avewright/proofwiki-math huggingface dataset:
Theorem: Identity Mapping is Frame Isomorphism

Let $L = \struct{S, \preceq}$ be a frame.
Let $\operatorname{id}_S$ denote the identity mapping on $S$.
Then:
:$\operatorname{id}_S$ is a frame isomorphism of $L$ to $L$

Proof:
From Identity Mapping is Bijection and Identity Mapping is Frame Homomorphism:
:$\operatorname{id}_S$ is a bijective frame homomorphism
Hence $\operatorname{id}_S$ is a frame isomorphism by definition.
{{qed}}
Category:Frame Isomorphisms
h9ytnyjd9b6yltuwrq43smfjxj4lwlx
Theorem: Convergent Sequence with Finite Number of Terms Inserted is Convergent

Let $\struct {X, d}$ be a metric space.
Let $\sequence {x_k}$ be a sequence in $X$.
Let $\sequence {x_k}$ be convergent.
Let a finite number of terms be inserted into $\sequence {x_k}$.
Then the resulting sequence is convergent.

Proof:
Let $I$ be the index of the last inserted term.
Let $N$ be the length of the sequence of terms to be inserted.
By Convergent Sequence with Finite Number of Terms Deleted is Convergent, we may delete the first $I$ terms from the original sequence $\sequence {x_k}$.
Let $\sequence {f_k}$ be the sequence of those $I$ terms, with the $N$ terms inserted.
So $\sequence {f_k}$ has $I + N$ terms, a finite amount.
By Convergent Sequence with Finite Elements Prepended is Convergent Sequence, we may prepend the modified $\sequence {x_k}$ with the terms in $\sequence {f_k}$.
The resulting sequence is $\sequence {x_k}$ with the finite amount of terms inserted.
And it is convergent.<!-- There's probably a shorter way to do this. The details obscure the handwave: Inserting can be done by deleting and prepending -->
{{Qed}}
bczct075bzd12odco2pkcpvd9pcay8i
Theorem: Transfer Operator on Lipshitz Functions on Shift Space is Linear Bounded Operator

Let $\struct {X ^+, \sigma}$ be a one-sided shift of finite type.
Let $F_\theta^+$ be the space of Lipschitz functions with the Lipschitz norm $\norm \cdot_\theta$.
Let $f \in F_\theta^+$.
Let $\LL_f : F_\theta^+ \to F_\theta^+$ be the transfer operator
Then $\LL_f$ is a bounded linear operator.

