hcju/proofwikiqa huggingface dataset:
Suppose $\left \langle {x_n}\right \rangle_{n \in \N}$ converges to a limit $l$. Consider the open set in $\R$: :$U := \left({l - \dfrac 1 2 \,.\,.\, l + \dfrac 1 2}\right)$ Then $\forall x \in \Z: x \in U \implies x = l$ It follows by definition of convergence that: :$\exists k \in \N: \forall m \in \N: m > k: x_m = x_k$ where $x_k = l$. Now suppose that: :$\exists k \in \N: \forall m \in \N: m > k: x_m = x_k$ Then trivially $\left \langle {x_n}\right \rangle_{n \in \N}$ converges to the limit $x_k$. {{qed}} Category:Sequences Category:Convergence 3xlxzoi9mutdaomqia6slko422gsgqr
From the definition of perfectly normal space, it is necessary to prove that $T$ is a $T_1$ space and that any closed set is $G_\delta$. From $T_2$ Space is $T_1$ Space and Sorgenfrey Line is Hausdorff: :the Sorgenfrey line is a $T_1$ space. From Complement of $F_\sigma$ Set is $G_\delta$ Set it is sufficient to prove that an open set of $T$ is $F_\sigma$. Let $W$ be any open set in $T$. Let $O \subseteq W$ be the interior of $W$ with respect to the metric space: :$\R = \struct {\R, d}$ where $d$ is the usual metric on $\R$. From the definition of $T$, for each $x \in W \setminus O$, we can choose $h_x \in W$ such that $\hointr x {h_x} \subseteq W$. Suppose $\hointr x {h_x} \cap \hointr y {h_y} \ne \O$ for some distinct points $x, y \in W \setminus O$. Then either $x < y < h_x$ or $y < x < h_y$. If $x < y < h_x$, then $y \in \openint x {h_x} \subseteq O$, contradicting $y \in W \setminus O$. Similarly, if $y < x < h_y$, then $x \in \openint y {h_y}) \subseteq O$, contradicting $x \in W \setminus O$. Thus $\family {\hointr x {h_x} }_{x \mathop \in W \setminus O}$ is a pairwise disjoint indexed family of open sets of $T$. From Sorgenfrey Line is Separable and Separable Space satisfies Countable Chain Condition: :$\set {\hointr x {h_x} : x \in W \setminus O}$ is countable and thus: :$W \setminus O$ is countable. From Metric Space is Perfectly T4: :$O$ is an $F_\sigma$ set in $\R$. Thus from Sorgenfrey Line is Expansion of Real Line: :$O$ is an $F_\sigma$ set in the Sorgenfrey line. Since $W \setminus O$ is a countable union of singletons and $T$ is a $T_1$ space: :$W \setminus O$ is an $F_\sigma$ set in $T$. Since $W = O \cup \left({W \setminus O}\right)$ and $F_\sigma$ sets are closed under unions: :$W$ is an $F_\sigma$ set in $T$. {{qed}} Category:Sorgenfrey Line Category:Perfectly Normal Spaces sxfiiej1t8wvwxi5jz5xw9wqkc7doyz
It is enough to prove that any open set in $\R$ is open in $T$. Let $a, b \in \R$. Then: :$\displaystyle \openint a b = \bigcup_{\epsilon \mathop > 0} \hointr {a + \epsilon} b$ Since $\hointr {a + \epsilon} b$ are open in $T$, $\openint a b$ is also open in $T$. {{qed}} Category:Sorgenfrey Line 5lkzxxu4seeyk0z1uuijn6e1df02vcl
We have Sorgenfrey Line is Perfectly Normal. The result follows from Sequence of Implications of Separation Axioms. {{qed}} Category:Sorgenfrey Line bf26hxhpvaaf8ewwt7bpfv4w79sbih9
