avewright/proofwiki-math huggingface dataset:
Theorem: Component of Complement of Jordan Curve has Curve as Boundary

Let $\phi : \closedint 0 1 \to \R^2$ be a Jordan curve.
Let $J = \phi \closedint 0 1$ be the image of $\phi$.
Suppose that $\R^2 \setminus J$ has at least two distinct components.
Then, for any component $U$ of $\R^2 \setminus J$:
:$\partial U = J$
where $\partial U$ denotes the boundary of $U$.

Proof:
By:
* Closed Real Interval is Compact Space
* Continuous Image of Compact Space is Compact
it follows that $J$ is compact.
Therefore, by Compact Subspace of Hausdorff Space is Closed:
:$J$ is closed in $\R^2$
so by definition of closed:
:$\R^2 \setminus J$ is open in $\R^2$
By definition of locally connected:
:Each component of $\R^2 \setminus J$ is open
Let $W$ be the union of all components of $\R^2 \setminus J$, excluding $U$.
By {{Open-set-axiom|1}}, $W$ is open.
Additionally:
:$U \cup W = \R^2 \setminus J$
:$U \cap W = \O$
By Open Set Disjoint from Set is Disjoint from Closure and the definition of closure:
:$\paren {U \cup \partial U} \cap W = \O$
By Intersection Distributes over Union, Set is Subset of Union, Subset of Empty Set:
:$\partial U \cap W = \O$
Combining the above with Set is Open iff Disjoint from Boundary, we get:
:$\partial U \cap \paren {\R^2 \setminus J} = \O$
That is:
:$\partial U \subseteq J$
Theorem: Catenary is Symmetric about Y-Axis

Consider a '''catenary''' $\CC$.
Let a cartesian plane be arranged so that the $y$-axis passes through the lowest point of the catenary.
$\CC$ exhibits reflectional symmetry in that $y$-axis.
</onlyinclude>

Proof:
From Cartesian Equation of Catenary: Formulation $2$, we have the equation of $\CC$:
:$y = \dfrac a 2 \paren {e^{x / a} + e^{-x / a} } = a \cosh \dfrac x a$
The result follows directly from Hyperbolic Cosine Function is Even.
{{qed}}
Theorem: Thales' Theorem/Converse

400px
Let $O$ be a circle.
Let $AOB$ be a diameter of $O$.
Then $\angle APB$ is a right angle {{iff}} $P$ lies on the circle $O$.

Proof:
=== Necessary Condition ===
Draw $OC \parallel PB$.
By Parallelism implies Equal Corresponding Angles:
:$\angle ACO = \angle APB$ and both are right angles.
$\triangle ACO$ and $\triangle APB$ share $\angle OAC$.
$\triangle ACO$ and $\triangle APB$ are both right triangles.
{{begin-eqn}}
{{eqn | l = \triangle ACO
      | o = \sim
      | r = \triangle APB
      | c = Triangles with Two Equal Angles are Similar
}}
{{eqn | l = AO : AB
      | r = AC : AP
      | c = {{Defof|Similar Triangles}}
}}
{{eqn | l = AO
      | r = \frac 1 2 AB
      | c = {{Defof|Radius of Circle}}
}}
{{eqn | l = AC
      | r = \frac 1 2 AP
      | c = {{Defof|Similar Triangles}}
}}
{{eqn | l = AC
      | r = CP
      | c = 
}}
{{end-eqn}}
Now draw $OP$.
$OC$ is shared.
{{begin-eqn}}
{{eqn | l = \angle ACO
      | r = \angle PCO
      | c = both are right angles
}}
{{eqn | l = \triangle ACO
      | o = \cong
      | r = \triangle PCO
      | c = Triangle Side-Angle-Side Congruence
}}
{{eqn | l = OP
      | r = OA
      | c = 
}}
{{end-eqn}}
But $OA$ is a radius of the circle.  
Therefore $OP$ is also a radius.
So $P$ must lie on the circle.
{{qed|lemma}}
Theorem: Cauchy's Convergence Criterion/General

