hcju/proofwikiqa huggingface dataset:
Let $z_1, z_2 \in \C_{\ne 0}$. Then by definition of complex number: :$z_1 = x_1 + i y_1, z_2 = x_2 + i y_2$ for some $x_1, y_1, x_2, y_2 \in \R$ such that: :$x_1 \ne 0$ or $y_1 \ne 0$ :$x_2 \ne 0$ or $y_2 \ne 0$ Expressing $z_1$ and $z_2$ in exponential form (although polar form is equally adequate): :$z_1 = r_1 e^{i \theta_1}$ and $z_2 = r_2 e^{i \theta_2}$ for some $r_1, r_2, \theta_1, \theta_2 \in \R$. Then by Product of Complex Numbers in Polar Form: :$z_1 \times z_2 = \paren {r_1 \times r_2} e^{i \paren {\theta_1 + \theta_2} }$ By definition of exponential form: :$r_1 = \sqrt {x_1^2 + y_1^2}$ :$r_2 = \sqrt {x_2^2 + y_2^2}$ Thus $r_1 > 0$ and $r_2 > 0$. Hence $r_1 \times r_2 > 0$ and so $z_1 \times z_2 \ne 0$. {{qed}}
Equivalently this is to say: :$z_1 z_2 = 0 \implies z_1 = 0 \lor z_2 = 0$ Let $z_1 z_2 = 0$. {{begin-eqn}} {{eqn | l = z_1       | r = \tuple {x_1, y_1}       | c = {{Defof|Complex Number|index = 2}}: for some $x_1, y_1 \in \R$ }} {{eqn | l = z_2       | r = \tuple {x_2, y_2}       | c = {{Defof|Complex Number|index = 2}}: for some $x_2, y_2 \in \R$ }} {{eqn | ll= \leadsto       | l = z_1 z_2       | r = \tuple {x_1 x_2 - y_1 y_2, x_1 y_2 + y_1 x_2}       | c = Definition of Complex Multiplication }} {{eqn | n = 1       | ll= \leadsto       | l = x_1 x_2       | r = y_1 y_2       | c =  }} {{eqn | n = 2       | lo= \land       | l = x_1 y_2       | r = -y_1 x_2       | c = as $z_1 z_2 = 0$ }} {{end-eqn}} {{WLOG}}, let $\tuple {x_2, y_2} \ne \tuple {0, 0}$. Suppose also that $\tuple {x_1, y_1} \ne \tuple {0, 0}$. Then: {{begin-eqn}} {{eqn | l = x_1       | r = \frac {y_1 y_2} {x_2}       | c = from $(1)$ }} {{eqn | ll= \leadsto       | l = \frac {y_1 y_2} {x_2} y_2       | r = -y_1 x_2       | c = substituting in $(2)$ }} {{eqn | ll= \leadsto       | l = y_2^2       | r = - x_2^2       | c =  }} {{eqn | ll= \leadsto       | l = x_2 = y_2       | r = 0       | c = as both $x_2$ and $y_2$ are real }} {{end-eqn}} From this contradiction it follows that: :$\tuple {x_1, y_1} = \tuple {0, 0}$ {{qed}}
Suppose $m$ is not prime. Then $m = r s$ for some $r, s \in \Z: 1 < r < m, 1 < s < m$. So $r, s \in \Z'_m$. But: :$r \times_m s \equiv 0 \pmod m$ and so $r \times_m s \notin \Z'_m$. So if $m$ is not prime, $\left({\Z_m, \times_m}\right)$ is not closed. The result follows from the Rule of Transposition. {{qed}}
By definition, a magma is an algebraic structure $\left({S, \circ}\right)$ where $\circ$ is closed. That is: :$\forall x, y \in S: x \circ y \in S$ From Set is Subset of Itself, $S \subseteq S$. By definition, $\left({T, \circ}\right)$ is a submagma of $S$ if: :$\forall x, y \in T: x \circ y \in T$ It follows from the above that $\left({S, \circ}\right)$ is, by definition, a submagma of itself. {{qed}}
By definition, a magma is an algebraic structure $\left({S, \circ}\right)$ where $\circ$ is closed. That is: :$\forall x, y \in S: x \circ y \in S$ By definition, $\left({T, \circ}\right)$ is a submagma of $S$ if: :$\forall x, y \in T: x \circ y \in T$ But: :$\not \exists x, y \in \varnothing: x \circ y \notin \varnothing$ it follows vacuously that: :$\forall x, y \in \varnothing: x \circ y \in \varnothing$ Hence the result. {{Qed}}
