hcju/proofwikiqa huggingface dataset:
Checking in turn each of the criteria for equivalence: === Reflexivity === Let $M$ be a topological space. From Identity Mapping on Metric Space is Isometry, the identity mapping $I_M: M \to M$ is an isometry. So $M \sim M$, and $\sim$ has been shown to be reflexive. {{qed|lemma}} === Symmetry === Let $M_1$ and $M_2$ be topological spaces such that $M_1 \sim M_2$. By definition, there exists an isometry $f: M_1 \to M_2$. From Inverse of Isometry of Metric Spaces is Isometry it follows that $f^{-1}: M_2 \to M_1$ is also an isometry. So $M_2 \sim M_1$, and $\sim$ has been shown to be symmetric. {{qed|lemma}} === Transitivity === Let $M_1, M_2, M_3$ be metric spaces such that $M_1 \sim M_2$ and $M_2 \sim M_3$. By definition, there exist isometries $f: M_1 \to M_2$ and $g: M_2 \to M_3$. From Composite of Isometries is Isometry it follows that $g \circ f: M_1 \to M_3$ is also an isometry. So $M_1 \sim M_3$, and $\sim$ has been shown to be transitive. {{qed|lemma}} $\sim$ has been shown to be reflexive, symmetric and transitive. Hence by definition it is an equivalence relation. {{qed}}
Checking in turn each of the criteria for equivalence: === Reflexivity === Let $M$ be a metric space. From Identity Mapping on Metric Space is Homeomorphism, the identity mapping $I_M: M \to M$ is a homeomorphism. So $M \sim M$, and $\sim$ has been shown to be reflexive. {{qed|lemma}} === Symmetry === Let $M_1$ and $M_2$ be topological spaces such that $M_1 \sim M_2$. By definition, there exists a homeomorphism $f: M_1 \to M_2$. From Inverse of Homeomorphism between Metric Spaces is Homeomorphism it follows that $f^{-1}: M_2 \to M_1$ is also a homeomorphism. So $M_2 \sim M_1$, and $\sim$ has been shown to be symmetric. {{qed|lemma}} === Transitivity === Let $M_1, M_2, M_3$ be metric spaces such that $M_1 \sim M_2$ and $M_2 \sim M_3$. By definition, there exist homeomorphisms $f: M_1 \to M_2$ and $g: M_2 \to M_3$. From Composite of Homeomorphisms between Metric Spaces is Homeomorphism it follows that $g \circ f: M_1 \to M_3$ is also a homeomorphism. So $M_1 \sim M_3$, and $\sim$ has been shown to be transitive. {{qed|lemma}} $\sim$ has been shown to be reflexive, symmetric and transitive. Hence by definition it is an equivalence relation. {{qed}}
By the definition of an isometry, $f$ is a bijection $f: A_1 \to A_2$ such that: : $\forall a, b \in A_1: d_1 \left({a, b}\right) = d_2 \left({\phi \left({a}\right), \phi \left({b}\right)}\right)$ By Isometry between Metric Spaces is Continuous, $f$ is a continuous mapping from $M_1$ to $M_2$. By the corollary to Isometry between Metric Spaces is Continuous, $f^{-1}$ is a continuous mapping from $M_2$ to $M_1$. Thus $f$ is a homeomorphism from $M_1$ to $M_2$. Hence the result. {{qed}} Category:Isometries Category:Homeomorphic Metric Spaces mygfttyy9exz01uppqpaf2h4p3jq0ky
From Isometry of Metric Spaces is Equivalence Relation, all isometries can be partitioned into equivalence classes. From Homeomorphism of Metric Spaces is Equivalence Relation, all homeomorphisms can be partitioned into equivalence classes. Let $\mathcal C_I$ be an equivalence class of isometries. Let $f: M_1 \to M_2$ be an element of $\mathcal C_I$ where $M_1$ and $M_2$ are some metric spaces. From Isometry of Metric Spaces is Homeomorphism, $\mathcal C_I$ is a homeomorphism. Thus $\mathcal C_I$ is an element of some equivalence class of homeomorphisms on the set of metric spaces. {{qed}}