Let $\theta_1 = \map \arg {z_1}, \theta_2 = \map \arg {z_2}$. Then the polar forms of $z_1, z_2$ are: :$z_1 = \cmod {z_1} \paren {\cos \theta_1 + i \sin \theta_1}$ :$z_2 = \cmod {z_2} \paren {\cos \theta_2 + i \sin \theta_2}$ By the definition of complex multiplication, factoring $\cmod {z_1} \cmod {z_2}$ from all terms, we have: :$z_1 z_2 = \cmod {z_1} \cmod {z_2} \paren {\paren {\cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2} + i \paren {\cos \theta_1 \sin \theta_2 + \sin \theta_1 \cos \theta_2} }$ Using Sine of Sum and Cosine of Sum, we have: :$z_1 z_2 = \cmod {z_1} \cmod {z_2} \paren {\map \cos {\theta_1 + \theta_2} + i \, \map \sin {\theta_1 + \theta_2} }$ The theorem follows from the definition of $\map arg z$, which says that $\map \arg {z_1 z_2}$ satisfies the equations: :$(1): \quad \dfrac {\cmod {z_1} \cmod {z_2} \map \cos {\theta_1 + \theta_2} } {\cmod {z_1} \cmod {z_2} } = \map \cos {\map \arg {z_1 z_2} }$ :$(2): \quad \dfrac {\cmod {z_1} \cmod {z_2} \map \sin {\theta_1 + \theta_2} } {\cmod {z_1} \cmod {z_2} } = \map \sin {\map \arg {z_1 z_2} }$  which in turn means that: :$\map \cos {\theta_1 + \theta_2} = \map \cos {\map \arg {z_1 z_2} }$ :$\map \sin {\theta_1 + \theta_2} = \map \sin {\map \arg {z_1 z_2} }$ There are $3$ possibilities for the size of $\theta_1 + \theta_2$: :$(1): \quad \theta_1 + \theta_2 > \pi$ Then: :$-\pi < \theta_1 + \theta_2 - 2 \pi \le \pi$ and we have: {{begin-eqn}} {{eqn | l = \map \cos {\theta_1 + \theta_2}       | r = \map \cos {\theta_1 + \theta_2 - 2 \pi}       | c = Cosine of Angle plus Full Angle }} {{eqn | l = \map \sin {\theta_1 + \theta_2}       | r = \map \sin {\theta_1 + \theta_2 - 2 \pi}       | c = Sine of Angle plus Full Angle }} {{end-eqn}} and so $\theta_1 + \theta_2 - 2 \pi$ is the argument of $z_1 z_2$ within its principal range. :$(2): \quad \theta_1 + \theta_2 \le -\pi$ Then: : $-\pi < \theta_1 + \theta_2 + 2 \pi \le \pi$ and we have: {{begin-eqn}} {{eqn | l = \map \cos {\theta_1 + \theta_2}       | r = \map \cos {\theta_1 + \theta_2 + 2 \pi}       | c = Cosine of Angle plus Full Angle }} {{eqn | l = \map \sin {\theta_1 + \theta_2}       | r = \map \sin {\theta_1 + \theta_2 + 2 \pi}       | c = Sine of Angle plus Full Angle }} {{end-eqn}} and so $\theta_1 + \theta_2 + 2 \pi$ is within the principal range of $z_1 z_2$. :$(3): \quad -\pi < \theta_1 + \theta_2 \le \pi$ Then $\theta_1 + \theta_2$ is already within the principal range of $z_1 z_2$. Therefore: :$\map \arg {z_1 z_2} = \theta_1 + \theta_2 = \map \arg {z_1} + \map \arg {z_2} + 2 k \pi$ where $k$ can be $0$, $1$ or $-1$. {{qed}}
Suppose $H = \left \langle {x_n}\right \rangle_{n \in \N}$ converges to a limit $L$. As $T$ is a discrete space, $\left\{{L}\right\}$ is an open set in $T$. Then trivially: :$\forall x \in H: x \in \left\{{L}\right\} \implies x = L$ So for $H$ to be convergent, it is necessary that: :$\exists k \in \N: \forall m \in \N: m > k: x_m = L$ It follows by definition of convergence that: :$\exists k \in \N: \forall m \in \N: m > k: x_m = x_k$ where $l = L$. Now suppose that: :$\exists k \in \N: \forall m \in \N: m > k: x_m = x_k$ Then trivially $\left \langle {x_n}\right \rangle_{n \in \N}$ converges to the limit $L$. {{qed}} Category:Sequences Category:Discrete Topology Category:Convergence 31dkxcqscqe5s3wd912otkzfhixnj47
