avewright/proofwiki-math huggingface dataset:
Theorem: Raw Moment of Erlang Distribution

Let $n, k$ be strictly positive integers. 
Let $\lambda$ be a strictly positive real number. 
Let $X$ have a continuous random variable with an Erlang distribution with parameters $k$ and $\lambda$.
Then the $n$th raw moment of $X$ is given by:
:$\ds \expect {X^n} = \frac 1 {\lambda^n} \prod_{m \mathop = 0}^{n - 1} \paren {k + m}$

Proof:
From the definition of the Erlang distribution, $X$ has probability density function:
:$\map {f_X} x = \dfrac {\lambda^k x^{k - 1} e^{- \lambda x} } {\map \Gamma k}$
From the definition of the expected value of a continuous random variable: 
:$\ds \expect {X^n} = \int_0^\infty x^n \map {f_X} x \rd x$
So:
{{begin-eqn}}
{{eqn	| l = \expect {X^n}
	| r = \frac {\lambda^k} {\map \Gamma k} \int_0^\infty x^{n + k - 1} e^{- \lambda x} \rd x
}}
{{eqn	| r = \frac {\lambda^k} {\lambda \map \Gamma k} \int_0^\infty \paren {\frac u \lambda}^{n + k - 1} e^{- u} \rd u
	| c = substituting $u = \lambda x$
}}
{{eqn	| r = \frac {\lambda^k} {\lambda^{n + k} \map \Gamma k} \int_0^\infty u^{n + k - 1} e^{- u} \rd u
}}
{{eqn	| r = \frac 1 {\lambda^n \map \Gamma k} \map \Gamma {n + k}
	| c = {{Defof|Gamma Function}}
}}
{{eqn	| r = \frac 1 {\lambda^n} \frac {\map \Gamma k} {\map \Gamma k} \prod_{m \mathop = 0}^{n - 1} \paren {k + m}
	| c = Gamma Difference Equation
}}
{{eqn	| r = \frac 1 {\lambda^n} \prod_{m \mathop = 0}^{n - 1} \paren {k + m}
}}
{{end-eqn}}
{{qed}}
Category:Raw Moments
Category:Erlang Distribution
6p9b6r03t5i1z5kkkkcsy830uif3iku
Theorem: Expectation of Erlang Distribution

Let $k$ be a strictly positive integer. 
Let $\lambda$ be a strictly positive real number. 
Let $X$ be a continuous random variable with an Erlang distribution with parameters $k$ and $\lambda$.
Then the expectation of $X$ is given by:
:$\expect X = \dfrac k \lambda$

Proof:
{{begin-eqn}}
{{eqn	| l = \expect X
	| r = \frac 1 {\lambda^1} \prod_{m \mathop = 0}^0 \paren {k + m}
	| c = Raw Moment of Erlang Distribution
}}
{{eqn	| r = \frac k \lambda
}}
{{end-eqn}}
{{qed}}
Category:Expectation
Category:Erlang Distribution
2mvhbomtjhf2h95radm04h4kd0kzyb2
Theorem: Variance of Erlang Distribution

Let $k$ be a strictly positive integer. 
Let $\lambda$ be a strictly positive real number. 
Let $X$ be a continuous random variable with an Erlang distribution with parameters $k$ and $\lambda$.
Then the variance of $X$ is given by: 
:$\var X = \dfrac k {\lambda^2}$

Proof:
By Variance as Expectation of Square minus Square of Expectation, we have: 
:$\var X = \expect {X^2} - \paren {\expect X}^2$
By Expectation of Erlang Distribution, we have: 
:$\expect X = \dfrac k \lambda$
We also have: 
{{begin-eqn}}
{{eqn	| l = \expect {X^2}
	| r = \frac 1 {\lambda^2} \prod_{m \mathop = 0}^1 \paren {k + m}
}}
{{eqn	| r = \frac {k \paren {k + 1} } {\lambda^2} 
}}
{{eqn	| r = \frac {k^2 + k} {\lambda^2}
}}
{{end-eqn}}
So:
{{begin-eqn}}
{{eqn	| l = \var X
	| r = \frac {k^2 + k} {\lambda^2} - \paren {\frac k \lambda}^2
}}
{{eqn	| r = \frac {k^2 + k - k^2} {\lambda^2}
}}
{{eqn	| r = \frac k {\lambda^2}
}}
{{end-eqn}}
{{qed}}
Category:Variance
Category:Erlang Distribution
hoxci9wul05aoxweky745hp408u3g4j
Theorem: Roots of Complex Number/Examples/z^4 - 81 = 0

