hcju/proofwikiqa huggingface dataset:
{{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: :$\displaystyle \norm {\mathbf u}_p = \dfrac 1 {\norm {\mathbf x}_p} \paren {\sum_{n \mathop = 1}^\infty \size {x_n}^p}^{1/p} = 1$ 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}}
From the definition of symmetric difference, we have: :$R * S = \left({R \setminus S}\right) \cup \left({S \setminus R}\right)$ Then from Set Difference is Subset of Union of Differences, we have: : $R \setminus S \subseteq \left({R \setminus T}\right) \cup \left({T \setminus S}\right)$ : $S \setminus R \subseteq \left({S \setminus T}\right) \cup \left({T \setminus R}\right)$ Thus: {{begin-eqn}} {{eqn | l = \left({R \setminus S}\right) \cup \left({S \setminus R}\right)       | o = \subseteq       | r = \left({R \setminus T}\right) \cup \left({T \setminus S}\right) \cup \left({S \setminus T}\right) \cup \left({T \setminus R}\right)       | c = Set Union Preserves Subsets }} {{eqn | r = \left({R \setminus T}\right) \cup \left({T \setminus R}\right) \cup \left({S \setminus T}\right) \cup \left({T \setminus S}\right)       | c = Union is Commutative }} {{eqn | r = \left({R * T}\right) \cup \left({S * T}\right)       | c = {{Defof|Symmetric Difference}} }} {{end-eqn}} {{qed}} Category:Symmetric Difference Category:Set Union c9zji7vn26npz9naldzlpdrxwyyft32
Axiom $(1)$ of a $\sigma$-algebra grants $X \in \Sigma$. By axiom $(2)$ and Set Difference with Self is Empty Set, it follows that $\varnothing = X \setminus X \in \Sigma$. {{qed}}
Define $A_1 = A, A_2 = B$, and for $n \in \N, n \ge 2: A_n = \varnothing$. Then by Sigma-Algebra Contains Empty Set, axiom $(3)$ of a $\sigma$-algebra applies. Hence $\displaystyle \bigcup_{n \mathop \in \N} A_n = A \cup B \in \Sigma$. {{qed}}
{{begin-eqn}} {{eqn | ll= \forall n \in \N:       | l = E_n \in \Sigma       | o = \implies       | r = X \setminus E_n \in \Sigma       | c = Axiom $(2)$ for $\sigma$-algebras }} {{eqn | o = \implies       | r = \bigcup_{n \mathop \in \N} \left({X \setminus E_n}\right) \in \Sigma       | c = Axiom $(3)$ for $\sigma$-algebras }} {{eqn | o = \implies       | r = X \setminus \left({\bigcup_{n \mathop \in \N} \left({X \setminus E_n}\right) }\right) \in \Sigma       | c = Axiom $(2)$ for $\sigma$-algebras }} {{end-eqn}} From De Morgan's laws: Complement of Intersection: :$\displaystyle \bigcup_{n \mathop \in \N} \left({X \setminus E_n}\right) = X \setminus \left({\bigcap_{n \mathop \in \N} E_n}\right)$ Also, by Set Difference with Set Difference and Set Union Preserves Subsets: :$\displaystyle X \setminus \left({X \setminus \left({\bigcap_{n \mathop \in \N} E_n}\right) }\right) = \bigcap_{n \mathop \in \N} E_n$ Combining the previous equalities, it follows that: : $\displaystyle \bigcap_{n \mathop \in \N} E_n \in \Sigma$ {{qed}}