=== Sufficient Condition === Suppose that $f_n$ converges to $f$ uniformly on all compact subsets of $U$. Let $z \in U$. Since $U$ is open, there is an open ball $B_\varepsilon \left({z}\right)$, around $z$, of radius $\varepsilon$, with the closure of the ball, $\overline{B_\varepsilon \left({z}\right)}$, contained in $U$. {{proofread}} Since $\overline {B_\varepsilon \left({z}\right)}$ is closed and bounded, it is compact, by the Heine-Borel Theorem.  We then have by assumption that $f_n$ converges uniformly on $\overline{B_\varepsilon \left({z}\right)}$. {{proofread}} Thus $f_n$ converges locally uniformly on $U$. {{qed|lemma}} === Necessary Condition === Now suppose that $f_n$ converges to $f$ locally uniformly. That is, for every $z \in U$, there is an $r$ so that $B_r \left({z}\right) \subset U$ and $f_n$ converges uniformly on $B_r \left({z}\right)$. Let $K \subset U$ be an arbitrary compact subset. Let $\left\{ {B_{r_z} \left({z}\right)}\right\}_{z \mathop \in K}$ be an open cover of $K$ by open balls, one centered at each $z \in K$, such that $f_n$ converges uniformly on each ball. Since $K$ is compact, there is a finite subset $\left\{ {B_{r_i} \left({z_i}\right) }\right\}_{i \mathop = 1}^n$ which covers $K$. Let $\varepsilon > 0$, and $1 \le i \le n$.  Then there is an $N_i$ large enough so that: :$\displaystyle \sup_{z \mathop \in B_{r_i} \left({z_i}\right)} \left\vert{f_n \left({z}\right) - f \left({z}\right)}\right\vert < \varepsilon$ for all $n > N_i$, since the convergence is uniform. Let $N = \displaystyle \max_{1 \mathop \le i \mathop \le n} \left\{N_i\right\}$. Let $B = \displaystyle \bigcup_{i \mathop = 1}^n B_{r_i} \left( z_i \right)$. Then: : $\displaystyle \sup_{z \in B} \left\vert f_n \left({z}\right) - f \left({z}\right) \right\vert < \varepsilon$ for all $n > N$. Thus $f_n$ converges uniformly on $B$, and so also on $K$. {{qed}} Category:Complex Analysis Category:Uniform Convergence rkvkyq8spt2njdh00fbbgiyc18kj8lv
Let $g \in G$. Since $K$ is normal in $G$, and $P \subset K$, the conjugate $g \circ P \circ g^{-1}$ of $P$ is also a subset of $K$. From Inner Automorphism is Automorphism, $g \circ P \circ g^{-1}$ is a subgroup of $K$ of the same order as $P$. Thus $g \circ P \circ g^{-1}$ is also a Sylow $p$-subgroup of $K$. By Second Sylow Theorem: :$\exists k \in K: g \circ P \circ g^{-1} = k \circ P \circ k^{-1}$ Then: :$g^{-1} \circ k \circ P \circ k^{-1} \circ g = P$ and so by definition of normalizer: :$k^{-1} \circ g \in \map {N_G} P$ We can then write $g$ as :$g = k \circ \paren {k^{-1} \circ g} \in K \circ \map {N_G} P$ Since $K$ is normal: :$\map {N_G} P \circ K = K \circ \map {N_G} P$ by Subset Product with Normal Subgroup as Generator. {{qed}} {{Namedfor|Giovanni Frattini|cat = Frattini}}
Let $\chi_E$ be the characteristic function of $E$. Because $A$ and $E \setminus A$ are disjoint: :$A \cap \paren {E \setminus A} = \O$ By Characteristic Function of Union: :$\chi_E = \chi_A + \chi_{E \mathop \setminus A}$  Integrating $f$ over $E$ gives: {{begin-eqn}} {{eqn | l = \int_E f \rd \mu       | r = \int \chi_E \cdot f \rd \mu       | c = {{Defof|Integral of Integrable Function over Measurable Set}} }} {{eqn | r = \int \paren {\chi_A + \chi_{E \mathop \setminus A} } \cdot f \rd \mu       | c = by the above argument }} {{eqn | r = \int \paren {\chi_A \cdot f + \chi_{E \mathop \setminus A} \cdot f} \rd \mu       | c = Pointwise Operation on Distributive Structure is Distributive }} {{eqn | r = \int \chi_A \cdot f \rd \mu + \int \chi_{E \mathop \setminus A} \cdot f \rd \mu       | c = Integral of Integrable Function is Additive }} {{eqn | r = \int_A f \rd \mu + \int_{E \mathop \setminus A} f \rd \mu       | c = {{Defof|Integral of Integrable Function over Measurable Set}} }} {{end-eqn}} {{qed}}
