avewright/proofwiki-math huggingface dataset:
Theorem: Continuity Test using Sub-Basis

Let $\struct {X_1, \tau_1}$ and $\struct {X_2, \tau_2}$ be topological spaces.
Let $f: X_1 \to X_2$ be a mapping.
Let $\SS$ be an analytic sub-basis for $\tau_2$.
Suppose that:
:$\forall S \in \SS: f^{-1} \sqbrk S \in \tau_1$
where $f^{-1} \sqbrk S$ denotes the preimage of $S$ under $f$.
Then $f$ is continuous.

Proof:
Let $\tau$ be the final topology on $X_2$ with respect to $f$.
By hypothesis, $\SS \subseteq \tau$.
By Synthetic Sub-Basis and Analytic Sub-Basis are Compatible, we have that $\tau_2$ is the topology generated by the synthetic sub-basis $\SS$.
By the definition of the generated topology, we have $\tau_2 \subseteq \tau$.
By the definition of the final topology, it follows that $f$ is continuous.
{{qed}}
Theorem: Final Topology is Topology

Let $X$ be a set.
Let $I$ be an indexing set.
Let $\family {\struct {Y_i, \tau_i} }_{i \mathop \in I}$ be an $I$-indexed family of topological spaces.
Let $\family {f_i: Y_i \to X}_{i \mathop \in I}$ be an $I$-indexed family of mappings.
Let $\tau$ be the final topology on $X$ with respect to $\family {f_i}_{i \mathop \in I}$.
Then $\tau$ is a topology on $X$.

Proof:
Define:
:$\forall i \in I: \vartheta_i = \set {U \subseteq X: \map {f_i^{-1} } U \in \tau_i} \subseteq \powerset X$
Then, by the definition of intersection:
:$\ds \tau = \bigcap_{i \mathop \in I} \vartheta_i$
From the Intersection of Topologies is Topology, it suffices to show, for all $i \in I$, that $\vartheta_i$ is a topology on $X$.
We now verify the axioms for $\vartheta_i$ to be a topology on $X$.
Theorem: Subspace Topology is Initial Topology with respect to Inclusion Mapping

Let $\struct {X, \tau}$ be a topological space.
Let $Y$ be a non-empty subset of $X$.
Let $\iota: Y \to X$ be the inclusion mapping.
Let $\tau_Y$ be the initial topology on $Y$ with respect to $\iota$.
Then $\struct {Y, \tau_Y}$ is a topological subspace of $\struct {X, \tau}$.
That is:
:$\tau_Y = \set {U \cap Y: U \in \tau}$

Proof:
By Initial Topology with respect to Mapping equals Set of Preimages, it follows that:
:$\tau_Y = \set {\iota^{-1} \sqbrk U: U \in \tau}$
From Preimage of Subset under Inclusion Mapping, we have:
:$\forall S \subseteq X: \iota^{-1} \sqbrk S = S \cap Y$
Hence the result.
{{qed}}
Category:Topological Subspaces
Category:Inclusion Mappings
Category:Initial Topology
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Theorem: Topological Subspace is Topological Space

Let $\struct {X, \tau}$ be a topological space.
Let $H \subseteq X$ be a non-empty subset of $X$.
Let $\tau_H = \set {U \cap H: U \in \tau}$ be the subspace topology on $H$.
Then the topological subspace $\struct {H, \tau_H}$ is a topological space.