Let $\sequence {x_n}$ be a sequence in $\R$ or $\C$.
Then $\sequence {x_n}$ is a Cauchy sequence {{iff}}:
:$\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall r \in \N: r \ge N: \forall k > 0: \size {\ds \sum_{i \mathop = 1}^k x_{r + i} } < \epsilon$

Proof:
{{ProofWanted}}
{{Namedfor|Augustin Louis Cauchy}}
Theorem: Nth Root Test/Weak Form

Let $\ds \sum_{n \mathop = 1}^\infty a_n$ be a series of (strictly) positive real numbers $\R$.
Let the sequence $\sequence {a_n}$ be such that the limit $\ds \lim_{n \mathop \to \infty} \size {a_n}^{1/n} = l$.
Then:
:If $l > 1$, the series $\ds \sum_{n \mathop = 1}^\infty a_n$ diverges.
:If $l < 1$, the series $\ds \sum_{n \mathop = 1}^\infty a_n$ converges absolutely.

Proof:
{{ProofWanted|in due course}}
Theorem: Complex Numbers form Preordered Vector Space

Consider the complex numbers $\C$ as a vector space over itself.
Define the relation $\ge^\C$ by:
:$z \ge^\C w$
{{iff}}:
:$z - w \in \hointr 0 \infty$
for each $z, w \in \C$.
Then $\struct {\C, \ge^\C}$ is a preordered vector space.

Proof:
From Characterization of Preordered Vector Spaces, it is enough to show that $\hointr 0 \infty$ is a convex cone.
Let $x \in \hointr 0 \infty$ and $\alpha \in \R_{\ge 0}$. 
Then $\alpha x \ge 0$, so $\alpha x \in \hointr 0 \infty$.
So $\hointr 0 \infty$ is a cone.
Now let $x, y \in \hointr 0 \infty$. 
Then we have $x + y \ge 0$, so $x + y \in \hointr 0 \infty$.
So $\hointr 0 \infty$ is a convex cone.
{{qed}}
Category:Preordered Vector Spaces
fieubamcyb5odeygq13d6exs01sxd2r
Theorem: Element of *-Algebra Uniquely Decomposes into Hermitian Elements

Let $\struct {A, \ast}$ be a $\ast$-algebra over $\C$. 
Let $a \in A$.
Then there exists unique Hermitian elements $b, c \in A$ such that:
:$a = b + i c$
In particular, $b = \map \Re a$ and $c = \map \Im a$ where $b$ and $c$ are the real and imaginary parts of $a$ respectively.

Proof:
=== Proof of Existence === 
{{finish|fill in $\text C^\ast x$ with template}}
Let:
:$b = \map \Re a = \dfrac 1 2 \paren {a + a^\ast}$
and:
:$c = \map \Im a = \dfrac 1 {2 i} \paren {a - a^\ast}$
Then we have using $(\text C^\ast 2)$ and $(\text C^\ast 1)$: 
:$b^\ast = \dfrac 1 2 \paren {a^\ast + a^{\ast \ast} } = \dfrac 1 2 \paren {a + a^\ast} = b$
and:
{{begin-eqn}}
{{eqn	| l = c^\ast
		| r = -\frac 1 {2 i} \paren {a^\ast - a^{\ast \ast} }
		| c = $(\text C^\ast 4)$, $(\text C^\ast 2)$
}}
{{eqn	| r = -\frac 1 {2 i} \paren {a^\ast - a}
		| c = $(\text C^\ast 1)$
}}
{{eqn	| r = \frac 1 {2 i} \paren {a - a^\ast} 
}}
{{eqn	| r = c
}}
{{end-eqn}}
We therefore have: 
{{begin-eqn}}
{{eqn	| l = b + i c
		| r = \frac 1 2 \paren {a + a^\ast} + \frac i {2 i} \paren {a - a^\ast} 
}}
{{eqn	| r = \frac 1 2 \paren {a + a^\ast + a - a^\ast} 
}}
{{eqn	| r = a
}}
{{end-eqn}}
{{qed|lemma}}
Theorem: Product of Element in *-Star Algebra with its Star is Hermitian

Let $\struct {A, \ast}$ be a $\ast$-algebra.
Let $a \in A$.
Then $a^\ast a$ and $a a^\ast$ are Hermitian.

