avewright/proofwiki-math huggingface dataset:
Theorem: Doob's Maximal Inequality/Discrete Time

Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \mathop \ge 0}, \Pr}$ be a filtered probability space.
Let $\sequence {X_n}_{n \mathop \ge 0}$ be a non-negative $\sequence {\FF_n}_{n \mathop \ge 0}$-submartingale.
Let:
:$\ds X_n^\ast = \max_{0 \mathop \le k \mathop \le n} X_k$
where $\max$ is the pointwise maximum. 
Let $\lambda > 0$.
Then:
:$\lambda \map \Pr {X_n^\ast \ge \lambda} \le \expect {X_n}$

Proof:
Let $\lambda > 0 $ and $n \ge 0$.
Let:
:$E := \set {X^\ast _n \ge \lambda}$.
That is, $E$ is a disjoint union:
:$(1):\quad \ds E = \bigsqcup _{0 \mathop \le k \mathop \le n} E_k$
where:
:$\ds E_k := \set {X_k \ge \lambda} \cap \bigcap _{0 \mathop \le j \mathop \le k-1} \set {X_j < \lambda}$
By construction, we have:
:$\forall k \in \set {0, \ldots, n} : E_k \in \FF_k$
In particular:
{{begin-eqn}}
{{eqn | l = \expect {X_n \chi_{E_k} \mid \FF_k } 
      | r = \expect {X_n \mid \FF_k } \chi_{E_k}
      | c = Rule for Extracting Random Variable from Conditional Expectation of Product
}}
{{eqn | o = \ge
      | r = X_k \chi_{E_k}
      | c = {{Defof|Submartingale}}
      | n = 2
}}
{{end-eqn}}
where $\chi_A$ denotes the characteristic function of $A \subseteq \Omega$.
Therefore:
{{begin-eqn}}
{{eqn | l = \expect {X_n}
      | o = \ge
      | r = \expect {X_n \chi_E}
      | c = as $X_n \ge 0$
}}
{{eqn | r = \sum_{0 \mathop \le k \mathop \le n} \expect {X_n \chi_{E_k} }
      | c = by $(1)$
}}
{{eqn | r = \sum_{0 \mathop \le k \mathop \le n} \expect { \expect {X_n \chi_{E_k} \mid \FF_k } }
      | c = Tower Property of Conditional Expectation
}}
{{eqn | o = \ge
      | r = \sum_{0 \mathop \le k \mathop \le n} \expect { X_k \chi_{E_k} }
      | c = by $(2)$
}}
{{eqn | o = \ge
      | r = \sum_{0 \mathop \le k \mathop \le n} \expect { \lambda \chi_{E_k} }
      | c = as $X_k \ge \lambda$ on $E_k$
}}
{{eqn | r = \lambda \sum_{0 \mathop \le k \mathop \le n} \expect {\chi_{E_k} }
      | c = Expectation is Linear
}}
{{eqn | r = \lambda \sum_{0 \mathop \le k \mathop \le n} \map \Pr {E_k}
      | c = Integral of Characteristic Function
}}
{{eqn | r = \lambda \map \Pr E
      | c = by $(1)$
}}
{{end-eqn}}
{{qed}}
Theorem: Expectation is Monotone

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ and $Y$ be integrable random variables such that:
:$\forall \omega \in \Omega: \map X \omega \le \map Y \omega$
{{explain|Should we craft a link to one of the instances of pointwise inequality? We this for general functions, do we need one for random variables?}}
Then:
:$\expect X \le \expect Y$

Proof:
From the definition of expectation we have:
:$\ds \expect X = \int X \rd \Pr$
and:
:$\ds \expect Y = \int Y \rd \Pr$
The result follows directly from Integral of Integrable Function is Monotone.
{{qed}}
Category:Expectation
kgzhf3379pyvuq13d31whi6ighqubtt
Theorem: Necessary and Sufficient Conditions for Continuous Linear Transformation Space to be Banach Space

Let $\struct {X, \norm {\, \cdot \,}_X}$ and $\struct {Y, \norm {\, \cdot \,}_Y}$ be normed vector spaces.
Let $\struct{\map {CL} {X, Y}, \norm{\, \cdot \,}}$ be the continuous linear transformation space equipped with the supremum operator norm.
Then $\struct {\map {CL} {X, Y}, \norm{\, \cdot \,} }$ is a Banach Space {{iff}} $\struct {Y, \norm{\, \cdot \,}_Y}$ is a Banach Space.

