avewright/proofwiki-math huggingface dataset:
Theorem: Mapping whose Graph is Closed in Chebyshev Product is not necessarily Continuous

Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $\AA = A_1 \times A_2$ be the cartesian product of $A_1$ and $A_2$.
Let $d_\infty: \AA \times \AA \to \R$ be the Chebyshev distance on $\AA$:
:$\ds \map {d_\infty} {x, y} = \max \set {\map {d_1} {x_1, y_1}, \map {d_2} {x_2, y_2} }$
where $x = \tuple {x_1, x_2}, y = \tuple {y_1, y_2} \in \AA$.
Let $\Gamma_f$ be the graph of $f$.
Let $f: M_1 \to M_2$ be a mapping such that the $\Gamma_f$ is a closed set of $\struct {A_1 \times A_2, d}$.
Then it is not necessarily the case that $f$ is a continuous mapping.

Proof:
Consider the mapping $f: \R \to \R$ defined as:
:$\map f x = \begin {cases} \dfrac 1 x : & x > 0 \\ 0 : & x \le 0 \end {cases}$
It is seen that $\map f x$ is continuous everywhere except at $x = 0$.
Hence from Graph of Continuous Mapping between Metric Spaces is Closed in Chebyshev Product, $\Gamma_f$ contains all its limit points except perhaps at $x = 0$.
We have that:
:$\tuple {0, 0} \in \Gamma_f$
For a bounded, non-continuous real function, the limit point at one of the discontinuities would not lie in the graph.
However, this is not the case here.
The sequence $\sequence {x_n, \map f {x_n} }$ for which $\ds \lim_{n \mathop \to \infty} = 0$ has only one limit point, that is: $\tuple {0, 0}$.
But $\tuple {0, 0}$ lies in $\Gamma_f$.
{{AimForCont}} there exists another limit point.
This would imply that:
:$\forall \epsilon > 0: \exists N \in \N: n > N \implies \map d {\map f x, \map f {x_n} } < \epsilon$
But $\map f x$ is not bounded.
So for some $x < x_n$ we have $\map f {\map f x, \map f {x_n} } > \epsilon$ if $x, x_n > 0$.
This contradicts what was deduced earlier.
So any limit point $\tuple {x, \map f x}$ for which $x = 0$ is $\tuple {0, 0}$ which is in $\Gamma_f$.
Hence $\Gamma_f$ contains all its limit points.
Hence the result by definition of closed set.
{{qed}}
Theorem: Right Order Topology on Strictly Positive Integers is Topology

Let $\Z_{>0}$ be the set of strictly positive integers.
Let $\tau$ be the '''right order topology on $\Z_{>0}$'''.
Then $\tau$ forms a topology on $\Z_{>0}$.
That is:
:$T = \struct {\Z_{>0}, \tau}$ is a topological space.

Proof:
Let $S := \Z_{>0}$ to ease notational clutter.
First we note that:
:$m \le n \implies O_n \subseteq O_m$
where $O_n := \set {x \in \Z_{>0}: x \ge n}$.
By definition we have that:
:$\O \in \tau$
Then each of the open set axioms is examined in turn:
Theorem: Right Order Topology on Strictly Positive Integers is not Metrizable

Let $\Z_{>0}$ be the set of strictly positive integers.
Let $T = \struct {\Z_{>0}, \tau}$ denote the '''right order space on $\Z_{>0}$'''.
Then $T = \struct {\Z_{>0}, \tau}$ is not a metrizable space.