By the definition of the max operation, either: :$\max \left({x, y}\right) = x$ or :$\max \left({x, y}\right) = y$ So $\max$ is closed on $S$. Then we have that the max operation is associative: : $\forall x, y, z \in S: \max \left({x, \max \left({y, z}\right)}\right) = \max \left({\max \left({x, y}\right), z}\right)$ Hence the result, by definition of semigroup. {{qed}}
The Min Semigroup is Commutative and idempotent.  Hence the result, by definition of a semilattice. {{qed}}
Let $x, y \in S$.  We have that the min operation is commutative: : $\min \left({x, y}\right) = \min \left({y, x}\right)$  Hence the result, by definition of commutative semigroup. {{qed}}
The fact that $\struct {S, \min}$ is a semigroup is demonstrated in Min Operation on Toset is Semigroup. Then the min operation is idempotent: :$\forall x \in S: \min \set {x, x} = x$ The result follows by the definition of idempotent semigroup. {{qed}}
From Well-Ordering is Total Ordering, we have that $\struct {S, \preceq}$ is a totally ordered set. Thus, by Max Operation on Toset is Semigroup, $\struct {S, \max}$ is a semigroup. By definition, a well-ordered set has a smallest element. So: :$\exists m \in S: \forall x \in S: m \preceq x$ where $m$ is that smallest element. It follows by definition of the max operation that: :$\forall x \in S: \max \set {m, x} = x$ Thus $m$ is the identity element of $\struct {S, \max}$. Hence the result, by definition of monoid. {{qed}} Category:Max and Min Operations Category:Well-Orderings Category:Examples of Monoids 9k93zgqq1r9n7kp28j7h8ndq6a8hhdv
By the Well-Ordering Principle, $\N$ is a well-ordered set. The result follows from Max Operation on Woset is Monoid. {{qed}}
Let $\epsilon>0$. Let $z \in \C$ be a complex number satisfying $\left\vert{z - z_0}\right\vert < \epsilon$. By the Reverse Triangle Inequality: : $\left\vert{ \left\vert{z}\right\vert - \left\vert{z_0}\right\vert }\right\vert \le \left\vert{z - z_0}\right\vert < \epsilon$ Hence the result, by the $\epsilon$-$\delta$ definition of continuity (taking $\delta = \epsilon$). {{qed}} Category:Complex Modulus Category:Continuous Functions c5b85ilzan02knmwpgg6iyhjvy62lag
By the Well-Ordering Principle, $\N$ is a well-ordered set. By definition, every subset of a well-ordered set is also well-ordered. Thus $S$ is a well-ordered set. The result follows from Max Operation on Woset is Monoid. {{qed}}
The Euclidean space $\R^m$ is a vector space over $\R$. That the norm axioms are satisfied is proven in Euclidean Space is Normed Space. Then we have Euclidean Space is Complete Metric Space. The result follows by the definition of a Banach space. {{qed}}
