avewright/proofwiki-math huggingface dataset:
Theorem: Vertices in Locally Finite Graph

Let $G$ be a locally finite graph.
Then if $G$ is infinite, it contains an infinite number of vertices.

Proof:
Suppose $G = \struct {V, E}$ has a finite number of vertices.
Let $V = \set {v_1, v_2, \ldots, v_n}$ where $n = \card V$ is the cardinality of $V$.
As $G$ is locally finite, each element of $V$ has a finite number of incident edges.
For each $v_k \in V$, let $r_k$ be the degree of $v_k$.
Then:
:$\ds \card E \le \sum_{i \mathop = 1}^n r_k$
where $\card E$ is the cardinality of $E$, that is, the number of edges in $G$.
As $\ds \sum_{i \mathop = 1}^n r_k$ is the sum of a finite number of finite numbers, it is itself finite.
So $\card E$ is itself finite, and so $G$ has:
:a finite number of vertices
:a finite number of edges
and so is by definition a finite graph.
By transposition, if $G$ is infinite, it must contain an infinite number of vertices.
{{qed}}
Theorem: Cantor Set is Closed in Real Number Space

Let $\CC$ be the Cantor set.
Let $\struct {\R, \tau_d}$ be the real number space $\R$ under the Euclidean topology $\tau_d$.
Then $\CC$ is a closed subset of $\struct {\R, \tau_d}$.

Proof:
By definition, the Cantor set is the complement of a union of open sets relative to the closed interval $\closedint 0 1$.
By the definition of a topology, that union is itself open in $\R$.
The closed interval $\closedint 0 1$ is itself the complement of a union of open sets $\openint \gets 0 \cup \openint 1 \to$.
Hence the result.
{{qed}}
Theorem: Word Metric is Metric

Let $\struct {G, \circ}$ be a group.
Let $S$ be a generating set for $G$ which is closed under inverses (that is, $x^{-1} \in S \iff x \in S$).
Let $d_S$ be the associated word metric.
Then $d_S$ is a metric on $G$.

Proof:
Let $g, h \in G$.
It is given that $S$ is a generating set for $G$.
It follows that there exist $s_1, \ldots, s_n \in S$ such that $g^{-1} \circ h = s_1 \circ \cdots \circ s_n$.
Therefore $\map {d_S} {g, h} \le n$, establishing that $\R$ is a valid codomain for the mapping $d_S$ with domain $G \times G$.
This is the form a mapping must have to be able to be a metric.
Now checking the other defining properties for a metric in turn:
Theorem: Cantor Space is Compact

Let $\CC$ be the Cantor set.
Let $\struct {\R, \tau_d}$ be the real number space $\R$ under the Euclidean topology $\tau_d$.
Then $\CC$ is a compact subset of $\struct {\R, \tau_d}$.

Proof:
We have Cantor Set is Closed in Real Number Space.
Taking, for example, $0 \in \CC$ and $1 \in \R$ it is clear that:
:$\forall x \in \CC: \map d {0, x} \le 1$
and so $\CC$ is bounded.
Hence the result from the Heine-Borel Theorem.
{{qed}}
Theorem: Cantor Space is Complete Metric Space

Let $T = \struct {\CC, \tau_d}$ be the Cantor space.
Then $T$ is a complete metric space.

Proof:
We have that the Cantor space is a metric subspace of the real number space $\R$, and hence a metric space.
We also have Cantor Space is Compact.
The result follows from Compact Metric Space is Complete.
{{qed}}
Theorem: Cantor Space satisfies all Separation Axioms

Let $T = \struct {\CC, \tau_d}$ be the Cantor space.
Then $T$ satisfies all the separation axioms.

Proof:
We have that the Cantor space is a metric subspace of the real number space $\R$, and hence a metric space.
The result follows from Metric Space fulfils all Separation Axioms.
{{qed}}
Theorem: Cantor Space is Second-Countable

Let $T = \struct {\CC, \tau_d}$ be the Cantor space.
Then $T$ is a second-countable space.