Proof:
=== Necessary Condition ===
Let $Y$ be a Banach space.
Let $\sequence {T_n}_{n \mathop \in \N} \in \map {CL} {X, Y}$ be a Cauchy sequence.
Let $x \in X$.
Let $\norm {\, \cdot \,}$ be the supremum operator norm.
Theorem: Area under Curve

Let $f: \R \to \R$ be a real function which is defined and (Darboux) integrable on the closed interval $\closedint a b$.
Consider the graph of $f$ embedded in a Cartesian plane.
Let $\AA$ denote the area between the curve $\map f x$, the straight lines $x = a$ and $x = b$, and the $x$-axis.
Then:
:$\AA = \ds \int_a^b \map f x \rd x$
where $\ds \int_a^b \map f x \rd x$ denotes the (Darboux) definite integral of $f$ from $a$ to $b$.

Proof:
=== Overview of Proof ===
:500px
Let $x \in \closedint a b$.
Let $\delta x$ be an arbitrarily small positive real number  such that $x + \delta x \in \closedint a b$.
Consider the small strip between:
:the $x$-axis
:the vertical straight lines through $\tuple {x, 0}$ and $\tuple {x + \delta x, 0}$
:the curve $\map f x$.
The area of this strip is approximated by $y \rdelta x$.
Hence we create a model of the geometric interpretation of the definite integral.
{{qed|lemma}}
Theorem: Single Point Characterization of Simple Closed Contour

Let $C$ be a simple closed contour in the complex plane $\C$ with parameterization $\gamma: \closedint a b \to \C$.
Let $t_0 \in \openint a b$ such that $\gamma$ is complex-differentiable at $t_0$.
Let $S \in \set {-1,1}$ and $r \in \R_{>0}$ such that:
:for all $\epsilon \in \openint 0 r$, we have $\map \gamma {t_0} + \epsilon i S \map {\gamma '}{t_0} \in \Int C$
where $\Int C$ denotes the interior of $C$.
If $S = 1$, then $C$ is positively oriented.
If $S = -1$, then $C$ is negatively oriented.

Proof:
We show that for all $t_1 \in \openint a b$ where $\gamma$ is complex-differentiable at $t_1$, there exists $r \in \R_{>0}$ such that for all $\epsilon \in \openint 0 r$, it follows that:
:$\map v { t_1, \epsilon} \in \Int C$.
The result then follows by the definitions of positively oriented contour and negatively oriented contour.
By definition of parameterization of contour, there exists $N \in \N$ and a subdivision $\set { c_0 , \ldots , c_N }$ such that $\gamma$ is complex-differentiable at all $t \in \openint {c_k}{ c_{k+1} }$.
Let $k \in \set {1, \ldots, N}$ such that $t_0 \in \openint {c_{k-1} }{ c_{k} }$.
First, suppose $t_1 \in \openint {c_{k-1} }{ c_{k} }$.
From Lemma 1, it follows that there exists $r \in \R_{>0}$ such that for all $\epsilon \in \openint 0 r$:
:$\map v { t_1, \epsilon} \in \Int C$
Set $l := \begin {cases} k+1 & : k < N \\ 1 & : k = N \end {cases}$.
Suppose $t_1 \in \openint {c_{l-1} }{ c_{l} }$.
From Lemma 2, it follows that there exists $r \in \R_{>0}$ such that for all $\epsilon \in \openint 0 r$:
:$\map v { t_1, \epsilon} \in \Int C$
By repeated use of Lemma 2, it follows that for all $n \in \set {k+1, k+2, \ldots, N, 1, 2, \ldots, k-1}$ and for all $t \in \openint {c_{n-1} }{ c_{n} }$, there exists $r \in \R_{>0}$ that fulfills the condition.
{{qed}}
Category:Orientation of Complex Contour
Category:Single Point Characterization of Simple Closed Contour
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Theorem: Necessary and Sufficient Conditions for Continuous Linear Transformation Space to be Banach Space/Corollary 1