Let us verify the axioms of a $\sigma$-algebra in turn. === Axiom $(1)$ === By Relative Complement with Self is Empty Set: : $\complement_X \left({X}\right) = \varnothing$ In particular, it is countable. Hence $X$ is co-countable, and so $X \in \Sigma$. {{qed|lemma}} === Axiom $(2)$ === The relative complement of a countable set is by definition co-countable. The converse holds by Relative Complement of Relative Complement. Hence: : $E \in \Sigma \implies \complement_X \left({E}\right) \in \Sigma$ {{qed|lemma}} === Axiom $(3)$ === Let $\left({E_n}\right)_{n \mathop \in \N} \in \Sigma$ be a collection. Suppose that all the $E_n$ are countable. Then by Countable Union of Countable Sets is Countable, $\displaystyle \bigcup_{n \mathop \in \N} E_n$ is also countable, hence in $\Sigma$. Now suppose that at least one $E_n$ is only co-countable. Then by Set Complement inverts Subsets and Set is Subset of Union: General Result: :$\displaystyle \complement_X \left({\bigcup_{n \mathop \in \N} E_n}\right) \subseteq \complement_X \left({E_n}\right)$ By definition of co-countable, the latter is countable. Thus, by Subset of Countably Infinite Set is Countable, it follows that $\displaystyle \bigcup_{n \mathop \in \N} E_n$ is co-countable, hence in $\Sigma$. Hence the result, from Proof by Cases. {{qed}}
Verifying the axioms for a $\sigma$-algebra in turn: === Axiom $(1)$ === Have $E = X \cap E$ from Intersection with Subset is Subset. Hence, as $X \in \Sigma$ by axiom $(1)$, $E \in \Sigma_E$ by definition of trace $\sigma$-algebra. {{qed|lemma}} === Axiom $(2)$ === Suppose that $S \in \Sigma_E$. Then there is some $A \in \Sigma$ such that $S = E \cap A$, by definition of trace $\sigma$-algebra. Hence $X \setminus A \in \Sigma$ by axiom $(2)$ for $\sigma$-algebras. So $E \cap \left({X \setminus A}\right)$ in $\Sigma_E$. By Intersection with Set Difference is Set Difference with Intersection, this equals $\left({X \cap E}\right) \setminus A = E \setminus A$. Now $E \setminus A = E \setminus S$ by Set Difference with Intersection is Difference. Hence $E \setminus S \in \Sigma_E$. {{qed|lemma}} === Axiom $(3)$ === Let $\left({S_i}\right)_{i \in \N}$ be a sequence in $\Sigma_E$. For each $i \in \N$, let $A_i \in \Sigma$ such that $S_i = E \cap A_i$. Then observe: {{begin-eqn}} {{eqn|l = \bigcup_{i \mathop \in \N} S_i      |r = \bigcup_{i \mathop \in \N} E \cap A_i }} {{eqn|r = E \cap \bigcup_{i \mathop \in \N} A_i      |c = Intersection Distributes over Union: General Result }} {{end-eqn}} As $\Sigma$ is a $\sigma$-algebra, $\displaystyle \bigcup_{i \mathop \in \N} A_i \in \Sigma$. Hence $\displaystyle \bigcup_{i \mathop \in \N} S_i \in \Sigma_E$ by definition of trace $\sigma$-algebra. {{qed}}
Verify the axioms for a $\sigma$-algebra in turn: === Axiom $(1)$ === As $f$ is a mapping, it is immediate that $f^{-1} \left({X'}\right) = X$. Also $X' \in \Sigma'$ as $\Sigma'$ is a $\sigma$-algebra. Hence $X \in f^{-1} \left({\Sigma'}\right)$. {{qed|lemma}} === Axiom $(2)$ === Let $S = f^{-1} \left({E'}\right) \in f^{-1} \left({\Sigma'}\right)$.  