Proof:
We have that the Cantor space is a topological subspace of the real number space with the usual (Euclidean) topology $\struct {\R, \tau_d}$.
We also have that the Real Number Line is Second-Countable.
The result follows from Second-Countability is Hereditary.
{{qed}}
Theorem: Cantor Space is Dense-in-itself

Let $T = \struct {\CC, \tau_d}$ be the Cantor space.
Then $T$ is dense-in-itself.

Proof:
{{Recall|Dense-in-itself|dense-in-itself}}
{{:Definition:Dense-in-itself}}
Let $U \in \tau_d$ be open in $T$.
Let $p \in U$.
Then:
:$\exists x \in U: \exists \epsilon \in \R: \map d {x, p} < \epsilon$
Thus $p$ is not an isolated point of $\T$.
Hence the result by definition of dense-in-itself.
{{qed}}
Theorem: Cantor Space is not Scattered

Let $T = \struct {\CC, \tau_d}$ be the Cantor space.
Then $T$ is not scattered.

Proof:
{{Recall|Scattered Space|scattered space}}
{{:Definition:Scattered Space/Definition 1}}
We have that Cantor Space is Dense-in-itself.
Hence the result by definition of a scattered space.
{{qed}}
Theorem: Cantor Space is Perfect

Let $T = \struct {\CC, \tau_d}$ be the Cantor space.
Then $\CC$ is a perfect set of the real number space $\R$ under the usual (Euclidean) topology $\tau_d$.

Proof:
{{Recall|Perfect Set|perfect set}}
{{:Definition:Perfect Set/Definition 2}}
From Cantor Space is Dense-in-itself, $\CC$ contains no isolated points.
We also have that the Cantor Set is Closed in Real Number Space.
The result follows from the definition of perfect set.
{{qed}}
Theorem: Cantor Space is Nowhere Dense

Let $T = \struct {\CC, \tau_d}$ be the Cantor space.
Then $T$ is nowhere dense in $\closedint 0 1$.

Proof:
From Cantor Set is Closed in Real Number Space, $\CC$ is closed.
So from Closed Set equals its Closure:
:$\CC^- = \CC$
where $\CC^-$ denotes the closure of $\CC$.
Let $0 \le a < b \le 1$.
Then $I = \openint a b$ is an open interval of $\closedint 0 1$.
Let $\epsilon = b - a$.
Clearly $\epsilon > 0$.
Let $n \in \N$ such that $3^{-n} < \epsilon$.
{{explain|Need to fill in the details to show that between $a$ and $b$ there is an interval $3^{-\paren {n + 1} }$ wide that contains numbers with $1$ in its $n + 1$th digit in its ternary expansion (or something like that) - intuitively obvious but fiddly to prove.}}
So there exists an open interval of $\closedint 0 1$ which has been deleted from $\closedint 0 1$ during the process of creating $\CC$.
Thus no open interval of $\closedint 0 1$ is disjoint from ''all'' the open intervals deleted from $\closedint 0 1$.
So an open interval of $\closedint 0 1$ can not be a subset of $\CC = \CC^-$.
Hence the result, by definition of nowhere dense.
{{qed}}
Theorem: Cantor Space is Nowhere Dense

Let $T = \struct {\CC, \tau_d}$ be the Cantor space.
Then $T$ is nowhere dense in $\closedint 0 1$.