Let $\mathbb K = \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a normed vector space over $\mathbb K$.
Let $CL$ be the continuous linear transformation space.
Let $\norm {\, \cdot \,}$ be the supremum operator norm.
Then $\struct {\map {CL} {X, \mathbb K}, \norm {\, \cdot \,} }$ is a Banach space.

Proof:
:$X' := \map {CL} {X, \mathbb K}$ is the dual space of $X$
So this theorem is the same as Normed Dual Space is Banach Space.
{{improve|proper proof please}}
{{qed}}
Theorem: Necessary and Sufficient Conditions for Continuous Linear Transformation Space to be Banach Space/Corollary 2

Let $\struct {X, \norm {\, \cdot \,}_X}$ be a Banach space.
Let $CL$ be the continuous linear transformation space.
Let $\norm {\, \cdot \,}$ be the supremum operator norm.
Then $\struct {\map {CL} {X, X}, \norm{\, \cdot \,}}$ is a Banach space.

Proof:
Take $Y = X$ in Necessary and Sufficient Conditions for Continuous Linear Transformation Space to be Banach Space.
{{qed}}
Theorem: Differentiable Operator-Valued Function is Continuous

Let $\struct {X, \norm \cdot_X}$ be a normed vector space.
Let $f : I \to X$ be a map defined on an interval $I$.
Let $x_0 \in I$ such that $f$ is differentiable at $x_0$.
Then $f$ is continuous at $x_0$.

Proof:
We have {{hypothesis}} that the derivative $\map {f'} {x_0}$ of $f$ at $x_0$ exists.
Hence:
{{begin-eqn}}
{{eqn | l = \lim_{x \mathop \to x_0} \norm {\map f {x} - \map f {x_0} }_X
      | r = \lim_{h \mathop \to 0} \norm {\map f {x_0 + h} - \map f {x_0} }_X
      | c = re-write limit
}}
{{eqn | r = \lim_{h \mathop \to 0} \norm {h \cdot \dfrac {\map f {x_0 + h} - \map f {x_0} } h}_X
}}
{{eqn | r = \lim_{h \mathop \to 0} \size h \cdot \norm {\dfrac {\map f {x_0 + h} - \map f {x_0} } h}_X
      | c = {{NormAxiomVector|2}}
}}
{{eqn | r = \lim_{h \mathop \to 0} \size h \cdot \norm {\dfrac {\map f {x_0 + h} - \map f {x_0} } h - \map {f'} {x_0} + \map {f'} {x_0} }_X
}}
{{eqn | o = \le
      | r = \lim_{h \mathop \to 0} \size h \paren {\norm {\dfrac {\map f {x_0 + h} - \map f {x_0} } h - \map {f'} {x_0} }_X + \norm {\map {f'} {x_0} }_X }
      | c = {{NormAxiomVector|3}}
}}
{{eqn | r = 0 \cdot \norm {\map {f'} {x_0} }_X = 0
      | c = $f$ is differentiable at $x_0$ with derivative $\map {f'} {x_0}$ by assumption
}}
{{end-eqn}}
Thus:
:$\map f x \to \map f {x_0}$ as $x \to x_0$
The result follows by the definition of continuity in metric spaces and Metric Induced by Norm is Metric.
{{qed}}
Theorem: Reciprocal of 451