Proof:
Let $m, n \in \Z_{>0}$ such that $m < n$.
Let $O_m$ and $O_n$ be arbitrary non-empty open sets of $T$.
Then:
:$O_m \cap O_n = O_m$
As $O_m$ and $O_n$ are arbitrary , it follows that there exist no $O_m$ and $O_n$ in $\tau$ such that $O_m \cap O_n = \O$.
Hence $T$ is not a $T_2$ (Hausdorff) space.
The result follows from Metrizable Space is Hausdorff.
{{qed}}
Theorem: Euclidean Metric and Chebyshev Distance on Real Metric Space give rise to Same Topological Space

For $n \in \N$, let $\R^n$ be an Euclidean space.
Let $d_2$ be the Euclidean metric on $\R^n$.
Let $d_\infty$ be the Chebyshev distance on $\R^n$.
Let $T_2 = \struct {\R^n, \tau_2}$ denote the topological space which is induced by $d_2$.
Let $T_\infty = \struct {\R^n, \tau_\infty}$ denote the topological space which is induced by $d_\infty$.
Then $T_2$ and $T_\infty$ are the same.
That is:
:$\tau_2 = \tau_\infty$

Proof:
From P-Product Metrics on Real Vector Space are Topologically Equivalent, $\tau_2$ and $\tau_\infty$ are topologically equivalent metrics.
The result follows from Topologically Equivalent Metrics induce Equal Topologies.
{{qed}}
Theorem: Topologically Equivalent Metrics induce Equal Topologies

Let $M_1 = \struct {A, d_1}$ and $M_2 = \struct {A, d_2}$ be metric spaces on the same underlying set $A$.
Let $d_1$ and $d_2$ be topologically equivalent.
Let $\tau_1$ and $\tau_2$ denote the topologies on $A$ induced by $d_1$ and $d_2$ respectively.
Then $\tau_1$ and $\tau_2$ are equal.

Proof:
Let $d_1$ and $d_2$ be topologically equivalent by hypothesis.
By definition of topological equivalence:
$d_1$ and $d_2$ are '''topologically equivalent''' {{iff}}:
:$U \subseteq A$ is $d_1$-open {{iff}} $U \subseteq A$ is $d_2$-open.
By definition of the induced topology:
:The '''topology on the metric space $M = \struct {A, d}$ induced by (the metric) $d$''' is defined as the set $\tau$ of all open sets of $M$.
Hence it follows that:
:$U \subseteq A$ is open in the topological space $\struct {A, \tau_1}$
{{iff}}:
:$U \subseteq A$ is open in the topological space $\struct {A, \tau_2}$
That is:
:$U \in \tau_1 \iff U \in \tau_2$
and the result follows.
{{qed}}
Category:Topologically Equivalent Metrics
fcwbbhg7j6t1l5mfk385dvtamt9yhz4
Theorem: Right Order Topology on Strictly Positive Integers as Neighborhood Space

Let $\Z_{>0}$ denote the set of (strictly) positive integers.
Let $n \in \Z_{>0}$.
Let $U \subseteq \Z_{>0}$ be defined as being a neighborhood of $n$ {{iff}}
:$\forall m \in \Z: m \ge n \implies m \in U$
Then the set $\NN$ of all $U$ for all $n \in \Z_{>0}$ forms a neighborhood space which is the same as the right order space on $\Z_{>0}$.

Proof:
First it is noted that a neighborhood of $n$ is exactly an element of the right order topology on $\Z_{>0}$.
It remains to be shown that $\NN$ actually forms a neighborhood space.
Let $\NN_n$ denote the set of all neighborhood of a given $n \in \Z_{>0}$.
Checking the neighborhood space axioms in turn:
Theorem: Translation Mapping is Bijection

Let $\struct {G, +}$ be an abelian group.
Let $g \in G$.
Let $\tau_g: G \to G$ be the translation by $g$:
:$\forall h \in G: \map {\tau_g} h = h + \paren {-g}$
where $-g$ is the inverse of $g$ with respect to $+$ in $G$.
Then $\tau_g$ is a bijection.

