avewright/proofwiki-math huggingface dataset:
Theorem: Reverse Young's Inequality for Products

Let $p, q \in \R_{> 0}$ be strictly positive real numbers satisfying:
:$\dfrac 1 p - \dfrac 1 q = 1$
Let $a \in \R_{\ge 0}$ be a positive real number.
Let $b \in \R_{> 0}$ be a strictly positive real number.
Then:
:$a b \ge \dfrac {a^p} p - \dfrac {b^{-q} } q$

Proof:
Define:
{{begin-eqn}}
{{eqn | n = 1
      | l = u
      | r = \frac 1 p
}}
{{eqn | l = v
      | r = \frac q p
}}
{{eqn | l = x
      | r = \paren {a b}^p
}}
{{eqn | l = y
      | r = b^{-p}
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \frac 1 u + \frac 1 v
      | r = p + \frac p q
}}
{{eqn | r = p \paren {1 + \frac 1 q}
}}
{{eqn | r = p \paren {\frac 1 p}
      | c = because $\dfrac 1 p - \dfrac 1 q = 1$, {{hypothesis}}
}}
{{eqn | r = 1
}}
{{end-eqn}}
Thus Young's Inequality for Products can be applied:
{{begin-eqn}}
{{eqn | l = x y
      | o = \le
      | r = \frac {x^u} u + \frac {y^v} v
}}
{{eqn | ll= \leadsto
      | l = \paren {a b}^p b^{-p}
      | o = \le
      | r = \frac {\paren {\paren {a b}^p}^{1/p} } {1/p} + \frac {\paren {b^{-p} }^{q/p} } {q/p}
      | c = substituting for $u, v, x, y$ from $(1)$ above
}}
{{eqn | ll= \leadsto
      | l = a^p
      | o = \le
      | r = p a b + p \frac {b^{-q} } q
      | c = algebraic simplification
}}
{{eqn | ll= \leadsto
      | l = \frac {a^p} p
      | o = \le
      | r = a b + \frac {b^{-q} } q
}}
{{eqn | ll= \leadsto
      | l = a b
      | o = \ge
      | r = \frac {a^p} p - \frac {b^{-q} } q
}}
{{end-eqn}}
{{qed}}
Theorem: Reverse Hölder's Inequality for Sums

Let $p, q \in \R_{>0}$ be strictly positive real numbers such that $\dfrac 1 p - \dfrac 1 q = 1$.
Suppose that the sequences $\mathbf x = \sequence {x_n}$ and $\mathbf y = \sequence {y_n}$ in $\C$ (or $\R$) are such that the series
:$\ds \paren {\sum_{n \mathop = 1}^\infty \size {x_n}^p}^{1/p}$
and
:$\ds \paren {\sum_{n \mathop = 1}^\infty \size {y_n}^{-q} }^{-1/q}$
are convergent.
Denote these values by $\norm {\mathbf x}_p$ and $\norm {\mathbf y}_{-q}$, respectively.
Here, the notation $\norm {\mathbf x}_p$ does not denote a norm, but is instead just a convenient notation similar to that of the $p$-norm, which is only defined when $p \ge 1$.
Let $\norm {\mathbf {x y} }_1$ denote the $1$-norm of $\mathbf {x y}$, if $\mathbf {x y}$ is in the Lebesgue space $\ell^1$.
Then $\norm {\mathbf {x y} }_1 \ge \norm {\mathbf x}_p \norm {\mathbf y}_{-q}$.

Proof:
{{WLOG}}, assume that $\mathbf x$ and $\mathbf y$ are non-zero.
Let:
:$\mathbf u = \sequence {u_n} = \dfrac {\mathbf x} {\norm {\mathbf x}_p}$
and:
:$\mathbf v = \sequence {v_n} = \dfrac {\mathbf y} {\norm {\mathbf y}_{-q} }$
Then:
:$\ds \norm {\mathbf u}_p = \dfrac 1 {\norm {\mathbf x}_p} \paren {\sum_{n \mathop = 1}^\infty \size {x_n}^p}^{1/p} = 1$
Similarly:
:$\norm {\mathbf v}_{-q} = 1$
It then suffices to prove that:
:$\norm {\mathbf {u v} }_1 = \dfrac {\norm {\mathbf {x y} }_1} {\norm {\mathbf x}_p \norm {\mathbf y}_{-q} } \ge 1$
By Reverse Young's Inequality for Products:
:$\size {u_n v_n} \ge \dfrac 1 p \size {u_n}^p - \dfrac 1 q \size {v_n}^{-q}$
Summing over all $n \in \N$ gives:
:$\norm {\mathbf {u v} }_1 \ge \dfrac 1 p \norm {\mathbf u}_p - \dfrac 1 q \norm {\mathbf v}_{-q} = 1$
as desired.
{{qed}}
Theorem: Real Vector Space is Vector Space