Proof:
We have:
{{begin-eqn}}
{{eqn	| l = \paren {a^\ast a}^\ast
		| r = a^\ast \paren {a^\ast}^\ast
		| c = $(\text C^\ast 3)$
}}
{{eqn	| r = a^\ast a 
		| c = $(\text C^\ast 1)$
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn	| l = \paren {a a^\ast}^\ast
		| r = \paren {a^\ast}^\ast a^\ast
		| c = $(\text C^\ast 3)$
}}
{{eqn	| r = a a^\ast
		| c = $(\text C^\ast 1)$
}}
{{end-eqn}}
So $a^\ast a$ and $a a^\ast$ are both Hermitian.
{{qed}}
Theorem: Median of Cauchy Distribution

Let $X$ be a continuous random variable with a '''Cauchy distribution''':
:$\map {f_X} x = \dfrac 1 {\pi \lambda \paren {1 + \paren {\frac {x - \gamma} \lambda }^2} }$
for:
:$\lambda \in \R_{>0}$
:$\gamma \in \R$
The median of $X$ is $\gamma$.

Proof:
From the definition of the Cauchy distribution, $X$ has probability density function:
:$\map {f_X} x = \dfrac 1 {\pi \lambda \paren {1 + \paren {\frac {x - \gamma} \lambda }^2} }$
Note that $f_X$ is non-zero, sufficient to ensure a unique median.
By the definition of a median, to prove that $\gamma$ is the median of $X$ we must verify: 
:$\ds \map \Pr {X < \gamma} = \int_{-\infty}^\gamma \map {f_X} x \rd x = \frac 1 2$
We have:
{{begin-eqn}}
{{eqn | l = \int_{-\infty}^\gamma \map {f_X} x \rd x
      | r = \frac 1 {\pi \lambda}  \int_{-\infty}^\gamma \dfrac 1 {\paren {1 + \paren {\frac {x - \gamma} \lambda }^2} } \rd x
}}
{{eqn | r = \frac 1 \pi \int_{-\infty}^0 \dfrac 1 {1 + t^2} \rd t
      | c = substituting $t = \dfrac {x - \gamma} {\lambda}$
}}
{{eqn | r = \dfrac 1 \pi \bigintlimits {\arctan t} {-\infty} {0}
      | c = Primitive of $\dfrac 1 {1 + x^2}$
}}
{{eqn | r = \dfrac 1 \pi \paren {0 - \paren {-\dfrac \pi 2} }
      | c = {{Defof|Arctangent}} and Arctangent of Zero is Zero
}}
{{eqn | r = \frac 1 2
}}
{{end-eqn}}
{{qed}}
Theorem: Cauchy Distribution is Symmetric about Median

Let $X$ be a continuous random variable with a '''Cauchy distribution''':
:$\map {f_X} x = \dfrac 1 {\pi \lambda \paren {1 + \paren {\frac {x - \gamma} \lambda }^2} }$
for:
:$\lambda \in \R_{>0}$
:$\gamma \in \R$
$X$ is (reflectionally) symmetric about the vertical line through the median $\gamma$.

Proof:
Recall from Median of Cauchy Distribution that $\gamma$ is indeed the median of $X$.
{{begin-eqn}}
{{eqn | l = \map {f_X} {2 \gamma - x}
      | r = \frac 1 {\pi \lambda \paren {1 + \paren {\frac {(2 \gamma - x) - \gamma} \lambda}^2} }
}}
{{eqn | r = \frac 1 {\pi \lambda \paren {1 + \paren {\frac {\gamma - x} \lambda}^2} }
}}
{{eqn | r = \frac 1 {\pi \lambda \paren {1 + \paren {\frac {x - \gamma} \lambda}^2} }
}}
{{eqn | r = \map {f_X} x
}}
{{end-eqn}}
So $X$ is symmetric about the line $x = \gamma$.
{{qed}}
{{explain|needs a link to an analytic geometrical demonstration that plane reflection is indeed governed by that equation}}
Theorem: Duplicated Triangle Forms a Kite or a Parallelogram