=== Necessary Condition === Let $\struct {T, \circ}$ be closed. Let $x \in S$ be arbitrary. Then: {{begin-eqn}} {{eqn | lo= \forall f, g \in T^S:       | l = \map {f \oplus g} x       | r = \map f x \circ \map g x       | c = {{Defof|Pointwise Operation}} }} {{eqn | o = \in       | r = T       | c = as $\struct {T, \circ}$ is closed }} {{end-eqn}} So $\oplus$ is closed on $T^S$. {{qed|lemma}} === Sufficient Condition === Let $\oplus$ is closed on $T^S$. {{AimForCont}} $\struct {T, \circ}$ is not closed. Then: :$(1): \quad \exists s, t \in T: s \circ t \notin T$ By definition, $T^S$ is the set of all mappings from $S$ \to $T$. As $S \ne \O$ it follows that $\exists x \in S$. Thus, let $x \in S$ be arbitrary. Let $f, g \in T^S$ such that: :$(2): \quad \map f x = s, \map g x = t$ Then: {{begin-eqn}} {{eqn | l = \map {f \oplus g} x       | r = \map f x \circ \map g x       | c = {{Defof|Pointwise Operation}} }} {{eqn | r = s \circ t       | c = from $(2)$ }} {{eqn | o = \notin       | r = T       | c = from $(1)$ }} {{end-eqn}} That is, $\oplus$ is not closed on $T^S$. From that contradiction it follows that $\struct {T, \circ}$ is closed. {{qed}} Category:Pointwise Operations aawr0dnrannkhib34n0pcxs1d4xq9t9
Let $f, g, h: S \to T$ be elements of $T^S$. Let $x \in S$. Then: {{begin-eqn}} {{eqn | l = \left ({f \circ' \left({g +' h}\right)}\right) \left({x}\right)       | r = f \left({x}\right) \circ \left({g \left({x}\right) + h \left({x}\right)}\right)       | c = {{Defof|Induced Structure}} }} {{eqn | r = \left({f \left({x}\right) \circ g \left({x}\right)}\right) + \left({f \left({x}\right) \circ h \left({x}\right)}\right)       | c = $\circ$ is distributive over $+$ on $T$ }} {{eqn | r = \left ({\left ({f \circ' g}\right) +' \left ({f \circ' h}\right)}\right) \left({x}\right)       | c = {{Defof|Induced Structure}} }} {{end-eqn}} Similarly: :$\left ({\left({g +' h}\right) \circ' f}\right) \left({x}\right) = \left ({\left ({g \circ' f}\right) +' \left ({h \circ' f}\right)}\right) \left({x}\right)$ Hence the result. {{qed}} Category:Pointwise Operations Category:Distributive Operations i7v04c2cehmt02hfx4jsqykh9wx9boy
We have Identity Mapping is Bijection. We also have Identity Mapping is Continuous. Hence, by definition, $I_T$ is a homeomorphism. {{qed}} Category:Homeomorphic Metric Spaces Category:Topologically Equivalent Metrics Category:Identity Mappings sxh0koi2n9pjcxoqy02l6jki7ki9tly
By definition, a homeomorphism is a bijection such that both $f$ and $f^{-1}$ are continuous. As $f$ is a bijection then by Bijection iff Inverse is Bijection, so is $f^{-1}$. So by definition $f^{-1}$ is a bijection such that both $f^{-1}$ and $\left({f^{-1}}\right)^{-1}$ are continuous. The result follows from Inverse of Inverse of Bijection. {{qed}} Category:Homeomorphic Metric Spaces 2f2r3swyhcm2fzhrr2czygk18dmhpsd