The roots of the polynomial:
:$z^4 - 81$
are:
:$\set {3, 3 i, -3, -3 i}$

Proof:
From Factorisation of $z^n - a$:
:$z^4 - a = \ds \prod_{k \mathop = 0}^3 \paren {z - \alpha^k b}$
where:
:$\alpha$ is a primitive complex $4$th root of unity
:$b$ is any complex number such that $b^4 = a$.
Here we can take $b = 3$, as $81 = 3^4$.
Thus:
:$z = \set {3 \exp \dfrac {k i \pi} 2}$
{{begin-eqn}}
{{eqn | n = k = 0
      | l = z
      | r = 3 \paren {\cos \dfrac 0 \pi 2 + i \sin \dfrac 0 \pi 2}
      | c = 
}}
{{eqn | r = 3
      | c = 
}}
{{eqn | n = k = 1
      | l = z
      | r = \cos \dfrac {\pi} 2 + i \sin \dfrac {\pi} 2
      | c = 
}}
{{eqn | r = 3 i
      | c = 
}}
{{eqn | n = k = 2
      | l = z
      | r = \cos \dfrac {2 \pi} 2 + i \sin \dfrac {2 \pi} 2
      | c = 
}}
{{eqn | r = -3
      | c = 
}}
{{eqn | n = k = 3
      | l = z
      | r = \cos \dfrac {3 \pi} 2 + i \sin \dfrac {3 \pi} 2
      | c = 
}}
{{eqn | r = -3 i
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Covariance as Expectation of Product minus Product of Expectations

Let $X$ and $Y$ be random variables. 
Let the expectations of $X$ and $Y$ exist and be finite. 
Then the covariance of $X$ and $Y$ is given by: 
:$\cov {X, Y} = \expect {X Y} - \expect X \expect Y$

Proof:
{{begin-eqn}}
{{eqn | l = \cov {X, Y}
      | r = \expect {\paren {X - \expect X} \paren {Y - \expect Y} }
      | c = {{Defof|Covariance}}
}}
{{eqn | r = \expect {X Y - X \expect Y - Y \expect X + \expect X \expect Y}
}}
{{eqn | r = \expect {X Y} - \expect Y \expect X - \expect Y \expect X + \expect X \expect Y
      | c = Expectation is Linear, Expectation of Constant
}}
{{eqn | r = \expect {X Y} - \expect X \expect Y
}}
{{end-eqn}}
{{qed}}
Theorem: Covariance of Independent Random Variables is Zero

Let $X$ and $Y$ be independent random variables.
Let the expectations of $X$ and $Y$ exist and be finite. 
Then the covariance of $X$ and $Y$ is $0$.

Proof:
{{begin-eqn}}
{{eqn | l = \cov {X, Y}
      | r = \expect {X Y} - \expect X \expect Y
      | c = Covariance as Expectation of Product minus Product of Expectations
}}
{{eqn | r = \expect X \expect Y - \expect X \expect Y
      | c = Condition for Independence from Product of Expectations
}}
{{eqn | r = 0
}}
{{end-eqn}}
{{qed}}
Theorem: Variance of Linear Combination of Random Variables

Let $X$ and $Y$ be random variables. 
Let the variances of $X$ and $Y$ be finite. 
Let $a$ and $b$ be real numbers. 
Then the variance of $a X + b Y$ is given by: 
:$\var {a X + b Y} = a^2 \, \var X + b^2 \, \var Y + 2 a b \, \cov {X, Y}$
where $\cov {X, Y}$ is the covariance of $X$ and $Y$.