Proof:
Let $\mathbf A$ be the $k\times k$ logical matrix of $X ^+$.
Let $n \ge 1$.
Let $x, y \in X^+$ be such that:
:$\forall i \in \hointr 0 n : x_i = y_i$
Then:
{{begin-eqn}}
{{eqn | l = \cmod {\map {\LL_f g} x - \map {\LL_f g} y}
      | r = \cmod {\sum_{\map {\mathbf A} {i, x_0} = 1} e^{\map f {ix} } \map g {ix} - \sum_{\map {\mathbf A} {i, y_0} = 1} e^{\map f {iy} } \map g {iy} }
}}
{{eqn | r = \cmod {\sum_{\map {\mathbf A} {i, x_0} = 1} e^{\map f {ix} }  \paren {\map g {ix} - \map g {iy} } + \paren {e^{\map f {ix} } - e^{\map f {iy} } } \map g {ix} }
}}
{{eqn | o = \le
      | r = \sum_{\map {\mathbf A} {i, x_0} = 1} e^{\norm f_\infty}  \cmod {\map g {ix} - \map g {iy} } + \cmod {e^{\map f {ix} } - e^{\map f {iy} } } \norm g_\infty
}}
{{eqn | o = \le
      | r = \sum_{\map {\mathbf A} {i, x_0} = 1} e^{\norm f_\infty}  \cmod {\map g {ix} - \map g {iy} } + e^{\norm f_\infty} \cmod {\map f {ix} - \map f {iy} } \norm g_\infty
      | c = Mean Value Theorem
}}
{{eqn | o = \le
      | r = \sum_{\map {\mathbf A} {i, x_0} = 1} e^{\norm f_\infty} \map {\mathrm{var}_{n+1} } g + e^{\norm f_\infty} \map {\mathrm{var}_{n+1} } f \norm g_\infty
}}
{{eqn | o = \le
      | r =  k e^{\norm f_\infty} \paren {\map {\mathrm{var}_{n+1} } g + \map {\mathrm{var}_{n+1} } f \norm g_\infty}
}}
{{end-eqn}}
Thus for $\forall n \ge 1$:
{{begin-eqn}}
{{eqn | o = \le
      | l = \frac {\map {\mathrm{var}_n } {\LL_f g} }{\theta^n}
      | r = \theta k e^{\norm f_\infty} \paren {\frac {\map {\mathrm{var}_{n+1} } g} {\theta^{n+1} } + \frac {\map {\mathrm{var}_{n+1} } f} {\theta^{n+1} } \norm g_\infty}
}}
{{eqn | o = \le
      | r = \theta k e^{\norm f_\infty} \paren {\size g_\theta + \size f_\theta \norm g_\infty}
}}
{{eqn | o = \le
      | r = \theta k e^{\norm f_\infty} \paren {1 + \size f_\theta} \norm g_\theta
}}
{{end-eqn}}
Furthermore:
{{begin-eqn}}
{{eqn | l = \frac {\map {\mathrm{var}_0} {\LL_f g} }{\theta^0}
      | r = \sup_{x,y\in X^+} \cmod {\map {\LL_f g} x - \map {\LL_f g} y}
}}
{{eqn | o = \le
      | r = 2 \norm {\LL_f g}_\infty
}}
{{eqn | o = \le
      | r = 2 \norm {\LL_f} \norm g_\infty
      | c = Transfer Operator with respect to One-Sided Shift Space of Finite Type is Linear Bounded Operator
}}
{{eqn | o = \le
      | r =  \norm g_\theta
}}
{{end-eqn}}
where $\norm {\LL_f}$ denotes the operator norm of  $\LL_f : \map C {X^+} \to \map C {X^+}$.
Let:
:$c_f := 2 \norm {\LL_f} + \theta k e^{\norm f_\infty} \paren {1 + \size f_\theta}$
Then:
:$\forall g \in F_\theta^+ : \size {\LL_f g}_\theta \le c_f \norm g_\theta$
Therefore:
:$\forall g \in F_\theta^+ : \norm {\LL_f g}_\theta \le \paren {\norm {\LL_f} + c_f} \norm g_\theta$
{{Finish|details}}
Category:Ergodic Theory
Category:Functional Analysis
9b2vzuecmaar4ph9ri1mp43m3z5ab4b
Theorem: Borel-Cantelli Lemma in Probability

Let $\sequence {E_n}$ be an arbitrary countable sequence of events.
Let the sum of the probabilities of $\sequence {E_n}$ be finite.
Then the probability that infinitely many of the events occur is zero.

Proof:
{{ProofWanted}}
{{Namedfor|Émile Borel|name2 = Francesco Paolo Cantelli|cat = Borel, Émile|cat2 = Cantelli}}
Theorem: Contravariant Functor Induces Covariant Functor from Dual Category

Let $\mathbf C$ and $\mathbf D$ be metacategories.
Let $F : \mathbf C \to \mathbf D$ be a contravariant functor.
Let $\mathbf C^{\text {op}}$ denote the dual category of $\mathbf C$.
Then $F': \mathbf C^{\text {op}} \to \mathbf D$ defined by:
:* each object $C^{\text{op}}$ of $\mathbf C^{\text{op}}$ is assigned to the object $FC$ of $\mathbf D$, that is, $\map {F'} {C^{\text{op}}} = FC$
:* each arrow $f^{\text{op}} : Y^{\text{op}} \to X^{\text{op}}$ of $\mathbf C^{\text{op}}$ is assigned to the arrow $Ff : FY \to FX$ of $\mathbf D$, that is, $\map {F'} {f^{\text{op}}} = Ff$
is a covariant functor.

