avewright/proofwiki-math huggingface dataset:
Theorem: Precisely One Function in terms of And, Or and Not

Let $\map P {A, B, C}$ denote the precisely one function on the statements $A$, $B$ and $C$.
Then:
:$\map P {A, B, C} \dashv \vdash \paren {A \land \neg B \land \neg C} \lor \paren {\neg A \land B \land \neg C} \lor \paren {\neg A \land \neg B \land C}$
where:
:$\land$ denotes conjunction
:$\lor$ denotes disjunction
:$\neg$ denotes negation

Proof:
We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|cccc||ccccc|} \hline
P & (A & B & C) & (((A & \land & \neg & B) & \land & \neg & C) & \lor & ((\neg & A & \land & B) & \land & \neg & C)) & \lor & ((\neg & A & \land & \neg & B) & \land & C) \\
\hline
F & F & F & F & F & F & T & F & F & T & F & F & T & F & F & F & F & T & F & F & T & F & T & T & F & F & F \\
T & F & F & T & F & F & T & F & F & F & T & F & T & F & F & F & F & F & T & T & T & F & T & T & F & T & T \\
T & F & T & F & F & F & F & T & F & T & F & T & T & F & T & T & T & T & F & T & T & F & F & F & T & F & F \\
F & F & T & T & F & F & F & T & F & F & T & F & T & F & T & T & F & F & T & F & T & F & F & F & T & F & T \\
T & T & F & F & T & T & T & F & T & T & F & T & F & T & F & F & F & T & F & T & F & T & F & T & F & F & F \\
F & T & F & T & T & T & T & F & F & F & T & F & F & T & F & F & F & F & T & F & F & T & F & T & F & F & T \\
F & T & T & F & T & F & F & T & F & T & F & F & F & T & F & T & F & T & F & F & F & T & F & F & T & F & F \\
F & T & T & T & T & F & F & T & F & F & T & F & F & T & F & T & F & F & T & F & F & T & F & F & T & F & T \\
\hline
\end{array}$
{{qed}}
Theorem: Lévy's Continuity Theorem

Let $\sequence {X_n}_{n \mathop \ge 1}$ be a sequence of discrete random variables with characteristic functions $\map {\phi_n} t := E \sqbrk {e^{i t X_n} }$.
Let the sequence $\sequence {\phi_n}$ converge to some real function $\phi$:
:$\forall t \in \R: \map {\phi_n} t \to \map \phi t$.
Then the following statements are equivalent:
:$(1): \quad$ The $\sequence {X_n}$ converges in distribution to some random variable $X$:
::$X_n \stackrel {dist} {\to} X$ with characteristic function $\map {\phi_X} t := \map \phi t$
:$(2): \quad$ The sequence $\sequence {X_n}$ is tight, that is:
::$\ds \lim_{M \mathop \to \infty} \sup_{n \mathop \ge 1} P \sqbrk {\size {X_n} \ge M} = 0$

Proof:
{{ProofWanted}}
{{Namedfor|Paul Pierre Lévy|cat = Lévy}}
Theorem: Composition of Three Mappings which form Identity Mapping

Let $A$, $B$ and $C$ be non-empty sets.
Let $f: A \to B$, $g: B \to C$ and $h: C \to A$ be mappings.
Let the following hold:
{{begin-eqn}}
{{eqn | l = h \circ g \circ f
      | r = I_A
}}
{{eqn | l = f \circ h \circ g
      | r = I_B
}}
{{eqn | l = g \circ f \circ h
      | r = I_C
}}
{{end-eqn}}
where:
:$g \circ f$ (and so on) denote composition of mappings
:$I_A$ (and so on) denote the identity mappings.
Then each of $f$, $g$ and $h$ are bijections, and:
{{begin-eqn}}
{{eqn | l = f^{-1}
      | r = h \circ g
}}
{{eqn | l = g^{-1}
      | r = f \circ h
}}
{{eqn | l = h^{-1}
      | r = g \circ f
}}
{{end-eqn}}
where $f^{-1}$ (and so on) denote the inverse mappings.