:$\dfrac 1 {451} = 0 \cdotp \dot 00221 \, 7294 \dot 9$

Proof:
Performing the calculation using long division:
<pre>
    0.0022172949002...
   -------------------
451)1.0000000000000
      902
      ---
       980
       902
       ---
        780
        451
        ---
        3290
        3157
        ----
         1330
          902
         ----
          4280
          4059
          ----
           2210
           1804
           ----
            4060
            4059
            ----
               1000
                902
               ----
               ....
</pre>
{{qed}}
Theorem: Composition of Continuous Mapping on Compact Space Preserves Uniform Convergence

Let $\struct{ X, d_X }$ be a compact metric space.
Let $\struct{ Y, d_Y }, \struct{ Z, d_Z }$ be metric spaces.
Let $\phi : Y \to Z$ be a continuous mapping.
For all $n \in \N$, let $f_n : X \to Y$ be a continuous mapping.
Let $f: X \to Y$ be a mapping such that $\sequence {f_n}_{n \in \N}$ converges to $f$ uniformly on $X$.
Then the sequence $\sequence {\phi \circ f_n}_{n \in \N}$ converges to $\phi \circ f$ uniformly on $X$.

Proof:
{{AimForCont}} that $\sequence {\phi \circ f_n}_{n \in \N}$ do not converge to $\phi \circ f$ uniformly.
This implies that there exists $\epsilon \in \R_{>0}$ such that for all $n \in \N$, there exist $N_n \ge n$ and $x_n \in X$ such that:
:$\map {d_Z}{ \map { \phi \circ f_{N_n} }{ x_n } , \map { \phi \circ f }{ x_n} } \ge \epsilon$
From Compact Subspace of Metric Space is Sequentially Compact in Itself, it follows that $\struct{ X, d_X }$ is sequentially compact in itself.
This implies that there exist $x' \in X$ and a subsequence $\sequence {x_{n_m} }_{m \in \N}$ such that $x_{n_m} \to x'$ for $m \to \infty$.
We calculate:
{{begin-eqn}}
{{eqn | o = \le
      | l = \epsilon
      | r = \lim_{m \mathop \to \infty} \map {d_Z}{ \map { \phi \circ f_{N_{n_m} } }{ x_{n_m} } , \map { \phi \circ f }{ x_{n_m} } }
}}
{{eqn | r = \map {d_Z}{ \lim_{m \mathop \to \infty} \map { \phi \circ f_{N_{n_m} } }{ x_{n_m} } , \lim_{m \mathop \to \infty} \map { \phi \circ f }{ x_{n_m} } }
      | c = by Metric is Continous Mapping and definition of continuity
}}
{{eqn | r = \map {d_Z}{ \map { \phi \circ f }{ x' } , \lim_{m \mathop \to \infty} \map { \phi \circ f }{ x_{n_m} } }
      | c = by Uniformly Convergent Sequence Evaluated on Convergent Sequence
}}
{{eqn | r = \map {d_Z}{ \map { \phi \circ f }{ x' } , \map { \phi \circ f }{ x' } }
      | c = by Composite of Continuous Mappings between Metric Spaces is Continuous
}}
{{eqn | r = 0
}}
{{end-eqn}}
which is a contradiction, as $\epsilon > 0$.
{{qed}}
Category:Uniform Convergence
Category:Sequentially Compact Spaces
mvd3tefjeqmiqwqp7oenb1uxcbhwwuv
Theorem: Primitive of Reciprocal of Root of x squared plus Constant

:$\ds \int \frac {\d x} {\sqrt {x^2 + k} } = \ln \size {x + \sqrt {x^2 + k} } + C$

Proof:
=== Positive Constant ===
Let $k > 0$.
Then $k = a^2$ for some $a \in \R$.
Hence from Primitive of $\dfrac 1 {\sqrt {x^2 + a^2} }$: Logarithm Form:
{{:Primitive of Reciprocal of Root of x squared plus a squared/Logarithm Form}}
from which the result follows.
{{qed|lemma}}
Theorem: Unique Tree of Order 2

There exists exactly one tree of order $2$:
:200px

Proof:
By definition, a graph of order $2$ has exactly $2$ nodes.
By Finite Connected Simple Graph is Tree iff Size is One Less than Order, such a tree has one edge.
There is trivially only one way to connect $2$ nodes with one edge.
Hence the result.
{{qed}}
Category:Tree Theory
tsygc5fl4tlqba2xegmme4svi2p7uuk
Theorem: Path Homotopy is Equivalence Relation

Let $X$ be a topological space.
Let $p, q \in X$.
Then path homotopy on the set of all paths in $X$ from $p$ to $q$ is an equivalence relation.