Proof:
We have:
{{begin-eqn}}
{{eqn | l = \map {F'} {g^{\text {op} } \circ f^{\text {op} } }
      | r = \map {F'} {\paren{f \circ g }^{\text {op} } } 
      | c = {{Defof|Dual Category}}
}}
{{eqn | r = \map F {f \circ g}
      | c = Definition of $F'$
}}
{{eqn | r = \map F g \circ \map F f
      | c = {{Defof|Contravariant Functor}}
}}
{{eqn | r = \map {F'} {g^{\text {op} } } \circ \map {F'} {f^{\text {op} } }
      | c = Definition of $F'$
}}
{{end-eqn}}
and
{{begin-eqn}}
{{eqn | l = \map {F'} {\operatorname{id}_{C^{\text{op} } } }
      | r = \map {F'} {\paren{\operatorname{id}_C}^{\text{op} } }
      | c = {{Defof|Dual Category}}
}}
{{eqn | r = \map F {\operatorname{id}_C}
      | c = Definition of $F'$
}}
{{eqn | r = \operatorname{id}_{FC}
      | c = {{Defof|Contravariant Functor}}
}}
{{eqn | r = \operatorname{id}_{\map {F'} {C^{\text{op} } } }
      | c = Definition of $F'$
}}
{{end-eqn}}
It follows that $F'$ is a covariant functor by definition.
{{qed}}
Category:Functors
Category:Dual Categories
cyvv27267t22yd5cyb81gpwxq64neub
Theorem: Contravariant Functor Induces Covariant Functor to Dual Category

Let $\mathbf C$ and $\mathbf D$ be metacategories.
Let $F : \mathbf C \to \mathbf D$ be a contravariant functor.
Let $\mathbf D^{\text {op}}$ denote the dual category of $\mathbf D$.
Then $F': \mathbf C \to \mathbf D^{\text {op}}$ defined by:
:* each object $C$ of $\mathbf C$ is assigned to the object $\paren{FC}^{\text {op}}$ of $\mathbf D^{\text {op}}$, that is, $\map {F'} C = \paren{FC}^{\text {op}}$.
:* each arrow $f : X \to Y$ of $\mathbf C$ is assigned to the arrow $\paren{Ff}^{\text {op}} : \paren{FX}^{\text {op}} \to \paren{FY}^{\text {op}}$ of $\mathbf D^{\text {op}}$, that is, $\map {F'} f = \paren{Ff}^{\text {op}}$
is a covariant functor.

Proof:
We have:
{{begin-eqn}}
{{eqn | l = \map {F'} {g \circ f}
      | r = \paren{\map F {g \circ f } }^{\text {op} }
      | c = Definition of $F'$
}}
{{eqn | r = \paren{Ff \circ Fg}^{\text {op} }
      | c = {{Defof|Contravariant Functor}}
}}
{{eqn | r = \paren{Fg}^{\text {op} } \circ \paren{Ff}^{\text {op} }
      | c = {{Defof|Dual Category}}
}}
{{eqn | r = \map {F'} g \circ \map {F'} f
      | c = Definition of $F'$
}}
{{end-eqn}}
and
{{begin-eqn}}
{{eqn | l = \map {F'} {\operatorname{id}_C}
      | r = \paren{\map F {\operatorname{id}_C} }^{\text {op} }
      | c = Definition of $F'$
}}
{{eqn | r = \paren{\operatorname{id}_{FC} }^{\text {op} }
      | c = {{Defof|Contravariant Functor}}
}}
{{eqn | r = \operatorname{id}_{\paren{FC}^{\text {op} } }
      | c = {{Defof|Dual Category}}
}}
{{eqn | r = \operatorname{id}_{\map {F'} C}
      | c = Definition of $F'$
}}
{{end-eqn}}
It follows that $F'$ is a covariant functor by definition.
{{qed}}
Category:Functors
Category:Dual Categories
0z0thdvjjzi7qm87ezbf3jadbnl0scc
Theorem: Continuous Mapping Induced by Frame Homomorphism is Continuous

