hcju/proofwikiqa huggingface dataset:
Suppose $T$ is a countable particular point space. Let $\CC$ be the open cover of $T$ defined as: :$\CC = \set {\set {x, p}: x \in S}$ $\CC$ is countable and has no open refinement except $\CC$ itself. But $\CC$ is not point finite because $\forall U \in \CC: p \in U$, and $\CC$ is (countably) infinite. Now suppose $T$ is an uncountable particular point space. Let $S' \subseteq S$ be a countable subset of $S$. Let $\CC$ be the open cover of $T$ defined as: :$\CC = \set {\set {x, p}: x \in S'} \cup \set {S \setminus S' \cup \set p}$ $\CC$ is also countable and has no open refinement except $\CC$ itself. And similarly, $\CC$ is not point finite because $\forall U \in \CC: p \in U$, and $\CC$ is (countably) infinite. Hence the result by definition of countably metacompact. {{qed}}
From the Rule of Exportation: :$\paren {p \land q} \implies r \dashv \vdash p \implies \paren {q \implies r}$ Then by Self-Distributive Law for Conditional: :$p \implies \paren {q \implies r} \dashv \vdash \paren {p \implies q} \implies \paren {p \implies r}$ {{qed}} Category:Implication Category:Conjunction oc5bpk4w3hco5d9sobteg96jn3b38gs
We prove the contrapositive. The main idea is that if such a theory $T$ is incomplete, we can construct size $\kappa$ models which disagree on a sentence. Suppose $T$ is not complete. By the definition of complete, this means that there is some sentence $\phi$ such that both $T \not \models \phi$ and $T \not \models \neg \phi$. This in turn means that both $T \cup \left\{{\neg \phi}\right\}$ and $T \cup \left\{{\phi}\right\}$ have models. Since $T$ has no finite models, this means that $T \cup \left\{{\neg \phi}\right\}$ and $T \cup \left\{{\phi}\right\}$ both have infinite models. We have that $\kappa$ is infinite and greater than the cardinality of the language. We also have that these theories have infinite models. From the Upward Löwenheim-Skolem Theorem one can prove that there are size $\kappa$ models $\mathcal M_{\neg \phi}$ and $\mathcal M_{\phi}$ of $T \cup \left\{{\neg \phi}\right\}$ and $T \cup \left\{{\phi}\right\}$ respectively. In particular, $\mathcal M_{\neg\phi}$ and $\mathcal M_\phi$ are models of $T$ which disagree about the sentence $\phi$. Such models cannot be isomorphic since isomorphisms preserve the truth of sentences. Thus, $T$ is not $\kappa$-categorical. {{qed}} {{Namedfor|Jerzy Maria Michał Łoś|name2 = Robert Lawson Vaught|cat = Łoś|cat2 = Vaught}} This result is also known as Vaught's Test. Category:Model Theory Category:Mathematical Logic 0cehvfrmryvn8jtame1htqwwd9mm8o6
A Sierpiński space is a particular point space by definition. A Particular Point Space is Irreducible. {{qed}}
The only closed sets of $T$ are $\O, \set 1$ and $\set {0, 1}$. $\set 1$ and $\set {0, 1}$ are not disjoint. Hence the result by definition of ultraconnected. {{qed}}
A Sierpiński space is a particular point space by definition. A Particular Point Space is Path-Connected. {{qed}}
A Sierpiński space is a particular point space by definition. A Particular Point Space is not Arc-Connected. {{qed}}
The only closed sets in $T$ are $\O$, $\set 1$ and $\set {0, 1}$. So there are no two separated sets $A, B \subseteq \set {0, 1}$. So $T$ is a $T_5$ space vacuously. {{qed}}
We have that the Sierpiński Space is $T_5$. Then we have that a $T_5$ Space is $T_4$. {{qed}}