By the definition of Fortissimo space, any $A \subset S \setminus \left\{{p}\right\}$ is open in $S$, because $p \notin A$. Thus by definition of topological subspace, $A \subset S \setminus \left\{{p}\right\}$ is open in $S \setminus \left\{{p}\right\}$. The result follows by the definition of discrete space. {{qed}} Category:Fortissimo Space Category:Discrete Topology lvele0fmw6hgzfeibimfmewtfwl2i3m
From Finite Space Satisfies All Compactness Properties, for every $p \in S$, $\left\{{p}\right\}$ is compact. Then $\displaystyle S = \bigcup_{p \mathop \in S} \left\{{p}\right\}$ is a countable union of compact sets. Thus by definition $T$ is $\sigma$-compact. {{qed}} Category:Sigma-Compact Spaces 916oy2s4pz8ar082wzktb4juasfaeyu
By definition of a Fortissimo space, $U \subseteq S$ is open  in $T$ {{iff}}: :$p \in \relcomp S U$ or :$\relcomp S U$ is countable or both. The result follows from the definition of closed set. {{qed}} Category:Closed Sets Category:Fortissimo Space 565de7e0p1d6wni18ezbm0minu19t5d
First let $A = \set {\tuple {0, 0} }$. From the definition of Arens-Fort space, $\set {\tuple {0, 0} }$ is closed because $S \setminus \set {\tuple {0, 0} }$ is open. From Closed Set Equals its Closure: :$\set {\tuple {0, 0} }^- = \set {\tuple {0, 0} }$ where $\set {\tuple {0, 0} }^-$ denotes the closure of $\set {\tuple {0, 0} }$. From the definition of Arens-Fort space, $\set {\tuple {0, 0} }$ is not open. Therefore the smallest open set contained in $\set {\tuple {0, 0} }$ is $\O$. Hence: :$\set {\tuple {0, 0} }^\circ = \O$  where $\set {\tuple {0, 0} }^\circ$ denotes the interior of $\set {\tuple {0, 0} }$. Thus we have that: :$\paren {\set {\tuple {0, 0} }^-}^\circ = \O$ and so by definition $\set {\tuple {0, 0} }$ is nowhere dense in $T$. From Union of Singleton it trivially follows that $\set {\tuple {0, 0} }$ is the union of a countable set of subsets of $S$ (that is: just the one) which are nowhere dense in $S$. Hence by definition $\set {\tuple {0, 0} }$ is meager. {{qed|lemma}} Now assume $A \ne \set {\tuple {0, 0} }$. Then $\exists x \in A: x \ne \tuple {0, 0}$.  By definition of the Arens-Fort space, $\set x$ is open in $T$.  Thus, from Interior of Open Set $\set x^\circ = \set x$. Hence $\set x \subseteq \paren {\set x^-}^\circ$ and by definition is not nowhere dense in $T$. So $A$ is not the union of a countable set of subsets of $S$ which are nowhere dense in $S$. So, by definition, $A$ is not meager. {{qed}} Category:Meager Spaces Category:Arens-Fort Space 07ys17kihokqi6du8kytry68idscq0f
We have the result Center is Intersection of Centralizers. That is, $\map Z G$ is the intersection of all the centralizers of $G$. All of these are subgroups of $G$ by Centralizer of Group Element is Subgroup. Thus from Intersection of Subgroups is Subgroup, $\map Z G$ is also a subgroup of $G$. {{qed}}