Let $L_1 = \struct{S_1, \preceq_1}$ and $L_2 = \struct{S_2, \preceq_2}$ be frames.
Let $f : L_1 \to L_2$ be a frame homomorphism.
Let:
:$\map {\operatorname{Sp}} f : \map {\operatorname{Sp}} {L_2} \to \map {\operatorname{Sp}} {L_1}$ denote the continuous mapping induced by $f$
where:
:considered as locales, $\map {\operatorname{Sp}} {L_1}$ and $\map {\operatorname{Sp}} {L_2}$ denote the spectra as completely prime filters of $L_1$ and $L_2$, respectively.
Then:
:$\map {\operatorname{Sp}} f$ is a continuous mapping

Proof:
By definition of spectra as completely prime filters:
:$\map {\operatorname{Sp}} {L_1}$ is the topological space $\struct{\map {\operatorname{pt}} {L_1}, \set{\Sigma^{\paren 1}_{a_1} : a_1 \in L_1}}$ where:
::*$\map {\operatorname{pt}} {L_1}$ denotes the set of points as completely prime filters of $L_1$
::*$\forall a_1 \in L_1 : \Sigma^{\paren 1}_{a_1} = \set{p_1 \in \map {\operatorname{pt}} {L_1} : a_1 \in p_1}$
Similarly:
:$\map {\operatorname{Sp}} {L_2}$ is the topological space $\struct{\map {\operatorname{pt}} {L_2}, \set{\Sigma^{\paren 2}_{a_2} : a_2 \in L_2}}$ where:
::*$\map {\operatorname{pt}} {L_2}$ denotes the set of points as completely prime filters of $L_2$
::*$\forall a_2 \in L_2 : \Sigma^{\paren 2}_{a_2} = \set{p_2 \in \map {\operatorname{pt}} {L_2} : a_2 \in p_2}$
By definition of continuous mapping induced by frame homomorphism:
:$\forall p_2 \in \map {\operatorname{pt}} {L_2} : \map {\map {\operatorname{Sp}} f} {p_2} = f^{-1} \sqbrk {p_2}$
From Inverse Image Mapping of Frame Homomorphism Preserves Completely Prime Filter:
:$\map {\operatorname{Sp}} f$ is well-defined
We have:
{{begin-eqn}}
{{eqn | q = \forall a_2 \in L_2
      | l = \map {\operatorname{Sp} } f^{-1} \sqbrk {\Sigma^{\paren 2}_{a_2} }
      | r = \set{p_1 \in \map {\operatorname{pt} } {L_1} : \map {\map {\operatorname{Sp} } f} {p_1} \in \Sigma^{\paren 2}_{a_2} }
      | c = {{Defof|Inverse Image}}
}}
{{eqn | r = \set{p_1 \in \map {\operatorname{pt} } {L_1} : f^{-1} \sqbrk {p_1} \in \Sigma^{\paren 2}_{a_2} }
      | c = {{Defof|Continuous Mapping Induced by Frame Homomorphism}}
}}
{{eqn | r = \set{p_1 \in \map {\operatorname{pt} } {L_1} : a_2 \in f^{-1} \sqbrk {p_1} }
      | c = Definition of $\Sigma^{\paren 2}_{a_2}$
}}
{{eqn | r = \set{p_1 \in \map {\operatorname{pt} } {L_1} : \map f {a_2} \in {p_1} }
      | c = {{Defof|Inverse Image}}
}}
{{eqn | r = \Sigma^{\paren 1}_{\map f {a_2} }
      | c = Definition of $\Sigma^{\paren 1}_{\map f {a_2} }$
}}
{{end-eqn}}
It has been shown that the inverse image of every open set of the spectrum of $L_2$ is an open set of the spectrum of $L_1$ under the mapping $\map {\operatorname{Sp}} f$.
It follows that $\map {\operatorname{Sp}} f$ is continuous by definition.
{{qed}}
Theorem: Brouwer's Fixed Point Theorem/Two-Dimensional Version

Let $D \subseteq \R^2$ be the closed disk defined as:
:$D = \set {\tuple {x, y} \in \R^2: x^2 + y^2 \le 1}$
Let $f: D \to D$ be a mapping which is continuous on $D$.
Then:
:$\exists \xi \in D: \map f \xi = \xi$