Let $\R$ be the set of real numbers.
Then the real vector space $\R^n$ is a vector space.

Proof:
=== Construction of Real Vector Space ===
From the definition, a vector space is a unitary module whose scalar ring is a field.
In order to call attention to the precise scope of the operators, let real addition and real multiplication be expressed as $+_\R$ and $\times_\R$ respectively.
Then we can express the field of real numbers as $\struct {\R, +_\R, \times_\R}$.
From Real Numbers under Addition form Group, $\struct {\R, +}$ is a group.
Again, in order to call attention to the precise scope of the operator, let real addition be expressed on $\struct {\R, +}$ as $+_G$.
That is, the group under consideration is $\struct {\R, +_G}$.
Consider the cartesian product:
:$\ds \R^n = \prod_{i \mathop = 1}^n \struct {\R, +_G} = \underbrace {\struct {\R, +_G} \times \cdots \times \struct {\R, +_G} }_{n \text{ copies} }$
Let:
:$\mathbf a = \tuple {a_1, a_2, \ldots, a_n}$
:$\mathbf b = \tuple {b_1, b_2, \ldots, b_n}$
be arbitrary elements of $\R^n$.
Let $\lambda$ be an arbitrary element of $\R$.
Let $+$ be the binary operation defined on $\R^n$ as:
:$\mathbf a + \mathbf b = \tuple {a_1 +_G b_1, a_2 +_G b_2, \ldots, a_n +_G b_n}$
Also let $\cdot$ be the binary operation defined on $\R \times \R^n$ as:
:$\lambda \cdot \mathbf a = \tuple {\lambda \times_\R a_1, \lambda \times_\R a_2, \ldots, \lambda \times_\R a_n}$
In this context, $\lambda \times_\R a_j$ is defined as real multiplication, as is appropriate (both $\lambda$ and $a_j$ are real numbers).
With this set of definitions, the structure $\struct {\R^n, +, \cdot}$ is a vector space, as is shown in Proof of Real Vector Space below.
Theorem: Symmetric Difference is Subset of Union of Symmetric Differences

Let $R, S, T$ be sets.
Then:
:$R \symdif S \subseteq \paren {R \symdif T} \cup \paren {S \symdif T}$
where $R \symdif S$ denotes the symmetric difference between $R$ and $S$.

Proof:
From the definition of symmetric difference, we have:
:$R \symdif S = \paren {R \setminus S} \cup \paren {S \setminus R}$
Then from Set Difference is Subset of Union of Differences, we have:
:$R \setminus S \subseteq \paren {R \setminus T} \cup \paren {T \setminus S}$
:$S \setminus R \subseteq \paren {S \setminus T} \cup \paren {T \setminus R}$
Thus:
{{begin-eqn}}
{{eqn | l = \paren {R \setminus S} \cup \paren {S \setminus R}
      | o = \subseteq
      | r = \paren {R \setminus T} \cup \paren {T \setminus S} \cup \paren {S \setminus T} \cup \paren {T \setminus R}
      | c = Set Union Preserves Subsets
}}
{{eqn | r = \paren {R \setminus T} \cup \paren {T \setminus R} \cup \paren {S \setminus T} \cup \paren {T \setminus S}
      | c = Union is Commutative
}}
{{eqn | r = \paren {R \symdif T} \cup \paren {S \symdif T}
      | c = {{Defof|Symmetric Difference}}
}}
{{end-eqn}}
{{qed}}
Category:Symmetric Difference
Category:Set Union
56944bqs6949sz09i2xmcr8xpngh2bi
Theorem: Sigma-Algebra Contains Empty Set