Proof:
First note that from Composition of Mappings is Associative:
:$\paren {h \circ g} \circ f = h \circ \paren {g \circ f}$
and so on.
However, while there is no need to use parenthesis to establish the order of composition of mappings, in the following the technique will be used in order to clarify what is being done.
We have that:
:$\paren {h \circ g} \circ f = I_A$
It follows from Injection iff Left Inverse that $f$ is an injection.
We also have that:
:$f \circ \paren {h \circ g} = I_B$
It follows from Surjection iff Right Inverse that $f$ is a surjection.
So $f$ is a bijection.
It follows from {{Corollary|Bijection iff Left and Right Inverse}} that $h \circ g$ is also a bijection.
Thus we have that $h \circ g$ is both a left inverse and a right inverse of $f$.
It follows by definition that $h \circ g$ is the inverse of $f$:
:$f^{-1} = h \circ g$
The same argument, {{Mutatis}}, can be used to show that:
:$g$ and $h$ are bijections
:the inverse of $g$ is $f \circ h$
:the inverse of $h$ is $g \circ f$.
{{qed}}
Theorem: Composition of Product Mappings on Natural Numbers

Let $a \in \N$ be a natural number.
Let $\mu_a: \N \to \N$ be the mapping defined as:
:$\forall x \in \N: \map {\mu_a} x = x a$
Then:
:$\mu_{a b} = \mu_b \circ \mu_a$

Proof:
{{begin-eqn}}
{{eqn | l = \mu_{a b}
      | r = x \paren {a b}
      | c = Definition of $\mu$
}}
{{eqn | r = \paren {x a} b
      | c = 
}}
{{eqn | r = \paren {\map {\mu_a} x} b
      | c = Definition of $\mu$
}}
{{eqn | r = \map {\mu_b} {\map {\mu_a} x}
      | c = Definition of $\mu$
}}
{{eqn | r = \map {\paren {\mu_b \circ \mu_a} } x
      | c = {{Defof|Composition of Mappings}}
}}
{{end-eqn}}
{{qed}}
Theorem: Linear Combination of Normal Random Variables

Let $X_1, X_2, X_3, \ldots, X_n$ be independent random variables. 
Let $\sequence {\alpha_i}_{1 \mathop \le i \mathop \le n}$ and $\sequence {\mu_i}_{1 \mathop \le i \mathop \le n}$ be sequences of real numbers.
Let $\sequence {\sigma_i}_{1 \mathop \le i \mathop \le n}$ be a sequence of positive real numbers.
Let $X_i \sim \Gaussian {\mu_i} {\sigma^2_i}$ for $1 \le i \le n$, where $\Gaussian {\mu_i} {\sigma^2_i}$ is the normal distribution with parameters $\mu_i$ and $\sigma^2_i$.
Then: 
:$\ds \sum_{i \mathop = 1}^n \alpha_i X_i \sim \Gaussian {\sum_{i \mathop = 1}^n \alpha_i \mu_i} {\sum_{i \mathop = 1}^n \alpha^2_i \sigma^2_i}$