Proof:
{{ProofWanted}}
{{Namedfor|Luitzen Egbertus Jan Brouwer}}
Theorem: Injectivity of Fourier transform over Lebesgue integrable functions

Fourier transform is injective over the space $\map {\mathrm L^1} {\R^N}$
{{explain|what is $\map {\mathrm L^1} {\R^N}$?}}

Proof:
Let $\FF$ denote the Fourier transform over the space of Lebesgue integrable functions $\map {\mathrm L^1} {\R^N}$.
The map $\FF$ is linear, so it is sufficient to prove that its kernel is trivial in order to show the injectivity (Linear Transformation is Injective iff Kernel Contains Only Zero).
Let $N \in \N$ and $f \in \map {\mathrm L^1} {\R^N}$ such as $\map \FF f = 0$.
As $f$ and $0$ are Lebesgue integrable over $\R^N$, we can apply the inverse Fourier transform to the equality $\map \FF f = 0$.
We then get $f = 0$ almost everywhere.
{{qed}}
Theorem: Buffon's Needle/General

Let a horizontal plane be divided into strips by a series of parallel lines a fixed distance $d$ apart.
Let a needle $N$ of length $l$ such that $l < d$ be dropped onto the plane randomly from a random height.
Then the probability that $N$ across one of the parallel lines is $\dfrac {2 l} {\pi d}$.

Proof:
Let the real number plane $\R^2$ divided into strips by the lines $x = k d$ for each integer $k$.
Define $\theta \in \hointr {-\dfrac \pi 2} {\dfrac \pi 2}$ as the angle between $N$ and the $x$-axis.
Then the $x$-component of the length of $N$ is $l \cos \theta$ for each $\theta$.
Let:
:$E$ be the event where $N$ falls across a vertical line
:$\Theta_\theta$ be the event where the angle between $N$ and the $x$-axis is $\theta$.
Let $N$ drop.
{{WLOG}}, let the end with the larger $x$-coordinate have $x$-coordinate $0 \le x_n < 1$.
Then for each $\theta$, $N$ falls across the line $x = 0$ exactly when $x_n \le l \cos \theta$.
Therefore the probability that this happens is:
:$\condprob E {\Theta_\theta} = \dfrac {l \cos \theta} d$
By considering $\theta$ as a continuous random variable:
{{begin-eqn}}
{{eqn | l = \map \Pr E
      | r = \sum_{\theta \mathop \in \hointr {-\pi / 2} {\pi / 2} } \condprob E {\Theta_\theta} \map \Pr {\Theta_\theta}
      | c = Total Probability Theorem
}}
{{eqn | r = \int_{-\pi / 2}^{\pi / 2} \dfrac {l \cos \theta \rd \theta} {\pi d}
}}
{{eqn | r = \intlimits {\frac l {d \pi} \sin \theta} {-\pi / 2} {\pi / 2}
      | c = Primitive of Cosine Function
}}
{{eqn | r = \frac l {d \pi} \paren {1 - \paren {-1} }
}}
{{eqn | r = \frac {2 l} {d \pi}
}}
{{end-eqn}}
{{qed}}
Theorem: Largest Number not Expressible as Sum of Multiples of Coprime Integers/Existence

Let $a_1,\ldots ,a_n \in \N_{>0}$.
Let $d = \gcd \set {a_1,\ldots ,a_n}$ be the greatest common divisor.
Let $d \N = \set { d k : k \in \N}$.
Let:
:$\map \N {a_1,\ldots ,a_n} = \set { k_1 a_1 + \cdots + k_n a_n : k_1,\ldots ,k_n \in \N}$
Then:
:$d \N \setminus \map \N {a_1,\ldots ,a_n}$
is a finite set.
In particular, it has a largest element.

Proof:
{{ProofWanted}}
Category:Largest Number not Expressible as Sum of Multiples of Coprime Integers
Category:Integer Combinations
gfdzjo6lspdspqc60se0f6a9jatnxdj
Theorem: PDF of Cantor Distribution does not Exist

The probability density function of the Cantor distribution $C$ does not exist.