By definition, Euclidean norm is the same as $p$-norm with $p = 2$. Let $\sequence {\mathbf x_n}_{n \mathop \in \N} = \sequence {\tuple {x_n^{\paren 1}, x_n^{\paren 2}, \ldots, x_n^{\paren m}}}_{n \mathop \in \N} $ be a Cauchy sequence in $\R^m$. Let $k \in \N_{> 0} : k \le m$. Then: {{begin-eqn}} {{eqn | l = \size {x_n^{\paren k} - x_m^{\paren k} }       | r = \paren {\paren {x_n^{\paren k} - x_m^{\paren k} }^2}^{\frac 1 2} }} {{eqn | o = \le       | r = \paren {\sum_{k \mathop = 0}^m \paren {x_n^{\paren k} - x_m^{\paren k} }^2}^{\frac 1 2} }} {{eqn | r = \norm { {\bf x}_n - {\bf x}_m}_2 }} {{end-eqn}} Hence, $\sequence {x_n^{\paren k} }_{n \mathop \in \N}$ is a Cauchy sequence in $\R$. Then: :$\displaystyle \lim_{n \mathop \to \infty} x_n^{\paren k} = L^{\paren k}$ Let $\mathbf L = \tuple {L^{\paren 1}, \ldots, L^{\paren m}} \in \R^m$ We have that for all $n > N$: {{begin-eqn}} {{eqn | l = \norm {\mathbf x_n - \mathbf L}       | r = \paren {\sum_{k \mathop = 1}^m \size {x^{\paren k}_n - x^{\paren k} }^2}^{\frac 1 2} }} {{eqn | o = \le       | r = \paren {\sum_{k \mathop = 1}^m \frac {\epsilon^2} m}^{\frac 1 2} }} {{eqn | r = \epsilon }} {{end-eqn}} Therefore, $\sequence {\mathbf x_n}_{n \mathop \in \N}$ converges to $\mathbf L$ in $\struct {\R^m, \norm {\, \cdot \,}_2}$. {{qed}}
Let $z = x + iy$ be a complex number, where $x, y \in \R$. Now, we can identify the complex number $z$ with the ordered pair $\left( x, y \right) \in \R^2$. The norm on $\C$ given by the complex modulus is then identical to the Euclidean norm on $\R^2$. Therefore, metric on $\C$ induced by the norm given by the complex modulus is identical to the Euclidean metric on $\R^2$ induced by the Euclidean norm. By Euclidean Space is Complete Metric Space, $\C$ is a complete metric space. {{qed}}
Let $\sequence {\tuple {x_{n,1}, x_{n,2}, \ldots, x_{n,m} } }_{n \mathop \in \N} $ be a Cauchy sequence in $\R^m$. Let $\epsilon > 0$. Then $\dfrac \epsilon m > 0$. We have that Real Number Line is Complete Metric Space. Therefore: :$\exists y_1, y_2, \ldots, y_m \in \R$ and $N_1, N_2, \ldots, N_m \in \N$ (depending on $\epsilon$) such that: :$\forall k \in \N: 1 \le k \le m: \forall n_k > N_k: \sequence {x_{n,k} – y_k}  < \dfrac \epsilon m$ From Euclidean Space is Normed Space: :$\displaystyle \norm {\tuple {x_{n,1}, x_{n,2}, \ldots, x_{n,m} } - \tuple {y_1, y_2, \ldots, y_m} } \le \sum_{k \mathop = 1}^m \size {x_{n, k} – y_k} < \epsilon$ Hence the Euclidean space is a complete metric space. {{qed}}
The complex plane $\C$ is a vector space over $\C$. That the norm axioms are satisfied is proven in Complex Modulus is Norm. Then we have Complex Plane is Complete Metric Space. Hence the result. {{qed}}
Proof by Counterexample: First note that Rational Numbers form Metric Space. It remains to be shown that $\struct {\Q, \tau_d}$ is not complete. Consider the sequence $\sequence {a_n}$ given by: :$a_n := \dfrac {f_{n + 1} } {f_n}$ where $\sequence {f_n}$ is the sequence of Fibonacci numbers. By Ratio of Consecutive Fibonacci Numbers: :$\displaystyle \lim_{n \mathop \to \infty} a_n = \phi := \dfrac {1 + \sqrt 5} 2$ But Square Root of Prime is Irrational. That means $\dfrac {1 + \sqrt 5} 2$ is likewise irrational. Thus $\sequence {a_n}$ is a Cauchy sequence of rational numbers which converges to a number which is not in $\Q$. The result follows by definition of complete metric space. {{qed}} In fact, any sequence of rational numbers that converges to an irrational number (in the metric space $\R$) is a Cauchy sequence that does not converge in $\Q$.