For brevity, suppress the symbol for the group operation (which may be $\circ$, or $+$). Apply the Two-Step Subgroup Test: === Condition $(1)$ === By the definition of identity, $e g = g e = g$ for all $g \in G$. So, $e \in \map Z G$, meaning $\map Z G$ is non-empty. {{qed|lemma}} === Condition $(2)$ === Suppose $a, b \in \map Z G$. Using the associative property and the definition of center, we have: :$\forall g \in G: \paren {a b} g = a \paren {b g} = a \paren {g b} = \paren {a g} b = \paren {g a} b = g \paren {a b}$ Thus, $a b \in \map Z G$. {{qed|lemma}} === Condition $(3)$ === Suppose $c \in \map Z G$. Then: {{begin-eqn}} {{eqn | l = c       | o = \in       | r = \map Z G }} {{eqn | ll= \leadsto       | l = \forall g \in G: c g       | r = g c       | c = Definition of center }} {{eqn | ll= \leadsto       | l = c^{-1} \paren {c g} c^{-1}       | r = c^{-1} \paren {g c} c^{-1} }} {{eqn | ll= \leadsto       | l = \paren {c^{-1} c} g c^{-1}       | r = c^{-1} g \paren {c c^{-1} } }} {{eqn | ll= \leadsto       | l = e g c^{-1}       | r = c^{-1} g e }} {{eqn | ll= \leadsto       | l = g c^{-1}       | r = c^{-1} g }} {{eqn | ll= \leadsto       | l = c^{-1}       | o = \in       | r = \map Z G }} {{end-eqn}} {{qed|lemma}} Therefore: : $\map Z G \le G$ {{qed}}
By Center of Group is Subgroup, $Z \left({G}\right)$ is a subgroup of $G$. The definition of the center $Z \left({G}\right)$ grants that all elements of $Z \left({G}\right)$ commute with all elements of $G$. In particular, all elements of $Z \left({G}\right)$ commute with all elements of $Z \left({G}\right)$ as $Z \left({G}\right) \subseteq G$. Therefore $Z \left({G}\right)$ is abelian. {{qed}}
By definition, the '''dihedral group''' $D_n$ of order $2 n$ is the group of symmetries of the regular $n$-gon. So, let $P$ denote a regular polygon with $n$ sides. Let $\alpha$ be a rotation of $P$ by $\dfrac {2 \pi} n$. Let $\beta$ be a reflection $P$ whose axis of reflection is the $y$ axis. It takes $n$ rotations by $\dfrac {2 \pi} n$ to return $P$ to its original position. Thus $\alpha$ is of order $n$, and so: :$\alpha^n = e$ Two reflections of $P$ in the same axis of reflection also returns $P$ to its original position. Thus $\beta$ is of order $2$, and so: :$\beta^2 = e$ It remains to be shown that $\beta \alpha \beta = \alpha^{-1}$. Let the vertices of $P$ be labelled in order anticlockwise from the $y$-axis: $A_1, A_2, \ldots, A_n$. Let $P$ be arranged so that $A_1$ is on the $y$-axis. From here, $\beta$ on $P$ exchanges the places of $A_2$ and $A_n$. From here, $\alpha$ sends $A_2$ to the position of $A_1$, $A_1$ to the original position of $A_2$, and $A_3$ to the original position of $A_n$. From here, $\beta$ sends exchanges the positions of $A_3$ and $A_1$. So $\beta \alpha \beta$ on $P$ results in $A_1$ moving to the original position of $A_n$, $A_2$ moving to the original position of $A_1$ and $A_3$ moving to the original position of $A_2$. That is, $\beta \alpha \beta$ is a rotation of $P$ by $-\dfrac {2 \pi} n$, or $\alpha^{-1}$. {{qed}}