Proof:
Let: 
:$\ds Z = \sum_{i \mathop = 1}^n \alpha_i X_i$ 
Let $M_Z$ be the moment generating function of $Z$. 
We aim to show that: 
:$\ds Z \sim \Gaussian {\sum_{i \mathop = 1}^n \alpha_i \mu_i} {\sum_{i \mathop = 1}^n \alpha^2_i \sigma^2_i}$
By Moment Generating Function of Normal Distribution and Moment Generating Function is Unique, it is therefore sufficient to show that: 
:$\ds \map {M_Z} t = \map \exp {\paren {\sum_{i \mathop = 1}^n \alpha_i \mu_i} t + \frac 1 2 \paren {\sum_{i \mathop = 1}^n \alpha^2_i \sigma^2_i} t^2}$
We also have, by Moment Generating Function of Normal Distribution, that the moment generating function of $X_i$, $M_{X_i}$, is given by: 
:$\map {M_{X_i}} t = \map \exp {\mu_i t + \dfrac 1 2 \sigma^2_i t^2}$
We have:
{{begin-eqn}}
{{eqn | l = \map {M_Z} t
      | r = \prod_{i \mathop = 1}^n \map {M_{X_i} } {\alpha_i t}
      | c = Moment Generating Function of Linear Combination of Independent Random Variables
}}
{{eqn | r = \prod_{i \mathop = 1}^n \map \exp {\mu_i \alpha_i t + \frac 1 2 \sigma^2_i \alpha^2_i t^2}
}}
{{eqn | r = \map \exp {\sum_{i \mathop = 1}^n \paren {\mu_i \alpha_i t + \frac 1 2 \sigma^2_i \alpha^2_i t^2} }
      | c = Exponential of Sum
}}
{{eqn | r = \map \exp {\paren {\sum_{i \mathop = 1}^n \alpha_i \mu_i} t + \frac 1 2 \paren {\sum_{i \mathop = 1}^n \alpha^2_i \sigma^2_i} t^2}
}}
{{end-eqn}}
{{qed}}
Category:Normal Distribution
tpdmj8pxjk5gd73n4c2zgy6p7s4a448
Theorem: Moment Generating Function of Linear Transformation of Random Variable

Let $X$ be a random variable. 
Let $\alpha$ and $\beta$ be real numbers.
Let $Z = \alpha X + \beta$.
Let $M_X$ be the moment generating function of $X$. 
Then the moment generating function of $Z$, $M_Z$, is given by: 
:$\map {M_Z} t = e^{\beta t} \map {M_X} {\alpha t}$

Proof:
{{begin-eqn}}
{{eqn | l = \map {M_Z} t
      | r = \expect {\map \exp {t Z} }
      | c = {{Defof|Moment Generating Function}}
}}
{{eqn | r = \expect {\map \exp {t \paren {\alpha X + \beta} } }
}}
{{eqn | r = \expect {\map \exp {\paren {\alpha t} X} \map \exp {\beta t} }
      | c = Exponential of Sum
}}
{{eqn | r = \map \exp {\beta t} \expect {\map \exp {\paren {\alpha t} X} }
      | c = Expectation is Linear
}}
{{eqn | r = e^{\beta t} \map {M_X} {\alpha t}
      | c = {{Defof|Moment Generating Function}}
}}
{{end-eqn}}
{{qed}}
Category:Moment Generating Functions
icxg9cpjlc0qq0jl0et2h72d795nd8f
Theorem: Linear Transformation of Normal Random Variable

Let $\mu$, $\alpha$ and $\beta$ be real numbers. 
Let $\sigma$ be a positive real number. 
Let $X \sim \Gaussian \mu {\sigma^2}$ where $\Gaussian \mu {\sigma^2}$ is the normal distribution with parameters $\mu$ and $\sigma^2$. 
Then: 
:$\alpha X + \beta \sim \Gaussian {\alpha \mu + \beta} {\alpha^2 \sigma^2}$

Proof:
Let $Z = \alpha X + \beta$. 
Let $M_Z$ be the moment generating function of $Z$. 
We aim to show that: 
:$Z \sim \Gaussian {\alpha \mu + \beta} {\alpha^2 \sigma^2}$
By Moment Generating Function of Normal Distribution and Moment Generating Function is Unique, it is sufficient to show that:
:$\map {M_Z} t = \map \exp {\paren {\alpha \mu + \beta} t + \dfrac 1 2 \alpha^2 \sigma^2 t^2}$
We also have, by Moment Generating Function of Normal Distribution, that the moment generating function of $X$, $M_X$, is given by: 
:$\map {M_X} t = \map \exp {\mu t + \dfrac 1 2 \sigma^2 t^2}$
We have:
{{begin-eqn}}
{{eqn | l = \map {M_Z} t
      | r = e^{\beta t} \map {M_X} {\alpha t}
      | c = Moment Generating Function of Linear Combination of Independent Random Variables
}}
{{eqn | r = e^{\beta t} \map \exp {\alpha \mu t + \frac 1 2 \sigma^2 \paren {\alpha t}^2}
}}
{{eqn | r = \map \exp {\paren {\alpha \mu + \beta} t + \frac 1 2 \sigma^2 \alpha^2 t^2}
      | c = Exponential of Sum
}}
{{end-eqn}}
{{qed}}
Category:Normal Distribution
5atpcina4lnlm8r3ovfhrifrhrqlmoq
Theorem: Standard Normal Random Variable as Transformation of Normal Random Variable