Proof:
By definition, the probability density function of a probability distribution is the derivative of its cumulative distribution function.
By definition, the cumulative distribution function of $C$ is the Cantor function.
From Cantor Function is Differentiable Almost Everywhere, the Cantor function is $0$ almost everywhere.
Hence the probability density function of the Cantor distribution would be $0$ almost everywhere.
Hence it cannot exist.
{{qed}}
Theorem: Midpoint Characterization of Extreme Point of Convex Set

Let $\GF \in \set {\R, \C}$. 
Let $X$ be a vector space over $\GF$. 
Let $K$ be a convex subset of $X$. 
Let $x \in K$.
Then $x$ is an extreme point of $K$ {{iff}}:
:whenever $\ds x = \frac {y + z} 2$ for $y, z \in K$, we have $x = y = z$.

Proof:
=== Necessary Condition ===
Suppose first that $K$ is a convex subset of $X$. 
Then:
:for each $t \in \openint 0 1$, if: 
::$x = t y + \paren {1 - t} z$ for some $y, z \in K$
:then $x = y = z$. 
In particular, taking $t = 1/2$, it holds that:
:whenever $\ds x = \frac {y + z} 2$ for $y, z \in K$, we have $x = y = z$. 
{{qed|lemma}}
Theorem: Characterization of Characters on Space of Complex-Valued Continuous Functions on Compact Hausdorff Space

Let $K$ be a compact Hausdorff space.
Let $\struct {\map \CC K, \overline \cdot, \norm {\, \cdot \,} }$ be the $\text C^\ast$-algebra of complex-valued continuous functions on $K$.
For each $x \in K$, define:
:$\map {\delta_x} f = \map f x$
for each $f \in \map \CC K$. 
Let $\phi : \map \CC K \to \C$ be a algebra homomorphism on $\map \CC K$. 
Then $\phi$ is a character {{iff}} there exists $x \in K$ with $\phi = \delta_x$.

Proof:
=== Sufficient Condition ===
Let $x \in K$.
We show that $\delta_x$ is an algebra homomorphism. 
Let $f, g \in \map \CC K$ and $\lambda \in \C$. 
Then we have:
{{begin-eqn}}
{{eqn	| l = \map {\delta_x} {\lambda f + g}
		| r = \map {\paren {\lambda f + g} } x
}}
{{eqn	| r = \map {\paren {\lambda f} } x + \map g x
}}
{{eqn	| r = \lambda \map f x + \map g x
}}
{{eqn	| r = \lambda \map {\delta_x} f + \map {\delta_x} g
}}
{{end-eqn}}
Hence $\delta_x$ is a linear functional.
Further:
{{begin-eqn}}
{{eqn	| l = \map {\delta_x} {f g}
		| r = \map {\paren {f g} } x
}}
{{eqn	| r = \map f x \map g x
}}
{{eqn	| r = \map {\delta_x} f \map {\delta_x} g
}}
{{end-eqn}}
So $\delta_x$ is an algebra homomorphism.
Now, take: 
:$\map f k = 1$ for each $k \in K$. 
From Constant Function is Continuous, we have $f \in \map \CC K$.
Hence we have $\map {\delta_x} f = \map f x = 1 \ne 0$. 
Hence $\delta_x$ is a non-zero algebra homomorphism.
{{qed|lemma}}
Theorem: Hermitian Element of Space of Complex-Valued Continuous Functions on Compact Hausdorff Space

Let $K$ be a compact Hausdorff space.
Let $\struct {\map \CC K, \overline \cdot, \norm {\, \cdot \,} }$ be the $\text C^\ast$-algebra of continuous functions on $K$.
Let $f \in \map \CC K$.
Then $f$ is Hermitian {{iff}} $\map f x \in \R$ for each $x \in K$.