Proof:
Suppose $X$ is a topological space.
Let $p, q \in X$.
Let $\sim$ denote the path homotopy on the set of all paths in $X$ from $p$ to $q$.
{{ProofWanted}}
Checking in turn each of the criteria for equivalence:
Theorem: Mertens' Convergence Theorem

Let $\sequence {a_n}_{n \mathop \ge 0}$ and $\sequence {b_n}_{n \mathop \ge 0}$ be real or complex  sequences.
Let:
:$\ds \sum_{n \mathop = 0}^\infty a_n$ converge to $A$
:$\ds \sum_{n \mathop = 0}^\infty b_n$ converge to $B$.
Let either $\ds \sum_{n \mathop = 0}^\infty a_n$ or $\ds \sum_{n \mathop = 0}^\infty b_n$ converge absolutely.
Then:
:$\ds \paren {\sum_{n \mathop = 0}^\infty a_n} \paren {\sum_{n \mathop = 0}^\infty b_n} = \sum_n \paren {\sum_{j \mathop + k \mathop = n} a_j b_k}$
If both $\ds \sum_{n \mathop = 0}^\infty a_n$ and $\ds \sum_{n \mathop = 0}^\infty b_n$ converge absolutely, then so also does the Cauchy product.

Proof:
{{ProofWanted}}
{{Namedfor|Franz Mertens|cat = Mertens}}
Theorem: Gauss's Hypergeometric Theorem

:$\map F {a, b; c; 1} = \dfrac {\map \Gamma c \map \Gamma {c - a - b} } {\map \Gamma {c - a} \map \Gamma {c - b} }$

Proof:
From Euler's Integral Representation of Hypergeometric Function, we have:
:$\ds \map F {a, b; c; x} = \dfrac {\map \Gamma c } {\map \Gamma b \map \Gamma {c - b} } \int_0^1 t^{b - 1} \paren {1 - t}^{c - b - 1} \paren {1 - x t}^{- a} \rd t$
Where $a, b, c \in \C$.
and $\size x < 1$
and $\map \Re c > \map \Re b > 0$.
Since Euler's Integral Representation only applies where $\size x < 1$, we will determine the limit of the integral as $x \to 1$.
Therefore:
{{begin-eqn}}
{{eqn | l = \map F {a, b; c; x}
      | r = \dfrac {\map \Gamma c } {\map \Gamma b \map \Gamma {c - b} } \int_0^1 t^{b - 1} \paren {1 - t}^{c - b - 1} \paren {1 - \paren {1} t}^{- a} \rd t
      | c = 
}}
{{eqn | ll= \leadsto
      | r = \dfrac {\map \Gamma c } {\map \Gamma b \map \Gamma {c - b} } \int_0^1 t^{b - 1} \paren {1 - t}^{c - b - a - 1} \rd t
      | c = simplifying and Product of Powers
}}
{{eqn | ll= \leadsto
      | r = \dfrac {\map \Gamma c } {\map \Gamma b \map \Gamma {c - b} } \dfrac {\map \Gamma b \map \Gamma {c - a - b } } {\map \Gamma {c - a } }
      | c = {{Defof|Beta Function}}
}}
{{eqn | ll= \leadsto
      | r = \dfrac {\map \Gamma c \map \Gamma {c - a - b} } {\map \Gamma {c - a} \map \Gamma {c - b} }
      | c = simplifying and canceling $\map \Gamma b$
}}
{{end-eqn}}
{{qed}}
Theorem: Closed Ball is Convex Set

Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Let $\map { {B_r}^-} x$ be a closed ball in $X$ with radius $r \in \R_{>0}$ and center $x \in X$.
Then $\map { {B_r}^-} x$ is convex.