Let a triangle be copied either by rotation or reflection.
Then if you mung them together by corresponding sides what you get is either a parallelogram or a kite.
{{mistake|you might get a dart}}

Proof:
{{tidy|including about using definitions with / in them}}
Let $\triangle ABC$ be any triangle.  We can form a congruent triangle by reflection or translation.
Theorem: Quadrilateral is Cyclic iff Opposite Angles sum to Two Right Angles

500px
Given $\Box ABCD$ with $A, B$ and $D$ on a circle.
Let $\angle ABC$ and $\angle ADC$ add to two right angles.
Then $C$ lies on the circle, and $\Box ABCD$ is a cyclic quadrilateral.

Proof:
=== Sufficient Condition ===
By Sum of Internal Angles of Polygon, since $\Box ABCD$ is a quadrilateral:
:$\angle ABC + \angle BCD + \angle BAD + \angle ADC = 360$
Therefore $\angle BAD$ and $\angle BCD$ are also supplementary angles.
Suppose $C$ does not lie on the circle, but lies internal.  
Extend $DC$ to meet the circle at $C'$.
By Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles:
:$\angle BC'D$ is supplementary to $\angle BAD$.
It follows that:
:$\angle BC'D = \angle BCD$
By External Angle of Triangle equals Sum of other Internal Angles:
:$\angle BCD = \angle BC'D + \angle C'BC$
Hence $BCD > BC'D$.  
But this is a contradiction.
Suppose $C$ lies outside the circle.  
Draw $DC$ and find the point $C'$ where $DC$ meets the circle.
By External Angle of Triangle equals Sum of other Internal Angles:
:$\angle BC'D = \angle BCD + \angle C'BC$
Hence $BC'D > BCD$.  
But this is a contradiction.
Since $C$ is neither internal to the circle nor external to it, it must lie on the circle.
{{qed|lemma}}
Theorem: Quotient of Normal Distributions has Cauchy Distribution/Corollary

Let $X$ and $Y$ be independent continuous random variables each with a '''normal distribution''' with:
:zero expectation
:the same variance $\sigma$:
{{begin-eqn}}
{{eqn | l = X
      | o = \sim
      | r = \Gaussian 0 {\sigma^2}
}}
{{eqn | l = Y
      | o = \sim
      | r = \Gaussian 0 {\sigma^2}
}}
{{end-eqn}}
Let $U$ be the continuous random variable defined as:
:$U = \dfrac X Y$
Then $U$ has the '''Cauchy distribution''':
:$U \sim \Cauchy 0 1$
and so does $\dfrac 1 U$:
:$\dfrac 1 U \sim \Cauchy 0 1$

Proof:
From Quotient of Normal Distributions has Cauchy Distribution:
:$U \sim \Cauchy 0 \lambda$
where:
:$\lambda = \dfrac {\sigma_x} {\sigma_y}$
and such that:
:$\sigma_x = \sigma_y = \sigma$
Hence $\lambda = 1$ and the result follows.
Similarly we have:
:$\dfrac 1 U = \dfrac Y X$
and again the result follows.
{{qed}}
Theorem: Tangent Points of Incircle in Terms of Semiperimeter

Let $\triangle ABC$ be a triangle whose sides are $a$, $b$ and $c$ opposite vertices $A$, $B$ and $C$ respectively.
Let $s$ denote the semiperimeter of $\triangle ABC$.
Then the distance from $A$ to a point tangent to the incircle is equal to $s - a$.