Proof:
=== Proof of Injectivity ===
{{begin-eqn}}
{{eqn | q = \forall h_1, h_2 \in G
      | l = \map {\tau_g} {h_1}
      | r = \map {\tau_g} {h_2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = h_1 + \paren {-g}
      | r = h_2 + \paren {-g}
      | c = Definition of $\tau_g$
}}
{{eqn | ll= \leadsto
      | l = h_1
      | r = h_2
      | c = Cancellation Laws
}}
{{end-eqn}}
{{qed|lemma}}
Theorem: Translation Mapping is Isometry

Let $\Gamma = \R^n$ denote the real Euclidean space of $n$ dimensions.
Let $\tau_\mathbf x$ be a translation on $\Gamma$:
:$\forall \mathbf y \in \R^n: \map {\tau_\mathbf x} {\mathbf y} = \mathbf y - \mathbf x$
where $\mathbf x$ is a vector in $\R^n$.
Then $\tau_\mathbf x$ is an isometry.

Proof:
From Translation Mapping is Bijection, $\tau_\mathbf x$ is a bijection.
From Euclidean Metric on Real Number Space is Translation Invariant, $\tau_\mathbf x$ is distance-preserving on $\Gamma$.
The result follows by definition of isometry.
{{qed}}
Theorem: Existence of Translation between Each Pair of Points in Euclidean Space

Let $\R^n$ denote the real Euclidean space of $n$ dimensions.
Let $\mathbf a = \tuple {a_1, a_2, \ldots, a_n}$ and $\mathbf b = \tuple {b_1, b_2, \ldots, b_n}$ be points in $\R^n$.
There exists an isometry $f: \R^n \to \R^n$ such that $\map f {\mathbf a} = b$.

Proof:
Let $\mathbf t = \mathbf a - \mathbf b$.
Then the translation $\tau_\mathbf t$ is such an isometry.
We have that:
:$\map {\tau_\mathbf t} {\mathbf a} = \mathbf a - \paren {\mathbf a - \mathbf b} = \mathbf b$
The result follows from Translation Mapping is Isometry.
{{qed}}
Theorem: Mapping is Continuous iff Inverse Images of Open Sets are Open

Let $X$ and $Y$ be normed vector spaces.
Let $f : X \to Y$ be a mapping.
Then:
:$f$ is continuous on $X$
{{iff}}:
:for every $V$ open in $Y$, $\map {f^{-1}} V$ is open in $X$.

Proof:
=== Sufficient Condition ===
Let $c \in X$.
Let $\epsilon \in \R_{\mathop > 0}$.
Let $V := \map {B_\epsilon} {\map f c}$ be an open ball in $Y$.
By Open Ball is Open Set in Normed Vector Space, $V$ is an open set in $Y$.
Let $\map {f^{-1}} V = \map {f^{-1}} {\map {B_\epsilon} {\map f c}}$ be an open set in $X$.
Note that $\map f c \in \map {B_\epsilon} {\map f c}$.
Hence, $c \in \map {f^{-1}} {\map {B_\epsilon} {\map f c}}$.
By definition of open set:
:$\exists \delta > 0 : \map {B_\delta} c \subseteq \map {f^{-1}} {\map {B_\epsilon} {\map f c}}$
In other words:
:$\forall x \in X : \norm {x - c} < \delta \implies x \in \map {f^{-1}} {\map {B_\epsilon} {\map f c}}$
or
:$\map f x \in \map {B_\epsilon} {\map f c}$
By definition of open ball:
:$\norm {\map f x - \map f c} < \epsilon$
But $\epsilon$ was arbitrary.
Hence:
:$\forall \epsilon \in \R_{\mathop > 0} : \exists \delta > 0 : \norm {x - c} < \delta \implies \norm {\map f x - \map f c} < \epsilon$
Therefore, $f$ is continuous at $c$.
However, $c \in X$ was arbitrary.
Hence, $f$ is continuous on $X$.
{{qed|lemma}}
Theorem: Mapping is Continuous iff Inverse Images of Open Sets are Open/Corollary

Let $X$ and $Y$ be normed vector spaces.
Let $f : X \to Y$ be a mapping.
Then:
:$f$ is continuous on $X$
{{iff}}:
:for every $F$ closed in $Y$, $\map {f^{-1}} F$ is closed in $X$.