By Preimage of Set Difference under Mapping and $f^{-1} \left({X'}\right) = X$, have: :$X \setminus S = f^{-1} \left({X'}\right) \setminus f^{-1} \left({E'}\right) = f^{-1} \left({X' \setminus E'}\right)$ As $X' \setminus E' \in \Sigma'$, this means $X \setminus S \in f^{-1} \left({\Sigma'}\right)$. {{qed|lemma}} === Axiom $(3)$ === Let $\left({S_i}\right)_{i \in \N}$ be a sequence in $f^{-1} \left({\Sigma'}\right)$. Let, for all $i \in \N$, $E'_i \in \Sigma'$ such that $S_i = f^{-1} \left({E'}\right)$. By Preimage of Union under Mapping: General Result, have: :$\displaystyle \bigcup_{i \mathop \in \N} S_i = \bigcup_{i \mathop \in \N} f^{-1} \left({E'_i}\right) = f^{-1} \left({\bigcup_{i \mathop \in \N} E'_i}\right)$ Now $\displaystyle \bigcup_{i \mathop \in \N} E'_i \in \Sigma'$ as $\Sigma'$ is a $\sigma$-algebra. Hence $\displaystyle \bigcup_{i \mathop \in \N} S_i \in f^{-1} \left({\Sigma'}\right)$. {{qed}}
Verify the axioms for a $\sigma$-algebra in turn: === Axiom $(1)$ === As all the $\Sigma_i$ are $\sigma$-algebras, $X \in \Sigma_i$ for all $i \in I$. Hence $X \in \Sigma$ by definition of intersection. {{qed|lemma}} === Axiom $(2)$ === Let $E \in \Sigma$.  Then for all $i \in I$, $E \in \Sigma_i$, by definition of intersection. As all the $\Sigma_i$ are $\sigma$-algebras, also, for all $i \in I$: :$X \setminus E \in \Sigma_i$ Hence $X \setminus E \in \Sigma$ by definition of intersection. {{qed|lemma}} === Axiom $(3)$ === Let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence in $\Sigma$. Then it is also a sequence in every $\Sigma_i$. Hence, for all $i \in I$, as the $\Sigma_i$ are all $\sigma$-algebras: :$\displaystyle \bigcup_{n \mathop \in \N} E_n \in \Sigma_i$ By definition of intersection, it follows that: :$\displaystyle \bigcup_{n \mathop \in \N} E_n \in \Sigma$ {{qed}}
By definition of $\sigma \left({\mathcal G}\right)$, $\mathcal G \subseteq \sigma \left({\mathcal G}\right)$. It follows that also $\mathcal F \subseteq \sigma \left({\mathcal G}\right)$. Hence, by definition of $\sigma \left({\mathcal F}\right)$, $\sigma \left({\mathcal F}\right) \subseteq \sigma \left({\mathcal G}\right)$. {{qed}}
By definition of Borel $\sigma$-algebra, $\mathcal B \left({\R^n}\right) = \sigma \left({\mathcal{O}^n}\right)$. The rest of the proof will be split in proving the following equalities: :$(1): \quad \sigma \left({\mathcal{O}^n}\right) = \sigma \left({\mathcal{C}^n}\right)$ :$(2): \quad \sigma \left({\mathcal{C}^n}\right) = \sigma \left({\mathcal{K}^n}\right)$ :$(3): \quad \sigma \left({\mathcal{O}^n}\right) = \sigma \left({\mathcal{J}_{ho}^n}\right)$ :$(4): \quad \sigma \left({\mathcal{J}_{ho}^n}\right) = \sigma \left({\mathcal{J}^n_{ho, \text{rat}}}\right)$ === Proof of $(1)$ === {{:Characterization of Euclidean Borel Sigma-Algebra/Open equals Closed}} === Proof of $(2)$ === {{:Characterization of Euclidean Borel Sigma-Algebra/Closed equals Compact}} === Proof of $(3)$ === {{:Characterization of Euclidean Borel Sigma-Algebra/Open equals Rectangle}} === Proof of $(4)$ === {{:Characterization of Euclidean Borel Sigma-Algebra/Rectangle equals Rational Rectangle}}