Let $\epsilon > 0$. As $\displaystyle \lim_{n \mathop \to \infty} x_n = l$, it follows that: :$\exists N: \forall n > N: \size {x_n - l} < \epsilon$ Now let $R = N$. Then from Strictly Increasing Sequence of Natural Numbers: :$\forall r > R: n_r \ge r$ Thus $n_r > N$ and so: :$\size {x_n - l} < \epsilon$ The result follows. {{qed}}
Let $d_\infty: \paren{A \times A}\times \paren{A \times A} \to \R$ be the metric on $A \times A$ defined by: :$\map {d_\infty} {\tuple{x, y}, \tuple{x', y'} } = \max \set{\map d {x, x'}, \map d {y, y'} }$ By P-Product Metric Induces Product Topology, $\tau$ is the topology on $A \times A$ induced by $d_\infty$. Let $\epsilon \in \R_{>0}$ be a strictly positive real number. Let $\tuple {x_0, y_0} \in A \times A$. Suppose that $\tuple {x, y} \in A \times A$ and $\map {d_\infty} {\tuple{x, y}, \tuple{x_0, y_0} } < \dfrac 1 2 \epsilon$. Then: {{begin-eqn}} {{eqn | l = \size {\map d {x, y} - \map d {x_0, y_0} }       | o = \le       | r = \size {\map d {x, y} - \map d {x_0, y} } + \size {\map d {x_0, y} - \map d {x_0, y_0} }       | c = Triangle Inequality for Real Numbers }} {{eqn | o = \le       | r = \map d {x, x_0} + \map d {y, y_0}       | c = Reverse Triangle Inequality }} {{eqn | o = <       | r = \epsilon       | c = by the definition of $d_\infty$ }} {{end-eqn}} The result follows from the definition of a continuous mapping. {{qed}} Category:Continuous Mappings Category:Metric Spaces Category:Product Spaces 6bu7lenh5xonjrmuql24zb8x2595eq4
Suppose $l > 0$. From the definition of convergence to a limit: :$\forall \epsilon > 0: \exists N: \forall n > N: \size {x_n - l} < \epsilon$ That is, $l - \epsilon < x_n < l + \epsilon$. As this is true for ''all'' $\epsilon > 0$, it is also true for $\epsilon = \dfrac l 2$ for some value of $N$. Thus: :$\exists N: \forall n > N: x_n > \dfrac l 2$ as required. Now suppose $l < 0$. By a similar argument: :$\forall \epsilon > 0: \exists N: \forall n > N: l - \epsilon < x_n < l + \epsilon$ Thus it is also true for $\epsilon = -\dfrac l 2$ for some value of $N$. Thus: :$\exists N: \forall n > N: x_n < \dfrac l 2$ as required. {{qed}}
Suppose $l > 0$. Let us choose $N$ such that: :$\forall n > N: \cmod {z_n - l} < \dfrac {\cmod l} 2$ Then: {{begin-eqn}} {{eqn | l = \cmod {z_n - l}       | o = <       | r = \frac {\cmod l} 2       | c =  }} {{eqn | ll= \leadsto       | l = \cmod l - \cmod {z_n}       | o = \le       | r = \cmod {z_n - l}       | c = Reverse Triangle Inequality }} {{eqn | o = <       | r = \frac {\cmod l} 2       | c =  }} {{eqn | ll= \leadsto       | l = \cmod {z_n}       | o = >       | r = \cmod l - \frac {\cmod l} 2       | c =  }} {{eqn | r = \frac {\cmod l} 2       | c =  }} {{end-eqn}} {{qed}} Category:Limits of Sequences 3ttfog1p4psqtbv9iperj3dyw9e0edy
We want to show that $\paren {n + 1}^{1 / \paren {n + 1} } \le n^{1/n}$. Thus: {{begin-eqn}} {{eqn | l = \paren {n + 1}^{1 / \paren {n + 1} }       | o = \le       | r = n^{1/n}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \paren {n + 1}^n       | o = \le       | r = n^{n + 1}       | c = raising both sides to the power of $n \paren {n + 1}$ }} {{eqn | ll= \leadstoandfrom       | l = \paren {n \paren {1 + \frac 1 n} }^n       | o = \le       | r = n^{n + 1}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \paren {1 + \frac 1 n}^n       | o = \le       | r = \frac {n^{n + 1} } {n^n} = n       | c =  }} {{end-eqn}} But from One Plus Reciprocal to the Nth: :$\paren {1 + \dfrac 1 n}^n < 3$ Thus the reversible chain of implication can be invoked and we see that $\paren {n + 1}^{1 / \paren {n + 1} } \le n^{1/n}$ when $n \ge 3$. So $\sequence {n^{1 / n} }$ is decreasing for $n \ge 3$. {{qed}}