By definition of homeomorphism, $f$ and $g$ are both bijections. From Composite of Bijections is Bijection it follows that $g \circ f$ is also a bijection. Similarly, also by definition of homeomorphism, $f$ and $g$ are both continuous mappings. From Composite of Continuous Mappings on Metric Spaces is Continuous it follows that $g \circ f$ is also a continuous mapping. Hence the result, from definition of homeomorphism. {{qed}} Category:Homeomorphic Metric Spaces pzwqca5p5a19hs7lun1ktkbdqz5vqg8
By definition, $T$ is homeomorphic to a topological space $\struct {S, \tau_d}$ such that $\tau_d$ is the topology induced by a metric $d$ on $S$. From Metric Space is Hausdorff, $\struct {S, d}$ is a $T_2$ (Hausdorff) space. As the open sets of $\struct {S, d}$ are the same as the open sets of $\struct {S, \tau_d}$, it follows that $\struct {S, \tau_d}$ is a $T_2$ (Hausdorff) space. From $T_2$ Space is Preserved under Homeomorphism it follows that $T$ is also a $T_2$ (Hausdorff) space. {{qed}}
Since $N$ is a neighborhood of $x$, there exists an open set $U \subseteq N$ containing $x$ by definition. As $\struct {S, \tau}$ is metrizable, the set $U$ is open with respect to some metric space $\struct {S, d}$. Hence: :$\exists \epsilon > 0: \map {\mathcal B_\epsilon} x \subseteq U$ where $\map {\mathcal B_\epsilon} x$ denotes the open $\epsilon$-ball of $x$. Taking $\zeta = \dfrac \epsilon 2$, we have that the closed $\zeta$-ball of $x$ is contained in $\map {\mathcal B_\epsilon} x$, and hence is a subset of $U \subseteq N$.  This closed ball is a closed neighborhood of $x$ in $\struct {S, \tau}$, and is a subset of $N$. Hence the result. {{qed}}
From the definition of closed set, $U$ is open in $T = \struct {S, \tau}$ {{iff}} $S \setminus U$ is closed in $T$. From Empty Set is Element of Topology, $\varnothing$ is open in $T$. From Set Difference with Empty Set is Self, we have $S \setminus \varnothing = S$, so $S$ is closed in $T$. {{qed}}
Let $x \in S$. Then $S$ itself is a neighborhood of $x$. {{qed}}
By definition of neighborhood: :$\exists U \in \tau: x \in U \subseteq N \subseteq S$ where $U$ is an open set of $T$. By Subset Relation is Transitive: :$U \subseteq N'$ The result follows by definition of neighborhood of $x$. {{qed}}
By definition of neighborhood: : $\exists U_1 \in \tau: x \in U_1 \subseteq M$ where $U_1$ is an open set of $T$. : $\exists U_2 \in \tau: x \in U_2 \subseteq N$ where $U_2$ is an open set of $T$. Thus by Set Intersection Preserves Subsets: : $U \subseteq M \cap N$ where $U = U_1 \cap U_2$ The result follows by definition of neighborhood of $a$. {{qed}}
By definition of neighborhood: :$\exists U \in \tau: x \in U \subseteq N \subseteq S$ where $U$ is an open set of $T$. By Set is Open iff Neighborhood of all its Points, $N' = U$ fulfils the conditions of the statement. {{qed}}
Suppose $\varnothing$ were not an open set of $\struct {S, \mathcal N}$. Then $\exists x \in \varnothing$ such that $\varnothing$ is not a neighborhood of $\varnothing$. By definition of empty set, such an $x$ does not exist. Hence the result. {{qed}}
Let $x \in S$. Then by neighborhood space axiom $\text N 1$ there exists a neighborhood $N$ of $x$. As $N \subseteq S$ it follows from neighborhood space axiom $\text N 3$ that $S$ is a neighborhood of $x$. As this holds for all $x \in S$ it follows that $S$ is an open set of $\struct {S, \NN}$. {{qed}}