Let $\mu$ be a real number. 
Let $\sigma$ be a positive real number. 
Let $X \sim \Gaussian \mu {\sigma^2}$ where $\Gaussian \mu {\sigma^2}$ is the normal distribution with parameters $\mu$ and $\sigma^2$. 
Then: 
:$\dfrac {X - \mu} \sigma \sim \Gaussian 0 1$
where $\Gaussian 0 1$ is the standard normal distribution.

Proof:
{{begin-eqn}}
{{eqn | l = \frac {X - \mu} \sigma
      | r = \frac 1 \sigma X - \frac \mu \sigma
}}
{{eqn | o = \sim
      | r = \Gaussian {\frac \mu \sigma - \frac \mu \sigma} {\paren {\frac 1 \sigma}^2 \sigma^2}
      | c = Linear Transformation of Normal Random Variable
}}
{{eqn | r = \Gaussian 0 1
}}
{{end-eqn}}
{{qed}}
Category:Normal Distribution
208kud8zlgxszdmo2rkcax8lw7jqxvo
Theorem: Composition of Addition Mappings on Natural Numbers

Let $a \in \N$ be a natural number.
Let $\alpha_a: \N \to \N$ be the mapping defined as:
:$\forall x \in \N: \map {\alpha_a} x = x + a$
Then:
:$\alpha_{a + b} = \alpha_b \circ \alpha_a$

Proof:
{{begin-eqn}}
{{eqn | l = \alpha_{a + b}
      | r = x + \paren {a + b}
      | c = Definition of $\alpha$
}}
{{eqn | r = \paren {x + a} + b
      | c = 
}}
{{eqn | r = \paren {\map {\alpha_a} x} + b
      | c = Definition of $\alpha$
}}
{{eqn | r = \map {\alpha_b} {\map {\alpha_a} x}
      | c = Definition of $\alpha$
}}
{{eqn | r = \map {\paren {\alpha_b \circ \alpha_a} } x
      | c = {{Defof|Composition of Mappings}}
}}
{{end-eqn}}
{{qed}}
Theorem: Reciprocal of Random Variable with Snedecor's F-Distribution has Snedecor's F-Distribution

Let $n, m$ be strictly positive integers. 
Let $X \sim F_{n, m}$ where $F_{n, m}$ is the Snedecor's $F$-distribution with $\tuple {n, m}$ degrees of freedom.
Then: 
:$\dfrac 1 X \sim F_{m, n}$

Proof:
Let $Y \sim \chi^2_n$ where $\chi^2_n$ is the chi-squared distribution with $n$ degrees of freedom. 
Let $Z \sim \chi^2_m$ where $\chi^2_m$ is the chi-squared distribution with $m$ degrees of freedom.
Then, by Quotient of Independent Random Variables with Chi-Squared Distribution Divided by Degrees of Freedom has Snedecor's F-Distribution:
:$X = \dfrac {Y / n} {Z / m} \sim F_{n, m}$
Applying Quotient of Independent Random Variables with Chi-Squared Distribution Divided by Degrees of Freedom has Snedecor's F-Distribution again, we have: 
:$\dfrac 1 X = \dfrac {Z / m} {Y / n} \sim F_{m, n}$
{{qed}}
Category:Snedecor's F-Distribution
plfn5kg734azb28xyu5fsmaftdl1jgk
Theorem: Quotient of Independent Random Variables with Chi-Squared Distribution Divided by Degrees of Freedom has Snedecor's F-Distribution

Let $n$ and $m$ be strictly positive integers.
Let $X$ and $Y$ be independent random variables. 
Let $X \sim \chi_n^2$ where $\chi_n^2$ is the chi-squared distribution with $n$ degrees of freedom. 
Let $Y \sim \chi_m^2$ where $\chi_m^2$ is the chi-squared distribution with $m$ degrees of freedom.
Then: 
:$\dfrac {X / n} {Y / m} \sim F_{n, m}$
where $F_{n, m}$ is Snedecor's $F$-distribution with $\tuple {n, m}$ degrees of freedom.