Proof:
Let $S_n$ and $C_n$ be as in the definition of the Cantor set as a limit of a decreasing sequence.
Then the length of every interval in $S_n$ is seen to be $\dfrac 1 {3^n} = 3^{-n}$.
Let $0 \le a < b \le 1$.
Then $\openint a b \subseteq \closedint 0 1$ is an open interval.
Let $n \in \N$ such that $3^{-n} < b - a$, so that the length of every interval in $S_n$ is $3^{-n} < b - a$.
Therefore, as the intervals in $S_n$ do not overlap, no interval of length $b - a$ is contained in $C_n = \ds \bigcup S_n$. 
Consequently, as $\CC \subseteq C_n$, no interval of length $b - a$ is contained in $\CC$.
Since the interval $\openint a b$ was of arbitrary length, there do not exist any open intervals in $\CC$.
Hence the result, by definition of nowhere dense.
{{qed}}
Theorem: Cantor Space is Meager in Closed Unit Interval

Let $T = \struct {\CC, \tau_d}$ be the Cantor space.
Then $T$ is meager in $\closedint 0 1$.

Proof:
From Cantor Space is Nowhere Dense, $T$ is nowhere dense in $\closedint 0 1$.
So, trivially, $\CC$ is the union of nowhere dense subsets of $\closedint 0 1$.
Hence the result from definition of meager.
{{qed}}
Theorem: Cantor Space is Non-Meager in Itself

Let $T = \struct {\CC, \tau_d}$ be the Cantor space.
Then $T$ is non-meager in itself.

Proof:
We have that the Cantor Space is Complete Metric Space.
By Baire Category Theorem, a complete metric space is also a Baire space.
The result then follows by Baire Space is Non-Meager.
{{qed}}
Theorem: Cantor Set is Uncountable

The Cantor set $\CC$ is uncountable.

Proof:
From the definition as a ternary representation, $\CC$ consists of all the elements of $\closedint 0 1$ which can be written without using a $1$.
So let $x \in \CC$. Then in base $3$ notation, we have (as $0 \le x \le 1$):
:$\ds x = \sum_{i \mathop = 1}^\infty r_j 3^{-j}$
From the definition of the Cantor set, we have $\forall j: r_j \in \set {0, 2}$.
Furthermore, from Representation of Ternary Expansions, the $r_j$ are unique.
Now define the following function:
:$\ds f: \CC \to \closedint 0 1, \quad \map f {\sum_{i \mathop = 1}^\infty r_j 3^{-j} } = \sum_{i \mathop = 1}^\infty \frac {r_j} 2 2^{-j}$
Observe that $\forall j: \dfrac {r_j} 2 \in \set {0, 1}$.
That the {{RHS}} expression is in fact an element of $\closedint 0 1$ now follows from binary notation.
Furthermore by Existence of Base-N Representation, any element $y$ of $\closedint 0 1$ may be written in the following form (where $\forall j: b_j \in \set {0, 1}$):
:$\ds y = \sum_{i \mathop = 1}^\infty b_j 2^{-j}$
Obviously, $y = \map f x$, where $x \in \CC$ is defined as follows:
:$\ds x = \sum_{i \mathop = 1}^\infty 2 b_j 3^{-j}$
It follows that $f$ is surjective.
From Closed Interval in Reals is Uncountable, the closed interval $\closedint 0 1$ is uncountable.
The result follows.
{{explain|it is obvious, but most likely there is some result that could be referred to}}
{{qed}}
Theorem: Cantor Set is Uncountable

The Cantor set $\CC$ is uncountable.

Proof:
It follows from Representation of Ternary Expansions that every string of the form $0.nnnnn \ldots$ where $n \in \set {0, 2}$ is an element of $\CC$.
We also have that every string of the form $0.nnnnn \ldots$ where $n \in \set {0, 1}$ is an element of $\closedint 0 1 \subset \R$ expressed in binary notation.
Let $f: \CC \to \closedint 0 1$ be the function defined by:
:$\forall x \in \CC: \map f x = \text{ the number obtained by replacing every } 2 \text { in } x \text { with a } 1$
where $x$ is expressed in base $3$ notation.
It is clear from the above that $f$ is a surjection.
{{qed}}
Theorem: Equality is Symmetric