By definition of set equality, it suffices to prove: :$X \cap \paren {Y \circ Z} \subseteq Y \circ \paren {X \cap Z}$ and: :$Y \circ \paren {X \cap Z} \subseteq X \cap \paren {Y \circ Z}$ === $X \cap \paren {Y \circ Z}$ is contained in $Y \circ \paren {X \cap Z}$ === Let $s \in X \cap \paren {Y \circ Z}$. {{begin-eqn}} {{eqn | l = s       | o = \in       | r = X \cap \paren {Y \circ Z}       | c =  }} {{eqn | lll= \leadsto       | l = s       | o = \in       | r = X       | c = Definition of Set Intersection }} {{eqn | lo= \land       | l = s       | o = \in       | r = Y \circ Z       | c = Definition of Set Intersection }} {{eqn | lll= \leadsto       | ll= \exists y \in Y, z \in Z:       | l = s       | r = y \circ z       | c = Definition of Subset Product }} {{eqn | n = 1       | lll= \leadsto       | l = z       | r = y^{-1} \circ s       | c = Group has Latin Square Property }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = Y       | o = \subseteq       | r = X       | c =  }} {{eqn | lll=\leadsto       | l = y       | o = \in       | r = X       | c = {{Defof|Subset}} }} {{end-eqn}} So we have: {{begin-eqn}} {{eqn | l = y       | o = \in       | r = X       | c =  }} {{eqn | lo= \land       | l = s       | o = \in       | r = X       | c = {{Defof|Subset}} }} {{eqn | lll=\leadsto       | l = y^{-1} \circ s       | o = \in       | r = X       | c = One-Step Subgroup Test }} {{eqn | lll=\leadsto       | l = z       | o = \in       | r = X       | c = from $(1)$ }} {{end-eqn}} Thus by definition of set intersection: : $z \in X \cap Z$ So: :$s = y \circ z \in Y \circ \paren {X \cap Z}$ By definition of subset: : $X \cap \paren {Y \circ Z} \subseteq Y \circ \paren {X \cap Z}$ {{qed|lemma}} === $Y \circ \paren {X \cap Z}$ is contained in $X \cap \paren {Y \circ Z}$ === {{begin-eqn}} {{eqn | l = Y \circ X       | o = \subseteq       | r = X \circ X       | c = Subset of Subset Product }} {{eqn | n = 1       | ll= \leadsto       | l = Y \circ X       | o = \subseteq       | r = X       | c = Product of Subgroup with Itself }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = Y \circ \paren {X \cap Z}       | o = \subseteq       | r = \paren {Y \circ X} \cap \paren {Y \circ Z}       | c = Product of Subset with Intersection }} {{eqn | o = \subseteq       | r = X \cap \paren {Y \circ Z}       | c = from $(1)$ }} {{end-eqn}} {{qed|lemma}} We have established that: : $x \in Y \circ \paren {X \cap Z} \iff x \in X \cap \paren {Y \circ Z}$ From the definition of set equality: : $Y \circ \paren {X \cap Z} = X \cap \paren {Y \circ Z}$ {{qed}}