{{WLOG}}, consider the right inverse of $\mathbf D$. Suppose none of the diagonal elements are zero.  Then by the definition of inverse, our assertion is that the matrix product of the two matrices in question is the unit matrix of order $n$. Now, observe that: {{begin-eqn}} {{eqn | l = \begin{bmatrix} a_{11} & 0 & \cdots & 0 \\ 0 & a_{22} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a_{nn} \\ \end{bmatrix} \begin{bmatrix} \frac 1 {a_{11} } & 0 & \cdots & 0 \\ 0 & \frac 1 {a_{22} } & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \frac 1 {a_{nn} } \\ \end{bmatrix}       | r = \begin{bmatrix} \frac{a_{11} }{a_{11} } & 0 & \cdots & 0 \\ 0 & \frac{a_{22} }{a_{22} } & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \frac{a_{nn} }{a_{nn} } \\ \end{bmatrix}       | c =  }} {{eqn | r = \begin{bmatrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \\ \end{bmatrix}       | c =  }} {{eqn | r = \mathbf I       | c =  }} {{end-eqn}} {{qed|lemma}} Now suppose one of the diagonal elements is zero. Then $\map \det {\mathbf D} = 0$, where $\det$ indicates the determinant of $\mathbf D$. From Determinant of Inverse Matrix it would follow that: :$\map \det {\mathbf D^{-1} } = \dfrac 1 {\map \det {\mathbf D} }$ But this equation has no solution, and so $\mathbf D$ cannot admit an inverse. {{qed}} Category:Diagonal Matrices Category:Inverse Matrices gh1d57u7f1tiwpqjlan4hfxa3atfd4x
This will be proved by demonstrating the contrapositive: :$S$ is not complete {{iff}} $S$ is not closed. === Necessary Condition === Suppose that $S$ is not complete. Then there exists a Cauchy sequence $\sequence {x_n}$ in $S$ such that the limit $\displaystyle x = \lim_{n \mathop \to \infty} x_n$, which exists in the complete metric space $M$, is not a member of $S$. For all $\epsilon > 0$, there exists an $N \in \N$ such that for all $n \ge N$: :$\map d {x, x_n} < \epsilon$ Hence $M \setminus S$ is not open. Therefore, $S$ is not closed. {{qed|lemma}} === Sufficient Condition === Suppose that $S$ is not closed. Then $M \setminus S$ is not open. Therefore, there exists a $x \in M \setminus S$ such that for all $\epsilon > 0$, there exists a $y \in S$ such that $\map d {x, y} < \epsilon$. So there exists a sequence $\sequence {y_n}$ in $S$ such that for all $n \in \N$: :$\map d {x, y_n} < \dfrac 1 n$ Now, we show that $\sequence {y_n}$ is a Cauchy sequence. Let $N \in \N$ be such that for all $n \ge N$: :$\map d {x, y_n} < \dfrac \epsilon 2$ Let $m, n \ge N$. Then, by the triangle inequality: :$\map d {y_m, y_n} \le \map d {x, y_m} + \map d {x, y_n} < \epsilon$ Hence $\sequence {y_n}$ is a Cauchy sequence. Because $\struct {M, d}$ is a complete metric space by assumption, the limit $\displaystyle \lim_{n \mathop \to \infty} y_n$ exists and is in $M$. Denote this limit by $y$. By the definition of $\sequence {y_n}$: :$\displaystyle \lim_{n \mathop \to \infty} \map d {x, y_n} = 0$ From Distance Function of Metric Space is Continuous and Continuity of Composite Mapping: :$\map d {x, y} = 0$ {{explain|Which two mappings are being composed here?}} By definition of a metric, this implies that $x = y$. Since $y \notin S$, $S$ is not complete. {{qed}} Category:Metric Subspaces Category:Complete Metric Spaces tsfu0pbjo82jwtsxn4b3m4i6dzxdk7w