Proof:
We verify the open set axioms for $\tau_H$ to be a topology on $H$.
=== {{Open-set-axiom|1|nolink}} ===
Let $\AA \subseteq \tau_H$.
It is to be shown that:
:$\ds \bigcup \AA \in \tau_H$
Define:
:$\ds \AA' = \set {V \in \tau: V \cap H \subseteq \bigcup \AA} \subseteq \tau$
Let:
:$\ds U = \bigcup \AA'$
By the definition of a topology, we have $U \in \tau$.
Then, from Intersection Distributes over Union and by Union is Smallest Superset: Family of Sets:
:$\ds U \cap H = \bigcup_{V \mathop \in \AA'} \paren {V \cap H} \subseteq \bigcup \AA$
By the definition of $\tau_H$ and by Set is Subset of Union: General Result, we have:
:$\ds \forall S \in \AA: \exists V \in \tau: S = V \cap H \subseteq \bigcup \AA$
That is:
:$\forall S \in \AA: \exists V \in \AA': S = V \cap H$
By Set is Subset of Union: General Result, we have:
:$\ds \forall V \in \AA': V \subseteq U$
Since intersection preserves subsets, it follows that:
:$\forall S \in \AA: S \subseteq U \cap H$
By Union is Smallest Superset: General Result, we conclude that:
:$\ds \bigcup \AA \subseteq U \cap H$
Hence, by definition of set equality:
:$\ds \bigcup \AA = U \cap H \in \tau_H$
{{qed|lemma}}
=== {{Open-set-axiom|2|nolink}} ===
Let $A, B \in \tau_H$.
Let $U, V \in \tau$ be such that $A = U \cap H$ and $B = V \cap H$.
By the definition of a topology, we have $U \cap V \in \tau$.
From Set Intersection is Self-Distributive:
:$A \cap B = \paren {U \cap V} \cap H \in \tau_H$
{{qed|lemma}}
=== {{Open-set-axiom|3|nolink}} ===
By the definition of a topology, we have $X \in \tau$.
By Intersection with Subset is Subset, it follows that $H = X \cap H \in \tau_H$.
{{qed|lemma}}
All the open set axioms are fulfilled, and the result follows.
{{qed}}
Theorem: Topological Subspace is Topological Space

Let $\struct {X, \tau}$ be a topological space.
Let $H \subseteq X$ be a non-empty subset of $X$.
Let $\tau_H = \set {U \cap H: U \in \tau}$ be the subspace topology on $H$.
Then the topological subspace $\struct {H, \tau_H}$ is a topological space.

Proof:
Follows directly from Subspace Topology is Initial Topology with respect to Inclusion Mapping.
{{qed}}
Theorem: Initial Topology with respect to Mapping equals Set of Preimages

Let $X$ be a set.
Let $\struct {Y, \tau_Y}$ be a topological space.
Let $f: X \to Y$ be a mapping.
Let $\tau_X$ be the initial topology on $X$ {{WRT}} $f$.
Then:
:$\tau_X = \set {f^{-1} \sqbrk U: U \in \tau_Y}$

Proof:
Define:
:$\tau = \set {f^{-1} \sqbrk U: U \in \tau_Y}$
By definition, $\tau_X$ is the topology generated by $\tau$.
Therefore:
:$\tau \subseteq \tau_X$
If $\tau$ is a topology on $X$, then it follows from the definition of the generated topology that:
:$\tau_X \subseteq \tau$
By definition of set equality:
:$\tau_X = \tau$
Hence, it suffices to prove that $\tau$ is a topology on $X$.
We now verify the open set axioms for $\tau$ to be a topology on $X$.
Theorem: Sommerfeld-Watson Transform

Let $\map f z$ be a mapping with isolated poles.
{{explain|Explain the context in which this theorem is placed. For example, what is the domain and range of $f$? One supposes $\C$ but it needs to be made clear.}}
Let $f$ go to zero faster than $\dfrac 1 {\size z}$ as $\size z \to \infty$.
{{explain|"faster"}}
Let $C$ be a contour that is deformed such that all poles of $\map f z$ are contained in $C$.
Then:
:$\ds \sum \limits_{n \mathop = -\infty}^\infty \paren {-1}^n \map f n = \frac 1 {2 i} \oint_C \frac {\map f z} {\sin \pi z} \rd z$