Let $U$ and $V$ be open sets of $\left({S, \mathcal N}\right)$. Let $x \in S$ such that $x \in U \cap V$. Then $x \in U$ and $x \in V$, both of which are neighborhoods of $x$ by definition of open set. By neighborhood space axiom $N 4$ it follows that $U \cap V$ is a neighborhood of $x$. As $x$ is arbitrary, it follows that the above is true for all $x \in U \cap V$. It follows by definition that $U \cap V$ is an open set of $\left({S, \mathcal N}\right)$. {{qed}}
Let $\ds x \in \bigcup_{\alpha \mathop \in I} U_\alpha$. Then $x \in U_\beta$ for some $\beta \in I$. By definition of open set, $U_\beta$ is a neighborhood of $x$. But from Set is Subset of Union: :$\ds U_\beta \subseteq x \in \bigcup_{\alpha \mathop \in I} U_\alpha$ By neighborhood space axiom $N 3$ it follows that $\ds \bigcup_{\alpha \mathop \in I} U_\alpha$ is a neighborhood of $x$. As $x$ is arbitrary, it follows that the above is true for all $\ds x \in \bigcup_{\alpha \mathop \in I} U_\alpha$. It follows by definition that $\ds \bigcup_{\alpha \mathop \in I} U_\alpha$ is an open set of $\struct {S, \NN}$. {{qed}}
Each of the open set axioms is examined in turn: === $O1$: Union of Open Sets === From Union of Open Sets of Neighborhood Space is Open, it follows that open set axiom $O1$ is fulfilled. {{qed|lemma}} === $O2$: Intersection of Open Sets === From Intersection of two Open Sets of Neighborhood Space is Open, it follows that open set axiom $O2$ is fulfilled. {{qed|lemma}} === $O3$: Underlying Set === From Whole Space is Open in Neighborhood Space, it follows that open set axiom $O3$ is fulfilled. {{qed|lemma}} All the open set axioms are fulfilled, and the result follows. {{qed}}
=== Necessary Condition === Let $N$ be a neighborhood of $x$. Then by neighborhood space axiom $N 5$, $N$ contains a neighborhood $U$ of $x$ such that $U$ is a neighborhood of each of its points. By neighborhood space axiom $N 2$, $x \in U$. {{qed|lemma}} === Sufficient Condition === Let $N$ be such that: :$\exists U \in \mathcal N: x \in U \subseteq N$ By definition of open set, $U$ is a neighborhood of $x$. By neighborhood space axiom $N 3$, $N$ is then a neighborhood of $x$. {{qed}}
Let $x \in S$. Let $\mathcal N_x$ be the neighborhood filter of $x$. From Basic Properties of Neighborhood in Topological Space, $\mathcal N_x$ fulfils the neighborhood space axioms. Hence the result. {{qed}} Category:Neighborhood Spaces hpgov4auvakl03167go55u84y7pcjmg
From Neighborhood Space is Topological Space, $\left({S, \mathcal N}\right)$ is a topological space. Consequently, the open sets of $\left({S, \tau}\right)$ are exactly the open sets of $\left({S, \mathcal N}\right)$. {{qed}} Category:Neighborhood Spaces n5jwkuh9ttiaatf3sqat7cdai2am1q8
Let $U \in \tau$ be an open set of $\left({S, \tau}\right)$. By Set is Open iff Neighborhood of all its Points, $U$ is a neighborhood of each of its points. By definition, $U$ is an open set of $\left({S, \mathcal N}\right)$. Thus by definition of neighborhood space induced by $\tau$ on $S$: :$U \in \mathcal N$ Then, by definition of the topological space induced by $\mathcal N$ on $S$: : $U \in \tau'$. Thus $\tau \subseteq \tau'$. Now let $U \in \tau'$ be an open set of $\left({S, \tau'}\right)$. Then, by definition, in the neighborhood space $\left({S, \mathcal N}\right)$, $U$ is an open set of $\left({S, \mathcal N}\right)$. That is, $U$ is a neighborhood in $\left({S, \mathcal N}\right)$ of each of its points. But the neighborhoods of $\left({S, \mathcal N}\right)$ are the neighborhoods of $\left({S, \tau}\right)$. Thus by Set is Open iff Neighborhood of all its Points, $U$ is an open set of $\left({S, \tau}\right)$. That is: : $U \in \tau$. It follows by definition of set equality that $\tau = \tau'$. {{qed}}