=== Proof of $(1)$ === For $x, y \in V$, compute: {{begin-eqn}} {{eqn | l = \left\Vert{x + y}\right\Vert^2       | r = \left\langle{x + y, x + y}\right\rangle       | c = Definition of $\left\Vert{\cdot}\right\Vert$ }} {{eqn | r = \left\langle{x, x}\right\rangle + \left\langle{x, y}\right\rangle + \left\langle{y, x}\right\rangle + \left\langle{y, y}\right\rangle       | c = Linearity of $\left\langle{\cdot, \cdot}\right\rangle$ }} {{eqn | o = \le       | r = \left\Vert{x}\right\Vert^2 + \sqrt{\left\langle{x, x}\right\rangle \left\langle{y, y}\right\rangle} + \sqrt{\left\langle{y, y}\right\rangle \left\langle{x, x}\right\rangle} + \left\Vert{y}\right\Vert^2       | c = Cauchy-Bunyakovsky-Schwarz Inequality }} {{eqn | r = \left\Vert{x}\right\Vert^2 + 2 \left\Vert{x}\right\Vert \left\Vert{y}\right\Vert + \left\Vert{y}\right\Vert^2       | c = Definition of $\left\Vert{\cdot}\right\Vert$ }} {{eqn | r = \left({\left\Vert{x}\right\Vert + \left\Vert{y}\right\Vert}\right)^2 }} {{end-eqn}} Taking square roots on either side gives the result. {{qed|lemma}} === Proof of $(2)$ === For $x \in V$, $a \in \Bbb F$, compute: {{begin-eqn}} {{eqn | l = \left\Vert{a x}\right\Vert^2       | r = \left\langle{a x, a x}\right\rangle       | c = Definition of $\left\Vert{\cdot}\right\Vert$ }} {{eqn | r = a \left\langle{x, a x}\right\rangle       | c = Linearity of $\left\langle{\cdot, \cdot}\right\rangle$ }} {{eqn | r = a \overline{\left\langle{a x, x}\right\rangle}       | c = Conjugate symmetry of $\left\langle{\cdot, \cdot}\right\rangle$ }} {{eqn | r = a \overline a \overline{\left\langle{x, x}\right\rangle}       | c = Linearity of $\left\langle{\cdot, \cdot}\right\rangle$ }} {{eqn | r = \left\vert{a}\right\vert^2 \left\Vert{x}\right\Vert^2       | c = Modulus in Terms of Conjugate }} {{end-eqn}} Taking square roots on either side gives the result. {{qed}}
Let us verify the norm axioms in turn. === Axiom $(\text N 1)$ === {{begin-eqn}} {{eqn | o =        | r = x = 0_V       | c =  }} {{eqn | ll= \leadstoandfrom       | o =        | r = \innerprod x x = 0       | c = Property $(5)$ of Inner Product }} {{eqn | ll= \leadstoandfrom       | o =        | r = \innerprod x x^{1 / 2} = 0 }} {{eqn | ll= \leadstoandfrom       | o =        | r = \norm x = 0       | c = {{Defof|Inner Product Norm}} }} {{end-eqn}} {{Qed|lemma}} === Axiom $(\text N 2)$ === Part $(2)$ of Properties of Semi-Inner Product. {{Qed|lemma}} === Axiom $(\text N 3)$ === Part $(3)$ of Properties of Semi-Inner Product. {{Qed|lemma}} Hence all the properties of a norm have been shown to hold. {{Qed}}
=== Proof of $\text M 1$ and $\text M 4$ === Let $x, y \in V$.  Then: :$\map d {x, y} = \norm {x - y} \ge 0$ and furthermore: {{begin-eqn}} {{eqn | l = \map d {x, y}       | r = 0 }} {{eqn | ll= \leadstoandfrom       | l = \norm {x - y}       | r = 0 }} {{eqn | ll= \leadstoandfrom       | l = x - y       | r = \mathbf 0_V       | c = Norm Axioms: Axiom $\text N 1$ }} {{eqn | ll= \leadstoandfrom       | l = x       | r = y }} {{end-eqn}} {{qed|lemma}} === Proof of $\text M 2$ === Let $x, y, z \in V$. Then: {{begin-eqn}} {{eqn | l = \map d {x, z}       | r = \norm {x - z} }} {{eqn | r = \norm {x - y + y - z}       | c =  }} {{eqn | o = \le       | r = \norm {x - y} + \norm {y - z}       | c = Norm Axioms: Axiom $\text N 3$ }} {{eqn | r = \map d {x, y} + \map d {y, z} }} {{end-eqn}} {{qed|lemma}} === Proof