Proof:
From Cauchy's Residue Theorem:
{{begin-eqn}}
{{eqn | l = \oint_C \map f z \rd z
      | r = 2 \pi i \, \sum \limits_{z_k} R_k(z_k)
      | c = 
}}
{{eqn | r = 2 \pi i \, \sum_{z_k} \lim_{z \mathop \to z_k} \paren {\paren {z - z_k} \frac {\map f z} {\sin \pi z} }
      | c = 
}}
{{end-eqn}}
This is for poles $z_k$ at order $N = 1$ because we say that simple poles exist for $\dfrac {\map f z} {\sin \pi z}$.
{{explain|Rewrite the above comprehensibly}}
Using l'Hôpital's rule:
{{begin-eqn}}
{{eqn | l = \oint_C \map f z \rd z
      | r = 2 \pi i \sum_{z_k} \lim_{z \mathop \to z_k} \paren {\frac {\partial_z \paren {z - z_k} \map f z} {\partial_z \sin \pi z} }
      | c = 
}}
{{eqn | r = 2 \pi i \sum_{z_k} \lim_{z \mathop \to z_k} \paren {\frac {\map f z + \paren {z - z_k} \map {f'} z} {\pi \cos \pi z} }
      | c = 
}}
{{end-eqn}}
{{explain|$\partial_z$}}
But $\sin \pi z$ has poles at $z_k = n$ for some $n \in \Z$ which implies:
{{begin-eqn}}
{{eqn | l = \oint_C \map f z \rd z
      | r = 2 \pi i \sum_{n \mathop = -\infty}^\infty \lim_{z \mathop \to n} \paren {\frac {\map f z + \paren {z - n} \map {f'} z} {\pi \cos \pi z} }
      | c = 
}}
{{eqn | r = 2 i \sum_{n \mathop = -\infty}^\infty \paren {\frac {\map f n} {\cos \pi n} }
      | c = 
}}
{{end-eqn}}
Finally:
:$\dfrac 1 {\cos \pi n} = \cos \pi n = \paren {-1}^n$
Therefore:
:$\ds \frac 1 {2 i} \oint_C \map f z \rd z = \sum_{n \mathop = -\infty}^\infty \paren {-1}^n \map f n$
{{qed}}
{{Namedfor|Arnold Johannes Wilhelm Sommerfeld|name2 = George Neville Watson|cat = Sommerfeld|cat2 = Watson}}
Category:Complex Analysis
blgk7kkx9jou79aols4np58m5ih7zfn
Theorem: Linear Transformation of Arithmetic Mean

Let $D = \set {x_0, x_1, x_2, \ldots, x_n}$ be a set of real data describing a quantitative variable.
Let $\overline x$ be the arithmetic mean of the data in $D$.
Let $T: \R \to \R$ be a linear transformation such that:
:$\forall i \in \set {0, 1, \ldots, n}: \map T {x_i} = \lambda x_i + \gamma$
Let $T \sqbrk D$ be the image of $D$ under $T$.
Then the arithmetic mean of the data in $T \sqbrk D$ is given by:
:$\map T {\overline x} = \lambda \overline x + \gamma$

Proof:
Follows from the definition of arithmetic mean and from Summation is Linear.
{{qed}}
Theorem: Linear Transformation of Arithmetic Mean

Let $D = \set {x_0, x_1, x_2, \ldots, x_n}$ be a set of real data describing a quantitative variable.
Let $\overline x$ be the arithmetic mean of the data in $D$.
Let $T: \R \to \R$ be a linear transformation such that:
:$\forall i \in \set {0, 1, \ldots, n}: \map T {x_i} = \lambda x_i + \gamma$
Let $T \sqbrk D$ be the image of $D$ under $T$.
Then the arithmetic mean of the data in $T \sqbrk D$ is given by:
:$\map T {\overline x} = \lambda \overline x + \gamma$

Proof:
This is a direct application of Expectation is Linear.
{{qed}}
Theorem: Category of Pointed Sets is Category

Let $\mathbf{Set}_*$ be the category of pointed sets.
Then $\mathbf{Set}_*$ is a metacategory.

Proof:
The axioms $(\text C 1)$ to $(\text C 3)$ are checked for a metacategory.
Pick any two morphisms $f : \struct {A, a} \to \struct {B, b}$ and $g : \struct {B, b} \to \struct {C, c}$ from $\mathbf{Set}_*$.
By the definition of composition in the category of pointed sets:
:$\map {\paren {g \circ f} } a = \map g {\map f a} = \map g b = c$
whence $g \circ f$ is a pointed mapping from $\struct {A, a}$ to $\struct {C, c}$.
That composition of pointed mappings is associative follows from Composition of Mappings is Associative.
For any object $\struct {A, a}$, the identity mapping $\operatorname{id}_A$ induces a pointed map $\operatorname{id}_{\struct {A, a} }: \struct {A, a} \to \struct {A, a}$, as $\map {\operatorname{id}_A} a = a$.
By Identity Mapping is Left Identity and Identity Mapping is Right Identity, this is the identity morphism for $\struct {A, a}$.
{{qed}}
Category:Category Theory
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Theorem: Closed Set in Topological Subspace/Corollary

Let $H$ be closed in $T$.
Then $V \subseteq H$ is closed in $T'$ {{iff}} $V$ is closed in $T$.