Proof:
{{ProofWanted}}
Category:Chi-Squared Distribution
Category:Snedecor's F-Distribution
8gywnfks3vlnwzzbkg6kpmj4thftam6
Theorem: Square of Random Variable with t-Distribution has Snedecor's F-Distribution

Let $k$ be a strictly positive integer. 
Let $X \sim t_k$ where $t_k$ is the $t$-distribution with $k$ degrees of freedom.
Then: 
:$X^2 \sim F_{1, k}$
where $F_{1, k}$ is Snedecor's $F$-distribution with $\tuple {1, k}$ degrees of freedom.

Proof:
Let $Y \sim F_{1, k}$. 
We aim to show that: 
:$\map \Pr {Y < x^2} = \map \Pr {\size X < x}$
for all $x \ge 0$. 
That is: 
:$\map \Pr {Y < x^2} = \map \Pr {-x < X < x}$
for all $x \ge 0$. 
We have: 
{{begin-eqn}}
{{eqn | l = \map \Pr {Y < x^2}
      | r = \int_0^{x^2} \frac {k^{k/2} 1^{1/2} u^{\paren {1/2} - 1} } {\paren {k + u}^{\paren {1 + k}/2} \map \Beta {1/2, k/2} } \rd u
      | c = {{Defof|Snedecor's F-Distribution|Snedecor's $F$-Distribution}}
}}
{{eqn | r = \frac {k^{k/2} } {k^{k/2} \sqrt k \map \Beta {1/2, k/2} } \int_0^{x^2} \frac 1 {\sqrt u} \paren {1 + \frac u k}^{-\paren {1 + k}/2} \rd u
      | c = extracting a factor of $k^{-\paren {1 + k}/2}$, rewriting
}}
{{eqn | r = \frac {\map \Gamma {\frac {k + 1} 2} } {\sqrt k \map \Gamma {\frac 1 2} \map \Gamma {\frac k 2} } \int_0^{x^2} \frac 1 {\sqrt u} \paren {1 + \frac u k}^{-\paren {1 + k}/2} \rd u
      | c = {{Defof|Beta Function}}
}}
{{eqn | r = \frac {\map \Gamma {\frac {k + 1} 2} } {\sqrt {\pi k} \map \Gamma {\frac k 2} } \int_0^{x^2} \frac 1 {\sqrt u} \paren {1 + \frac u k}^{-\paren {1 + k}/2} \rd u
      | c = Gamma Function of One Half
}}
{{eqn | r = \frac {\map \Gamma {\frac {k + 1} 2} } {\sqrt {\pi k} \map \Gamma {\frac k 2} } \int_0^x \frac {2 v} {\sqrt {v^2} } \paren {1 + \frac {v^2} k}^{-\paren {1 + k}/2} \rd v
      | c = substituting $u = v^2$
}}
{{eqn | r = \frac {\map \Gamma {\frac {k + 1} 2} } {\sqrt {\pi k} \map \Gamma {\frac k 2} } \int_0^x 2 \paren {1 + \frac {v^2} k}^{-\paren {1 + k}/2} \rd v
}}
{{eqn | r = \frac {\map \Gamma {\frac {k + 1} 2} } {\sqrt {\pi k} \map \Gamma {\frac k 2} } \int_{-x}^x \paren {1 + \frac {v^2} k}^{-\paren {1 + k}/2} \rd v
      | c = Definite Integral of Even Function
}}
{{eqn | r = \map \Pr {-x < X < x}
      | c = {{Defof|Student's t-Distribution}}
}}
{{end-eqn}}
{{qed}}
Theorem: Union of Connected Sets with Common Point is Connected

Let $T = \struct {S, \tau}$ be a topological space.
Let $\family {B_\alpha}_{\alpha \mathop \in A}$ be a family of connected sets of $T$.
Let $\exists x \in \ds \bigcap \family {B_\alpha}_{\alpha \mathop \in A}$.
Then
:$\ds \bigcup \family {B_\alpha}_{\alpha \mathop \in A}$ is a connected set of $T$.

