letting $d$ tend to zero.
== Sources ==
* : Chapter $\text {B}.1$: The Pythagorean Theorem: Appendix: The Formulas of Heron and Brahmagupta
* : The Area Enclosed Against The Seashore: $31$
Category:Heron's Formula
Heron's Formula/Proof 3

== Theorem ==
== Proof ==
:
Let $\AA$ be the area of $\triangle ABC$.
Construct the incircle of $\triangle ABC$.
Let the incenter of $\triangle ABC$ be $M$.
Let the inradius of $\triangle ABC$ be $r$.
$\triangle ABC$ is made up of three triangles: $\triangle AMB$, $\triangle BMC$ and $\triangle CMA$.
From Area of Triangle in Terms of Side and Altitude, the areas of $\triangle AMB$, $\triangle BMC$ and $\triangle CMA$ are given by:
:$\Area \paren {\triangle AMB} = \dfrac {r c} 2$
:$\Area \paren {\triangle BMC} = \dfrac {r a} 2$
:$\Area \paren {\triangle CMA} = \dfrac {r b} 2$
Thus:
:$(1): \quad \AA = \dfrac {r \paren {c + a + b} } 2 = r s$
where $s$ is the semiperimeter of $\triangle ABC$.
Construct the excircle of $\triangle ABC$ with excenter $N$ tangent to $AB$, and to $AC$ and $BC$ produced at $D$ and $E$ respectively.
By Length of Tangent to Excircle:
:$s = CD = CE$
Therefore:
:$DA = s - b$
:$EB = s - a$
Note that:
:$AF + DA = BG + EB$
and:
:$AF + BG = C$
We need to show that $\triangle MAF \sim \triangle DNA$.
Note that:
:$AM$ bisects $\angle BAC$
:$AN$ bisects $\angle BAD$
But $\angle BAC$ and $\angle BAD$ are supplementary angles.
Hence the half-angles $\angle MAF$ and $\angle DAN$ are complementary angles.
It follows that:
:$\angle MAF = \angle DNA$
Since $\triangle MAF$ and $\triangle DNA$ are both right triangles:
:$\triangle DNA \sim \triangle MAF$
As right triangles sharing an acute angle:
:$\triangle NDC \sim \triangle MFC$
from which:
== Sources ==
* : Chapter $\text {B}.1$: The Pythagorean Theorem: Appendix: The Formulas of Heron and Brahmagupta
Category:Heron's Formula
Heron's Principle of Reflection

== Physical Law ==
The angle of incidence of a ray of light is equal to the angle of reflecion when that ray is reflected in a (plane) mirror.
== Proof ==
:
Let a ray of light $L$ travel from $A$ to $B$ by way of a mirror $M$.
Let $B'$ be the mirror image of $B$ so that $M$ is the perpendicular bisector of $BB'$.
Let $P$ be the point where $L$ is reflected from $M$.
The total length of the path of $L$ is $AP + PB = AP + PB'$.
From Fermat's Principle of Least Time, this length is required to be a minimum.
It is to be demonstrated that $P$ is the point on $M$ such that $APB'$ is a straight line.
 $L$ went through any point $P'$, for example, which is not on the straight line $AB'$.
Then $AP'B'$ is a triangle.
But $AP' + P'B$ is longer than $APB$.
Hence $L$ does not pass along the line $AP'B'$.
The result follows by Proof by Contradiction.
== Examples ==
== Historical Note ==
== Sources ==
* : Chapter $\text {A}.7$: Heron (first century A.D.)
* : Light Reflected off a Mirror
Category:Optics
Category:Heron's Principle of Reflection
Heron's Principle of Reflection/Examples/Journey to the River