Let $X$ be a set.
Let $\Sigma$ be a $\sigma$-algebra on $X$.
Then:
:$\O \in \Sigma$

Proof:
Axiom $(1)$ of a $\sigma$-algebra grants:
:$X \in \Sigma$
By axiom $(2)$ and Set Difference with Self is Empty Set, it follows that:
:$\O = X \setminus X \in \Sigma$
{{qed}}
Theorem: Sigma-Algebra Closed under Union

Let $X$ be a set, and let $\Sigma$ be a $\sigma$-algebra on $X$.
Let $A, B \in \Sigma$ be measurable sets.
Then $A \cup B \in \Sigma$, where $\cup$ denotes set union.

Proof:
Define $A_1 = A, A_2 = B$, and for $n \in \N, n \ge 2: A_n = \O$.
Then by Sigma-Algebra Contains Empty Set, axiom $(3)$ of a $\sigma$-algebra applies.
Hence:
:$\ds \bigcup_{n \mathop \in \N} A_n = A \cup B \in \Sigma$
{{qed}}
Theorem: Sigma-Algebra Closed under Countable Intersection

Let $X$ be a set, and let $\Sigma$ be a $\sigma$-algebra on $X$.
Suppose that $\sequence {E_n}_{n \mathop \in \N} \in \Sigma$ is a collection of measurable sets.
Then:
:$\ds \bigcap_{n \mathop \in \N} E_n \in \Sigma$,
where $\ds \bigcap$ denotes set intersection.

Proof:
{{begin-eqn}}
{{eqn | q = \forall n \in \N
      | l = E_n
      | o = \in
      | r = \Sigma
}}
{{eqn | ll= \leadsto
      | l = X \setminus E_n
      | o = \in
      | r = \Sigma
      | c = Axiom $(2)$ for $\sigma$-algebras
}}
{{eqn | ll= \leadsto
      | l = \bigcup_{n \mathop \in \N} \paren {X \setminus E_n}
      | o = \in
      | r = \Sigma
      | c = Axiom $(3)$ for $\sigma$-algebras
}}
{{eqn | ll= \leadsto
      | l = X \setminus \paren {\bigcup_{n \mathop \in \N} \paren {X \setminus E_n} }
      | o = \in
      | r = \Sigma
      | c = Axiom $(2)$ for $\sigma$-algebras
}}
{{end-eqn}}
From De Morgan's laws: Complement of Intersection:
:$\ds \bigcup_{n \mathop \in \N} \paren {X \setminus E_n} = X \setminus \paren {\bigcap_{n \mathop \in \N} E_n}$
Also, by Set Difference with Set Difference and Set Union Preserves Subsets:
:$\ds X \setminus \paren {X \setminus \paren {\bigcap_{n \mathop \in \N} E_n} } = \bigcap_{n \mathop \in \N} E_n$
Combining the previous equalities, it follows that:
:$\ds \bigcap_{n \mathop \in \N} E_n \in \Sigma$
{{qed}}
Theorem: Sigma-Algebra of Countable Sets

Let $X$ be a set.
Let $\Sigma$ be the set of countable and co-countable subsets of $X$.
Then $\Sigma$ is a $\sigma$-algebra.

Proof:
Let us verify in turn the axioms of a $\sigma$-algebra.
Theorem: Trace Sigma-Algebra is Sigma-Algebra

Let $X$ be a set.
Let $\Sigma$ be a $\sigma$-algebra on $X$.
Let $E \subseteq X$ be a subset of $X$.
Then the trace $\sigma$-algebra $\Sigma_E$ is a $\sigma$-algebra on $E$.

Proof:
Verifying the axioms for a $\sigma$-algebra in turn:
Theorem: Pre-Image Sigma-Algebra on Domain is Sigma-Algebra

Let $X, X'$ be sets, and let $f: X \to X'$ be a mapping.
Let $\Sigma'$ be a $\sigma$-algebra on $X'$.
Then $f^{-1} \sqbrk {\Sigma'}$, the pre-image $\sigma$-algebra on the domain of $f$, is a $\sigma$-algebra on $X$.