Proof:
400px
Let $D$, $E$ and $F$ be the points where the incircle is tangent to the sides $AC$, $AB$ and $BC$ respectively.
Then:
{{begin-eqn}}
{{eqn | l = AD
      | r = AE
      | c = Tangents to Circle from Point are of Equal Length
}}
{{eqn | l = CD
      | r = CF
      | c = Tangents to Circle from Point are of Equal Length
}}
{{eqn | l = BE
      | r = BF
      | c = Tangents to Circle from Point are of Equal Length
}}
{{eqn | ll= \leadsto
      | l = BF + FC + AD
      | r = s
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a + AD
      | r = s
      | c = 
}}
{{eqn | ll= \leadsto
      | l = AD
      | r = s - a
      | c = 
}}
{{end-eqn}}
The above argument can be used {{mutatis}} to show that:
{{begin-eqn}}
{{eqn | l = BE
      | r = s - b
}}
{{eqn | l = CF
      | r = s - c
}}
{{end-eqn}}
{{qed}}
Category:Incircles of Triangles
f5e2b2mdb9rapifzdky3dcozx1s2a0p
Theorem: Length of Tangent to Excircle

The excircle on one side of a triangle is tangent to that side and divides it into the same two lengths as the tangent to the incircle, but in opposite order.
{{rename|because the existing name does not match what is being proved}}

Proof:
<onlyinclude>
:700px
Let the sides opposite vertex $A, B, C$ be $a,b,c$.
Let the semiperimeter of $\triangle ABC$ be $s$.
Construct the incircle of $\triangle ABC$ with center $M$.
Let $\triangle ABC$ be tangent to the incircle at $H$.
Construct the excircle of $\triangle ABC$ on side $c$ with center $N$.
Let $\triangle ABC$ be tangent to the excircle at $T$.
By Tangent Points of Incircle in Terms of Semiperimeter:
:$AF = AH = s - a$
:$CF = CG = s - c$
:$BG = BH = s - b$
By addition:
:$AB = AH + BH = \paren {s - a} + \paren {s - b}$
:$AB = AT + BT = \paren {s - a} + \paren {s - b}$
By Tangents to Circle from Point are of Equal Length:
:$AD = AT$
:$BE = BT$
By addition:
:$AT + BT = c$
Substituting:
:$AD + BE = c$
Rearranging:
:$BE = c - AD$
By Tangents to Circle from Point are of Equal Length, the tangents drawn to the excircle from $C$ are equal:
:$AD + b = BE + a$
By addition:
:$AD + b = c - AD + a$
:$2AD = a + c - b$
:$AD = \dfrac {a + c - b} 2$
By the definition of $s$ above:
:$\paren {s - b} = \dfrac {a + c + b} 2 - b = \dfrac {a + c - b} 2$
Combining results:
:$AD = s - b$
We find one pair of tangents to the incircle by substitution:
:$AT = AD = s - b$
Since $c = \paren {s - a} + \paren {s - b} = AD + BE$:
:$BE = s - a$
We find a second pair of tangents to the incircle by substitution:
:$BT = BE = s - a$
The third pair of tangents $CD$ and $CE$ are
:$CD = CE = 3 s - \paren {a + b + c} = s$
{{qed}}
Category:Incircles of Triangles
Category:Excircles of Triangles
5dnei9qar4jtdogpbx6yv9c3r21wab9
Theorem: Student's t-Distribution with One Degree of Freedom is Standard Cauchy Distribution

The Student's $t$-distribution with one degree of freedom is a special case of a standard Cauchy distribution.