:$\forall a, b: a = b \implies b = a$

Proof:
{{begin-eqn}}
{{eqn | l = a
      | r = b
      | c = 
}}
{{eqn | ll= \vdash
      | l = \map P a
      | o = \iff
      | r = \map P b
      | c = Leibniz's Law
}}
{{eqn | ll= \vdash
      | l = \map P b
      | o = \iff
      | r = \map P a
      | c = Biconditional is Commutative
}}
{{eqn | ll= \vdash
      | l = b
      | r = a
      | c = Leibniz's Law
}}
{{end-eqn}}
{{qed}}
Theorem: Derivative of Hyperbolic Sine Function

:$\map {\dfrac \d {\d x} } {\sinh u} = \cosh u \dfrac {\d u} {\d x}$

Proof:
{{begin-eqn}}
{{eqn | l = \map {\frac \d {\d x} } {\sinh u}
      | r = \map {\frac \d {\d u} } {\sinh u} \frac {\d u} {\d x}
      | c = Chain Rule for Derivatives
}}
{{eqn | r = \cosh u \frac {\d u} {\d x}
      | c = Derivative of Hyperbolic Sine
}}
{{end-eqn}}
{{qed}}
Theorem: Derivative of Hyperbolic Cosine Function

:$\map {\dfrac \d {\d x} } {\cosh u} = \sinh u \dfrac {\d u} {\d x}$

Proof:
{{begin-eqn}}
{{eqn | l = \map {\frac \d {\d x} } {\cosh u}
      | r = \map {\frac \d {\d u} } {\cosh u} \frac {\d u} {\d x}
      | c = Chain Rule for Derivatives
}}
{{eqn | r = \sinh u \frac {\d u} {\d x}
      | c = Derivative of Hyperbolic Cosine
}}
{{end-eqn}}
{{qed}}
Theorem: Cantor Space is Totally Separated

Let $T = \struct {\CC, \tau_d}$ be the Cantor space.
Then $T$ is totally separated.

Proof:
Let $a, b \in \CC$ such that $a < b$.
Then $b - a = \epsilon$.
Consider $n \in \N$ such that $3^{-n} < \epsilon$.
{{wtd|To be proved that between $a + 3^{-\paren {n + 1} }$ and $a + 2 \times 3^{-\paren {n + 1} }$ there exists an interval $I$ which is excluded from $\CC$. Or something like that.<br/>Then we pick some $r \in I$ and continue as follows.}}
So $\exists r \in \R: a < r < b, r \notin \CC$.
Let $A = \CC \cap \hointr 0 r$ and $B = \CC \cap \hointl r 1$.
Thus $A \mid B$ is a separation of $\CC$ such that $a \in A, b \in B$.
Such a separation can be found for any two distinct $a, b \in \CC$.
Hence the result by definition of totally separated.
{{qed}}
Theorem: Cantor Space is not Extremally Disconnected

Let $T = \struct {\CC, \tau_d}$ be the Cantor space.
Then $T$ is not extremally disconnected.

Proof:
{{Recall|Extremally Disconnected Space|extremally disconnected space}}
{{:Definition:Extremally Disconnected Space/Definition 3}}
Consider the real number $\dfrac 1 4 = 0.020202 \ldots_3$.
We have that:
:$C_1 := \CC \cap \hointr 0 {\dfrac 1 4}$
:$C_2 := \CC \cap \hointl {\dfrac 1 4} 1$
are disjoint sets both of which are open sets of $T$.
However, as $\dfrac 1 4 \in \CC$ we have that:
:$\dfrac 1 4 \in C_1^-$
and
:$\dfrac 1 4 \in C_2^-$
where $C_1^-$ and $C_2^-$ are the closures of $C_1$ and $C_2$ in $T$.
Thus $C_1^- \cap C_2^- \ne \O$.
The result follows by definition.
{{qed}}
Theorem: Cantor Space as Countably Infinite Product

Let $A_n = \struct {\set {0, 2}, \tau_n}$ be the discrete space of the two points $0$ and $2$.
Let $\ds A = \prod_{n \mathop = 1}^\infty A_n$.
Let $\struct {A, \tau}$ be the product space where $\tau$ is the product topology on $A$.
Then $A$ is homeomorphic to the Cantor space.