Let $D = \bigcup \mathcal C$.  Observe that any $C \in \mathcal C$ is a chain in $P$, hence $C \subseteq P$. Therefore, $D \subseteq P$, by Set Union Preserves Subsets. Now suppose that $a, b \in D$. By definition of union, there exist $A, B \in \mathcal C$ such that $a \in A$ and $b \in B$.  Since $\mathcal C$ is a chain in $\left({\operatorname{ch} \left({P}\right), \subseteq}\right)$, either $A \subseteq B$ or $B \subseteq A$. Hence either $a \in B$ or $b \in A$. Both $A$ and $B$ are chains in $P$, and thence $a \leq b$ or $b \leq a$. Therefore, $D$ is totally ordered by $\le$. That is, it is a chain in $P$. {{qed}} Category:Set Theory Category:Order Theory bqwyzf61gost9w8p44l40rgcb9flhyx
For real $p \ne 0$, the Hölder mean is defined as: :$\displaystyle M_p \left({x_1, x_2, \ldots, x_n}\right) = \left({\frac 1 n \sum_{k \mathop = 1}^n x_k^p}\right)^{1/p}$ whenever the above expression is defined. Consider the function: : $\phi : \R_{\ge 0} \to \R_{\ge 0}$ defined by $\phi \left({x}\right) = x^{q/p}$ {{begin-eqn}} {{eqn| l=D_x \left({x^{q/p} }\right)      | r=\dfrac{q}{p} x^{q/p-1}      | c=power rule }} {{end-eqn}} From Real Function is Strictly Convex iff Derivative is Strictly Increasing that $\phi$ is strictly convex if $q > 0$ Similarly, from Real Function is Strictly Concave iff Derivative is Strictly Decreasing and strictly concave if $q < 0$. {{explain|The above needs to be amplified, as it is far from clear why.}} Now apply Jensen's inequality to $x_1^p, x_2^p, \ldots, x_n^p$. For $q > 0$, this gives: :$\displaystyle \left({ \frac 1 n \sum_{k \mathop = 1}^n x_k^p }\right)^{q/p} \le \frac 1 n \sum_{k \mathop = 1}^n x_k^q$ For $q < 0$, the reverse inequality holds. Also by Jensen's inequality, equality holds if and only if $x_1 = x_2 = \cdots = x_n$. In either case, the result follows. {{qed}} Category:Real Analysis Category:Inequalities Category:Measures of Central Tendency ssitrtujj0bf4izv2svida7swzg7uw0
We use the Two-Step Vector Subspace Test. $\left\{{\mathbf 0}\right\}$ is not empty, because it contains $\mathbf{0}$. $\left\{{\mathbf 0}\right\}$ is closed under $+$ because: :$\forall \mathbf{x},\mathbf{y} \in \left\{{\mathbf 0}\right\}, \mathbf{x} + \mathbf{y} = \mathbf{0} + \mathbf{0} = \mathbf{0} \in \left\{{\mathbf 0}\right\}$ $\left\{{\mathbf 0}\right\}$ is closed under multiplication because: :$\forall \lambda \in K, \mathbf{x}\in \left\{{\mathbf 0}\right\}: \lambda\mathbf{x} = \lambda\mathbf{0} = \mathbf{0}\in \left\{{\mathbf 0}\right\}$ Hence the result, from the Two-Step Vector Subspace Test. {{qed}}
This is simply a special case of Cartesian Product of Countable Sets is Countable. {{qed}}
=== Proof for $p = 2$ === $p = 2$ is an easily proved special case: {{:Minkowski's Inequality for Sums/Index 2}} {{qed|lemma}} === Proof for $p > 1$ === {{:Minkowski's Inequality for Sums/Index Greater than 1}} {{qed|lemma}} === Proof for $p < 1$, $p \ne 0$ === {{:Minkowski's Inequality for Sums/Index Less than 1}} {{qed}} {{namedfor|Hermann Minkowski|cat=Minkowski}} Category:Algebra Category:Analysis Category:Inequalities 602bzuaclyhirrgrph9d48773r6rc2m
=== $(1)$ implies $(2)$ === Follows directly from the definition of characteristic function. {{qed|lemma}} === $(2)$ implies $(1)$ === Let $s \in S$. Suppose that $s \in A$. Then by assumption, $f \left({s}\right) = 1$. Also, by definition of characteristic function, $\chi_A \left({s}\right) = 1$. Next, suppose that $s \notin A$. Then $f \left({s}\right) \ne 1$ by assumption. As $f \left({s}\right) \in \left\{{0, 1}\right\}$, it follows that $f \left({s}\right) = 0$. Again, by definition of characteristic function, also have $\chi_A \left({s}\right) = 0$. Hence, for all $s \in S$, have $f \left({s}\right) = \chi_A \left({s}\right)$. By Equality of Mappings, it follows that $f = \chi_A$. {{qed}} Category:Characteristic Functions 07z7f5u39h69xemwks9f4667lhgeey0