Proof:
Let $X$ be a continuous random variables with a Student's $t$-distribution with one degree of freedom.
Then $X$  has probability density function: 
:$\map {f_X} x = \dfrac {\map \Gamma {\frac {k + 1} 2} } {\sqrt {\pi k} \map \Gamma {\frac k 2} } \paren {1 + \dfrac {x^2} k}^{-\frac {k + 1} 2}$
where $k = 1$.
Hence:
{{begin-eqn}}
{{eqn | l = \map {f_X} x
      | r = \dfrac {\map \Gamma {\frac {1 + 1} 2} } {\sqrt {\pi \times 1} \map \Gamma {\frac 1 2} } \paren {1 + \dfrac {x^2} 1}^{-\frac {1 + 1} 2}
      | c = setting $k = 1$
}}
{{eqn | r = \dfrac {\map \Gamma 1} {\sqrt \pi \paren {1 + x^2} \map \Gamma {\frac 1 2} }
      | c = simplifying
}}
{{eqn | r = \dfrac 1 {\sqrt \pi \paren {1 + x^2} \map \Gamma {\frac 1 2} }
      | c = Gamma Function Extends Factorial: $\map \Gamma 1 = 1$
}}
{{eqn | r = \dfrac 1 {\pi \paren {1 + x^2} }
      | c = Gamma Function of One Half: $\map \Gamma {\frac 1 2} = \sqrt \pi$
}}
{{end-eqn}}
which is exactly the standard Cauchy distribution.
{{qed}}
Theorem: Set of Invertible Continuous Transformations is Open Subset of Continuous Linear Transformations in Supremum Operator Norm Topology

Let $X$ be a Banach space.
Let $\map {CL} X$ be the continuous linear operator space on $X$.
Let $\map {GL} X$ denote the set of all invertible continuous linear operators on $X$.
Then $\map {GL} X \subseteq \map {CL} X$ in the supremum operator norm topology.

Proof:
Let $T_0 \in \map {GL} X$.
By definition:
:$T_0^{-1} \in \map {CL} X$.
Let $T \in \map {B_\epsilon} {T_0}$ where $\map {B_\epsilon} x$ is an open ball in $\struct {\map {GL} X, \norm {\, \cdot \,} }$ topology.
By definition:
:$\norm {T - T_0} < \epsilon$
We also have that:
{{begin-eqn}}
{{eqn | l = \norm {\paren {T - T_0} \circ T_0^{-1} } 
      | o = \le 
      | r = \norm {T - T_0} \norm {T_0^{-1} }
      | c = Supremum Operator Norm on Continuous Linear Transformation Space is Submultiplicative
}}
{{eqn | o = < 
      | r = \epsilon \norm {T_0^{-1} } 
}}
{{end-eqn}}
Zero operator is not invertible.
Hence, $T_0^{-1} \ne \mathbf 0$ and $\norm {T_0^{-1} } \ne 0$.
Choose $\epsilon$ such that $\epsilon \norm {T_0^{-1} } < 1$, for example, $\epsilon = \dfrac 1 {2 \norm {T_0^{-1} } }$.
Then:
{{begin-eqn}}
{{eqn | l = \norm {\paren {T - T_0} \circ T_0^{-1} } 
      | o = < 
      | r = \frac 1 2 
}}
{{eqn | o = <
      | r = 1
}}
{{end-eqn}}
By Neumann series theorem:
:$I + \paren {T - T_0} \circ T_0^{-1} \in \map {GL} X$
However, $T_0 \in \map {GL} X$ too.
Hence:
{{begin-eqn}}
{{eqn | l = T
      | r = T_0 + \paren {T - T_0} 
}}
{{eqn | r = \paren {I + \paren {T - T_0}\circ T_0^{-1} } \circ T_0
}}
{{end-eqn}}
We have that $T_0 \in \map {GL} X$ and $I + \paren {T - T_0}\circ T_0^{-1} \in \map {GL} X$.
Also, Composite of Bijections is Bijection and Bijection is Invertible.
Hence, their composition is also in $\map {GL} X$:
:$T = \paren {I + \paren {T - T_0}\circ T_0^{-1} } \circ T_0 \in \map {GL} X$
$T$ was arbitrary.
So we have the implication that:
:$\forall T \in \text{topology of } \struct {\map {GL} X, \norm {\, \cdot \,} } : \paren {T \in \map {GL} X} \implies \paren {T \in \map {CL} X}$
or in other words:
:$\map {GL} X \subseteq \map {CL} X$
in the supremum operator norm topology.
{{qed}}
Theorem: Quotient Space of Compact Space is Compact

Let $T = \struct {X, \tau}$ be a compact topological space.
Let $\RR \subseteq X \times X$ be an equivalence relation on $X$.
Then, the quotient space:
:$T / \RR$
is compact.