Proof:
Let $B = \ds \bigcup_{\alpha \mathop \in A} B_\alpha$.
{{Recall|Connected Set (Topology)|connected set|index = 3}}
{{:Definition:Connected Set (Topology)/Definition 3}}
Thus $B$ is connected in $T$ {{iff}} the subspace $\struct {B, \tau_B}$ is a connected space.
{{Recall|Connected Topological Space|connected space|index = 4}}
{{:Definition:Connected Topological Space/Definition 4}}
Hence $\struct {B, \tau_B}$ is connected {{iff}} the only clopen sets in $\struct {B, \tau_B}$ are $B$ and $\O$.
Let $U$ be any clopen set of the subspace of $\struct {B, \tau_B}$.
Let $V = B \setminus U$.
From Complement of Clopen Set is Clopen, $V$ is also clopen.
Hence $U, V$ are disjoint clopen sets such that $B = U \cup V$. 
{{WLOG}} assume $x \in U$.
From Set is Subset of Union:
:$\ds \forall \alpha \in A : B_\alpha \subseteq B = U \cup V$
From Connected Subset of Union of Disjoint Open Sets:
:$\forall \alpha \in A : B_\alpha \subseteq U$
From Union is Smallest Superset:
:$B \subseteq U$
Since $U \subseteq B$ then $U = B$.
From Set Difference with Self is Empty Set,
:$V = B \setminus U = B \setminus B = \O$
Hence the only clopen sets in $\struct {B, \tau_B}$ are $B$ and $\O$.
The result follows. 
{{qed}}
Theorem: Union of Connected Sets with Common Point is Connected

Let $T = \struct {S, \tau}$ be a topological space.
Let $\family {B_\alpha}_{\alpha \mathop \in A}$ be a family of connected sets of $T$.
Let $\exists x \in \ds \bigcap \family {B_\alpha}_{\alpha \mathop \in A}$.
Then
:$\ds \bigcup \family {B_\alpha}_{\alpha \mathop \in A}$ is a connected set of $T$.

Proof:
Follows immediately from Union of Connected Sets with Non-Empty Intersections is Connected.
{{qed}}
Theorem: Connected Subset of Union of Disjoint Open Sets

Let $T = \struct{S, \tau}$ be a topological space.
Let $A$ be a connected set of $T$.
Let $U, V$ be disjoint open sets.
Let $A \subseteq U \cup V$.
Then
:either $A \subseteq U$ or $A \subseteq V$.

Proof:
Let $U' = A \cap U$ and $V' = A \cap V$.
By definition $U'$ and $V'$ are open sets in the subspace $\struct{A, \tau_A}$. 
From Intersection is Empty Implies Intersection of Subsets is Empty $U'$ and $V'$ are disjoint.
Hence $U'$ and $V'$ are separated sets by definition.
Now
{{begin-eqn}}
{{eqn | l = A
      | r = A \cap \paren {U \cup V}
      | c = Intersection with Subset is Subset
}}
{{eqn | r = \paren {A \cap U} \cup \paren {A \cap V}
      | c = Intersection Distributes over Union
}}
{{eqn | r = U' \cup V'
}}
{{end-eqn}}
Since $A$ is connected then one of $U'$ or $V'$ is empty.
{{WLOG}} assume that $V' = \O$.
Then
{{begin-eqn}}
{{eqn | l = A
      | r = U' \cup V'
}}
{{eqn | r = U' \cup \O
}}
{{eqn | r = U'
      | c = Union with Empty Set
}}
{{eqn | r = A \cap U
}}
{{eqn | ll= \leadsto
      | l = A
      | o = \subseteq
      | r = U
      | c = Intersection with Subset is Subset
}}
{{end-eqn}}
{{qed}}
Theorem: Axiom of Archimedes/Variant