Proceed by induction on $n$, that is, on the number of generators. === Basis for the Induction === The case $n = 0$ is verified as follows: :$\sigma \left({\varnothing}\right) = \left\{{\varnothing, X}\right\}$ from Sigma-Algebra Generated by Empty Set. Thus $\sigma \left({\varnothing}\right)$ is finite. This is the basis for the induction. === Induction Hypothesis === Fix $n \in \N$. Assume that $\sigma \left({\left\{{A_1, \ldots, A_n}\right\}}\right)$ is finite. This is the induction hypothesis. === Induction Step === This is the induction step: For brevity, write $\Sigma_n$ for $\sigma \left({A_1, \ldots, A_n}\right)$. Observe that: :$\left\{{A_1, \ldots, A_{n+1}}\right\} \subseteq \Sigma_n \cup \left\{{A_{n+1}}\right\} \subseteq \Sigma_{n+1}$ Hence Condition on Equality of Generated Sigma-Algebras applies to yield: :$\Sigma_{n+1} = \sigma \left({\Sigma_n \cup \left\{{A_{n+1}}\right\}}\right)$ From Sigma-Algebra Extended by Single Set, it follows that: :$\sigma \left({\Sigma_n \cup \left\{{A_{n+1}}\right\}}\right) = \left\{{\left({E_1 \cap A_{n+1}}\right) \cup \left({E_2 \cap A_{n+1}^c}\right): E_1, E_2 \in \Sigma_n}\right\}$ Thus there is a surjection $f: \Sigma_n \times \Sigma_n \to \Sigma_{n+1}$, defined by: :$f \left({E_1, E_2}\right) := \left({E_1 \cap A_{n+1}}\right) \cup \left({E_2 \cap A_{n+1}^c}\right)$ By Cardinality of Surjection, it now follows that: :$\left\vert{\Sigma_{n+1}}\right\vert \le \left\vert{\Sigma_n \times \Sigma_n}\right\vert$ and the latter equals $\left\vert{\Sigma_n}\right\vert^2$ by Cardinality of Cartesian Product. By the induction hypothesis, this is necessarily finite. Therefore, $\Sigma_{n+1}$ is also finite. The result follows by the Principle of Mathematical Induction. {{qed}}
We have Set of all Self-Maps is 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}}
$0$ is not invertible in $k$ since $0 a = 0$ for all $a \in k$. Thus $0 \notin k^\times$. Consider now $0 \ne a \in k$. From the axioms of fields it follows that there exists $a^{-1}$. Thus $a \in k^\times$. {{qed}} Category:Field Theory i7zr3adupmhtbqdhthmn6w0x117qiuw
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}}
{{begin-eqn}} {{eqn | l = \frac \d {\d x} \left({\mathbf r \left({x}\right) \cdot \mathbf q \left({x}\right) }\right)       | r = \frac \d {\d x} \left({\sum_{i \mathop = 1}^n r_i \left({x}\right) q_i \left({x}\right)}\right)       | c = {{Defof|Dot Product}} }} {{eqn | r = \sum_{i \mathop = 1}^n \frac \d {\d x} \left({r_i \left({x}\right) q_i \left({x}\right)}\right)       | c = Sum Rule for Derivatives }} {{eqn | r = \sum_{i \mathop = 1}^n \left({ {r_i}' \left({x}\right) q_i \left({x}\right) + r_i \left({x}\right) {q_i}' \left({x}\right)}\right)       | c = Product Rule for Derivatives }} {{eqn | r = \sum_{i \mathop = 1}^n {r_i}' \left({x}\right) q_i \left({x}\right) + \sum_{i \mathop = 1}^n r_i \left({x}\right) {q_i}' \left({x}\right)        | c = Summation is Linear }} {{eqn | r = \mathbf r'\left({x}\right) \cdot \mathbf q \left({x}\right) + \mathbf r \left({x}\right) \cdot \mathbf q' \left({x}\right)       | c = {{Defof|Dot Product}} }} {{end-eqn}} {{qed}}
By definition of Borel $\sigma$-algebra, it holds that: :$\mathcal B \left({X, \tau}\right) = \sigma \left({\tau}\right)$ and also, by definition of subspace topology: :$\tau_A = A \cap \tau = \left\{{A \cap U: U \in \tau}\right\}$ Thus, it follows that: :$\mathcal B \left({A, \tau_A}\right) = \sigma \left({A \cap \tau}\right)$ Thereby, the desired equality: :$\mathcal B \left({A, \tau_A}\right) = \mathcal B \left({X, \tau}\right)_A$ follows directly from applying Trace Sigma-Algebra of Generated Sigma-Algebra with $\mathcal G = \tau$. {{qed}}