The idea is: :to extend the language by adding $\kappa$ many new constants and: :to extend the theory by adding sentences asserting that these constants  are distinct. It is shown that this new theory is finitely satisfiable using an infinite model of $T$. Compactness then implies that the new theory has a model. Some care needs to be taken to ensure that we  construct a model of exactly size $\kappa$. Let $\LL^*$ be the language formed by adding new constants $\set {c_\alpha: \alpha < \kappa}$ to $\LL$. Let $T^*$ be the $\LL^*$-theory formed by adding the sentences $\set {c_\alpha \ne c_\beta: \alpha, \beta < \kappa, \ \alpha \ne \beta}$ to $T$. We show that $T^*$ is finitely satisfiable: Let $\Delta$ be a finite subset of $T^*$. Then $\Delta$ contains: :finitely  many sentences from $T$ along with: :finitely many sentences of the form $c_\alpha \ne c_\beta$ for the new constant symbols. Since $T$ has an infinite model, it must have a model $\MM$ of cardinality at most $\card \LL + \aleph_0$. This model already satisfies everything in $T$. So, since we can find arbitrarily many distinct elements in it,  it can also be used as a model of $\Delta$ by interpreting the finitely many new constant symbols in $\Delta$ as distinct elements of $\MM$. Since $T^*$ is finitely satisfiable, it follows by the Compactness Theorem that $T^*$ itself is satisfiable. Since $T^*$ ensures the existence of $\kappa$ many distinct elements, this means it has models of size at least $\kappa$. It can be proved separately or observed from the ultraproduct proof of the compactness theorem that $T^*$ then has a model $\MM^*$ of exactly size $\kappa$. {{explain|That proof needs to be proved, and / or a link needs to be provided to that ultraproduct proof and its implications explained.}} Since $T^*$ contains $T$, $\MM^*$ is a model of $T$ of size $\kappa$. {{qed}} {{Namedfor|Leopold Löwenheim|name2 = Thoralf Albert Skolem|cat = Löwenheim|cat2 = Skolem}}
By definition, $T$ is finitely satisfiable means that every finite subset of $T$ is satisfiable. Since the direction :$T$ satisfiable implies $T$ finitely satisfiable is trivial, the proofs below all justify the converse :$T$ finitely satisfiable implies $T$ satisfiable.
By definition: :$\tau^*_p = \set {U \cup \set p: U \in \tau} \cup \set \O$ We have that $\varnothing \in \tau^*_p$ by definition. We also have that $S \in \tau$ so $S \cup \set p \in \tau^*_p$. Now let $U_1, U_2 \in \tau^*_p$. Then $U_1^* = U_1 \setminus \set p \in \tau, U_2^* = U_2 \setminus \set p \in \tau$. Then $p \in U_1$ and $p \in U_2$ and so $p \in U_1 \cap U_2$. So $U_1 \cap U_2 = \paren {U_1^* \cap U_2^*} \cup \set p \in \tau^*_p$.  Finally consider $\UU \subseteq \tau^*_p$. :$\displaystyle \bigcup \UU = \bigcup_{U \mathop \in \UU} \paren {U \setminus \set p} \cup \set p$ So $\displaystyle \bigcup \UU \in \tau^*_p$. So $\tau^*_p$ is a topology on $S \cup \set p$. {{qed}}
The proof is done by induction on complexity of formulas. Note that since interpretations of terms with parameters from $\MM$ are preserved when passing to superstructures, we have that $\map {t^\MM} {\bar a} = \map {t^\NN} {\bar a}$ whenever $t$ is an $\LL$-term and $\bar a$ is in $\MM$. First, we verify the theorem for atomic formulas.  This is the base for the inductive proof. Suppose $\phi$ is $t_1 = t_2$ for terms $t_1$ and $t_2$: We have $\MM \models \map \phi {\bar a}$ {{iff}} $\MM \models \map {t_1} {\bar a} = \map {t_2} {\bar a}$. The {{RHS}} of this holds {{iff}} $\map {t_1^\MM} {\bar a} = \map {t_2^\MM} {\bar a}$. By the note above, this in turn holds {{iff}} $\map {t_1^\NN} {\bar a} = \map {t_2^\NN} {\bar a}$, which is equivalent to $\NN \models \map \phi {\bar a}$. Suppose $\phi$ is $R(t_1, \dots, t_n)$ where $R$ is a relation symbol and each $t_i$ is a term: We