From the construction of: :the neighborhood space induced by $\tau$ on $S$ :the topological space induced by $\mathcal N$ on $S$ the mappings $\phi$ and $\psi$ are well-defined mappings. From Topological Space induced by Neighborhood Space induced by Topological Space, $\phi$ is a bijection. From Neighborhood Space induced by Topological Space induced by Neighborhood Space, $\psi$ is a bijection. Hence the result. {{qed}}
Let $x \in S$. Let $\mathcal N_x$ be the set of all neighborhoods of $x$. Let $N \in \mathcal N_x$ be a neighborhood of $x$. From Subset in Neighborhood Space is Neighborhood iff it contains Open Set, $N$ is the superset of some open set $U$ in $\left({S, \mathcal N}\right)$. By Neighborhood Space is Topological Space we have that $U$ is an open set of $\left({S, \tau}\right)$. Thus, by definition, $N$ is a neighborhood of $x$ in the context of the topological space $\left({S, \tau}\right)$. Thus by definition of the neighborhood space induced by $\tau$ on $S$, it follows that: : $N \in \mathcal N'_x$ where $\mathcal N'_x$ is the set of all neighborhoods of $x$ in $\left({S, \mathcal N'}\right)$. Thus $\mathcal N_x \subseteq \mathcal N'_x$. {{qed|lemma}} Now suppose that $N \in \mathcal N'_x$. Then, by definition of neighborhood, $N$ is the superset of some open set $U$ in $\left({S, \tau}\right)$ such that $x \in U$. Since $U \in \tau$ it follows that $U$ is an open setin $\left({S, \mathcal N}\right)$. So by Subset in Neighborhood Space is Neighborhood iff it contains Open Set, $N$ is a neighborhood of $x$ in $\left({S, \mathcal N}\right)$. That is: : $N \in \mathcal N_x$ Thus it follows that $\mathcal N_x \subseteq \mathcal N'_x$ {{qed|lemma}} It follows by definition of set equality that $\mathcal N_x = \mathcal N'_x$. As $x$ is arbitrary, it applies to all $x \in S$. Hence the result. {{qed}}
=== Definition 1 implies Definition 2 === Follows directly from Cardinality of Set of Subsets. {{qed|lemma}} === Definition 1 implies Definition 3 === This is proved in the Binomial Theorem. {{qed}}
From Form of Geometric Sequence of Integers the $j$th term of $G_n$ is given by: :$a_j = k p^{n - j} q^j$ where the common ratio is $\dfrac q p$. Thus: :$a_0 = k p^n$ :$a_n = k q^n$ From Geometric Sequence in Lowest Terms has Coprime Extremes it follows that $k = 1$. Hence the result. {{qed}} Category:Geometric Sequences of Integers cwfob7tfrj3rzjzscijwag34jsinvz6
Let $r$ be the common ratio of $Q_n$. Let the elements of $Q_n$ be the smallest positive integers such that $Q_n$ has common ratio $r$. From Geometric Sequence with Coprime Extremes is in Lowest Terms, the elements of $Q_n$ are the smallest positive integers such that $Q_n$ has common ratio $r$. From Form of Geometric Sequence of Integers in Lowest Terms the $j$th term of $P$ is given by: :$a_j = q^j p^{n - j}$ where $r = \dfrac p q$. Hence the result. {{qed}} Category:Geometric Sequences of Integers m62ikkujr35cgadosyuwq1eegs6vevx