Proof:
=== Sufficient Condition ===
Suppose that for every closed $F$ in $Y$, $\map {f^{-1}} F$ is closed in $X$.
Let $V$ be open in $Y$.
By definition, $Y \setminus V$ is closed in $Y$.
By assumption, $\map {f^{-1}} {Y \setminus V}$ is closed in $X$.
We have that:
{{begin-eqn}}
{{eqn | l = \map {f^{-1} } {Y \setminus V} 
      | r = \map {f^{-1} } Y \setminus \map {f^{-1} } V 
      | c = Preimage of Set Difference under Mapping
}}
{{eqn | r = X \setminus \map {f^{-1} } V
      | c = Definition of $f$
}}
{{end-eqn}}
Thus, $X \setminus \map {f^{-1} } V$ is closed in $X$.
By definition, $X \setminus \map {f^{-1} } V$ is closed iff $\map {f^{-1} } V$ is open.
Hence, for every $V$ open in $Y$, $ \map {f^{-1}} V$ is open in $X$.
By Mapping is Continuous iff Inverse Images of Open Sets are Open, $f$ is continuous on $X$.
{{qed|lemma}}
Theorem: Set of Isometries in Euclidean Space under Composition forms Group

Let $\struct {\R^n, d}$ be a real Euclidean space of $n$ dimensions.
Let $S$ be the set of all mappings $f: \R^n \to \R^n$ which preserve distance:
That is:
:$\map d {\map f a, \map f b} = \map d {a, b}$
Let $\struct {S, \circ}$ be the algebraic structure formed from $S$ and the composition operation $\circ$.
Then $\struct {S, \circ}$ is a group.

Proof:
From Euclidean Metric on Real Vector Space is Metric, $\R^n$ is a metric space.
Hence it is seen that a complex function $f: \C \to \C$ which preserves distance is in fact an isometry on $\C$.
Taking the group axioms in turn:
Theorem: Composite of Continuous Mappings between Normed Vector Spaces is Continuous

Let $X, Y, Z$ be normed vector spaces.
Let $f : X \to Y$ and $g : Y \to Z$ be continuous mappings on $X$ and $Y$ respectively.
Let $g \circ f : X \to Z$ be a composite mapping where:
:$\forall x \in X : \map {\paren {g \circ f} } x := \map g {\map f x}$
Then $g \circ f$ is continuous on $X$.

Proof:
Let $W$ be open in $Z$.
$g$ is continuous on $Y$.
By Mapping is Continuous iff Inverse Images of Open Sets are Open, $\map {g^{-1} } W$ is open in $Y$.
$f$ is continuous on $X$.
By Mapping is Continuous iff Inverse Images of Open Sets are Open, $\map {f^{-1} } {\map {g^{-1} } W}$ is open in $X$.
By Inverse of Composite Relation:
:$\map {\paren {g \circ f}^{-1} } W = \map {f^{-1} } {\map {g^{-1} } W}$
Hence, $\map {\paren {g \circ f}^{-1} } W$ is continuous on $X$.
By Mapping is Continuous iff Inverse Images of Open Sets are Open, $g \circ f$ is continuous on $X$.
{{qed}}
Theorem: Dot Product of Unit Vectors

Let $\mathbf a$ and $\mathbf b$ be unit vectors.
Then their dot product $\mathbf a \cdot \mathbf b$ is:
:$\mathbf a \cdot \mathbf b = \cos \theta$
where $\theta$ is the angle between $\mathbf a$ and $\mathbf b$.