Proof:
{{begin-eqn}}
{{eqn | l = \var {a X + b Y}
      | r = \expect {\paren {a X + b Y - \expect {a X + b Y} }^2}
      | c = {{Defof|Variance}}
}}
{{eqn | r = \expect {\paren {a X + b Y - a \, \expect X - b \, \expect Y}^2}
      | c = Expectation is Linear
}}
{{eqn | r = \expect {\paren {a \paren {X - \expect X} + b \paren {Y - \expect Y} }^2}
}}
{{eqn | r = \expect {a^2 \paren {X - \expect X}^2 + b^2 \paren {Y - \expect Y}^2 + 2 a b \paren {X - \expect X} \paren {Y - \expect Y} }
}}
{{eqn | r = a^2 \, \expect {\paren {X - \expect X}^2} + b^2 \, \expect {\paren {Y - \expect Y}^2} + 2 a b \, \expect {\paren {X - \expect X} \paren {Y - \expect Y} }
      | c = Expectation is Linear
}}
{{eqn | r = a^2 \, \var X + b^2 \, \var Y + 2 a b \, \cov {X, Y}
      | c = {{Defof|Variance}}, {{Defof|Covariance}}
}}
{{end-eqn}}
{{qed}}
Category:Variance
5d0czeyn78w162wxrh068zxo5itb019
Theorem: Square of Expectation of Product is Less Than or Equal to Product of Expectation of Squares

Let $X$ and $Y$ be random variables. 
Let the expectation of $X Y$, $\expect {X Y}$, exist and be finite. 
Then: 
:$\paren {\expect {X Y} }^2 \le \expect {X^2} \expect {Y^2}$

Proof:
Note that: 
:$\map \Pr {Y^2 \ge 0} = 1$
so we have by Expectation of Non-Negative Random Variable is Non-Negative: 
:$\expect {Y^2} \ge 0$
First, take $\expect {Y^2} > 0$. 
Let $Z$ be a random variable with: 
:$Z = X - Y \dfrac {\expect {X Y} } {\expect {Y^2} }$
Note that we have: 
:$\map \Pr {Z^2 \ge 0} = 1$
so again applying Expectation of Non-Negative Random Variable is Non-Negative, we have: 
:$\expect {Z^2} \ge 0$
That is: 
{{begin-eqn}}
{{eqn | l = 0
      | o = \le
      | r = \expect {Z^2}
}}
{{eqn | r = \expect {\paren {X - Y \frac {\expect {X Y} } {\expect {Y^2} } }^2}
}}
{{eqn | r = \expect {X^2 - 2 X Y \frac {\expect {X Y} } {\expect {Y^2} } + Y^2 \frac {\paren {\expect {X Y} }^2} {\paren {\expect {Y^2} }^2} }
      | c = Square of Sum
}}
{{eqn | r = \expect {X^2} - 2 \expect {X Y} \frac {\expect {X Y} } {\expect {Y^2} } + \expect {Y^2} \frac {\paren {\expect {X Y} }^2} {\paren {\expect {Y^2} }^2}
      | c = Expectation is Linear
}}
{{eqn | r = \expect {X^2} - \frac {\paren {\expect {X Y} }^2 } {\expect {Y^2} }
}}
{{end-eqn}}
We therefore have: 
:$\dfrac {\paren {\expect {X Y} }^2} {\expect {Y^2} } \le \expect {X^2}$
giving: 
:$\paren {\expect {X Y} }^2 \le \expect {X^2} \expect {Y^2}$
as required. 
It remains to address the case $\expect {Y^2} = 0$. 
Note that since $\map \Pr {Y^2 \ge 0} = 1$, from Condition for Expectation of Non-Negative Random Variable to be Zero we necessarily have: 
:$\map \Pr {Y^2 = 0} = 1$
That is: 
:$\map \Pr {Y = 0} = 1$
This implies that the random variable $X Y$ has: 
:$\map \Pr {X Y = 0} = 1$
From which, we have that: 
:$\map \Pr {X Y \ge 0} = 1$
So, applying Expectation of Non-Negative Random Variable is Non-Negative again we have: 
:$\expect {X Y} = 0$
With that, we have: 
:$\paren {\expect {X Y} }^2 = 0$
and: 
:$\expect {X^2} \expect {Y^2} = \expect {X^2} \times 0 = 0$
So the inequality: 
:$\paren {\expect {X Y} }^2 \le \expect {X^2} \expect {Y^2}$
also holds in the case $\expect {Y^2} = 0$, completing the proof.
{{qed}}
Theorem: Square of Covariance is Less Than or Equal to Product of Variances

Let $X$ and $Y$ be random variables. 
Let the variances of $X$ and $Y$ exist and be finite. 
Then: 
:$\paren {\cov {X, Y} }^2 \le \var X \, \var Y$
where $\cov {X, Y}$ denotes the covariance of $X$ and $Y$.