Proof:
Let $V \subseteq H$ be closed in $T'$.
Then, from Closed Set in Topological Subspace, $V = H \cap V$ is closed in $T'$.
If $V$ is closed in $T'$ then $V = H \cap W$ where $W$ is closed in $T$.
Since $H$ is closed in $T$, it follows by Topology Defined by Closed Sets that $V$ is closed in $T$.
{{qed}}
Theorem: Unit Interval is Path-Connected in Real Numbers

Let $\R$ be the real number line with the usual (Euclidean} metric.
The closed unit interval $\I = \closedint 0 1$ is a path-connected metric subspace of $\R$.

Proof:
Follows directly from Subset of Real Numbers is Path-Connected iff Interval.
{{qed}}
Theorem: Subset of Real Numbers is Path-Connected iff Interval

Let $\R$ be the real number line considered as an Euclidean space.
Let $S \subseteq \R$ be a subset of $\R$.
Then $S$ is a path-connected metric subspace of $\R$ {{iff}} $S$ is a real interval.

Proof:
=== Necessary Condition ===
Let $S$ be a path-connected metric subspace of $\R$.
From Path-Connected Space is Connected, it follows that $S$ is connected.
From Subset of Real Numbers is Interval iff Connected, it follows that $S$ is a real interval.
{{qed|lemma}}
Theorem: Size of Linearly Independent Subset is at Most Size of Finite Generator

Let $R$ be a division ring.
Let $V$ be an $R$-vector space.
Let $F \subseteq V$ be a finite generator of $V$ over $R$.
Let $L \subseteq V$ be linearly independent over $R$.
Then:
:$\size L \le \size F$

Proof:
Let $S \subseteq \N$ be the set of all natural numbers $n \in \N$ such that:
:For any finite generator $F$ of $V$ over $R$, if $\card {F \cap L} \ge n$, then $\card L \le \card F$.
It is to be demonstrated that $S = \N$.
That is, that $\card {F \cap L} \ge n \implies \card L \le \card F$ for all $n \in \N$.
By Intersection is Subset and Cardinality of Subset of Finite Set, we have $\card {F \cap L} \le \card F$.
Hence, it is vacuously true that $\card F + 1 \in S$.
Therefore, $S$ is non-empty.
From the well-ordering principle, $S$ has a smallest element $N$.
If $N = 0$, the theorem immediately follows.
{{AimForCont}} $N \ge 1$.
Let $\card {F \cap L} \ge N - 1$.
If $L \subseteq F$, the theorem follows from Cardinality of Subset of Finite Set.
Otherwise, there exists a $v \in L$ such that $v \notin F$.
Let $F' = F \cup \set v$.
By Intersection is Subset, we have $F' \cap L \subseteq L$, so it follows by Subset of Linearly Independent Set is Linearly Independent that $F' \cap L$ is linearly independent over $R$.
Also, by Intersection is Subset:
:$F' \cap L \subseteq F'$
We have that $F'$ is a generator of $V$ over $R$.
By Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set, there exists a basis $B$ for $V$ such that:
:$F' \cap L \subseteq B \subseteq F'$
Since $v \notin F$ is a linear combination of $F$, it follows that $F'$ is linearly dependent over $R$.
By the definition of a basis:
:$B \subsetneq F'$
By Cardinality of Subset of Finite Set and Cardinality is Additive Function:
:$\card B < \card {F'} = \card F + 1$
Hence:
:$\card B \le \card F$
We have that:
{{begin-eqn}}
{{eqn | l = \card {B \cap L}
      | o = \ge
      | r = \card {\paren {F' \cap L} \cap L}
      | c = Set Intersection Preserves Subsets and Cardinality of Subset of Finite Set
}}
{{eqn | r = \card {F' \cap L}
      | c = Intersection is Associative and Set Intersection is Idempotent
}}
{{eqn | r = \card {\paren {F \cap L} \cup \paren {\set v \cap L} }
      | c = Intersection Distributes over Union
}}
{{eqn | r = \card {\paren {F \cap L} \cup \set v}
      | c = Intersection with Subset is Subset
}}
{{eqn | o = \ge
      | r = N
      | c = Cardinality is Additive Function
}}
{{end-eqn}}
Since $N \in S$, it follows by the definition of a basis that:
:$\card L \le \card B \le \card F$
Hence:
:$N - 1 \in S$
But this contradicts the assumption that $N$ is the smallest element of $S$.
Therefore:
:$N = 0$
Hence the result.
{{qed}}
Theorem: Size of Linearly Independent Subset is at Most Size of Finite Generator