Proof:
We have by definition of dot product :
:$\mathbf a \cdot \mathbf b = \norm {\mathbf a} \, \norm {\mathbf b} \cos \theta$
where $\norm {\mathbf a}$ denotes the length of $\mathbf a$.
From Length of Unit Vector is 1:
:$\norm {\mathbf a} = \norm {\mathbf b} = 1$
from which the result follows.
{{qed}}
Theorem: Square of Sum of Vectors

Let $\mathbf a$ and $\mathbf b$ be vector quantities.
Then:
:$\paren {\mathbf a + \mathbf b}^2 = \mathbf a^2 + 2 \mathbf a \cdot \mathbf b + \mathbf b^2$
where:
:$\mathbf a \cdot \mathbf b$ denotes dot product
:$\mathbf a^2$ denotes the square of $\mathbf a$, that is: $\mathbf a \cdot \mathbf a$.

Proof:
{{begin-eqn}}
{{eqn | l = \paren {\mathbf a + \mathbf b}^2
      | r = \paren {\mathbf a + \mathbf b} \cdot \paren {\mathbf a + \mathbf b}
      | c = {{Defof|Square of Vector Quantity}}
}}
{{eqn | r = \mathbf a \cdot \paren {\mathbf a + \mathbf b} + \mathbf b \cdot \paren {\mathbf a + \mathbf b}
      | c = Dot Product Distributes over Addition
}}
{{eqn | r = \mathbf a \cdot \mathbf a + \mathbf a \cdot \mathbf b + \mathbf b \cdot \mathbf a + \mathbf b \cdot \mathbf b
      | c = Dot Product Distributes over Addition
}}
{{eqn | r = \mathbf a \cdot \mathbf a + 2 \mathbf a \cdot \mathbf b + \mathbf b \cdot \mathbf b
      | c = Dot Product Operator is Commutative
}}
{{eqn | r = \mathbf a^2 + 2 \mathbf a \cdot \mathbf b + \mathbf b^2
      | c = {{Defof|Square of Vector Quantity}}
}}
{{end-eqn}}
{{qed}}
Theorem: Dot Product of Sum with Difference of Vectors

Let $\mathbf a$ and $\mathbf b$ be vector quantities.
Then:
:$\paren {\mathbf a + \mathbf b} \cdot \paren {\mathbf a - \mathbf b} = \mathbf a^2 - \mathbf b^2$
where:
:$\cdot$ denotes dot product
:$\mathbf a^2$ denotes the square of $\mathbf a$, that is: $\mathbf a \cdot \mathbf a$.

Proof:
{{begin-eqn}}
{{eqn | l = \paren {\mathbf a + \mathbf b} \cdot \paren {\mathbf a - \mathbf b}
      | r = \mathbf a \cdot \paren {\mathbf a - \mathbf b} + \mathbf b \cdot \paren {\mathbf a - \mathbf b}
      | c = Dot Product Distributes over Addition
}}
{{eqn | r = \mathbf a \cdot \mathbf a - \mathbf a \cdot \mathbf b + \mathbf b \cdot \mathbf a - \mathbf b \cdot \mathbf b
      | c = Dot Product Distributes over Addition
}}
{{eqn | r = \mathbf a \cdot \mathbf a - \mathbf b \cdot \mathbf b
      | c = Dot Product Operator is Commutative
}}
{{eqn | r = \mathbf a^2 + - \mathbf b^2
      | c = {{Defof|Square of Vector Quantity}}
}}
{{end-eqn}}
{{qed}}
Theorem: Square of Vector Quantity in Coordinate Form

Let $\mathbf a$ be a vector in a vector space $\mathbf V$ of $n$ dimensions:
$\ds \mathbf a = \sum_{k \mathop = 1}^n a_k \mathbf e_k$
where $\tuple {\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}$ is the standard ordered basis of $\mathbf V$.
Then:
:$\ds \mathbf a^2 = \sum_{k \mathop = 1}^n a_k^2$
where $\mathbf a^2$ denotes the square of $\mathbf a$.