Proof:
Verify the axioms for a $\sigma$-algebra in turn:
Theorem: Intersection of Sigma-Algebras

Let $X$ be a set.
Let $\family {\Sigma_i}_{i \mathop \in I}$ be a indexed set of $\sigma$-algebras on $X$.
Then $\Sigma := \ds \bigcap_{i \mathop \in I} \Sigma_i$ is also a $\sigma$-algebra on $X$.
Here, $\ds \bigcap$ denotes set intersection.

Proof:
Verify the axioms for a $\sigma$-algebra in turn:
Theorem: Generated Sigma-Algebra Preserves Subset

Let $X$ be a set.
Let $\FF, \GG \subseteq \powerset X$ be collections of subsets of $X$.
Suppose that:
:$\FF \subseteq \GG$
Then:
:$\map \sigma \FF \subseteq \map \sigma \GG$
where $\map \sigma \GG$ denotes the $\sigma$-algebra generated by $\GG$

Proof:
By definition of $\sigma$-algebra generated by $\GG$:
:$\GG \subseteq \map \sigma \GG$
It follows that also:
:$\FF \subseteq \map \sigma \GG$
By definition of $\sigma$-algebra generated by $\FF$:
:$\FF \subseteq \map \sigma \FF$
We are given that:
:$\FF \subseteq \GG$.
Hence:
:$\map \sigma \FF \subseteq \map \sigma \GG$
{{qed}}
Theorem: Characterization of Euclidean Borel Sigma-Algebra

Let $\OO^n$, $\CC^n$ and $\KK^n$ be the collections of open, closed and compact subsets of the Euclidean space $\struct {\R^n, \tau}$, respectively.
Let $\JJ_{ho}^n$ be the collection of half-open rectangles in $\R^n$.
Let $\JJ^n_{ho, \text {rat} }$ be the collection of half-open rectangles in $\R^n$ with rational endpoints.
Then the Borel $\sigma$-algebra $\map \BB {\R^n}$ satisfies:
:$\map \BB {\R^n} = \map \sigma {\OO^n} = \map \sigma {\CC^n} = \map \sigma {\KK^n} = \map \sigma {\JJ_{ho}^n} = \map \sigma {\JJ^n_{ho, \text {rat} } }$
where $\sigma$ denotes generated $\sigma$-algebra.

Proof:
By definition of Borel $\sigma$-algebra, $\map \BB {\R^n} = \map \sigma {\OO^n}$.
The rest of the proof will be split in proving the following equalities:
:$(1): \quad \map \sigma {\OO^n} = \map \sigma {\CC^n}$
:$(2): \quad \map \sigma {\CC^n} = \map \sigma {\KK^n}$
:$(3): \quad \map \sigma {\OO^n} = \map \sigma {\JJ_{ho}^n}$
:$(4): \quad \map \sigma {\JJ_{ho}^n} = \map \sigma {\JJ^n_{ho, \text {rat} } }$
Theorem: Generated Sigma-Algebra Preserves Finiteness

Let $X$ be a set, and let $A_1, \ldots, A_n \subseteq X$ be subsets of $X$.
Then $\map \sigma {\set {A_1, \ldots, A_n} }$ is a finite set, where $\sigma$ denotes generated $\sigma$-algebra.

Proof:
Proceed by induction on $n$, that is, on the number of generators.
Theorem: Set of Invertible Mappings forms Symmetric Group

Let $S$ be a set.
Let $\GG$ be the set of all invertible mappings from $S$ to $S$.
Then $\struct {\GG, \circ}$ is the symmetric group on $S$.

Proof:
Let $\struct {S^S, \circ}$ be the algebraic structure formed from the set of all mappings from $S$ to itself.
From Set of all Self-Maps under Composition forms Monoid, $\struct {S^S, \circ}$ is a monoid.
By Inverse of Permutation is Permutation, if $f$ is a permutation of $S$, then so is its inverse $f^{-1}$.
By Bijection iff Inverse is Bijection, it follows that all the invertible elements of $S^S$ are exactly the permutations on $S$.
The result follows from Invertible Elements of Monoid form Subgroup of Cancellable Elements.
{{qed}}
Theorem: Group of Units of Field

Let $k^\times$ denote the group of units of a field $\struct {k, +, \times}$.
Then:
:$k^\times = k \setminus \set 0$
where:
:$\setminus$ denotes set difference
:$\set 0$ denotes the set containing only the zero of $k$.
That is, $0$ is the only element of $k$ which does not have a multiplicative inverse in $k$.