Proof:
Let $y \in \map { {B_1}^-} {\mathbf 0}$.
From {{NormAxiomVector|2}}, it follows that:
:$\norm {r y} = r \norm y$
It follows that:
:$y \in \map { {B_1}^-} {\mathbf 0}$, {{iff}} $r y \in \map { {B_r}^-} {\mathbf 0}$
As $\norm {r y - \mathbf 0} = \norm {\paren {r y + x} - x}$, it follows that:
:$r y \in \map { {B_r}^-} {\mathbf 0}$
{{iff}}:
:$r y + x \in \map { {B_r}^-} {\mathbf x}$
It follows that:
:$\map { {B_r}^-} {\mathbf x} = r \map { {B_1}^-} {\mathbf 0} + x$
From Closed Unit Ball is Convex Set, it follows that $\map { {B_1}^-} {\mathbf 0}$ is convex.
From Dilation of Convex Set in Vector Space is Convex, it follows that $r \map { {B_1}^-} {\mathbf 0}$ is convex.
From Translation of Convex Set in Vector Space is Convex, it follows that $r \map { {B_1}^-} {\mathbf 0} + x$ is convex.
{{qed}}
Theorem: Paths in Interior and Exterior of Jordan Curve

Let $\gamma : \closedint 0 1 \to \R^2$ be a Jordan curve.
Let $t_1 , t_2 \in \closedint 0 1$ with $t_1 < t_2$.
Let $h \in \R_{>0}$.
Let $\Img \gamma$, $\Int \gamma$ and $\Ext \gamma$ denote the image, interior, and exterior of $\gamma$.
Then there exists $r \in \R_{>0}$ such that $r \le h$, and for all $p_1 \in \map {B_r}{ \map \gamma {t_1} }$ and $p_2 \in \map {B_r}{ \map \gamma {t_2} }$:
* if $p_1, p_2 \in \Int \gamma$, then there exists a path $\sigma : \closedint 0 1 \to \Int \gamma$ with $\map \sigma 0 = p_1, \map \sigma 1 = p_2$ such that $\ds \max_{s \mathop \in \closedint 0 1} \map d {\gamma \sqbrk{\closedint {t_1}{t_2} }, \map \sigma s} < h$;
* if $p_1, p_2 \in \Ext \gamma$, then there exists a path $\sigma : \closedint 0 1 \to \Ext \gamma$ with $\map \sigma 0 = p_1, \map \sigma 1 = p_2$ such that $\ds \max_{s \mathop \in \closedint 0 1} \map d {\gamma \sqbrk{\closedint {t_1}{t_2} }, \map \sigma s} < h$.
Here $\map {B_r} x$ denotes the open ball with radius $r$ and center $x$, and $\map d {X,y}$ denotes the distance between the set $X$ and $y \in \R^2$.

Proof:
Let $\phi : \R^2 \to \R^2$ be the homeomorphism defined in the Jordan-Schönflies Theorem such that:
{{begin-eqn}}
{{eqn | l = \phi \sqbrk {\Img \gamma} 
      | r = \mathbb S_1
}}
{{eqn | l = \phi \sqbrk {\Int \gamma} 
      | r = \map {B_1}{ \mathbf 0 }
}}
{{eqn | l = \phi \sqbrk {\Ext \gamma} 
      | r = \R^2 \setminus \map {B_1^-} { \mathbf 0 }
}}
{{end-eqn}}
where $\mathbb S_1$ denotes the unit circle in $\R^2$, and $\map {B_1^- }{ \mathbf 0 }$ denotes a closed ball.
For the main part of this proof, we will assume that $p_1 , p_2 \in \Int \gamma$.
We will construct a path between $\map \phi {p_1}$ and $\map \phi {p_2}$ in the interior of $\phi \circ \gamma$.
This path will be a concatenation of three segments $\sigma_{\epsilon_\sigma, \operatorname {Int} }$, $\overline \sigma_1$, and $\overline \sigma_2$.
Theorem: Interior and Exterior of Partially Disjoint Jordan Curves