Proof:
By definition of square of $\mathbf a$:
:$\mathbf a^2 = \mathbf a \cdot \mathbf a$
By definition of dot product:
:$\ds \mathbf a \cdot \mathbf a = a_1 a_1 + a_2 a_2 + \cdots + a_n a_n = \sum_{k \mathop = 1}^n a_k^2$
{{qed}}
Theorem: Continuous Composition of Measurable Functions into Second Countable Space is Measurable

Let $\struct {X, \Sigma}$ be a measurable space.
Let $\struct {X_i, \tau_i}$ for $i = 1, \ldots, n$ and $\struct {Y, \tau_Y}$ be topological spaces such that $X_1, \ldots, X_n$ are second countable. 
Let $f_i: \struct {X, \Sigma} \to \struct {X_i, \map \BB {X_i, \tau_i} }$, $i = 1, \ldots, n$ be measurable functions where $\map \BB {X_i, \tau_i}$ denotes the Borel-sigma algebra.
Let $F: \struct {\ds \prod_{i \mathop = 1}^n X_i, \tau} \to \struct {Y, \tau_Y}$ be continuous where $\tau$ denotes the product topology on the finite Cartesian product $\ds \prod_{i \mathop = 1}^n X_i$. 
Then the composition $h: \struct {X, \Sigma} \to \struct {Y, \map \BB {Y, \tau_Y} }$, $x \mapsto \map F {\map {f_1} x, \ldots, \map {f_n} x}$ is measurable.

Proof:
By Mapping Measurable iff Measurable on Generator, it suffices to check that $h$ is measurable on open sets.
Thus let $U \in \tau_Y$ be given.
As $F$ is continuous by assumption, the pre-image $F^{-1} \sqbrk U$ is in $\tau$. 
By:
:definition of basis
:Countable Product of Second-Countable Spaces is Second-Countable
there exists an at most countable indexing set $J \subseteq \N$ and indexed family of open sets $\family {U^i_j}_{j \mathop \in J} \subseteq \tau_i$, $i = 1, \ldots, n$ such that:
:$F^{-1} \sqbrk U = \ds \bigcup_{j \mathop \in J} \prod_{i \mathop = 1}^n U^i_j$
Putting things together:
{{begin-eqn}}
{{eqn | l = h^{-1} \sqbrk U
      | r = \set {x \in X : \map F {\map {f_1} x ,\ldots, \map {f_n} x} \in U}
}}
{{eqn | r = \set {x \in X : \tuple {\map {f_1} x, \ldots, \map {f_n} x} \in F^{-1} \sqbrk U}
}}
{{eqn | r = \set {x \in X : \tuple {\map {f_1} x, \ldots, \map {f_n} x} \in \bigcup_{j \mathop \in J} \prod_{i \mathop = 1}^n U^i_j }
}}
{{eqn | r = \bigcup_{j \mathop \in J} \set {x \in X : \tuple { \map {f_1} x, \ldots , \map {f_n} x} \in \prod_{i \mathop = 1}^n U^i_j }
}}
{{eqn | r = \bigcup_{j \mathop \in J} \bigcap_{i \mathop = 1}^n \set {x \in X : \map {f_i} x \in U^i_j }
}}
{{eqn | r = \bigcup_{j \mathop \in J} \bigcap_{i \mathop = 1}^n f_i^{-1} \sqbrk {U^i_j} 
}}
{{end-eqn}}
By assumption each $f_i$ is measurable.
By definition of the Borel-sigma algebra:
:$\tau_i \subseteq \map \BB {X_i, \tau_i}$
Hence it follows that every $f_i^{-1} \sqbrk {U^i_j}$ is in $\Sigma$.
Therefore by Sigma-Algebra Closed under Countable Intersection:
:$\ds \bigcap_{i \mathop = 1}^n f_i^{-1} \sqbrk {U^i_j}$
is also in $\Sigma$.
Finally, we note that by definition of $\sigma$-algebra, $\Sigma$ contains at most countable unions of measurable sets.
Hence it follows that:
:$h^{-1} \sqbrk U = \ds \bigcup_{j \mathop \in J} \bigcap_{i \mathop = 1}^n f_i^{-1} \sqbrk {U^i_j}$ is in $\Sigma$
This concludes the proof. 
{{qed}}
Theorem: Point dividing Line Segment between Two Points in Given Ratio