Proof:
$0$ is not invertible in $k$ since $0 a = 0$ for all $a \in k$.
Thus $0 \notin k^\times$.
Consider now $a \in k$ such that $a \ne 0$.
From {{Field-axiom|M4}} it follows that there exists $a^{-1}$ such that:
:$a \times a^{-1} = 1 = a^{-1} \times a$
where $1$ is the unity of $k$.
So:
:$a \in k^\times$
Hence we have demonstrated that $k^\times$ equals $k$ without $0$, that is:
:$k^\times = k \setminus 0$
{{qed}}
Category:Field Theory
fmwvh2q2cab2ozi3uel1tn6cj0ib9ba
Theorem: Automorphism Group is Subgroup of Symmetric Group

Let $\struct {S, *}$ be an algebraic structure.
Let $\Aut S$ be the automorphism group of $\struct {S, *}$.
Then $\Aut S$ is a subgroup of the symmetric group  $\struct {\Gamma \paren S, \circ}$ on $S$.

Proof:
An automorphism is an isomorphism $\phi: S \to S$ from an algebraic structure $S$ to itself.
The Identity Mapping is Automorphism, so $\Aut S$ is not empty.
The composite of isomorphisms is itself an isomorphism, as demonstrated on Isomorphism is Equivalence Relation.
So:
:$\phi_1, \phi_2 \in \Aut S \implies \phi_1 \circ \phi_2 \in \Aut S$
demonstrating closure.
If $\phi \in \Aut S$, then $\phi$ is bijective and an isomorphism.
Hence from Inverse of Algebraic Structure Isomorphism is Isomorphism, $\phi^{-1}$ is also bijective and an isomorphism.
So $\phi^{-1} \in \Aut S$.
The result follows by the Two-Step Subgroup Test.
{{Qed}}
Theorem: Differentiation of Vector-Valued Function Componentwise

Let:
:$\mathbf r: t \mapsto \tuple {\map {r_1} t, \map {r_2} t, \ldots, \map {r_n} t}$ 
be a differentiable vector-valued function.
The derivative of a vector-valued function can be calculated by differentiating each of its component functions:
:$\dfrac {\d \map {\mathbf r} t} {\d t} = \tuple {\dfrac \d {\d t} \map {r_1} t, \dfrac \d {\d t} \map {r_2} t, \ldots, \dfrac \d {\d t} \map {r_n} t}$

Proof:
{{begin-eqn}}
{{eqn | l = \frac {\d \map {\mathbf r} t} {\d t}
      | r = \lim_{\Delta t \mathop \to 0} \frac {\map {\mathbf r} {t + \Delta t} - \map {\mathbf r} t} {\Delta t}
      | c = {{Defof|Derivative of Vector-Valued Function}}
}}
{{eqn | r = <nowiki>\begin {bmatrix}
\ds \lim_{\Delta t \mathop \to 0} \dfrac {\map {r_1} {t + \Delta t} - \map {r_1} t} {\Delta t} \\ \ds \lim_{\Delta t \mathop \to 0} \dfrac {\map {r_2} {t + \Delta t} - \map {r_2} t} {\Delta t} \\ \vdots \\ \ds \lim_{\Delta t \mathop \to 0} \dfrac {\map {r_n} {t + \Delta t} - \map {r_n} t} {\Delta t}
\end {bmatrix}</nowiki>
      | c = {{Defof|Limit of Vector-Valued Function}}
}}
{{eqn | r = \begin {bmatrix} \dfrac \d {\d t} \map {r_1} t \\ \dfrac \d {\d t} \map {r_2} t \\ \vdots \\ \dfrac \d {\d t} \map {r_n} t \end {bmatrix}
      | c = {{Defof|Derivative of Real Function}}
}}
{{end-eqn}}
{{qed}}
Theorem: Derivative of Dot Product of Vector-Valued Functions