Let $R$ be a division ring.
Let $V$ be an $R$-vector space.
Let $F \subseteq V$ be a finite generator of $V$ over $R$.
Let $L \subseteq V$ be linearly independent over $R$.
Then:
:$\size L \le \size F$

Proof:
Let $\alpha_1, \alpha_2, \ldots, \alpha_n$ be a generator of $V$.
Let $\xi_1, \xi_2, \ldots, \xi_r$ be a linearly independent set of elements of $V$.
Hence the sequence $\sequence {\xi_1, \alpha_1, \alpha_2, \ldots, \alpha_n}$ is a linearly dependent sequence of elements of $V$.
One of these elements, which cannot be $\xi_1$, is a linear combination of the preceding elements.
Let this elements be $\alpha_i$.
So we can omit $\alpha_i$ from that sequence, and the remaining set is still a generator of $V$.
Therefore $\xi_2$ is a linear combination of these.
Thus $\sequence {\xi_2, \xi_1, \alpha_1, \alpha_2, \ldots, \alpha_{i - 1}, \alpha_{i + 1}, \ldots, \alpha_n}$ is a linearly dependent sequence of elements of $V$.
Again, one of them is a linear combination of the preceding elements.
This cannot be $\xi_2$, as $\xi_2$ has no preceding elements.
Neither can it be $\xi_1$, as $\xi_1$ and $\xi_2$ are linearly independent.
Thus we can omit whichever $\alpha_j$ it is, and we have a new set which is a generator of $V$.
This consists of $\xi_1$, $\xi_2$ and whichever $n - 2$ of the remaining elements of $\set {\alpha_1, \alpha_2, \ldots, \alpha_n}$.
After $p$ such steps, we have a set which is a generator of $V$ which consists of:
:$\xi_1, \xi_2, \ldots, \xi_p$ and $n - p$ of the elements of $\set {\alpha_1, \alpha_2, \ldots, \alpha_n}$.
{{AimForCont}} suppose $n < r$.
Then when $p = n$, the remaining set which is a generator of $V$ consists of:
:$\xi_1, \xi_2, \ldots, \xi_n$
and there is at least one more element $\xi_{n + 1}$.
This is a linear combination of $\set {\xi_1, \xi_2, \ldots, \xi_n}$.
But this contradicts the supposition that $\set {\xi_1, \xi_2, \ldots, \xi_n, \xi_{n + 1} }$ is a linearly independent set.
Hence, by Proof by Contradiction, $n \ge r$.
The result follows.
{{qed}}
Theorem: Intermediate Value Theorem/Corollary

Let $0 \in \R$ lie between $\map f a$ and $\map f b$.
That is, either:
:$\map f a < 0 < \map f b$
or:
:$\map f b < 0 < \map f a$
Then $f$ has a root in $\openint a b$.

Proof:
Follows directly from the Intermediate Value Theorem and from the definition of root.
{{qed}}
Theorem: Real Number Line with Point Removed is Not Path-Connected

Let $\R$ be the real number line considered as an Euclidean space.
Let $x \in \R$ be a real number.
Then $\R \setminus \set x$, where $\setminus$ denotes set difference, is not path-connected.

Proof:
We have that $x - 1$ and $x + 1$ are both real numbers, so:
:$x - 1 \in \R \setminus \set x$
:$x + 1 \in \R \setminus \set x$
Let $\I := \closedint 0 1$ be the closed unit interval.
{{AimForCont}} there exists a path $f: \I \to \R \setminus \set x$ from $x - 1$ to $x + 1$.
Then by Image of Real Interval under Continuous Real Function is Real Interval, it follows that $x = \map f s$ for some $s \in \I$.
But $x \notin \R \setminus \set x$ by definition of set difference.
This contradicts our assertion that such an $f$ exists.
Hence $f$ cannot exist.
The result follows by definition of path-connected.
{{qed}}
Theorem: Euclidean Space is Path-Connected

Let $\R^n$ be the $n$-dimensional Euclidean space for $n \in \N$ a natural number.
Then $\R^n$ is path-connected.