Let $A$ and $B$ be points whose position vectors relative to an origin $O$ of a Euclidean space are $\mathbf a$ and $\mathbf b$.
Let $\mathbf r$ be the position vector of a point $R$ on $AB$ which divides $AB$ in the ratio $m : n$.
:320px
Then:
:$\mathbf r = \dfrac {n \mathbf a + m \mathbf b} {m + n}$

Proof:
Let the coordinates of $A$ be $\tuple {x_1, y_1}$.
Let the coordinates of $B$ be $\tuple {x_2, y_2}$.
Let the coordinates of $R$ be $\tuple {X, Y}$.
Then we have:
:$\dfrac {x_2 - X} {X - x_1} = \dfrac n m$
and so:
:$X \paren {m + n} = m x_2 + n x_1$
Similarly for $Y$, giving $R$ as:
:$\tuple {\dfrac {m x_2 + n x_1} {m + n}, \dfrac {m y_2 + n y_1} {m + n} }$
The result follows.
{{qed}}
Theorem: Centroid of Weighted Pair of Points

Let $A$ and $B$ be two points in Euclidean space.
Let $A$ and $B$ have weights $p$ and $q$ respectively.
Let $G$ be the centroid of $A$ and $B$.
Then $G$ divides the line $AB$ in the ratio $q : p$.
That is:
:$AG = \dfrac q {p + q} AB$
:$BG = \dfrac p {p + q} AB$

Proof:
Let the position vectors of $A$ and $B$ be given by $\mathbf a$ and $\mathbf b$ repectively.
By definition of centroid:
:$\vec {O G} = \dfrac {p \mathbf a + q \mathbf b} {p + q}$
The result follows from Point dividing Line Segment between Two Points in Given Ratio.
{{qed}}
Theorem: Centroid of Weighted Set of Points is Independent of Origin

Let $O'$ be a point whose position vector from $O$ is $\mathbf l$.
Let $S = \set {A_1, A_2, \ldots, A_n}$ be a set of $n$ points in Euclidean space whose position vectors are given by $\mathbf a_1, \mathbf a_2, \dotsc, \mathbf a_n$ repectively.
Let $W: S \to \R$ be a weight function on $S$.
Let $G$ be the centroid of $S$ with weight function $W$:
:$\vec {OG} = \dfrac {w_1 \mathbf a_1 + w_2 \mathbf a_2 + \dotsb + w_n \mathbf a_n} {w_1 + w_2 + \dotsb + w_n}$
where $w_i = \map W {A_i}$ for each $i$.
Let $\vec {OG'}$ be the centroid of $S$ with weight function $W$ with respect to $O'$.
Then:
:$\vec {OG} - \mathbf l = \vec {OG'}$
That is, the centroid of $S$ with weight function $W$ is independent of the position of the origin.

Proof:
The position vectors of the elements of $S$ are given by:
:$\mathbf a_1 - \mathbf l, \mathbf a_2 - \mathbf l, \dotsc, \mathbf a_n - \mathbf l$
Hence the centroid of $S$ with weight function $W$ with respect to $O'$ ias:
:$\vec {OG'} = \dfrac {w_1 \paren {\mathbf a_1 - \mathbf l} + w_2 \paren {\mathbf a_2 - \mathbf l} + \dotsb + w_n \paren {\mathbf a_n - \mathbf l} } {w_1 + w_2 + \dotsb + w_n}$
{{begin-eqn}}
{{eqn | l = \vec {OG'}
      | r = \dfrac {w_1 \paren {\mathbf a_1 - \mathbf l} + w_2 \paren {\mathbf a_2 - \mathbf l} + \dotsb + w_n \paren {\mathbf a_n - \mathbf l} } {w_1 + w_2 + \dotsb + w_n}
      | c = 
}}
{{eqn | r = \dfrac {w_1 \mathbf a_1 + w_2 \mathbf a_2 + \dotsb + w_n \mathbf a_n} {w_1 + w_2 + \dotsb + w_n} - \mathbf l
      | c = 
}}
{{eqn | r = \vec {OG} - \mathbf l
      | c = 
}}
{{end-eqn}}
{{Qed}}
Theorem: Centroid of Combined Systems of Weighted Points