Proof:
In $\CC$, basis elements are sets of the form $\set {y: \size {x - y} < \epsilon}$ for $x \in \CC$ and some $\epsilon \in \R_{>0}$.
In $\ds \prod_{n \mathop = 1}^\infty A_n$, sets of the form $\set {\sequence {a_i} \in \prod A_n: a_i \text { is fixed for } 1 \le i \le n}$ forms a basis for the product topology.
Consider the function $f$ taking each point from $\tuple {a_1, a_2, \ldots, a_n, \ldots}$ in $\prod A_n$ to the point $0 \cdotp a_1 a_2 \ldots a_n \ldots_3$ in $\CC$.
As $f$ takes basis elements to basis elements, both $f$ and $f^{-1}$ are seen to be continuous.
{{qed}}
Theorem: Cantor Space as Countably Infinite Product

Let $A_n = \struct {\set {0, 2}, \tau_n}$ be the discrete space of the two points $0$ and $2$.
Let $\ds A = \prod_{n \mathop = 1}^\infty A_n$.
Let $\struct {A, \tau}$ be the product space where $\tau$ is the product topology on $A$.
Then $A$ is homeomorphic to the Cantor space.

Proof:
Since $\CC$ is a metric space, $\CC$ is Hausdorff. By Tychonoff's Theorem, $\ds A = \prod_{n \mathop = 1}^\infty A_n$ is compact.
Consider the function $f$ taking each point from $\tuple {a_1, a_2, \ldots, a_n, \ldots}$ in $\prod A_n$ to the point $0 \cdotp a_1 a_2 \ldots a_n \ldots_3$ in $\CC$.
$f$ is a bijection.
In $\CC$, the basis elements are sets of the form $\ds B_{N,x} = \set {y \in \CC: \size {x - y} < 3^{-N}}$, where $N \in \N_{>0}$ and $x \in \CC$.
Let $\ds x=\sum_{j \mathop = 1}^\infty a_j 3^{-j}$
where $\forall j: a_j \in \set{0,2}$
$\ds f^{-1} \sqbrk {B_{N,x}} = \set{\tuple {b_1, b_2, \ldots, b_n, \ldots} \in A: \size {\sum_{j \mathop = 1}^\infty (a_j-b_j) 3^{-j}} < 3^{-N}  }$
The equation $\ds \size {\sum_{j \mathop = 1}^\infty (a_j-b_j) 3^{-j}} < 3^{-N}$, where $\forall j: a_j, b_j \in \set{0,2}$ holds true if and only if $a_j = b_j$ for all $j \in \set{1,2, \ldots, N}$ and it is not the case $a_j-b_j = 2$ for all $ j >N$ and it is not the case $b_j-a_j =2$ for all $ j >N$.
If $\ds x=\sum_{j \mathop = 1}^N a_j 3^{-j}$,
$\ds f^{-1} \sqbrk {B_{N,x}} = \set{\tuple {b_1, b_2, \ldots,b_N, b_{N+1}, \ldots} \in A: a_j = b_j,  \forall j \in \set{1,2,...,N} } \setminus \set{\tuple {a_1, a_2, \ldots, a_N, 2,2,2, \ldots} }$
If $\ds x=3^{-N} + \sum_{j \mathop = 1}^N a_j 3^{-j}$,
$\ds f^{-1} \sqbrk {B_{N,x}} = \set{\tuple {b_1, b_2, \ldots,b_N, b_{N+1}, \ldots} \in A: a_j = b_j,  \forall j \in \set{1,2,...,N} } \setminus \set{\tuple {a_1, a_2, \ldots, a_N, 0,0,0, \ldots} }$
Else,
$\ds f^{-1} \sqbrk {B_{N,x}} = \set{\tuple {b_1, b_2, \ldots,b_N, b_{N+1}, \ldots} \in A: a_j = b_j,  \forall j \in \set{1,2,...,N} } $
In all cases, $f^{-1} \sqbrk {B_{N,x}}$ is open since from Product Space is T2 iff Factor Spaces are T2, singleton sets are closed.
{{explain|Why does that follow?}}
Thus $f$ is continuous.
Since $f$ is continuous, it follows from Continuous Bijection from Compact to Hausdorff is Homeomorphism that $f$ is a homeomorphism.
{{qed}}
Theorem: Cantor Space is not Locally Connected