Let $\mathbf a: \R \to \R^n$ and $\mathbf b: \R \to \R^n$ be differentiable vector-valued functions.
The derivative of their dot product is given by:
:$\map {\dfrac \d {\d x} } {\mathbf a \cdot \mathbf b} = \dfrac {\d \mathbf a} {\d x} \cdot \mathbf b + \mathbf a \cdot \dfrac {\d \mathbf b} {\d x}$

Proof:
Let:
:$\mathbf a: x \mapsto \tuple {\map {a_1} x, \map {a_2} x, \ldots, \map {a_n} x}$
:$\mathbf b: x \mapsto \tuple {\map {b_1} x, \map {b_2} x, \ldots, \map {b_n} x}$
Then:
{{begin-eqn}}
{{eqn | l = \map {\frac \d {\d x} } {\mathbf a \cdot \mathbf b}
      | r = \map {\frac \d {\d x} } {\sum_{i \mathop = 1}^n a_i b_i}
      | c = {{Defof|Dot Product}}
}}
{{eqn | r = \sum_{i \mathop = 1}^n \map {\frac \d {\d x} } {a_i b_i}
      | c = Sum Rule for Derivatives
}}
{{eqn | r = \sum_{i \mathop = 1}^n \paren {\map {\frac \d {\d x} } {a_i} b_i + a_i \map {\frac \d {\d x} } {b_i} }
      | c = Product Rule for Derivatives
}}
{{eqn | r = \sum_{i \mathop = 1}^n \map {\frac \d {\d x} } {a_i} b_i + \sum_{i \mathop = 1}^n a_i \map {\frac \d {\d x} } {b_i}
      | c = Summation is Linear
}}
{{eqn | r = \dfrac {\d \mathbf a} {\d x} \cdot \mathbf b + \mathbf a \cdot \dfrac {\d \mathbf b} {\d x}
      | c = {{Defof|Dot Product}}
}}
{{end-eqn}}
{{qed}}
Theorem: Derivative of Dot Product of Vector-Valued Functions

Let $\mathbf a: \R \to \R^n$ and $\mathbf b: \R \to \R^n$ be differentiable vector-valued functions.
The derivative of their dot product is given by:
:$\map {\dfrac \d {\d x} } {\mathbf a \cdot \mathbf b} = \dfrac {\d \mathbf a} {\d x} \cdot \mathbf b + \mathbf a \cdot \dfrac {\d \mathbf b} {\d x}$

Proof:
Let $v = \mathbf a \cdot \mathbf b$.
Then:
{{begin-eqn}}
{{eqn | l = \dfrac {\d v} {\d x}
      | r = \lim_{h \mathop \to 0} \dfrac {\map v {x + h} - \map v x} h
      | c = {{Defof|Derivative of Real Function}}
}}
{{eqn | r = \lim_{h \mathop \to 0} \dfrac {\map {\mathbf a} {x + h} \cdot \map {\mathbf b} {x + h} - \map {\mathbf a} x \cdot \map {\mathbf b} x} h
      | c = Definition of $\mathbf v$
}}
{{eqn | r = \lim_{h \mathop \to 0} \dfrac {\map {\mathbf a} {x + h} \cdot \map {\mathbf b} {x + h} - \map {\mathbf a} {x + h} \cdot \map {\mathbf b} x + \map {\mathbf a} {x + h} \cdot \map {\mathbf b} x - \map {\mathbf a} x \cdot \map {\mathbf b} x} h
      | c = rearranging
}}
{{eqn | r = \lim_{h \mathop \to 0} \map {\mathbf a} {x + h} \cdot \dfrac {\map {\mathbf b} {x + h} - \map {\mathbf b} x} h + \lim_{h \mathop \to 0} \dfrac {\map {\mathbf a} {x + h} - \map {\mathbf a} x} h \cdot \map {\mathbf b} x
      | c = rearranging
}}
{{eqn | r = \mathbf a \cdot \dfrac {\d \mathbf b} {\d x} + \dfrac {\d \mathbf a} {\d x} \cdot \mathbf b
      | c = {{Defof|Derivative of Vector-Valued Function}}
}}
{{end-eqn}}
{{qed}}
Theorem: Borel Sigma-Algebra of Subset is Trace Sigma-Algebra