Let $O'$ be a point whose position vector from $O$ is $\mathbf l$.
Let $S = \set {A_1, A_2, \ldots, A_n}$ be a set of $n$ points in Euclidean space.
Let $W_S: S \to \R$ be a weight function on $S$.
Let $T = \set {B_1, B_2, \ldots, B_m}$ be a set of $m$ points in Euclidean space.
Let $W_T: T \to \R$ be a weight function on $T$.
Let $H$ be the centroid of $S$ with weight function $W_S$.
Let $H'$ be the centroid of $T$ with weight function $W_T$.
Then the centroid of $S \cup T$ with weight function $W_S \cup W_T$ is the centroid of the two points $H$ and $H'$ with weights $\ds \sum_{i \mathop = 1}^n \map {W_S} {A_i}$ and $\ds \sum_{i \mathop = 1}^m \map {W_T} {B_m}$ respectively.

Proof:
Let the position vectors of the points in $S = \set {A_1, A_2, \ldots, A_n}$ be given by $\mathbf a_1, \mathbf a_2, \dotsc, \mathbf a_n$ repectively.
Let the position vectors of the points in $T = \set {B_1, B_2, \ldots, B_m}$ be given by $\mathbf b_1, \mathbf b_2, \dotsc, \mathbf b_n$ repectively.
We have that:
:$\vec OH = \dfrac {\ds \sum_{i \mathop = 1}^n \map {W_S} {A_i} \mathbf a_i} {\ds \sum_{i \mathop = 1}^n \map {W_S} {A_i} }$
and:
:$\vec OH' = \dfrac {\ds \sum_{i \mathop = 1}^m \map {W_T} {B_i} \mathbf b_i} {\ds \sum_{i \mathop = 1}^m \map {W_T} {B_i} }$
Let $\vec OG$ denote the centroid of the two points $H$ and $H'$ with weights $\ds \sum_{i \mathop = 1}^n \map {W_S} {A_i}$ and $\ds \sum_{i \mathop = 1}^m \map {W_T} {B_m}$ respectively.
Then:
{{begin-eqn}}
{{eqn | l = \vec OG
      | r = \dfrac {\paren {\ds \sum_{i \mathop = 1}^n \map {W_S} {A_i} } \vec {OH} + \paren {\ds \sum_{i \mathop = 1}^m \map {W_T} {B_i} } \vec {OH'} } {\ds \sum_{i \mathop = 1}^n \map {W_S} {A_i} + \ds \sum_{i \mathop = 1}^n \map {W_S} {A_i} }
      | c = 
}}
{{eqn | r = \dfrac {\ds \sum_{i \mathop = 1}^n \map {W_S} {A_i} \mathbf a_i + \ds \sum_{i \mathop = 1}^m \map {W_T} {B_i} \mathbf b_i} {\ds \sum_{i \mathop = 1}^n \map {W_S} {A_i} + \ds \sum_{i \mathop = 1}^n \map {W_S} {A_i} }
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Unit Vector in Direction of Vector

Let $\mathbf v$ be a vector quantity.
The '''unit vector''' $\mathbf {\hat v}$ in the direction of $\mathbf v$ is:
:$\mathbf {\hat v} = \dfrac {\mathbf v} {\norm {\mathbf v} }$
where $\norm {\mathbf v}$ is the magnitude of $\mathbf v$.

Proof:
From Vector Quantity as Scalar Product of Unit Vector Quantity:
:$\mathbf v = \norm {\mathbf v} \mathbf {\hat v}$
whence the result.
{{qed}}