Let $\gamma_0 , \gamma_1 , \gamma_2 : \closedint 0 1 \to \R^2$ be Jordan arcs such that:
:$ \map {\gamma_0} 0 = \map {\gamma_1} 1 = \map {\gamma_2} 1$
:$\map {\gamma_0} 1 = \map {\gamma_1} 0 = \map {\gamma_2} 0$
Let $\gamma_0 \sqbrk { \openint 0 1 }$ and $\gamma_1 \sqbrk { \openint 0 1 }$ be disjoint sets, and let $\gamma_0 \sqbrk { \openint 0 1 }$ and $\gamma_2 \sqbrk { \openint 0 1 }$ be disjoint sets.
Let $\mathbf c \in \R^2$ and $r_0 \in \R_{>0}$ such that:
:$\Img {\gamma_1} \cup \Img {\gamma_2} \subseteq \map {B_{r_0} }{ \mathbf c }$
where $\Img {\gamma_1}$ denotes the image of $\gamma_1$, and $\map {B_{r_0} }{ \mathbf c }$ denotes the open ball with radius $r_0$ and center $\mathbf c$.
Let $t \in \openint 0 1$ such that $\map {\gamma_0} t \notin \map {B_{r_0} }{ \mathbf c }$.
Then there exists $r \in \R_{>0}$ such that for all $p \in \map {B_r}{ \map {\gamma_0} t }$:
:$p \in \Int { \gamma_0 * \gamma_1 }$, {{Iff}} $p \in \Int { \gamma_0 * \gamma_2 }$
:$p \in \Ext { \gamma_0 * \gamma_1 }$, {{Iff}} $p \in \Ext { \gamma_0 * \gamma_2 }$
where $\Int { \gamma_0 * \gamma_1 }$ and $\Ext { \gamma_0 * \gamma_1 }$ denote the interior, respectively the exterior of the Jordan curve created by concatenating $\gamma_0$ and $\gamma_1$.

Proof:
By definition of Jordan curve, it follows that $\gamma_0 * \gamma_1$ and $\gamma_0 * \gamma_2$ are Jordan curves.
Theorem: Sequences of Projections in 2-Sequence Space Converges in Strong Operator Topology

Let $\struct {\ell^2, \norm {\, \cdot \,}_2}$ be the $p$-sequence normed vector space.
Let $\map {CL} {\ell^2} := \map {CL} {\ell^2, \ell^2}$ be the continuous linear transformation space.
For $n \in \N$ let $P_n \in \map {CL} {\ell^2}$ be the projection operator over $\ell^2$.
Let $\sequence {P_n}_{n \mathop \in \N}$ be a sequence.
Let $I \in \map {CL} {\ell^2}$ be the identity operator.
Then $\sequence {P_n}_{n \mathop \in \N}$ converges to $I$ in the strong operator topology.

Proof:
Let $\mathbf a = \tuple {a_1, a_2, \ldots} \in \ell^2$.
Then:
{{begin-eqn}}
{{eqn | l = \norm {I \mathbf a - P_n \mathbf a}^2_2
      | r = \norm {\ldots, 0, a_{n+1}, a_{n + 2}, \ldots}^2_2
      | c = {{Defof|Projection Operator over 2-Sequence Spaces}}
}}
{{eqn | r = \sum_{k \mathop = n + 1}^\infty \size {a_k}^2
}}
{{eqn | ll = \leadsto
      | l = \lim_{n \mathop \to \infty } \norm {I \mathbf a - P_n \mathbf a}^2_2
      | r = \sum_{k \mathop = n + 1}^\infty \size {a_k}^2
}}
{{eqn | r = 0
      | c = {{Defof|P-Sequence Space}}, Tail of Convergent Series tends to Zero
}}
{{end-eqn}}
Hence:
:$\ds \lim_{n \mathop \to \infty } \norm {I \mathbf a - P_n \mathbf a}_2 = 0$
By definition, $\sequence {P_n}_{n \mathop \in \N}$ converges to $I$ in the strong operator topology.
{{qed}}
Theorem: Sequences of Projections in 2-Sequence Space do not Converge in Uniform Operator Topology