Let $x$ and $y$ be a natural numbers.
Then there exists a natural number $n$ such that:
:$n x \ge y$

Proof:
{{AimForCont}} there exists $x, y \in \N$ such that $n x < y$ for every natural number $n$.
Consider the set $S$, defined as:
:$S := \set {y - n x: n \in \N}$
We have {{hypothesis}} that $S$ contains only natural numbers.
By the Well-Ordering Principle, $S$ contains a smallest element, $y - m x$ for example.
But $y - \paren {m + 1} x$ is also in $S$, as $S$ contains all natural numbers of this form.
Furthermore, we have:
:$y - \paren {m + 1} x = \paren {y - m x} - x < y - m x$
But this contradicts our assertion that $y - m x$ is the smallest element of $S$.
This assertion originated from the assumption that there exists $x, y \in \N$ such that $n x < y$ for every natural number $n$.
Hence that assumption cannot hold.
Hence the result by Proof by Contradiction.
{{qed}}
{{Namedfor|Archimedes of Syracuse|cat = Archimedes}}
Theorem: Set Intersection Preserves Subsets/Families of Sets/Intersection is Empty Implies Intersection of Subsets is Empty

Let $I$ be an indexing set.
Let $\family {A_\alpha}_{\alpha \mathop \in I}$ and $\family {B_\alpha}_{\alpha \mathop \in I}$ be indexed families of subsets of a set $S$.
Let:
:$\forall \beta \in I: A_\beta \subseteq B_\beta$
Then:
:$\ds \bigcap_{\alpha \mathop \in I} B_\alpha = \O \implies \bigcap_{\alpha \mathop \in I} A_\alpha = \O$

Proof:
Let $\ds \bigcap_{\alpha \mathop \in I} B_\alpha = \O$.
From Set Intersection Preserves Subsets/Families of Sets:
:$\ds \bigcap_{\alpha \mathop \in I} A_\alpha \subseteq \bigcap_{\alpha \mathop \in I} B_\alpha = \O$
From Subset of Empty Set:
:$\ds \bigcap_{\alpha \mathop \in I} A_\alpha = \O$
{{qed}}
Category:Set Intersection Preserves Subsets
Category:Indexed Families
g9h0lyq7idgca39apd16knz8fybcoi5
Theorem: Union of Path-Connected Sets with Common Point is Path-Connected

Let $T = \struct {S, \tau}$ be a topological space.
Let $\family {B_\alpha}_{\alpha \mathop \in A}$ be a family of path-connected sets of $T$.
Let $\exists x \in \ds \bigcap \family {B_\alpha}_{\alpha \mathop \in A}$.
Then
:$\ds \bigcup \family {B_\alpha}_{\alpha \mathop \in A}$ is a path-connected set of $T$.

Proof:
Let $B = \ds \bigcup \family {B_\alpha}_{\alpha \mathop \in A}$.
Let $a, b \in B$.
Then 
:$\exists \alpha, \beta \in A: a \in B_\alpha \land b \in B_\beta$.
As $B_\alpha$ is a path-connected set in $T$ then $a$ and $x$ are path-connected points.
Similarly, $x$ and $b$ are path-connected points.
From Joining Paths makes Another Path, $a$ and $b$ are path-connected points.
Since $a$ and $b$ were arbitrary points then $B$ is a path-connected set of $T$.
{{qed}}
Theorem: Sum of Powers of 2

Let $n \in \N_{>0}$ be a (strictly positive) natural number.
Then:
{{begin-eqn}}
{{eqn | n = 1
      | l = 2^n - 1
      | r = \sum_{j \mathop = 0}^{n - 1} 2^j
      | c = 
}}
{{eqn | r = 1 + 2 + 2^2 + 2^3 + \dotsb + 2^{n - 1}
      | c = 
}}
{{end-eqn}}