Proof:
We have, by the definition of variance, that both: 
:$\expect {\paren {X - \expect X}^2}$
and: 
:$\expect {\paren {Y - \expect Y}^2}$
exist and are finite. 
Therefore:
{{begin-eqn}}
{{eqn | l = \paren {\cov {X, Y} }^2
      | r = \paren {\expect {\paren {X - \expect X} \paren {Y - \expect Y} } }^2
      | c = {{Defof|Covariance}}
}}
{{eqn | o = \le 
      | r = \expect {\paren {X - \expect X}^2} \expect {\paren {Y - \expect Y}^2}
      | c = Square of Expectation of Product is Less Than or Equal to Product of Expectation of Squares
}}
{{eqn | r = \var X \, \var Y
      | c = {{Defof|Variance}}
}}
{{end-eqn}}
{{qed}}
Theorem: Absolute Value of Pearson Correlation Coefficient is Less Than or Equal to 1

Let $X$ and $Y$ be random variables. 
Let the variances of $X$ and $Y$ exist and be finite. 
Then: 
:$\size {\map \rho {X, Y} } \le 1$
where $\map \rho {X, Y}$ denotes the Pearson correlation coefficient of $X$ and $Y$.

Proof:
{{begin-eqn}}
{{eqn | l = \paren {\map \rho {X, Y} }^2
      | r = \paren {\frac {\map {\operatorname {Cov} } {X, Y} } {\sqrt {\var X \var Y} } }^2
      | c = {{Defof|Pearson Correlation Coefficient}}
}}
{{eqn | r = \frac {\paren {\map {\operatorname {Cov} } {X, Y} }^2} {\var X \var Y}
}}
{{eqn | o = \le
      | r = 1
      | c = Square of Covariance is Less Than or Equal to Product of Variances
}}
{{end-eqn}}
So:
:$\size {\map \rho {X, Y} } \le 1$
{{qed}}
{{expand|There's a proof available here that uses Cauchy's Inequality, but that means pulling apart the {{Defof|Pearson Correlation Coefficient}} and covariance and so on}}
Theorem: Equivalence of Definitions of Locally Connected Space/Definition 1 implies Definition 2

Let each point of $T$ have a local basis consisting entirely of connected sets in $T$.

Proof:
From Local Basis for Open Sets Implies Neighborhood Basis of Open Sets, it follows directly that:
:each point of $T$ has a neighborhood basis  consisting entirely of connected sets in $T$.
Theorem: Equivalence of Definitions of Locally Connected Space/Definition 2 implies Definition 1

Let $T$ be weakly locally connected at each point of $T$.
That is, each point of $T$ has a neighborhood basis consisting of connected sets of $T$.

Proof:
Let $x \in S$ and $x \in U \in \tau$.
Let $\BB_x = \set {W \in \tau : x \in W, W \text{ is connected in } T}$.
By definition of local basis, we have to show that there exists a connected open set $V \in \BB_x$ with $x \in V \subset U$.
Let $V = \map {\operatorname{Comp}_x} U$ denote the component of $x$ in $U$.
From Open Set in Open Subspace, it suffices to show that $V$ is open in $U$.
By Set is Open iff Neighborhood of all its Points, we may do this by showing that $V$ is a neighborhood in $U$ of all of its points.
Let $y \in V$.
By assumption, there exists a connected neighborhood $W$ of $y$ in $T$ with $W \subset U$.
By Neighborhood in Topological Subspace, $W$ is a neighborhood of $y$ in $U$.
By definition of component:
:$W \subseteq \map {\operatorname{Comp}_y} U = \map {\operatorname{Comp}_x} U = V$
By Neighborhood iff Contains Neighborhood, $V$ is a neighborhood of $y$ in $U$.
Because $y$ was arbitrary, $V$ is open in $U$.
From Open Set in Open Subspace, $V$ is open in $T$.
Hence $V \in \BB_x$.
Because $U$ was arbitrary, $\BB_x$ is a local basis of $x$ consisting of (open) connected sets.
Since $x$ was arbitrary, then each point of $T$ has a local basis consisting entirely of connected sets in $T$.
Theorem: Equivalence of Definitions of Locally Connected Space/Definition 1 implies Definition 3

Let each point $x$ of $T$ have a local basis $\DD_x$ consisting entirely of connected sets in $T$.