Proof:
We have that $f$ is Hermitian {{iff}} $f = \overline f$.
That is, {{iff}} $\map f x = \overline {\map f x}$ for each $x \in K$. 
From Complex Number equals Conjugate iff Wholly Real, this is equivalent to $\map f x \in \R$ for each $x \in K$. 
{{qed}}
Category:Space of Continuous Functions on Compact Hausdorff Space
dnuinsibsaq31qu8yu5lh0e5q0oc2s9
Theorem: Projection on Space of Complex-Valued Continuous Functions on Compact Hausdorff Space

Let $K$ be a compact Hausdorff space.
Let $\struct {\map \CC K, \overline \cdot, \norm {\, \cdot \,} }$ be the $\text C^\ast$-algebra of continuous functions on $K$.
Let $p \in \map \CC K$ be a projection.
Then there exists open sets (possibly empty) $U_0, U_1 \subseteq K$ such that:
:$U_0 \cap U_1 = \O$
and:
:$K = U_0 \cup U_1$
such that:
:$\ds \map p x = \begin{cases}1 & x \in U_1 \\ 0 & x \in U_0\end{cases}$

Proof:
Let $\map {B_{1/2} } 0$ and $\map {B_{1/2} } 1$ be the open balls of radius $1/2$ and centers $0$ and $1$ respectively.
Since $p$ is continuous, we have that:
:$p^{-1} \sqbrk {\map {B_{1/2} } 0}$ and $p^{-1} \sqbrk {\map {B_{1/2} } 1}$ are open in $K$. 
Let $U_0 = p^{-1} \sqbrk {\map {B_{1/2} } 0}$ and $U_1 = p^{-1} \sqbrk {\map {B_{1/2} } 1}$.
Since $\map {B_{1/2} } 0 \cap \map {B_{1/2} } 1 = \O$, we have that $U_0 \cap U_1 = \O$.
Since $p$ is a projection, we have:
:$\paren {\map p x}^2 = \map p x$ for each $x \in K$.
Hence:
:$\map p x \paren {\map p x - 1} = 0$
Hence for each $x \in K$:
:$\map p x = 0$ or $\map p x = 1$.
That is:
:$p^{-1} \sqbrk {\map {B_{1/2} } 0} \cup p^{-1} \sqbrk {\map {B_{1/2} } 1} = K$
and:
:$\ds \map p x = \begin{cases}1 & x \in U_1 \\ 0 & x \in U_0\end{cases}$
{{qed}}
Category:Space of Continuous Functions on Compact Hausdorff Space
elbue24i65d45m7hdp2sc8osh85e58p
Theorem: Image of Extreme Point under Injective Linear Isomorphism is Extreme Point

Let $\Bbb F \in \set {\R, \C}$.
Let $X$ and $Y$ be vector spaces over $\Bbb F$.
Let $K \subseteq X$ be convex.
Let $T : X \to Y$ be an injective linear transformation.
Let $x \in K$.
Then $x$ is an extreme point of $K$ {{iff}} $T x$ is an extreme point of $T \sqbrk K$.

Proof:
=== Necessary Condition ===
Suppose first that $x$ is an extreme point of $K$.
Let $y, z \in K$ be such that:
:$T x = t T y + \paren {1 - t} T z$ 
Then since $T$ is linear, we have:
:$T x = \map T {t y + \paren {1 - t} z}$
Since $T$ is injective, we have:
:$x = t y + \paren {1 - t} z$
Snce $x$ is an extreme point, we have $y = z$. 
Hence $T y = T z$. 
So $T x$ is an extreme point of $T \sqbrk K$.
{{qed|lemma}}
Theorem: Intersection of Closed Subsets satisfying Finite Intersection Property in Compact Space

Let $T = \struct {S, \tau}$ be a compact topological space.
Let $\FF = \set {V_\alpha}_{\alpha \mathop \in I} \subseteq \powerset S$ be a family of closed sets in $T$.
Suppose that $\FF$ satisfies the finite intersection property.
:That is, the intersection of every finite subset is non-empty.
Then the intersection of $\FF$ is non-empty.