Let $\struct {X, \tau}$ be a topological space, and let $A \subseteq X$ be a subset of $X$.
Let $\tau_A$ be the subspace topology on $A$.
Then the following equality of $\sigma$-algebras on $A$ holds:
:$\map \BB {A, \tau_A} = \map \BB {X, \tau}_A$
where $\BB$ signifies Borel $\sigma$-algebra, and $\map \BB {X, \tau}_A$ signifies trace $\sigma$-algebra.

Proof:
By definition of Borel $\sigma$-algebra, it holds that:
:$\map \BB {X, \tau} = \map \sigma \tau$
and also, by definition of subspace topology:
:$\tau_A = A \cap \tau = \set {A \cap U: U \in \tau}$
Thus, it follows that:
:$\map \BB {A, \tau_A} = \map \sigma {A \cap \tau}$
Thereby, the desired equality:
:$\map \BB {A, \tau_A} = \map \BB {X, \tau}_A$
follows directly from applying Trace Sigma-Algebra of Generated Sigma-Algebra with $\GG = \tau$.
{{qed}}
Theorem: Existence and Uniqueness of Monotone Class Generated by Collection of Subsets

Let $X$ be a set.
Let $\GG \subseteq \powerset X$ be a collection of subsets of $X$.
Then $\map {\mathfrak m} \GG$, the monotone class generated by $\GG$, exists and is unique.

Proof:
=== Existence ===
By Power Set is Monotone Class, there is at least one monotone class containing $\GG$.
Now let $\Bbb M$ be the collection of monotone classes containing $\GG$:
:$\Bbb M := \set {\mathfrak m': \GG \subseteq \mathfrak m', \mathfrak m' \text{ is a monotone class} }$
By Intersection of Monotone Classes is Monotone Class, $\mathfrak m := \bigcap \Bbb M$ is a monotone class.
Also, by Set Intersection Preserves Subsets:
:$\GG \subseteq \mathfrak m$
Now let $\mathfrak m'$ be a monotone class containing $\GG$.
Then by construction of $\mathfrak m$, and Intersection is Subset: General Result:
:$\mathfrak m \subseteq \mathfrak m'$
{{qed|lemma}}
Theorem: Derivative of Product of Real Function and Vector-Valued Function

Let:
:$\mathbf z:\R \to \R^n$
be a differentiable vector-valued function, where:
:$\mathbf z = \begin {bmatrix} z_1 \\ z_2 \\ \vdots \\ z_n \end {bmatrix}$
such that:
:$z_1, z_2, \cdots, z_n$
are differentiable real functions.
Let:
:$y: \R \to \R$
be a differentiable real function.
Then:
:$\map {D_x} {y \, \mathbf z} = \dfrac {\d y} {\d x} \mathbf z + y \dfrac {\d \mathbf z} {\d x}$

Proof:
{{begin-eqn}}
{{eqn | l = \map {D_x} {y \, \mathbf z}
      | r = \map {D_x} {\begin {bmatrix} y \ z_1 \\ y \ z_2 \\ \vdots \\ y \ z_n \end {bmatrix} }
}}
{{eqn | r = \begin {bmatrix} \map {D_x} {y \ z_1} \\ \map {D_x} {y \ z_2} \\ \vdots \\ \map {D_x} {y \ z_n} \end{bmatrix}
      | c = Differentiation of Vector-Valued Function Componentwise
}}
{{eqn | r = \begin{bmatrix} \dfrac {\d y} {\d x} z_1 + y \dfrac {\d z_1} {\d x} \\ \dfrac {\d y} {\d x}z_2 + y \dfrac {\d z_2} {\d x} \\ \vdots \\ \dfrac {\d y} {\d x} z_n + y \dfrac {\d z_n} {\d x} \end{bmatrix}
      | c = Product Rule for Derivatives of Real Functions
}}
{{eqn | r = \begin{bmatrix} \dfrac {\d y} {\d x} z_1 \\ \dfrac {\d y} {\d x} z_2 \\ \vdots \\ \dfrac {\d y} {\d x} z_n \end{bmatrix} + \begin{bmatrix} y \dfrac {\d z_1} {\d x} \\ y \dfrac {\d z_2} {\d x} \\ \vdots \\ y \dfrac {\d z_n} {\d x} \end{bmatrix}
}}
{{eqn | r = \dfrac {\d y} {\d x} \begin{bmatrix} z_1 \\ z_2 \\ \vdots \\ z_n \end{bmatrix} + y \begin{bmatrix} \dfrac {\d z_1} {\d x} \\ \dfrac {\d z_2} {\d x} \\ \vdots \\ \dfrac {\d z_n} {\d x} \end{bmatrix}
}}
{{eqn | r = \frac {\d y} {\d x} \mathbf z + y \frac {\d \mathbf z} {\d x}
      | c = Differentiation of Vector-Valued Function Componentwise
}}
{{end-eqn}}
{{qed}}
Theorem: Generated Sigma-Algebra by Generated Monotone Class