== Example of Use of Heron's Principle of Reflection ==
Mary, who is standing at a point $S$, wants to walk to the point $T$.
:
However, first she wants to take a drink of water from a river.
She wants to walk as short a distance as possible.
To what point on the riverbank should she walk?
== Solution ==
Imagine the point $T'$ which is $T$ reflected in the line taken by the river bank.
The point $P$ is that point on the river bank directly in line with $S$ and $T'$.
:
Or you can reflect $S$ in the line taken by the river bank to get point $S'$.
Then $P$ is the point on the river bank directly in line with $S'$ and $T$.
== Sources ==
* : Light Reflected off a Mirror: $18$
Category:Heron's Principle of Reflection
Heron of Alexandria/Problems/Pythagorean Triangle

== Problem ==
:In a Pythagorean triangle, the sum of the area and the perimeter is $280$.
:Find the sides and the area.
== Solution ==
The Pythagorean triangle has sides $20$, $21$ and $29$.
Its area is $210$.
== Proof ==
There is no simple way to solve this except by trial and error.
First we note that:
:the area of the $20-21-29$ Pythagorean triangle is $\dfrac {20 \times 21} 2 = 210$
and we have that:
:$210 + 20 + 21 + 29 = 280$
Then from Sequence of Solutions of Pythagorean Equation:
from which it can soon be seen that the $20-21-29$ triangle is the one required.
== Sources ==
* 
* : Light Reflected off a Mirror: $22$
Category:Examples of Algebra Problems
Heron of Alexandria/Problems/Rectangles

== Problem ==
:Find two rectangles with integral sides, such that:
::the area of the first is $3$ times the area of the second
::the perimeter of the second is $3$ times the perimeter of the first.
== Solution ==
The rectangles are:
:$53 \times 54$
and:
:$318 \times 3$
== Proof ==
As can be seen:
and:
and it is seen that the two rectangles fulfil the given conditions.
Let the given ratio be made general, that is, $n$ rather than $3$.
Let $u, v$ and $x, y$ be the sides of the rectangles.
Then we can write:
Thus the general solution is:
Setting $n = 3$ gives the solution.
== Sources ==
* 
* : Light Reflected off a Mirror: $21$
Category:Examples of Algebra Problems
Heronian Triangle is Similar to Integer Heronian Triangle

== Theorem ==
Let $\triangle {ABC}$ be a Heronian triangle.
Then there exists an integer Heronian triangle $\triangle {A'B'C'}$ such that $\triangle {ABC}$ and $\triangle {A'B'C'}$ are similar.
== Proof ==
Let $\triangle {ABC}$ have sides whose lengths are $a$, $b$ and $c$.
By definition of Heronian triangle, each of $a$, $b$ and $c$ are rational.
By definition of rational number, we can express:
:$a = \dfrac {p_a} {q_a}$, $b = \dfrac {p_b} {q_b}$ and $c = \dfrac {p_c} {q_c}$
where each of $p_a, q_a, p_b, q_b, p_c, q_c$ are integers.
Now let:
Let $\triangle {A'B'C'}$ be the triangle whose sides have lengths $a'$, $b'$ and $c'$.
By definition, $\triangle {ABC}$ and $\triangle {A'B'C'}$ are similar.
Each of $a'$, $b'$ and $c'$ are integers.
Consider the area of triangle $\triangle {A'B'C'}$
Let the area of $\triangle {ABC}$ be $A$.
Then the area $\triangle {A'B'C'}$ is $q_a q_b q_c A$, which is rational.
Hence $\triangle {A'B'C'}$ is an integer Heronian triangle.
== Sources ==
* : Bachet
Category:Heronian Triangles
Category:Similar Triangles
Heronian Triangle whose Altitude and Sides are Consecutive Integers