Proof:
By definition of local basis, each of these connected sets in $\DD_x$ is open in $T$.
Consider the set $\ds \DD = \bigcup_{x \mathop \in S} \DD_x$.
From Union of Local Bases is Basis, $\DD$ is a basis for the topology $\tau$.
Since each $\DD_x$ consists entirely of connected sets, $\DD$ also consists entirely of connected sets.
Theorem: Equivalence of Definitions of Locally Connected Space/Definition 3 implies Definition 1

Let $T$ have a basis $\BB$ consisting of connected sets in $T$.

Proof:
For each $x \in S$ we define:
:$\BB_x = \set {B \in \BB: x \in B}$
From Basis induces Local Basis, $\BB_x$ is a local basis.
As each element of $\BB_x$ is also an element of $\BB$, it follows that $\BB_x$ is also formed of connected sets.
Thus, for each point $x \in S$, there is a local basis which consists entirely of connected sets.
Theorem: Equivalence of Definitions of Locally Connected Space/Definition 3 implies Definition 4

Let $T$ have a basis consisting of connected sets in $T$.

Proof:
Let $U$ be an open subset of $T$.
From Open Set is Union of Elements of Basis, $U$ is a union of open connected sets in $T$.
From Open Set in Open Subspace and Connected Set in Subspace, $U$ is a union of open connected sets in $U$.
From Components are Open iff Union of Open Connected Sets, the components of $U$ are open in $U$.
From Open Set in Open Subspace then the components of $U$ are open in $T$.
Theorem: Equivalence of Definitions of Locally Connected Space/Definition 4 implies Definition 3

Let the components of the open sets of $T$ are also open in $T$.

Proof:
Let $\BB = \set {U \in \tau : U \text{ is connected in } T}$.
Let $U$ be open in $T$.
By assumption, the components of $U$ are open in $T$.
From Connected Set in Subspace, the components of $U$ are connected in $T$.
By the definition of the components of a topological space, $U$ is the union of its components.
Hence $U$ is the union of open connected sets in $T$.
By definition, $\BB$ is an basis for $T$.
Theorem: Connected Component is Closed

Let $T = \struct {S, \tau}$ be a topological space.
Then every connected component of $T$ is closed.

Proof:
Let $H$ be a connected component of $T$.
By the definition of connected component, $H$ is connected.
From Closure of Connected Set is Connected then the closure $H^-$ is connected.
By the definition of the closure, $H \subseteq H^-$.
By the definition of connected component, $H$ is a maximal connected set.
Hence $H = H^-$.
From Set is Closed iff Equals Topological Closure, $H$ is closed.
{{qed}}
Theorem: Constant Mapping is Non-Commutative

Let $S$ be a set whose cardinality is greater than one.
Let $f: S \to S$ and $g: S \to S$ be constant mappings on $S$.
Then:
:$f \circ g \ne g \circ f$
where $\circ$ denotes composition of mappings.

Proof:
First note that if $S$ is a singleton, then there exists only one constant mapping on $S$.
In such a circumstance, $f = g$ and so $f \circ g \ne g \circ f$.
{{qed|lemma}}
So, let $\card S > 1$.
Then there exist at least $2$ distinct elements $a$ and $b$ of $S$.
Thus, let $f$ and $g$ be defined as:
:$\forall x \in S: \map f x = a$
:$\forall x \in S: \map g x = b$
for some $a, b \in S$ such that $a \ne b$.
We have that:
{{begin-eqn}}
{{eqn | l = \map {\paren {f \circ g} } x
      | r = \map f {\map g x}
      | c = 
}}
{{eqn | r = \map f b
      | c = Definition of $\map g x$
}}
{{eqn | r = a
      | c = Definition of $\map f x$
}}
{{eqn | l = \map {\paren {g \circ f} } x
      | r = \map g {\map f x}
      | c = 
}}
{{eqn | r = \map g a
      | c = Definition of $\map f x$
}}
{{eqn | r = b
      | c = Definition of $\map g x$
}}
{{end-eqn}}
So $f \circ g \ne g \circ f$ as required.
{{qed}}
Theorem: Quotient Mapping is Injection iff Equality

Let $\RR$ be an equivalence relation on $S$.
Then the quotient mapping $q_\RR: S \to S / \RR$ is an injection {{iff}} $\RR$ is the equality relation.