{{begin-eqn}} {{eqn | l = \mathbf r \left({t}\right) \times \mathbf q \left({t}\right)       | r = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \times \begin{bmatrix} \chi \\ \gamma \\ \zeta \end{bmatrix} }} {{eqn | r = \begin{bmatrix} y \zeta - z \gamma \\ z \chi - x \zeta \\ x\gamma -y \chi \end{bmatrix}       | c = {{Defof|Vector Cross Product}} }} {{eqn | ll= \leadsto       | l = D_t \left({\mathbf r \left({t}\right) \times \mathbf q \left({t}\right)}\right)       | r = \begin {bmatrix} D_t\left({y \zeta - z \gamma}\right) \\ D_t\left({z \chi - x \zeta}\right) \\ D_t\left({x\gamma -y \chi}\right) \end{bmatrix}        | c = Differentiation of Vector-Valued Function Componentwise }} {{eqn | r = \begin {bmatrix} \dfrac{\d y} {\d t}\zeta + y \dfrac {\d \zeta} {\d t} - \dfrac {\d z} {\d t} \gamma - z \dfrac {\d \gamma} {\d t} \\ \dfrac {\d z} {\d t} \chi + z \dfrac {\d \chi} {\d t} - \dfrac {\d x} {\d t} \zeta - x \dfrac {\d \zeta} {\d t} \\ \dfrac {\d x} {\d t} \gamma + x \dfrac {\d \gamma} {\d t} - \dfrac {\d y} {\d t} \chi - y \dfrac {\d \chi} {\d t} \end{bmatrix}       | c = Product Rule for Derivatives, Linear Combination of Derivatives }} {{eqn | l = \mathbf r' \left({x}\right) \times \mathbf q \left({x}\right) + \mathbf r \left({x}\right) \times \mathbf q'\left({x}\right)       | r = \begin {bmatrix} \dfrac {\d x} {\d t} \\ \dfrac {\d y} {\d t} \\ \dfrac {\d z} {\d t} \end {bmatrix} \times \begin {bmatrix} \chi \\ \gamma \\ \zeta \end {bmatrix} + \begin {bmatrix} x \\ y \\ z \end {bmatrix} \times \begin {bmatrix} \dfrac {\d \chi} {\d t} \\ \dfrac {\d \gamma} {\d t} \\ \dfrac {\d \zeta} {\d t} \end{bmatrix} }} {{eqn | r = \begin {bmatrix} \dfrac {\d y} {\d t} \zeta - \dfrac {\d z} {\d t} \gamma \\ \dfrac {\d z} {\d t} \chi - \dfrac {\d x} {\d t} \zeta \\ \dfrac {\d x} {\d t} \gamma - \dfrac {\d y} {\d t} \chi \end {bmatrix} + \begin {bmatrix} y \dfrac {\d \zeta} {\d t} - z \dfrac {\d \gamma} {\d t} \\ z \dfrac {\d \chi} {\d t} - x \dfrac {\d \zeta} {\d t} \\ x \dfrac {\d \gamma} {\d t} - y \dfrac {\d \chi} {\d t} \end {bmatrix}       | c = {{Defof|Vector Cross Product}} }} {{eqn | r = \begin {bmatrix} \dfrac {\d y} {\d t} \zeta - \dfrac {\d z} {\d t} \gamma + y \dfrac {\d \zeta} {\d t} - z \dfrac {\d \gamma} {\d t} \\ \dfrac {\d z} {\d t} \chi - \dfrac {\d x} {\d t} \zeta + z \dfrac {\d \chi} {\d t} - x \dfrac {\d \zeta} {\d t} \\ \dfrac {\d x} {\d t} \gamma - \dfrac {\d y} {\d t} \chi + x \dfrac {\d \gamma} {\d t} - y \dfrac {\d \chi} {\d t} \end {bmatrix} }} {{eqn | r = \begin {bmatrix} \dfrac {\d y} {\d t} \zeta + y \dfrac {\d \zeta} {\d t} - \dfrac {\d z} {\d t} \gamma - z \dfrac {\d \gamma} {\d t} \\ \dfrac {\d z} {\d t} \chi + z \dfrac {\d \chi} {\d t} - \dfrac {\d x} {\d t} \zeta - x \dfrac {\d \zeta} {\d t} \\ \dfrac {\d x} {\d t} \gamma + x \dfrac {\d \gamma} {\d t} - \dfrac {\d y} {\d t} \chi - y \dfrac {\d \chi} {\d t} \end {bmatrix} }} {{eqn | r = D_t \left({\mathbf r \left({t}\right) \times \mathbf q \left({t}\right)}\right) }} {{end-eqn}} {{qed}}