Proof:
From Sum of Geometric Sequence:
:$\ds \sum_{j \mathop = 0}^{n - 1} x^j = \frac {x^n - 1} {x - 1}$
The result follows by setting $x = 2$.
{{Qed}}
Theorem: Sum of Powers of 2

Let $n \in \N_{>0}$ be a (strictly positive) natural number.
Then:
{{begin-eqn}}
{{eqn | n = 1
      | l = 2^n - 1
      | r = \sum_{j \mathop = 0}^{n - 1} 2^j
      | c = 
}}
{{eqn | r = 1 + 2 + 2^2 + 2^3 + \dotsb + 2^{n - 1}
      | c = 
}}
{{end-eqn}}

Proof:
Let $S \subseteq \N_{>0}$ denote the set of (strictly positive) natural numbers for which $(1)$ holds.
=== Basis for the Induction ===
We have:
{{begin-eqn}}
{{eqn | l = 2^1 - 1
      | r = 2 - 1
      | c = 
}}
{{eqn | r = 1
      | c = 
}}
{{eqn | r = 2^0
      | c = 
}}
{{eqn | r = \sum_{j \mathop = 0}^{1 - 1} 2^j
      | c = 
}}
{{end-eqn}}
So $1 \in S$.
This is our basis for the induction.
=== Induction Hypothesis ===
We now show that, if $k \in S$ is true, where $k \ge 1$, then it logically follows that $k + 1 \in S$.
So this is our induction hypothesis:
:$\ds \sum_{j \mathop = 0}^{k - 1} 2^j = 2^k - 1$
Then we need to show:
:$\ds \sum_{j \mathop = 0}^k 2^j = 2^{k + 1} - 1$
=== Induction Step ===
This is our induction step:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 0}^k 2^j
      | r = \sum_{j \mathop = 0}^{k - 1} 2^j + 2^k
      | c = 
}}
{{eqn | r = 2^k - 1 + 2^k
      | c = Induction Hypothesis
}}
{{eqn | r = 2 \times 2^k - 1
      | c = 
}}
{{eqn | r = 2^{k + 1} - 1
      | c = 
}}
{{end-eqn}}
So $k \in S \implies k + 1 \in S$.
It follows by the Principle of Finite Induction that $S = \N_{>0}$.
That is:
:$\ds \forall n \in \N_{> 0}: \sum_{j \mathop = 0}^{n - 1} 2^j = 2^n - 1$
{{qed}}
Theorem: Upper Bound for Lucas Number

Let $L_n$ denote the $n$th Lucas number.
Then:
:$L_n < \paren {\dfrac 7 4}^n$

Proof:
The proof proceeds by complete induction.
For all $n \in \Z_{\ge 1}$, let $\map P n$ be the proposition:
:$L_n < \paren {\dfrac 7 4}^n$
$\map P 1$ is the case:
{{begin-eqn}}
{{eqn | l = L_1
      | r = 1
      | c = 
}}
{{eqn | o = <
      | r = \dfrac 7 4
      | c = 
}}
{{end-eqn}}
Thus $\map P 1$ is seen to hold.
Theorem: Factorial Greater than Square for n Greater than 3

Let $n \in \Z$ be an integer such that $n > 3$.
Then $n! > n^2$.

Proof:
We note that:
{{begin-eqn}}
{{eqn | l = 1!
      | r = 1
      | c = 
}}
{{eqn | r = 1^2
      | c = 
}}
{{eqn | l = 2!
      | r = 2
      | c = 
}}
{{eqn | o = <
      | r = 4
      | c = 
}}
{{eqn | r = 2^2
      | c = 
}}
{{eqn | l = 3!
      | r = 6
      | c = 
}}
{{eqn | o = <
      | r = 9
      | c = 
}}
{{eqn | r = 3^2
      | c = 
}}
{{end-eqn}}
The proof then proceeds by induction.
For all $n \in \Z_{\ge 4}$, let $\map P n$ be the proposition:
:$n! > n^2$