== Theorem ==
There exists exactly one Heronian triangle one of whose altitudes and its sides are all consecutive integers.
This is the Heronian triangle whose sides are $\tuple {13, 14, 15}$ and which has an altitude $12$.
== Proof ==
We note that a Heronian triangle whose sides are all consecutive integers is also known as a Fleenor-Heronian triangle.
From Sequence of Fleenor-Heronian Triangles, we have that the smallest such triangles are as follows:
:$\tuple {1, 2, 3}$, which has an altitude of $0$
This is the degenerate case where the Heronian triangle is a straight line.
While $0, 1, 2, 3$ is a sequence of $4$ consecutive integers, this is not technically a triangle.
:$\tuple {3, 4, 5}$ with area $6$.
It has altitudes $3$, $4$ and $\dfrac {12} 5$.
:$\tuple {13, 14, 15}$
This can be constructed by placing the $2$ Pythagorean triangles $\tuple {5, 12, 13}$ and $\tuple {9, 12, 15}$ together along their common side $12$:
:
Thus the altitude and sides are:
:$\tuple {12, 13, 14, 15}$
and this is the Heronian triangle we seek.
It has area $84$.
The next largest Fleenor-Heronian triangle has sides $\tuple {51, 52, 53}$.
Using Heron's Formula, its area is given by:
:$\AA = \sqrt {78 \times 25 \times 26 \times 27} = 1170$
Hence its altitudes are:
:$45 \frac {45} {51}$, $45$, $44 \frac 8 {53}$
For still larger triangles, the altitudes are never within $1$ unit of the sides:
Consider the triangle with sides $\tuple {a - 1, a, a + 1}$.
Using Heron's Formula, its area is given by:
Its longest altitude is therefore:
and we have:
This shows that for $a \ge 22$, all altitudes of the triangle is less than $a - 2$.
Hence there are no more examples.
== Sources ==
* : Bachet: $109$
Category:Heronian Triangles
Hex Theorem/Corollary

== Theorem ==
Every game of Hex has exactly one winner.
== Proof ==
 there is a sequence of moves:
:$P_1, Q_1, P_2, Q_2, \dotsc$
such that neither player wins.
By the Pigeonhole Principle, every tile is marked after $n^2$ moves.
But, by the Hex Theorem, that board has exactly one winner.
This contradicts the assertion that neither player wins.
Thus, by Proof by Contradiction, every game of Hex has at least one winner.
 there is a game of Hex in which both players win.
Let $B$ be the board at the end of the game.
Let every unmarked tile of $B$ be marked with an arbitrary player symbol, by the Principle of Finite Choice.
Then, the resulting board $B'$ has every tile marked.
But, by definition of winning, each player has a finite sequence of adjacent tiles marked with their own symbol, which spans their opposite sides.
In particular, every tile in the sequence is marked in $B$, and so is marked the same in $B'$.
Thus, both players win in $B'$.
This contradicts the Hex Theorem.
Thus, by Proof by Contradiction, every game of Hex has no more than one winner.
The result follows.
Category:Hex Theorem
Hexadecimal Notation/Examples

== Examples of Hexadecimal Notation ==
=== Example: $4 \mathrm B$ ===
=== Example: $\mathrm C 8$ ===
=== Example: $12 \mathrm C$ ===
=== Example: $2 \mathrm C 8$ ===
=== Example: $6 \mathrm C 5$ ===
=== Example: $2 \mathrm B 7 \mathrm E$ ===
Category:Examples of Hexadecimal Notation
Hexadecimal Notation/Examples/12C

== Example of Hexadecimal Notation ==
The integer expressed in decimal as $300$ is expressed in hexadecimal as $12 \mathrm C$.
== Proof ==
== Sources ==
* : $\text {1-2}$ The Basis Representation Theorem: Example $\text {1-3}$
Category:Examples of Hexadecimal Notation
Hexadecimal Notation/Examples/C8

== Example of Hexadecimal Notation ==
The integer expressed in decimal as $200$ is expressed in hexadecimal as $\mathrm C 8$.
== Proof ==
== Sources ==
* : $\text {1-2}$ The Basis Representation Theorem: Example $\text {1-3}$
Category:Examples of Hexadecimal Notation
Hexagonal Number/Sequence

== Definition ==
The sequence of hexagonal numbers, for $n \in \Z_{\ge 0}$, begins:
:$0, 1, 6, 15, 28, 45, 66, 91, 120, 153, 190, 231, \ldots$
== Sources ==
* : $45$
* : Tables: $2$ The First $20$ Pentagonal, Hexagonal, Heptagonal and Octagonal Numbers
* : $45$
* : Tables: $2$ The First $20$ Pentagonal, Hexagonal, Heptagonal and Octagonal Numbers
Category:Hexagonal Numbers
Hexagonal Number as 4 times Triangular Number plus n