of $\text M 3$ === Let $x, y \in V$. Then:  {{begin-eqn}} {{eqn | l = \map d {x, y}       | r = \norm {x - y} }} {{eqn | r = \norm {-1 \paren {y - x} } }} {{eqn | r = \norm {-1} \times \norm {y - x}       | c = Norm Axioms: Axiom $\text N 2$ }} {{eqn | r = \norm {y - x} }} {{eqn | r = \map d {y, x} }} {{end-eqn}} {{qed|lemma}} As $d$ satisfies the metric space axioms, it is a metric. {{qed}} Category:Norm Theory Category:Metric Spaces 1ojmpq5wvlf7ez22c3n312r3utnvffc
Proof by induction: For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the proposition: : $\displaystyle \exists w, x, y, z \in \Z: \prod_{j \mathop = 1}^n \left({a_j^2 + b_j^2 + c_j^2 + d_j^2}\right) = w^2 + x^2 + y^2 + z^2$ $P \left({1}\right)$ is true, as this just says: : $\exists w, x, y, z \in \Z: a^2 + b^2 + c^2 + d^2 = w^2 + x^2 + y^2 + z^2$ which is trivially true. === Basis for the Induction === $P \left({2}\right)$ is the case: : $\exists w, x, y, z \in \Z: \left({a_1^2 + b_1^2 + c_1^2 + d_1^2}\right) \left({a_2^2 + b_2^2 + c_2^2 + d_2^2}\right) = w^2 + x^2 + y^2 + z^2$ which follows from Product of Sums of Four Squares. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true. So this is our induction hypothesis: : $\displaystyle \exists w, x, y, z \in \Z: \prod_{j \mathop = 1}^k \left({a_j^2 + b_j^2 + c_j^2 + d_j^2}\right) = w^2 + x^2 + y^2 + z^2$ Then we need to show that it directly implies: : $\displaystyle \exists w, x, y, z \in \Z: \prod_{j \mathop = 1}^{k+1} \left({a_j^2 + b_j^2 + c_j^2 + d_j^2}\right) = w^2 + x^2 + y^2 + z^2$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | o =        | r = \prod_{j \mathop = 1}^{k+1} \left({a_j^2 + b_j^2 + c_j^2 + d_j^2}\right)       | c =  }} {{eqn | r = \left({\prod_{j \mathop = 1}^k \left({a_j^2 + b_j^2 + c_j^2 + d_j^2}\right)}\right) \left({a_{k+1}^2 + b_{k+1}^2 + c_{k+1}^2 + d_{k+1}^2}\right)       | c =  }} {{eqn | r = \left({r^2 + s^2 + t^2 + u^2}\right) \left({a_{k+1}^2 + b_{k+1}^2 + c_{k+1}^2 + d_{k+1}^2}\right)       | c = from the induction hypothesis: for some $r, s, t, u, \in \Z$ }} {{eqn | r = w^2 + x^2 + y^2 + z^2       | c = from the basis for the induction: for some $w, x, y, z, \in \Z$ }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: : $\displaystyle \forall n \in \Z_{>0}: \exists w, x, y, z \in \Z: \prod_{j \mathop = 1}^n \left({a_j^2 + b_j^2 + c_j^2 + d_j^2}\right) = w^2 + x^2 + y^2 + z^2$ {{qed}} Category:Product of Sums of Four Squares Category:Proofs by Induction mqkxsxvflrnnxqfi8350kswpk0gnpj9
By definition of subset product: :$H \circ K = \left\{{h \circ k: h \in H, k \in K}\right\}$ So: {{begin-eqn}} {{eqn | l=x       | o=\in       | r=H       | c= }} {{eqn | ll=\implies       | l=x       | r=x \circ e       | c=as $e$ is the identity of $G$ }} {{eqn | o=\in       | r=H \circ K       | c=as $e \in K$ from Identity of Subgroup }} {{eqn | ll=\implies       | l=H       | o=\subseteq       | r=H \circ K       | c=Definition of subset }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l=y       | o=\in       | r=K       | c= }} {{eqn | ll=\implies       | l=y       | r=e \circ y       | c=as $e$ is the identity of $G$ }} {{eqn | o=\in       | r=H \circ K       | c=as $e \in H$ from Identity of Subgroup }} {{eqn | ll=\implies       | l=K       | o=\subseteq       | r=H \circ K       | c=Definition of subset }} {{end-eqn}} {{qed}} Category:Subset Products Category:Subgroups dd8alail40q7bzc449ly8bqi4ahdwac