have: :$\MM \models \map \phi {\bar a}$ iff $\paren {\map {t_1^\NN} {\bar a}, \dots, \map {t_n^\NN} {\bar a} } \in R^\MM$ Since $\MM$ is a substructure of $\NN$, the interpretation of $R$ in $\NN$ must extend $R^\MM$, so the right side of this statement is equiavlent to: :$\paren {\map {t_1^\NN} {\bar a}, \dots, \map {t_n^\NN} {\bar a} } \in R^\NN$ By the note above, this in turn is equivalent to: :$\paren {\map {t_1^\NN} {\bar a}, \dots, \map {t_n^\NN} {\bar a} } \in R^\NN$ which is the same as: :$\NN \models \map \phi {\bar a}$ Having verified the theorem for the atomic formulas, we proceed with the inductive step of the proof. Suppose the result holds for $\psi$, and consider $\neg \psi$: We have $\MM \models \neg \map \psi {\bar a}$ {{iff}} $\MM \not \models \map \psi {\bar a}$. By the inductive hypothesis, the right side of this statement is equivalent to $\NN \not\models \map \psi {\bar a}$, and so the result follows. Suppose the result holds for $\psi_1$ and $\psi_2$.# Consider $\psi_1 \wedge \psi_2$: We have $\MM \models \map {\psi_1} {\bar a} \wedge \map {\psi_2} {\bar a}$ {{iff}} $\MM \models \map {\psi_1} {\bar a}$ and $\MM \models \map {\psi_2} {\bar a}$. By the inductive hypothesis, the {{RHS}} of this is equivalent to $\NN \models \map {\psi_1} {\bar a}$ and $\NN \models \map {\psi_2} {\bar a}$. The result follows, completing the proof. {{qed}} {{MissingLinks}} Category:Model Theory Category:Mathematical Logic k9mvibx55zw5o2p5iik6700k6s5yzmr
We prove the contrapositive: that a Galois field cannot be algebraically closed. Let $F$ be Galois. Define the polynomial: :$\displaystyle \map f x = 1 + \prod_{a \mathop \in F} \paren {x - a}$ By definition, a field is a ring. Thus by Ring Product with Zero: :$\displaystyle \forall x \in F: \prod_{a \mathop \in F} \paren {x - a} = 0$ But: :$\forall a \in F: \map f a = 1 + 0 \ne 0$ Therefore $\map f x$ cannot have a root in $F$. Thus $F$ is not algebraically closed. The result follows by the Rule of Transposition. {{qed}} Category:Field Theory dxwe85ej6e1gfs3xfgm4kzqok4lswql
By the Łoś-Vaught Test, it suffices to show that $ACF_p$ is satisfiable, has no finite models, and is $\kappa$-categorical for some uncountable $\kappa$. === Satisfiability === $\C$ is an algebraically closed field of characteristic $0$. If $p$ is a prime, then the algebraic closure of $\Z / \Z_p$ is an algebraically closed field of characteristic $p$. Thus $ACF_p$ is satisfiable. {{explain|Needs to be made clear why what is above leads to that statement.}} {{qed|lemma}} === No Finite Models === From Algebraically Closed Field is Infinite: $ACF_p$ has no finite models {{qed|lemma}} === $\kappa$-Categorical === From: : Field of Uncountable Cardinality $\kappa$ has Transcendence Degree $\kappa$ and: : Algebraically Closed Fields are Isomorphic iff they have the same Characteristic and Transcendence Degree it follows that: : $ACF_p$ is $\kappa$-categorical for all uncountable $\kappa$ Hence: : $ACF_p$ is $\kappa$-categorical for some uncountable $\kappa$ as was to be proved. {{qed}} Category:Model Theory Category:Mathematical Logic q2tolc8ek8jowx6j72tpmdgmhgs5lmj