== Theorem ==
Let $H_n$ be the $n$th hexagonal number.
Then:
:$H_n = 4 T_{n - 1} + n$
where $T_{n - 1}$ is the $n - 1$th triangular number.
== Proof ==
Hence the result.
== Sources ==
* : $45$
* : $45$
Category:Hexagonal Numbers
Category:Triangular Numbers
Higher-Aleph Complement Topology is Topology

== Theorem ==
Let $T = \struct {S, \tau}$ be an $\aleph_n$ complement space.
Then $\tau$ is a topology on $T$.
== Proof ==
By definition, we have that $\O \in \tau$.
We also have that $S \in \tau$ as $\relcomp S S = \O$ which is trivially finite.
Now suppose $A, B \in \tau$.
Let $H = A \cap B$.
Then:
But as $A, B \in \tau$ it follows that $\relcomp S A$ and $\relcomp S B$ both have a cardinality strictly smaller than that of $S$.
Hence their union also has a cardinality strictly smaller than that of $S$.
So $\relcomp S H$ has a cardinality strictly smaller than that of $S$.
So $H = A \cap B \in \tau$ as its complement has a cardinality strictly smaller than that of $S$.
Now let $\UU \subseteq \tau$.
Then from De Morgan's laws: Complement of Union:
:$\ds \relcomp S {\bigcup \UU} = \bigcap_{U \mathop \in \UU} \relcomp S U$
But as:
:$\forall U \in \UU: \relcomp S U \in \tau$
each of the $\relcomp S U$ has a cardinality strictly smaller than that of $S$.
Hence so has their intersection.
So $\ds \relcomp S {\bigcup \UU}$ has a cardinality strictly smaller than that of $S$, which means:
:$\ds \bigcup \UU \in \tau$
So $\tau$ is a topology on $T$.
== Sources ==
* Believed original to Prime.mover
Category:Higher-Aleph Complement Topology
Higher Derivatives of Exponential Function

== Theorem ==
Let $\exp x$ be the exponential function.
Then:
:$\map {\dfrac {\d^n} {\d x^n} } {\exp x} = \exp x$
=== Corollary ===
== Proof ==
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\map {\dfrac {\d^n} {\d x^n} } {\exp x} = \exp x$
=== Basis for the Induction ===
$\map P 1$ is true, as this is the case proved in Derivative of Exponential Function:
:$\map {\dfrac \d {\d x} } {\exp x} = \exp x$
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show 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 our induction hypothesis:
:$\map {\dfrac {\d^k} {\d x^k} } {\exp x} = \exp x$
Then we need to show:
:$\map {\dfrac {\d^{k + 1} } {\d x^{k + 1} } } {\exp x} = \exp x$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \N_{>0}: \map {\dfrac {\d^n} {\d x^n} } {\exp x} = \exp x$
Category:Derivatives involving Exponential Function
Higher Derivatives of Exponential Function/Corollary

== Corollary to Higher Derivatives of Exponential Function ==
Let $\exp x$ be the exponential function.
Let $c$ be a constant.
Then:
:$\map {\dfrac {\d^n} {\d x^n} } {\map \exp {c x} } = c^n \map \exp {c x}$
== Proof ==
From Derivatives of Function of $a x + b$:
:$\map {\dfrac {\d^n} {\d x^n} } {\map f {a x + b} } = a^n \map {\dfrac {\d^n} {\d z^n} } {\map f z}$
where $z = a x + b$.
Here we set $a = c$ and $b = 0$ so that:
:$\map {\dfrac {\d^n} {\d x^n} } {\map f {c x} } = c^n \map {\dfrac {\d^n} {\d z^n} } {\map f z}$
where $z = c x$.
Then from Higher Derivatives of Exponential Function:
:$\map {\dfrac {\d^n} {\d z^n} } {\exp z} = \exp z$
Hence the result.
Category:Derivatives involving Exponential Function