Proof:
Let $\UU$ be an open cover of $T / \RR$.
By definition of quotient topology, for every $U \in \UU$:
:$q_\RR^{-1} \sqbrk U \in \tau$
where $q_\RR$ is the quotient mapping induced by $\RR$.
Therefore:
:$\VV = \set {q_\RR^{-1} \sqbrk U : U \in \UU}$
is a set of open sets of $T$.
Let $x \in X$ be arbitrary.
By definition of cover, there is some $U \in \UU$ such that:
:$\eqclass x \RR \in U$
Therefore:
:$x \in q_\RR^{-1} \sqbrk U$
As $x \in X$ was arbitrary, it follows that $\VV$ is a cover of $X$.
As $\VV$ is a cover consisting of open sets, it is by definition an open cover of $T$.
Thus, by definition of compact space, there is some finite subcover $\set {V_i}_{1 \le i \le n} \subseteq \VV$.
But by definition of $\VV$, for every $1 \le i \le n$, there is some open set $U_i \in \UU$ such that:
:$V_i = q_\RR^{-1} \sqbrk U_i$
It remains to show that:
:$\set {U_i}_{1 \le i \le n}$
is a cover of $X / \RR$, the quotient set of $X$ induced by $\RR$.
Let $y \in X / \RR$ be arbitrary.
By definition of quotient set, there is some $x \in X$ for which:
:$y = \eqclass x \RR$
By definition of cover there is some $1 \le i \le n$ such that:
:$x \in V_i$
By definition of preimage:
:$\eqclass x \RR = \map {q_\RR} x \in U_i$
As $y \in X / \RR$ was arbitrary, it follows that:
:$\set {U_i}_{1 \le i \le n} \subseteq \UU$
is a finite subcover of $\UU$.
As $\UU$ was arbitrary, $T / \RR$ is compact by definition.
{{qed}}
Theorem: Quotient Space of Compact Space is Compact

Let $T = \struct {X, \tau}$ be a compact topological space.
Let $\RR \subseteq X \times X$ be an equivalence relation on $X$.
Then, the quotient space:
:$T / \RR$
is compact.

Proof:
Let $q_\RR$ be the quotient mapping induced by $\RR$.
By the definition of the quotient topology, $q_\RR$ is continuous.
Further, $T / \RR = q_\RR \sqbrk T$. 
From Continuous Image of Compact Space is Compact, $q_\RR \sqbrk T$ is compact. 
Hence $T / \RR$ is compact.
{{qed}}
Theorem: Maximal Inequality for Positive Operators

Let $\struct {X, \BB, \mu}$ be a probability space.
Let $\map {L^1} \mu$ be a real-valued $L^1$ space with respect to $\mu$.
Let $U : \map {L^1} \mu \to \map {L^1} \mu$ be a positive linear operator, that is:
:$\forall f \in \map {L^1} \mu : f \ge 0 \implies U f \ge 0$
Suppose:
:$\norm U \le 1$
where $\norm \cdot$ denotes the operator norm.
Then, for all $N \in \N_{>0}$:
:$\ds \int_{\set {F_N > 0} } f \rd \mu \ge 0$
where:
:$F_N := \max \set {f_n : 1 \le n \le N}$
and:
:$f_n := f + U f + \cdots + U^{n-1} f$