We verify Peano's axioms in turn. First, suppose that $s \left({m}\right) = s \left({n}\right)$ for some $m, n \in S$. That is: :$m \circ 1 = n \circ 1$ By Axiom $(NO2)$, it follows that $m = n$. Hence Axiom $(P3)$ holds. Next, suppose that $s \left({n}\right) = 0$ for some $n \in S$. That is: :$n \circ 1 = 0$ By Sum with One is Immediate Successor in Naturally Ordered Semigroup, this would imply: :$n \prec 0$ However, this contradicts the definition of $0$. Consequently, Axiom $(P4)$ is satisfied. Finally, let $A \subseteq S$ be such that $0 \in A$ and, for all $n \in S$: :$n \in A \implies s \left({n}\right) \in A$ Define $T := S \setminus A$. Suppose that $T \ne \varnothing$. Then by Axiom $(NO3)$, $T$ has a minimal element $n$. By assumption, $0 \in A$, so that $n \ne 0$. Therefore, by definition of $1$: :$1 \preceq n$ By Ordering in terms of Addition, there exists a $p \in S$ such that: :$p \circ 1 = n$ since $\circ$ is commutative. By Sum with One is Immediate Successor in Naturally Ordered Semigroup: :$p \prec n$ Since $n$ is the minimal element of $T$, it follows that $p \notin T$. Therefore, by definition, $p \in A$. But by assumption on $A$, this means that $n = p \circ 1 \in A$. This is a contradiction, hence $T = \varnothing$. That is to say, $A = S$, and Axiom $(P5)$ holds as well. Having verified all the axioms, we conclude that $\left({S, 0, s}\right)$ is a Peano structure. {{qed}} Category:Naturally Ordered Semigroup Category:Peano's Axioms bzvkvibijvhiu90m46w9gnrd53xyz86
Let $a_j \divides a_{j + 1}$ for some $j \in \set {0, 1, \ldots, n - 1}$. Then by definition of integer divisibility: :$\exists r \in \Z: r a_j = a_{j + 1}$ Thus the common ratio of $Q_n$ is $r$. So by definition of geometric sequence: :$\forall j, k \in \set {0, 1, \ldots, n}, j < k: r^{k - j} a_j = a_k$ and so $a_j \divides a_k$. The converse is trivial. {{qed}} Category:Geometric Sequences of Integers 9wvw4e48rd7y1uyoccxqq5hzi2ue3tc
From Form of Geometric Sequence of Integers, the $j$th term of $Q_n$ is given by: :$a_j = k q^j p^{n - j}$ where: :the common ratio of $Q_n$ expressed in canonical form is $\dfrac q p$ :$k$ is an integer. As $a_0 = 1$ it follows that: :$1 = k p^{n - j}$ from which it follows that: :$k = 1$ :$p = 1$ and the common ratio of $Q_n$ is $q$. Thus: :$a_1 = q$ Setting $a = a_1$ yields the result as stated. {{qed}} Category:Geometric Sequences of Integers oe6s3lsppor1abhbcntfeh882woz8za
First it is necessary to establish that every element of the set $\set {1, p, p^2, \ldots, p^{n - 1}, p^n}$ is in fact a divisor of $p^n$. For any $j \in \set {1, 2, \ldots, n}$: :$p^n = p^j p^{n - j}$ and so each of $1, p, p^2, \ldots, p^{n - 1}, p^n$ is a divisor of $p^n$. {{qed|lemma}} Let: :$a \in \Z_{>0}: a \notin \set {1, p, p^2, \ldots, p^{n - 1}, p^n}$ Let $a = p^j$ where $j \in \Z: j > n$. Then: :$p^j = p^n p^{j - n}$ and so $p^n$ is a divisor of $p^j$. Hence $p_j \nmid p^n$. Now let: :$a \notin \set {p^k: k \in \Z_{>0} }$ Then: :$\exists q \in \Bbb P: q \divides a$ where: :$\Bbb P$ is the set of all prime numbers :$\divides$ denotes divisibility. {{AimForCont}} $a \divides p^n$. From Divisor Relation is Transitive it follows that $q \divides p^n$. From Euclid's Lemma for Prime Divisors: General Result it follows that: :$q \divides p$ As $p$ is a prime, by definition its only divisors are $1$ and $p$. This contradicts the supposition that $q$ is a divisor of $p^n$. Hence $a \nmid p^n$. {{qed}} Category:Prime Numbers 7rk579ji8oqt9bogdoui6eysst01x2q