Suppose that $\map {\chi_{A \mathop \setminus B} } s = 1$. Then by definition of characteristic function: :$s \in A \setminus B$ That is, by definition of set difference, $s \in A$ and $s \notin B$; in particular, $s \notin A \cap B$. Hence: :$\map {\chi_A} s = 1$ and $\map {\chi_{A \cap B} } s = 0$ Conclude that: :$\map {\chi_{A \mathop \setminus B} } s = 1 \implies \map {\chi_A} s - \map {\chi_{A \cap B} } s = 1$ Now conversely suppose $\map {\chi_A} s - \map {\chi_{A \cap B} } s = 1$. Considering that $\map {\chi_{A \cap B} } s \ge 0$, it must be that $\map {\chi_A} s = 1$ and $\map {\chi_{A \cap B} } s = 0$. Hence, $s \in A$, and $s \notin A \cap B$. By definition of set intersection, this means $s \notin B$. Hence $s \in A \setminus B$ by definition of set difference. Thus: :$\map {\chi_{A \mathop \setminus B} } s = 1$ Hence, it has been established that: :$\map {\chi_{A \mathop \setminus B} } s = 1 \iff \map {\chi_A} s - \map {\chi_{A \cap B} } s = 1$ which, by application of Characteristic Function Determined by 1-Fiber, gives the result. {{qed}}
By definition of symmetric difference: : $A * B = \left({A \cup B}\right) \setminus \left({A \cap B}\right)$ Thus: :$\chi_{A * B} = \chi_{A \cup B} - \chi_{\left({A \cup B}\right) \cap \left({A \cap B}\right)}$ by Characteristic Function of Set Difference. But by Intersection is Subset of Union and Intersection with Subset is Subset: : $\left({A \cup B}\right) \cap \left({A \cap B}\right) = A \cap B$ Hence it follows that: :$\chi_{A * B} = \chi_{A \cup B} - \chi_{A \cap B}$ which by Characteristic Function of Union: Variant 2 becomes: :$\chi_{A * B} = \chi_A + \chi_B - 2 \chi_{A \cap B}$ as desired. {{qed}}
{{ProofWanted}} === Proof of Corollary === Suppose $\mathbf 0 \in \mathbb W$. Then $\mathbb W$ contains an element and is non-empty. Suppose $\mathbb W$ contains an element $\mathbf x \in \R^n$. Then, by criterion $(3)$: :$0 \mathbf x \in \mathbb W$ where $0$ is the zero scalar. But $0 \mathbf x = \mathbf 0$ from Vector Scaled by Zero is Zero Vector, so $\mathbf 0 \in \mathbb W$. {{qed}}
Define: {{begin-eqn}} {{eqn|n = 1      |l = u      |r = \frac 1 p }} {{eqn|l = v      |r = \frac q p }} {{eqn|l = x      |r = \left({a b}\right)^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 \left({1 + \frac 1 q}\right) }} {{eqn|r = p \left({\frac 1 p}\right)      |c = because $\dfrac 1 p - \dfrac 1 q = 1$, by 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 = \implies      |l = \left({a b}\right)^p b^{-p}      |o = \le      |r = \frac {\left({\left({a b}\right)^p}\right)^{1/p} } {1/p} + \frac {\left({b^{-p} }\right)^{q/p} } {q/p}      |c = substituting for $u, v, x, y$ from $(1)$ above }} {{eqn|ll = \implies      |l = a^p      |o = \le      |r = pab + p \frac { b^{-q} } q      |c = algebraic simplification }} {{eqn|ll = \implies      |l = \frac {a^p} p      |o = \le      |r = ab + \frac { b^{-q} } q }} {{eqn|ll = \implies      |l = ab      |o = \ge      |r = \frac {a^p} p - \frac { b^{-q} } q }} {{end-eqn}} {{qed}} {{namedfor|William Henry Young|cat=Young}} Category:Analysis 47nxl9wd0fqa22sq8bin4oka1gbbbll