Let $T = \struct {\CC, \tau_d}$ be the Cantor space.
Then $T$ is not locally connected.

Proof:
Let $\BB$ be a basis of $T$.
Let $A \in \BB$.
By definition of $\BB$, $A$ is an open set of $T$.
But the Cantor Space is Totally Separated.
Therefore $A$ is not a connected set.
Hence the result from definition of a locally connected space.
{{qed}}
Theorem: Local Connectedness is not Preserved under Infinite Product

The property of local connectedness is not preserved under the operation of forming an infinite product space.

Proof:
Let $T = \struct {\CC, \tau_d}$ be the Cantor space.
Let $A_n = \struct {\set {0, 2}, \tau_n}$ be the discrete space of the two points $0$ and $2$.
Let $\ds A = \prod_{n \mathop = 1}^\infty A_n$.
Let $T' = \struct {A, \tau}$ be the product space where $\tau$ is the product topology on $A$.
From Cantor Space as Countably Infinite Product, $T'$ is homeomorphic to $T$.
From Totally Disconnected and Locally Connected Space is Discrete, we have that $A_n$ is locally connected.
But we also have that the Cantor Space is not Locally Connected.
Hence the result.
{{qed}}
Theorem: Derivative of Hyperbolic Tangent Function

:$\map {\dfrac \d {\d x} } {\tanh u} = \sech^2 u \dfrac {\d u} {\d x}$

Proof:
{{begin-eqn}}
{{eqn | l = \map {\frac \d {\d x} } {\tanh u}
      | r = \map {\frac \d {\d u} } {\tanh u} \frac {\d u} {\d x}
      | c = Chain Rule for Derivatives
}}
{{eqn | r = \sech^2 u \frac {\d u} {\d x}
      | c = Derivative of Hyperbolic Tangent
}}
{{end-eqn}}
{{qed}}
Theorem: Difference of Squares of Hyperbolic Cosine and Sine

:$\cosh^2 x - \sinh^2 x = 1$
where $\cosh$ and $\sinh$ are hyperbolic cosine and hyperbolic sine.

Proof:
{{begin-eqn}}
{{eqn | l = \cosh^2 x - \sinh^2 x
      | r = \paren {\frac {e^x + e^{-x} } 2}^2 - \paren {\frac {e^x - e^{-x} } 2}^2  
      | c = {{Defof|Hyperbolic Cosine}} and {{Defof|Hyperbolic Sine}}
}}
{{eqn | r = \paren {\frac {\paren {e^x}^2 + 2 \paren {e^x} \paren {e^{-x} } + \paren {e^{-x} }^2} 4} - \paren {\frac {\paren {e^x}^2 - 2 \paren {e^x} \paren {e^{-x} } + \paren {e^{-x} }^2} 4}  
      | c = Square of Sum
}}
{{eqn | r = \paren {\frac {e^{2 x} + 2 + e^{-2 x} } 4} - \paren {\frac {e^{2 x} - 2 + e^{-2 x} } 4}  
      | c = Exponential of Sum
}}
{{eqn | r = \frac {e^{2 x} - e^{2 x} + e^{-2 x} - e^{-2 x} + 2 + 2} 4
}}
{{eqn | r = 1
}}
{{end-eqn}}
{{qed}}
Theorem: Derivative of Arccosecant Function