=== Existence === By Power Set is Monotone Class, there is at least one monotone class containing $\mathcal G$. Now let $\Bbb M$ be the collection of monotone classes containing $\mathcal G$: :$\Bbb M := \left\{{\mathfrak{m}': \mathcal G \subseteq \mathfrak{m}', \text{$\mathfrak{m}'$ is a monotone class}}\right\}$ By Intersection of Monotone Classes is Monotone Class, $\mathfrak m := \bigcap \Bbb M$ is a monotone class. Also, by Set Intersection Preserves Subsets: : $\mathcal G \subseteq \mathfrak m$ Now let $\mathfrak{m}'$ be a monotone class containing $\mathcal G$. Then by construction of $\mathfrak m$, and Intersection is Subset: General Result: : $\mathfrak m \subseteq \mathfrak{m}'$ {{qed|lemma}} === Uniqueness === Suppose $\mathfrak{m}_1$ and $\mathfrak{m}_2$ are both monotone classes generated by $\mathcal G$. Then property $(2)$ for these monotone classes implies both $\mathfrak{m}_1 \subseteq \mathfrak{m}_2$ and $\mathfrak{m}_2 \subseteq \mathfrak{m}_1$. By definition of set equality: : $\mathfrak{m}_1 = \mathfrak{m}_2$ {{qed}}
{{begin-eqn}} {{eqn | l = D_x \left({y \, \mathbf{z} }\right)       | r = D_x \left({\begin{bmatrix} y \ z_1 \\ y \ z_2 \\ \vdots \\ y \ z_n \end{bmatrix} }\right) }} {{eqn | r = \begin{bmatrix} D_x \left({y \ z_1}\right) \\ D_x \left({y \ z_2 }\right) \\ \vdots \\ D_x \left({y \ z_n }\right) \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}} {{proofread}}
By Sigma-Algebra is Monotone Class, and the definition of generated monotone class, it follows that: :$\mathfrak m \left({\mathcal G}\right) \subseteq \sigma \left({\mathcal G}\right)$ Next, define $\Sigma$ by: :$\Sigma := \left\{{M \in \mathfrak m \left({\mathcal G}\right): X \setminus M \in \mathfrak m \left({\mathcal G}\right)}\right\}$ By $(1)$, it follows that $\mathcal G \subseteq \Sigma$. Next, we will show that $\Sigma$ is a $\sigma$-algebra. To this end, let $G \in \mathcal G$ be arbitrary. By $(1)$, also $X \setminus G \in \mathcal G$. Hence from the definition of monotone class: :$\varnothing = G \cap \left({X \setminus G}\right) \in \mathfrak m \left({\mathcal G}\right)$ :$X = G \cup \left({X \setminus G}\right) \in \mathfrak m \left({\mathcal G}\right)$ by virtue of Set Difference Intersection with Second Set is Empty Set and Set Difference and Intersection form Partition. Since $\varnothing = X \setminus X$, it follows that: :$X, \varnothing \in \Sigma$ Further, from Set Difference with Set Difference, it follows that: :$E \in \Sigma \implies X \setminus E \in \Sigma$ Lastly, for any sequence $\left({E_n}\right)_{n \in \N}$ in $\Sigma$, the definition of monotone class implies that: :$\displaystyle \bigcup_{n \mathop \in \N} E_n \in \mathfrak m \left({\mathcal G}\right)$ Now to ensure that it is in fact in $\Sigma$, compute: {{begin-eqn}} {{eqn|l = X \setminus \left({\bigcup_{n \mathop \in \N} E_n}\right)      |r = \bigcap_{n \mathop \in \N} \left({X \setminus E_n}\right)      |c = De Morgan's Laws: Difference with Union }} {{end-eqn}} All of the $X \setminus E_n$ are in $\mathfrak m \left({\mathcal G}\right)$ as each $E_n$ is in $\Sigma$. Hence, from the definition of monotone class, we conclude: :$\displaystyle \bigcap_{n \mathop \in \N} \left({X \setminus E_n}\right) \in \mathfrak m \left({\mathcal G}\right)$ which subsequently implies that: :$\displaystyle \bigcup_{n \mathop \in \N} E_n \in \Sigma$ Thus, having verified all three axioms, $\Sigma$ is a $\sigma$-algebra. Since $\mathcal G \subseteq \Sigma$, this means, by definition of generated $\sigma$-algebra, that: :$\sigma \left({\mathcal G}\right) \subseteq \Sigma \subseteq \mathfrak m \left({\mathcal G}\right)$ By definition of set equality: :$\mathfrak m \left({\mathcal G}\right) = \sigma \left({\mathcal G}\right)$ {{qed}}
Let $M = \struct {A, d}$ be totally bounded. Let $\epsilon = 1, \dfrac 1 2, \dfrac 1 3, \ldots$ As $M$ is totally bounded, for each $\epsilon$ there exists a finite $\epsilon$-net $\CC$ for $M$. From Net forms Basis for Metric Space, $\CC$ is a countable basis for $M$. That is, $M$ is second-countable. {{qed}}
Follows directly from: : Totally Bounded Metric Space is Separable : Separable Metric Space is Second-Countable {{qed}}
=== Necessary Condition === This follows directly from: * Compact Metric Space is Complete * Compact Metric Space is Totally Bounded {{qed|lemma}} === Sufficient Condition === This follows directly from: * Complete and Totally Bounded Metric Space is Sequentially Compact * Sequentially Compact Metric Space is Compact {{qed}} {{ACC|Complete and Totally Bounded Metric Space is Sequentially Compact}} {{namedfor|Heinrich Eduard Heine|name2=Émile Borel}}
Let $U \in \tau$ be an open set, and define $C$ by: :$C := X \setminus U$ hence $C$ is a closed set. Further, define, for all $n \in \N$: :$C_n := \displaystyle \bigcup_{c \mathop \in C} B \left({c; \frac 1 n}\right)$ where $B$ denotes open ball. The $C_n$ are open sets, being the union of open balls. It is clear that $C \subseteq C_n$ for all $n \in \N$. Conversely, as $U$ is open, for any $u \in U$ (that is, $u \notin C$), find $n \in \N$ such that: :$B \left({u; \dfrac 1 n}\right) \subseteq U$ as is possible from the definition of open set in a metric space. Thus, for all $c \in C = X \setminus U$, this means: :$d \left({u, c}\right) \ge \dfrac 1 n$ whence $u \notin C_n$. That is, we have established that: :$c \in C \iff \forall n \in \N: c \in C_n$ Phrased in terms of intersection, this means: :$C = \displaystyle \bigcap_{n \mathop \in \N} C_n$ Thus, since $C_n \in \tau \subseteq \mathfrak m \left({\tau}\right)$: :$C \in \mathfrak m \left({\tau}\right)$ Now define: :$\complement_X \left({\tau}\right) := \left\{{X \setminus U: U \in \tau}\right\}$ Then we have shown: :$\mathfrak m \left({\tau \cup \complement_X \left({\tau}\right)}\right) \subseteq \mathfrak m \left({\tau}\right)$ and the reverse inclusion (and hence equality) follows from Generated Monotone Class Preserves Subset. Now applying Generated Sigma-Algebra by Generated Monotone Class: Corollary: :$\sigma \left({\tau}\right) = \mathfrak m \left({\tau \cup \complement_X \left({\tau}\right)}\right)$ Combining these two equalities gives the result, by definition of Borel $\sigma$-algebra. {{qed}}
Combine Measure is Finitely Additive Function with Additive Function is Strongly Additive. {{qed}}