Proof:
Let $\eqclass x \RR, \eqclass y \RR \in S / \RR$
Theorem: Closure of Subset in Subspace

Let $T = \struct{S, \tau}$ be a topological space.
Let $H \subseteq S$ be an arbitrary subset of $S$.
Let $T_H = \struct {H, \tau_H}$ be the topological subspace on $H$.
Let $A \subseteq H$ be an arbitrary subset of $H$.
Then:
:$\map {\cl_H} A = H \cap \map \cl A$
where:
:$\map {\cl_H} A$ denotes the closure of $A$ in $T_H$
:$\map \cl A$ denotes the closure of $A$ in $T$.

Proof:
{{begin-eqn}}
{{eqn | l = \map {\cl_H} A
      | r = \bigcap \set {K \subseteq H: A \subseteq K, K \text{ is closed in } T_H}
      | c = {{Defof|Closure (Topology)}}
}}
{{eqn | r = \bigcap \set {N \cap H: A \subseteq N, N \text{ is closed in } T}
      | c = Closed Set in Topological Subspace
}}
{{eqn | r = H \cap \bigcap \set {N: A \subseteq N, N \text{ is closed in } T}
      | c = Set Intersection is Self-Distributive over Family of Sets
}}
{{eqn | r = H \cap \map \cl A
      | c = {{Defof|Closure (Topology)}}
}}
{{end-eqn}}
{{qed}}
Theorem: Equivalence of Definitions of Locally Path-Connected Space/Definition 1 implies Definition 2

Let each point of $T$ have a local basis consisting entirely of path-connected sets in $T$.

Proof:
From Local Basis for Open Sets Implies Neighborhood Basis of Open Sets, it follows directly that:
:each point of $T$ has a neighborhood basis  consisting entirely of path-connected sets in $T$.
Theorem: Equivalence of Definitions of Locally Path-Connected Space/Definition 2 implies Definition 1

Let each point of $T$ have a neighborhood basis consisting of path-connected sets in $T$.

Proof:
Let $x \in S$ and $x \in U \in \tau$.
Let $\BB_x = \set {W \in \tau : x \in W, W \text { is path-connected in } T}$.
By definition of local basis, we have to show that there exists a path-connected open set $V \in \tau$ with $x \in V \subset U$.
Let $V = \map {\operatorname {PC}_x} U$ denote the path component of $x$ in $U$.
From Open Set in Open Subspace, it suffices to show that $V$ is open in $U$.
By Set is Open iff Neighborhood of all its Points, we may do this by showing that $V$ is a neighborhood in $U$ of all of its points.
Let $y \in V$.
By assumption, there exists a path-connected neighborhood $W$ of $y$ in $T$ with $W \subset U$.
By Neighborhood in Topological Subspace, $W$ is a neighborhood of $y$ in $U$.
By definition of path component:
:$W \subset \map {\operatorname {PC}_y} U = \map {\operatorname {PC}_x} U = V$
By Neighborhood iff Contains Neighborhood, $V$ is a neighborhood of $y$ in $U$.
Because $y$ was arbitrary, $V$ is open in $U$.
From Open Set in Open Subspace, $V$ is open in $T$.
Hence $V \in \BB_x$.
Because $U$ was arbitrary, $\BB_x$ is a local basis of $x$ consisting of (open) path-connected sets.
Since $x$ was arbitrary, then each point of $T$ has a local basis consisting entirely of path-connected sets in $T$.
Theorem: Equivalence of Definitions of Locally Path-Connected Space/Definition 1 implies Definition 3

Let each point $x$ of $T$ have a local basis $\DD_x$ consisting entirely of path-connected sets in $T$.

Proof:
By definition of local basis, each of these path-connected sets in $\DD_x$ is open in $T$.
Consider the set $\ds \DD = \bigcup_{x \mathop \in S} \DD_x$.
From Union of Local Bases is Basis, $\DD$ is a basis for the topology $\tau$.
Since each $\DD_x$ consists entirely of path-connected sets, $\DD$ also consists entirely of path-connected sets.
Theorem: Equivalence of Definitions of Locally Path-Connected Space/Definition 3 implies Definition 1

Let $T$ have a basis $\BB$ consisting of path-connected sets in $T$.