Let $\struct {\ell^2, \norm {\, \cdot \,}_2}$ be the $p$-sequence normed vector space.
Let $\map {CL} {\ell^2} := \map {CL} {\ell^2, \ell^2}$ be the continuous linear transformation space.
Let $\norm {\, \cdot \,}$ be the supremum operator norm.
For $n \in \N$ let $P_n \in \map {CL} {\ell^2}$ be the projection operator over $\ell^2$.
Let $\sequence {P_n}_{n \mathop \in \N}$ be a sequence.
Let $I \in \map {CL} {\ell^2}$ be the identity operator.
Then $\sequence {P_n}_{n \mathop \in \N}$ does not converge to $I$ in the uniform operator topology.

Proof:
{{AimForCont}} $\sequence {P_n}_{n \mathop \in \N}$ converges to $I$ in the uniform operator topology.
By definition:
:$\forall \epsilon \in \R_{> 0} : \exists N \in \N : \forall n \in \N : n > N \implies \norm {I - P_n} < \epsilon$
Suppose $\mathbf e_{N + 1} \in \ell^2$ is such that:
:$\mathbf e_{N + 1} = \tuple {\underbrace {0, \ldots, 0}_N, 1, 0 \ldots}$
Then:
{{begin-eqn}}
{{eqn | l = 1
      | r = \norm {\mathbf e_{N + 1} }_2
}}
{{eqn | r = \norm {\paren {I - P_N} \mathbf e_{N + 1} }_2
}}
{{eqn | o = \le
      | r = \norm {I - P_N} \norm {\mathbf e_{N + 1} }_2
      | c = Supremum Operator Norm as Universal Upper Bound
}}
{{eqn | o = <
      | r = \epsilon \cdot 1
}}
{{eqn | r = \epsilon
}}
{{end-eqn}}
This has to hold for all $\epsilon > 0$.
However, for $\epsilon < 1$ this does not hold.
Hence, we have a contradiction.
Therefore, $\sequence {P_n}_{n \mathop \in \N}$ does not converge to $I$ in the uniform operator topology.
{{qed}}
Theorem: Addition of Numbers is not Distributive over Multiplication

Addition of numbers is not distributive over multiplication.
That is, for numbers $a$, $b$ and $c$ it is not necessarily the case that $a + \paren {b \times c} = \paren {a + b} \times \paren {a + c}$.

Proof:
Proof by Counterexample:
{{begin-eqn}}
{{eqn | l = 2 + \paren {3 \times 6}
      | r = 2 + 18
      | c = 
}}
{{eqn | r = 20
      | c = 
}}
{{eqn | l = \paren {2 + 3} \times \paren {2 + 6}
      | r = 6 + 8
      | c = 
}}
{{eqn | r = 48
      | c = 
}}
{{eqn | o = \ne
      | r = 2 + \paren {3 \times 6}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Number of Sides of Polygon equals Number of Interior Angles

Let $A$ be a polygon.
The number of sides of $A$ is equal to the number of interior angles of $A$.

Proof:
{{ProofWanted|Obvious, which makes it difficult}}
Theorem: Equilateral Polygon is not necessarily Equiangular

Let $P$ be an equilateral polygon with more than $3$ sides.
Then it is not necessarily the case that $P$ is also equiangular.

Proof:
;Proof by Counterexample
We take as an example the rhombus:
{{:Definition:Rhombus}}{{qed}}
Theorem: Equiangular Polygon is not necessarily Equilateral

Let $P$ be an equiangular polygon with more than $3$ sides.
Then it is not necessarily the case that $P$ is also equilateral.

Proof:
;Proof by Counterexample
We take as an example the rectangle:
{{:Definition:Rectangle}}{{qed}}