Proof:
Let $\mathbf x, \mathbf y \in \R^n$ be arbitrary points of $\R^n$.
Define $l: \closedint 0 1 \to \R^n$ by:
:$\map l t = \paren {1 - t} \mathbf x + t \mathbf y$
Then $\map l 0 = 1 \mathbf x + 0 \mathbf y = \mathbf x$, whereas $\map l 1 = 0 \mathbf x + 1 \mathbf y = \mathbf y$.
Finally, it remains to show that $l$ is continuous. 
Fix $\epsilon > 0$ and suppose that $t, t' \in \closedint 0 1$ are such that $\size {t - t'} < \dfrac {\epsilon} {1 + \norm {\mathbf x} + \norm {\mathbf y} }$.
Then:
{{begin-eqn}}
{{eqn | l = \map l t - \map l {t'}
      | r = \paren {1 - t} \mathbf x + t \mathbf y - \paren {\paren {1 - t'} \mathbf x + t' \mathbf y}
}}
{{eqn | r = \paren {\paren {1 - t} - \paren {1 - t'} } \mathbf x + \paren {t - t'} \mathbf y
}}
{{eqn | r = \paren {t' - t} \mathbf x + \paren {t - t'} \mathbf y
}}
{{end-eqn}}
We can now estimate the norm of this last expression:
{{begin-eqn}}
{{eqn | l = \norm {\paren {t' - t} \mathbf x + \paren {t - t'} \mathbf y}
      | o = \le
      | r = \norm {\paren {t' - t} \mathbf x} + \norm {\paren {t - t'} \mathbf y}
      | c = Triangle Inequality for Vectors in Euclidean Space
}}
{{eqn | r = \size {t' - t} \norm {\mathbf x} + \size {t - t'} \norm {\mathbf y}
      | c = {{NormAxiomVector|2}}
}}
{{eqn | r = \size {t' - t} \paren {\norm {\mathbf x} + \norm {\mathbf y} }
}}
{{eqn | o = <
      | r = \frac \epsilon {1 + \norm {\mathbf x} + \norm {\mathbf y} } \paren {\norm {\mathbf x} + \norm {\mathbf y} }
}}
{{eqn | o = <
      | r = \epsilon
}}
{{end-eqn}}
Since $\epsilon$ was arbitrary, we conclude that $l$ is continuous.
Therefore, it forms a path from $\mathbf x$ to $\mathbf y$.
Since $\mathbf x$ and $\mathbf y$ were arbitrary, it follows that $\R^n$ is path-connected.
{{qed}}
Theorem: Coequalizer is Epimorphism

Let $\mathbf C$ be a metacategory.
Let $q: D \to Q$ be the coequalizer of two morphisms $f, g: C \to D$.
Then $q$ is an epimorphism.

Proof:
Follows directly from Equalizer is Monomorphism and the Duality Principle.
{{qed}}
Theorem: Ordering on Multiindices is Partial Order

Let $Z$ be the set of multiindices indexed by a set $J$.
The ordering on $Z$ is a partial ordering.

Proof:
Let $\le$ denote the ordering on integers.
Let $\preceq$ denote the ordeing on multiindices.
Recall that if $k = \family {k_j}_{j \mathop \in J}$ and $\ell = \family {\ell_j}_{j \mathop \in J}$ are multiindices, then $k \le \ell$ if $k_j \preceq \ell_j$ for all $j \in J$.
Let $k$ and $\ell$ be as above, and let $m = \family {m_j}_{j \mathop \in J} \in Z$.
From Integers under Addition form Totally Ordered Group, the integers are totally ordered by $\le$.
Hence the ordering on the integers is reflexive, antisymmetric and transitive.
Theorem: Euclidean Space without Origin is Path-Connected

Let $n \in \Z: n \ge 2$.
Let $\R^n$ be the $n$-dimensional Euclidean space.
Let $\R^n \setminus \set {\mathbf 0}$ be $\R^n$ with the origin removed.
Then $\R^n \setminus \set {\mathbf 0}$ is path-connected.

Proof:
{{proof wanted|Another day, this is tedious.}}
Theorem: Multiindices under Addition form Commutative Monoid

Let $Z$ be the set of multiindices.
Let $+$ denote the addition of multiindices.
Then $\left({Z, +}\right)$ is a commutative monoid.