Proof:
For each $x \in S$ we define:
:$\BB_x = \set {B \in \BB: x \in B}$
From Basis induces Local Basis, $\BB_x$ is a local basis.
As each element of $\BB_x$ is also an element of $\BB$, it follows that $\BB_x$ is also formed of path-connected sets.
Thus, for each point $x \in S$, there is a local basis which consists entirely of path-connected sets.
Theorem: Equivalence of Definitions of Locally Path-Connected Space/Definition 3 implies Definition 4

Let $T$ have a basis consisting of path-connected sets in $T$.

Proof:
Let $U$ be an open subset of $T$.
From Open Set is Union of Elements of Basis, $U$ is a union of open path-connected sets in $T$.
From Open Set in Open Subspace and Path-Connected Set in Subspace, $U$ is a union of open path-connected sets in $U$.
From Path Components are Open iff Union of Open Path-Connected Sets, the path components of $U$ are open in $U$.
From Open Set in Open Subspace then the path components of $U$ are open in $T$.
Theorem: Equivalence of Definitions of Locally Path-Connected Space/Definition 4 implies Definition 3

Let the path components of open sets of $T$ be also open in $T$.

Proof:
Let $\BB = \set {U \in \tau : U \text{ is path-connected in } T}$.
Let $U$ be open in $T$.
By assumption, the path components of $U$ are open in $T$.
From Path-Connected Set in Subspace, the path components of $U$ are path-connected in $T$.
By the definition of the path components of a topological space, $U$ is the union of its path components.
Hence $U$ is the union of open path-connected sets in $T$.
By definition, $\BB$ is an basis for $T$.
Theorem: Binomial Coefficient 2 n Choose n is Divisible by All Primes between n and 2 n

Let $\dbinom {2 n} n$ denote a binomial coefficient.
Then for all prime numbers $p$ such that $n < p < 2 n$:
:$p \divides \dbinom {2 n} n$
where $\divides$ denotes divisibility.

Proof:
By definition of binomial coefficient:
{{begin-eqn}}
{{eqn | l = \dbinom {2 n} n
      | r = \dfrac {\paren {2 n}!} {\paren {n!}^2}
      | c = where $n!$ denotes the factorial of $n$
}}
{{eqn | ll= \leadsto
      | l = \dbinom {2 n} n \paren {n!}^2
      | r = \paren {2 n}!
      | c = 
}}
{{end-eqn}}
Let $\mathbb P$ denote the set of prime numbers.
Let $p \in \mathbb P$ such that $n < p < 2 n$.
By definition of factorial:
:$p \divides \paren {2 n}!$
That is:
:$p \divides \dbinom {2 n} n \paren {n!}^2$
From Prime iff Coprime to all Smaller Positive Integers:
:$p \nmid n!$
where $\nmid$ denotes non-divisibility.
and so:
:$p \nmid \paren {n!}^2$
But from Euclid's Lemma for Prime Divisors, $p$ is a divisor of either $\dbinom {2 n} n$ or $\paren {n!}^2$.
Hence it must be the case that:
:$p \divides \dbinom {2 n} n$
{{qed}}
Theorem: Path-Connected Set in Subspace

Let $T = \struct {S, \tau}$ be a topological space.
Let $A \subseteq B \subseteq S$.
Let $T_B = \struct {B, \tau_B}$ be the topological space where $\tau_B$ is the subspace topology on $B$.
Then
:$A$ is path-connected in $T_B$ {{iff}} $A$ is path-connected in $T$.

Proof:
Let $\tau_A$ be the subspace topology on $A$ induced by $\tau$.
Let $\tau'_A$ be the subspace topology on $A$ induced by $\tau_B$.
By the definition of a path-connected set, $A$ is path-connected in $T$ {{iff}} every two points in $A$ are path-connected in $\struct {A, \tau_A}$.
Similarly, $A$ is path-connected in $T_B$ {{iff}} every two points in $A$ are path-connected in $\struct {A, \tau'_A}$.
From Subspace of Subspace is Subspace, we have $\tau_A = \tau'_A$.
Hence every two points in $A$ are path-connected in $\struct {A, \tau_A}$ {{iff}} every two points in $A$ are path-connected in $\struct {A, \tau'_A}$.
{{qed}}
Category:Path-Connected Sets
Category:Path-Connected Spaces
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