:<nowiki>$\dfrac {\map \d {\arccsc x} } {\d x} = \dfrac {-1} {\size x \sqrt {x^2 - 1} } = \begin {cases} \dfrac {-1} {x \sqrt {x^2 - 1} } & : 0 < \arccsc x < \dfrac \pi 2 \ (\text {that is: $x > 1$}) \\
\dfrac {+1} {x \sqrt {x^2 - 1} } & : -\dfrac \pi 2 < \arccsc x < 0 \ (\text {that is: $x < -1$})  \\
\end{cases}$</nowiki>

Proof:
{{:Graph of Arccosecant Function}}
Let $y = \arccsc x$ where  $x < -1$ or $x > 1$.
Then:
{{begin-eqn}}
{{eqn | l = y
      | r = \arccsc x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | r = \csc y
      | c = where $y \in \closedint 0 \pi \land y \ne \dfrac pi 2$
}}
{{eqn | ll= \leadsto
      | l = \frac {\d x} {\d y}
      | r = -\csc y \cot y
      | c = Derivative of Cosecant Function
}}
{{eqn | ll= \leadsto
      | l = \frac {\d y} {\d x}
      | r = \frac {-1} {\csc y \cot y}
      | c = Derivative of Inverse Function
}}
{{eqn | ll= \leadsto
      | l = \paren {\frac {\d y} {\d x} }^2
      | r = \frac 1 {\csc^2 y \cot^2 y}
      | c = squaring both sides
}}
{{eqn | r = \frac 1 {\csc^2 y \paren {\csc^2 y - 1} }
      | c = Difference of Squares of Cosecant and Cotangent
}}
{{eqn | r = \frac 1 {x^2 \paren {x^2 - 1} }
      | c = Definition of $x$
}}
{{eqn | ll= \leadsto
      | l = \size {\dfrac {\d y} {\d x} }
      | r = \dfrac 1 {\size x \sqrt {x^2 - 1} }
      | c = squaring both sides
}}
{{end-eqn}}
Since $\dfrac {\d y} {\d x} = \dfrac {-1} {\csc y \cot y}$, the sign of $\dfrac {\d y} {\d x}$ is opposite to the sign of $\csc y \cot y$.
Writing $\csc y \cot y$ as $\dfrac {\cos y} {\sin^2 y}$, it is evident that the sign of $\dfrac {\d y} {\d x}$ is opposite to the sign of $\cos y$.
From Sine and Cosine are Periodic on Reals, $\cos y$ is never negative on its domain ($y \in \closedint {-\dfrac \pi 2} {\dfrac \pi 2} \land y \ne 0$).
However, by definition of the arccosecant of $x$:
:$0 < \arccsc x < \dfrac \pi 2 \implies x > 1$
:$-\dfrac \pi 2 < \arccsc x < 0 \implies x < -1$
Thus:
:<nowiki>$\dfrac {\map \d {\arccsc x} } {\d x} = \dfrac {-1} {\size x \sqrt {x^2 - 1} } = \begin {cases} \dfrac {-1} {x \sqrt {x^2 - 1} } & : 0 < \arccsc x < \dfrac \pi 2 \ (\text {that is: $x > 1$}) \\
\dfrac {+1} {x \sqrt {x^2 - 1} } & : -\dfrac \pi 2 < \arccsc x < 0 \ (\text {that is: $x < -1$}) \\
\end{cases}$</nowiki>
{{qed}}
Theorem: Derivative of Hyperbolic Secant Function

:$\map {\dfrac \d {\d x} } {\sech u} = -\sech u \tanh u \dfrac {\d u} {\d x}$

Proof:
{{begin-eqn}}
{{eqn | l = \map {\frac \d {\d x} } {\sech u}
      | r = \map {\frac \d {\d u} } {\sech u} \frac {\d u} {\d x}
      | c = Chain Rule for Derivatives
}}
{{eqn | r = -\sech u \tanh u \frac {\d u} {\d x}
      | c = Derivative of Hyperbolic Secant
}}
{{end-eqn}}
{{qed}}
Theorem: Power of Sum Modulo Prime