Proof:
{{AimForCont}} $\ds \bigcap_{\alpha \mathop \in I} V_\alpha = \O$.
Then, by De Morgan's Laws:
:$\ds \bigcup_{\alpha \mathop \in I} \relcomp S {V_\alpha} = \relcomp S \O = S$
Since each $V_\alpha$ is closed, $\relcomp S {V_\alpha}$ is open.
Hence, $\set {\relcomp S {V_\alpha}}_{\alpha \mathop \in I}$ forms an open cover of $T$.
So by definition of a compact space, it has a finite subcover:
:$\FF' = \set {\relcomp S {V_{\alpha_1} }, \relcomp S {V_{\alpha_2} }, \dots, \relcomp S {V_{\alpha_n} } }$
But then:
:$\ds \bigcup_{i \mathop = 1}^n \relcomp S {V_{\alpha_i} } = S$
and so by De Morgan's Laws:
:$\ds \relcomp S {\bigcap_{i \mathop = 1}^n V_{\alpha_i} } = S$
which is to say:
:$\ds \bigcap_{i \mathop = 1}^n V_{\alpha_i} = \O$
But that contradicts the hypothesis that $\FF$ satisfies the finite intersection property.
Hence, by Proof by Contradiction:
:$\ds \bigcap_{\alpha \mathop \in I} V_\alpha \ne \O$
{{qed}}
Theorem: Number of Spanning Trees on Complete Graph

Let $K_n$ denote the complete graph of order $n$.
The number of spanning trees on $K_n$ is equal to $n^{n - 2}$.

Proof:
{{ProofWanted|Induction may work here}}
Theorem: Closed Unit Ball of Space of Zero-Limit Sequences has no Extreme Points

Let $\struct {c_0, \norm {\, \cdot \,}_\infty}$ be the Banach space of zero-limit sequences.
Let $B_{c_0}^-$ be the closed unit ball of $\struct {c_0, \norm {\, \cdot \,}_\infty}$.
Then $B_{c_0}^-$ has no extreme points.

Proof:
From Space of Zero-Limit Sequences with Supremum Norm forms Banach Space, $\struct {c_0, \norm {\, \cdot \,}_\infty}$ is a Banach space.
From Closed Unit Ball is Convex Set, $B_{c_0}^-$ is convex and hence we can speak of its extreme points..
Let $\sequence {x_n}_{n \mathop \in \N} \in B_{c_0}^-$. 
Since $x_n \to 0$, there exists $N \in \N$ such that:
:$\cmod {x_N} < 1/4$
Define:
:$e_k = \begin{cases}1 & N = k \\ 0 & N \ne k\end{cases}$
Then $e_k = 0$ for $k > N$, and hence $e_k \to 0$.
So $e = \sequence {e_n}_{n \mathop \in \N} \in c_0$.
Since $c_0$ is a vector space, we have $x + e, x - e \in c_0$.
Define $y_\pm = x \pm e$.
We have $\paren {y_\pm}_k = x_k$ for $k \ne N$ and $\paren {y_\pm}_N = x_k \pm \frac 1 4$.
We therefore have, by the Triangle Inequality for Complex Numbers:
:$\ds \cmod {\paren {y_\pm}_N} \le \cmod {x_k} + \frac 1 4 < \frac 1 2$ 
while:
:$\ds \cmod {\paren {y_\pm}_k} < \frac 1 4$ for $k \ne N$.
Hence we have $y_\pm \in B_{c_0}^-$. 
However, we have:
:$\ds x = \frac 1 2 \paren {x + e + \paren {x - e} } = \frac 1 2 \paren {y_+ + y_-}$
with $y_\pm \ne x$.
Hence $x$ is not an extreme point.
Since $x \in B_{c_0}^-$ was arbitrary, $B_{c_0}^-$ has no extreme points.
{{qed}}
Category:Extreme Points of Convex Sets
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Theorem: Ceiling Function/Examples/Ceiling of 3.2

:$\ceiling {3 \cdotp 2} = 4$

Proof:
We have that:
:$3 < 3 \cdotp 2 \le 4$
Hence $4$ is the ceiling of $3 \cdotp 2$ by definition.
{{qed}}
Theorem: Ceiling Function/Examples/Ceiling of 5

:$\ceiling 5 = 5$

Proof:
We have that:
:$4 < 5 \le 5$
Hence $5$ is the ceiling of $5$ by definition.
{{qed}}