Proof:
Let:
:$G_N := \max \set {f_n : 0 \le n \le N}$
where $f_0 := 0$.
For all $0 \le n \le N$:
:$U G_N + f \ge U f_n + f = f_{n+1}$
In particular:
:$\paren 1 :\quad U G_N + f \ge F_N$
Let $x \in X$.
If $\map {F_N} x > 0$, then:
{{begin-eqn}}
{{eqn | l = \map {G_N} x
      | r = \max \set {\map {f_n} x : 0 \le n \le N}
}}
{{eqn | r = \max \set {\map {f_n} x : 1 \le n \le N}
}}
{{eqn | r = \map {F_N} x
}}
{{eqn | o = \le
      | r = \map {U G_N} x + \map f x
      | c = by $\paren 1$
      | n = 2
}}
{{end-eqn}}
Further, from $G_N \ge 0$ follows:
:$\paren 3 :\quad U G_N \ge 0$ 
Therefore:
{{begin-eqn}}
{{eqn | l = \int_{\set {F_N > 0} } f \rd \mu
      | o = \ge
      | r = \int_{\set {F_N > 0} } G_N - U G_N \rd \mu
      | c = by $\paren 2$
}}
{{eqn | r = \int_{\set {F_N > 0} } G_N \rd \mu - \int_{\set {F_N > 0} } U G_N \rd \mu
}}
{{eqn | r = \int G_N \rd \mu - \int_{\set {F_N > 0} } U G_N \rd \mu
      | c = as $\map {F_N} x \le 0$ implies $\map {G_N} x = 0$
}}
{{eqn | o = \ge
      | r = \int G_N \rd \mu - \int U G_N \rd \mu
      | c = by $\paren 3$
}}
{{eqn | r = \norm {G_N}_{\map {L^1} \mu} - \norm {U G_N}_{\map {L^1} \mu}
}}
{{eqn | o = \ge
      | r = \norm {G_N}_{\map {L^1} \mu} - \norm U \norm {G_N}_{\map {L^1} \mu}
      | c = {{Defof|Norm on Bounded Linear Transformation|$\norm U$}}
}}
{{eqn | o = \ge
      | r = 0
      | c = as $\norm U \le 1$
}}
{{end-eqn}}
{{qed}}
Theorem: Compact Convex Set with Nonempty Interior is Homeomorphic to Cone on Boundary

Let $n \in \N_{> 0}$.
Let $T \subseteq \R^n$ be a compact convex subset of real Euclidean $n$-space.
Suppose the interior of $T$ is non-empty.
Then, $T$ is homeomorphic to the cone on its boundary.

Proof:
Let $\bsx_0 \in T^\circ$ be an interior point of $T$.
Let $C \partial T$ denote the cone on the boundary of $T$.
Define $\phi : C \partial T \to T$ as:
:$\map \phi {\eqclass {\tuple {e, \bsx, t}} \RR} = t \bsx + \paren {1 - t} \bsx_0$
where:
:$e$ is the unique element of the trivial topological space used in the construction of the cone.
:$\RR$ is the equivalence relation used to define the join
For $\phi$ to be well-defined, it is necessary that if:
:$\tuple {\tuple {e, \bsx, t}, \tuple {e', \bsx', t'}} \in \RR$
then:
:$t \bsx + \paren {1 - t} \bsx_0 = t' \bsx' + \paren {1 - t'} \bsx_0$
By definition of join, the hypothesis implies that one of:
:$t = t' = 0$ and $e = e'$
:$t = t' \in \openint 0 1$ and $\tuple {e, \bsx, t} = \tuple {e', \bsx', t'}$
:$t = t' = 1$ and $\bsx = \bsx'$
is true.
If the first holds, then:
:$t \bsx + \paren {1 - t} \bsx_0 = \bsx_0 = t' \bsx' + \paren {1 - t'} \bsx_0$
If the second holds, then:
:$e = e'$
:$\bsx = \bsx'$
:$t = t'$
and the equality holds trivially.
If the third holds, then:
:$t \bsx + \paren {1 - t} \bsx_0 = \bsx = \bsx' = t' \bsx' + \paren {1 - t'} \bsx_0$
Thus, in every case, $\phi$ is well-defined.
Theorem: Hurwitz's Theorem (Normed Division Algebras)

The only normed division algebras over the real numbers are:
:the real numbers $\R$ themselves
:the complex numbers $\C$
:the quaternions $\H$
:the octonions $\mathbb O$

Proof:
{{ProofWanted}}
{{Namedfor|Adolf Hurwitz|cat = Hurwitz}}