=== Necessary Condition === Suppose $S_i$ is path-connected for each $i\in I$.  Let $x \ne y \in S$ be arbitrary. Since each $S_i$ is path-connected, we have a continuous mapping: :$f_i: \closedint {0} {1} \to S_i$ such that $\map {f_i} 0 = x_i$ and $\map {f_i} 1  = y_i$ for all $i \in I$. We then define: : $f : \closedint {0} {1} \to S$ by setting: :$\forall t \in \closedint {0} {1} : \map f t_i = \map {f_i} t$ We have that $f_i = \pr_i \circ f$ are all continuous. So, by Continuous Mapping to Product Space, it follows that $f$ is continuous. Also: : $\map f 0_i = \map {f_i} 0 = x_i$ and: : $\map f 1_i = \map {f_i} 1 = y_i$ Therefore $\map f 0 = x$ and $\map f 1 = y$. So there exists a continuous mapping: : $f: \closedint {0} {1} \to S$  such that $\map f 0 = x$ and $\map f 1 = y$. Since $x$ and $y$ were arbitrary, we have shown that $S$ is path-connected. {{qed|lemma}} === Sufficient Condition === Conversely, suppose $T$ is a path-connected space. Let $j$ be an arbitrary $j \in I$. Let $x_j$ and $y_j$ be points in $S_j$ such that $x_j \ne y_j$. We have that $\forall i \in I: S_i \ne \O$. Using the axiom of choice, a choice function can be set up to allow the choice of some $q_i \in S_i$ for each $i \ne j$. We then define: : $\forall s \in T: \map {\pr_i} s = q_i$ From Constant Mapping is Continuous, it follows that each of $\pr_i: T \to S_i$ where $i \ne j$ is continuous: Thus there exist points in $x, y \in S$ such that:  : $\map {\pr_i} x = q_i$ : $\map {\pr_i} y = q_i$ for all $i \ne j$, and: : $\map {\pr_j} x = x_j$ : $\map {\pr_j} y = y_j$ Since $T$ is path-connected, there exists a continuous mapping: : $f: \closedint {0} {1} \to S$  such that $\map f 0 = x$ and $\map f 1 = y$. By Continuous Mapping to Product Space: : $\pr_j \circ f: \closedint {0} {1} \to S_j$ is also continuous. Furthermore: : $\map {\pr_j \circ f} 0 = \map {\pr_j} x = x_j$ and: : $\map {\pr_j \circ f} 1 = \map {\pr_j} y = y_j$ Thus since $x_j$ and $y_j$ were arbitrary, we have shown that $S_j$ is path-connected.   Since $j$ was arbitrary, we have shown that all $S_i$ are path-connected. {{qed}} {{AoC}}
From Subset Product with Normal Subgroup is Subgroup: :$N H = H N$ is a subgroup of $G$. By definition of subset product all elements of $H N$ can be written in the form: :$h n \in H N$ where $h \in H, n \in N$. Let $h n \in H N$. Let $n_1 \in N$. From Inverse of Group Product: :$\paren {h n} n_1 \paren {h n}^{-1} = h n n_1 n^{-1} h^{-1}$ We have that: :$n n_1 n \in N$ :$h, h^{-1} \in G$. Then, since $N$ is a normal subgroup of $G$: :$\paren {h n} n_1 \paren {n^{-1} h^{-1} } \in N$ Thus: : $N \lhd N H$ {{qed}}