Proof:
We check each of the axioms in turn.
Let $k = \left \langle {k_j}\right \rangle_{j \in J}$, $\ell = \left \langle {\ell_j}\right \rangle_{j \in J}$ and $m = \left \langle {m_j}\right \rangle_{j \in J}$ be multiindices.
Theorem: Composite of Continuous Mappings is Continuous/Corollary

Let $T_1, T_2, T_3$ each be one of:
:metric spaces
:the complex plane
:the real number line
Let $f: T_1 \to T_2$ and $g: T_2 \to T_3$ be continuous mappings.
Then the composite mapping $g \circ f: T_1 \to T_3$ is continuous.

Proof:
These follow directly from:
* Real Number Line is Metric Space
* Complex Plane is Metric Space
* Metric Induces Topology
{{explain|How the claim follows from these links? Noting is proved about the continuity of the composition.}}
{{qed}}
Theorem: Binomial Theorem/Multiindex

Let $\alpha$ be a multiindex, indexed by $\set {1, \ldots, n}$ such that $\alpha_j \ge 0$ for $j = 1, \ldots, n$.
Let $x = \tuple {x_1, \ldots, x_n}$ and $y = \tuple {y_1, \ldots, y_n}$ be ordered tuples of real numbers.
Then:
:$\ds \paren {x + y}^\alpha = \sum_{0 \mathop \le \beta \mathop \le \alpha} \dbinom \alpha \beta x^\beta y^{\alpha - \beta}$
where $\dbinom n k$ is a binomial coefficient.

Proof:
First of all, by definition of multiindexed powers:
:$\ds \paren {x + y}^\alpha = \prod_{k \mathop = 1}^n \paren {x_k + y_k}^{\alpha_k}$
Then:
{{begin-eqn}}
{{eqn | l = \paren {x + y}^\alpha
      | r = \prod_{k \mathop = 1}^n \sum_{\beta_k \mathop = 0}^{\alpha_k} \dbinom {\alpha_k} {\beta_k} x_k^{\alpha_k - \beta_k} y_k^{\beta_k}
      | c = Binomial Theorem for Integral Index
}}
{{eqn | r = \sum_{\beta_1 \mathop = 0}^{\alpha_1} \cdots \sum_{\beta_n \mathop = 0}^{\alpha_n} \dbinom {\alpha_1} {\beta_1} \cdots \dbinom {\alpha_n} {\beta_n} x_1^{\alpha_1 - \beta_1} \cdots x_n^{\alpha_n - \beta_n} y_1^{\beta_1} \cdots y_n^{\beta_n}
      | c = expanding out the product
}}
{{end-eqn}}
On the other hand:
{{begin-eqn}}
{{eqn | l = \sum_{0 \mathop \le \beta \mathop \le \alpha} \dbinom \alpha \beta x^\beta y^{\alpha - \beta}
      | r = \sum_{0 \mathop \le \beta \mathop \le \alpha} \dbinom {\alpha_1} {\beta_1} \cdots \dbinom {\alpha_n} {\beta_n} x_1^{\alpha_1 - \beta_1} \cdots x_n^{\alpha_n - \beta_n} y_1^{\beta_1} \cdots y_n^{\beta_n}
      | c = {{Defof|Power (Algebra)/Multiindices|Power}} and {{Defof|Binomial Coefficient/Multiindices|Binomial Coefficient}}
}}
{{eqn | r = \sum_{\beta_1 \mathop = 0}^{\alpha_1} \cdots \sum_{\beta_n \mathop = 0}^{\alpha_n} \dbinom {\alpha_1} {\ell_1}  \cdots \dbinom {\alpha_n} {\ell_n} x_1^{\alpha_1 - \ell_1} \cdots x_n^{\alpha_n - \ell_n} y_1^{\ell_1} \cdots y_n^{\ell_n}
      | c = {{Defof|Ordering on Multiindices}}
}}
{{end-eqn}}
{{Mistake|In the second of the equations above, $\ell$ has not been introduced.}}
This shows that:
:$\ds \paren {x + y}^\alpha = \sum_{0 \mathop \le \beta \mathop \le \alpha} \dbinom \alpha \beta x^\beta y^{\alpha - \beta}$
as required.
{{Qed}}
{{Proofread}}
Category:Binomial Theorem
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