Let $X$ be a set, and let $\GG \subseteq \powerset X$ be a nonempty collection of subsets of $X$.
Suppose that $\GG$ satisfies the following condition:
:$(1):\quad A \in \GG \implies \relcomp X A \in \GG$
that is, $\GG$ is closed under complement in $X$.
Then:
:$\map {\mathfrak m} \GG = \map \sigma \GG$
where $\mathfrak m$ denotes generated monotone class, and $\sigma$ denotes generated $\sigma$-algebra.

Proof:
By Sigma-Algebra is Monotone Class, and the definition of generated monotone class, it follows that:
:$\map {\mathfrak m} \GG \subseteq \map \sigma \GG$
Next, define $\Sigma$ by:
:$\Sigma := \set {M \in \map {\mathfrak m} \GG: X \setminus M \in \map {\mathfrak m} \GG}$
By $(1)$, it follows that $\GG \subseteq \Sigma$.
Next, we will show that $\Sigma$ is a $\sigma$-algebra.
To this end, let $G \in \GG$ be arbitrary.
By $(1)$, also $X \setminus G \in \GG$.
Hence from the definition of monotone class:
:$\O = G \cap \paren {X \setminus G} \in \map {\mathfrak m} \GG$
:$X = G \cup \paren {X \setminus G} \in \map {\mathfrak m} \GG$
by virtue of Set Difference Intersection with Second Set is Empty Set and Set Difference and Intersection form Partition.
Since $\O = X \setminus X$, it follows that:
:$X, \O \in \Sigma$
Further, from Set Difference with Set Difference, it follows that:
:$E \in \Sigma \implies X \setminus E \in \Sigma$
Lastly, for any sequence $\sequence {E_n}_{n \mathop \in \N}$ in $\Sigma$, the definition of monotone class implies that:
:$\ds \bigcup_{n \mathop \in \N} E_n \in \map {\mathfrak m} \GG$
Now to ensure that it is in fact in $\Sigma$, compute:
{{begin-eqn}}
{{eqn|l = X \setminus \paren {\bigcup_{n \mathop \in \N} E_n}
     |r = \bigcap_{n \mathop \in \N} \paren {X \setminus E_n}
     |c = De Morgan's Laws: Difference with Union
}}
{{end-eqn}}
All of the $X \setminus E_n$ are in $\map {\mathfrak m} \GG$ as each $E_n$ is in $\Sigma$.
Hence, from the definition of monotone class, we conclude:
:$\ds \bigcap_{n \mathop \in \N} \paren {X \setminus E_n} \in \map {\mathfrak m} \GG$
which subsequently implies that:
:$\ds \bigcup_{n \mathop \in \N} E_n \in \Sigma$
Thus, having verified all three axioms, $\Sigma$ is a $\sigma$-algebra.
Since $\GG \subseteq \Sigma$, this means, by definition of generated $\sigma$-algebra, that:
:$\map \sigma \GG \subseteq \Sigma \subseteq \map {\mathfrak m} \GG$
By definition of set equality:
:$\map {\mathfrak m} \GG = \map \sigma \GG$
{{qed}}