Let $p$ be a prime number.
Then:
:$\paren {a + b}^p \equiv a^p + b^p \pmod p$

Proof:
From the Binomial Theorem:
:$\ds \paren {a + b}^p = \sum_{k \mathop = 0}^p \binom p k a^k b^{p - k}$
Also note that:
:$\ds \sum_{k \mathop = 0}^p \binom p k a^k b^{p-k} = a^p + \sum_{k \mathop = 1}^{p - 1} \binom p k a^k b^{p - k} + b^p$
So:
{{begin-eqn}}
{{eqn | q = \forall k: 0 < k < p
      | l = \binom p k
      | o = \equiv
      | r = 0
      | rr= \pmod p
      | c = Binomial Coefficient of Prime
}}
{{eqn | ll= \leadsto
      | l = \binom p k a^k b^{p - k}
      | o = \equiv
      | r = 0
      | rr= \pmod p
      | c = {{Defof|Modulo Multiplication}}
}}
{{eqn | ll= \leadsto
      | l = \sum_{k \mathop = 1}^{p - 1} \binom p k a^k b^{p - k}
      | o = \equiv
      | r = 0
      | rr= \pmod p
      | c = {{Defof|Modulo Addition}}
}}
{{eqn | ll= \leadsto
      | l = \sum_{k \mathop = 0}^p \binom p k a^k b^{p - k}
      | o = \equiv
      | r = a^p + b^p
      | rr= \pmod p
      | c = from above
}}
{{end-eqn}}
{{qed}}
Theorem: Volume of Right Circular Cone

The volume $V$ of a right circular cone is given by:
:$V = \dfrac 1 3 \pi r^2 h$
where:
:$r$ is the radius of the base
:$h$ is the height of the cone, that is, the distance between the apex and the center of the base.

Proof:
This proof utilizes the Method of Disks and thus is dependent on Volume of Right Circular Cylinder.
From the Method of Disks, the volume of the cone can be found by the definite integral:
:$\ds (1): \quad V = \pi \int_0^{AC} \paren {\map R x}^2 \rd x$
where $\map R x$ is the function describing the line which is to be rotated about the $x$-axis in order to create the required solid of revolution.
In this example, $\map R x$ describes the line segment $\overline {AB}$, and so:
:$\map R x = \dfrac r h x$
We have also defined:
:$\overline {AC}$ as the axis of the cone, whose length is $h$
:$A$ as the origin.
So the equation $(1)$ is interpreted as:
{{begin-eqn}}
{{eqn | l = V
      | r = \pi \int_0^h \paren {\frac r h x}^2 \rd x
      | c = 
}}
{{eqn | r = \intlimits {\pi \paren {\frac r h}^2 \frac {x^3} 3} {x \mathop = 0} {x \mathop = h}
      | c = Linear Combination of Definite Integrals, Integral of Power
}}
{{eqn | r = \frac 1 3 \pi r^2 h
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Intersection of Topologies is Topology

Let $\family {\tau_i}_{i \mathop \in I}$ be an arbitrary indexed family of topologies on a set $S$.
Then $\tau := \ds \bigcap_{i \mathop \in I} \tau_i$ is also a topology on $S$.

Proof:
Each of the open set axioms are examined in turn:
Theorem: Power Set is Closed under Set Difference

Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Then:
:$\forall A, B \in \powerset S: A \setminus B \in \powerset S$
where $A \setminus B$ denotes the set difference of $A$ and $B$.

Proof:
Let $A, B \in \powerset S$.
Then by the definition of power set, $A \subseteq S$ and $B \subseteq S$.
We also have $A \setminus B \subseteq A$ from Set Difference is Subset.
Thus by Subset Relation is Transitive, $A \setminus B \subseteq S$.
Thus $A \setminus B \in \powerset S$, and closure is proved.
{{qed}}