=== $(1)$ iff $(2)$ === From Theory of Algebraically Closed Fields of Characteristic $p$ is Complete: the theory $ACF_p$ of algebraically closed fields of characteristic $p$ is complete. That is, all such fields satisfy the exact same $\LL_r$ sentences. {{qed|lemma}} === $(2)$ implies $(4)$ === Let $\phi$ be true in some such field. Then: :$ACF_0 \models \phi$ By Gödel's Completeness Theorem and the finiteness of proofs, it follows that there is a finite subset $\Delta$ of $ACF_0$ such that $\Delta \models \phi$. Such a $\Delta$ can only make finitely many assertions about the character of its models. Hence, as long as $p$ is selected sufficiently large, an algebraically closed field of characteristic $p$ will satisfy $\phi$. {{qed|lemma}} === $(4)$ implies $(3)$ === We have that all sufficiently large $p$ work. Hence it follows that it is always possible to find arbitrarily large $p$ that work. {{qed|lemma}} === $(3)$ implies $(2)$ === We prove this by contraposition. Suppose there is no algebraically closed field of characteristic $0$ where $\phi$ is true. Then $\phi$ is false in algebraically closed fields of characteristic $0$ Since $ACF_0$ is complete, this means that $ACF_0 \models \neg \phi$. Similarly to the case of $(2)$ implies $(4)$, there must then be a finite subset $\Delta$ of $ACF_0$ such that $\Delta \models \neg \phi$. But then, for all sufficiently large $p$, we have that $\phi$ is false in the algebraically closed fields of characteristic $p$. Hence, it cannot be true for arbitrarily large $p$. By Rule of Transposition, $\phi$ is true in algebraically closed fields for arbitrarily large primes $p$ implies there exists an algebraically closed field of characteristic $0$ where $\phi$ is true. {{qed}} {{Namedfor|Solomon Lefschetz|cat = Lefschetz}} Category:Algebraic Geometry Category:Field Theory Category:Model Theory Category:Mathematical Logic Category:Proofs by Contraposition 2ztr54llb9zkvlktqc9933kmxbwp44x
By definition: :$\tau^*_p = \set {U \cup \set p: U \in \tau} \cup \set \O$ Let $V \subseteq S^*_p$ be closed in $T^*_p$. Then $S^*_p \setminus V$ is open in $T^*_p$. Then $\struct {S^*_p \setminus V} \setminus \set p$ is open in $T$. From Set Difference with Union we have: :$\struct {S^*_p \setminus V} \setminus \set p = S^*_p \setminus \struct {V \cup \set p}$ Hence the result. {{qed}}
Directly apparent from the definitions of particular point topology, discrete topological space and closed extension space. {{qed}}
By definition: :$\tau^*_p = \set {U \cup \set p: U \in \tau} \cup \set \O$ Let $x \in S^*_p, x \ne p$. Let $U = \set p$. Then $U \in \tau^*_p$ such that $p \in U, x \notin U$. But there is no $V \in \tau^*_p$ such that $x \in V, p \notin V$, by definition of the closed extension topology. Hence $T^*_p$ can not be a $T_1$ (Fréchet) space. {{qed}}
By definition, $\Bbb A$ is the subset of the complex numbers which consists of roots of polynomials with coefficients in $\Q$. We can prove the theorem by a cardinality argument, counting the number of such polynomials and roots. By Set of Polynomials over Infinite Set has Same Cardinality, the set $\Q \sqbrk x$ of polynomials over $\Q$ is countable. Next, note that $\Bbb A$ can be written as the union of the set of roots of each polynomial. That is: :$\Bbb A = \displaystyle \bigcup_{\map p x \mathop \in \Q \sqbrk x} \set {x \in \C: \map p x = 0}$ From Polynomial over Field has Finitely Many Roots, $\Bbb A$ is a union of countably many finite sets. From Countable Union of Countable Sets is Countable it follows that $\Bbb A$ is countable. {{qed}}
Every open set of $T^*_p = \struct {S^*_p, \tau^*_p}$ except $\O$ contains the point $p$ by definition. So every open set $U \in \tau^*_p$ such that $x \in U$ contains $p$. So by definition of the limit point of a point, $x$ is a limit point of $p$. {{qed}}
By definition, $\forall U \in \tau^*_p, u \ne \O: p \in U$. From Limit Points in Closed Extension Space, every point in $S^*_p$ is a limit point of $p$. So by definition of closure, every point in $S^*_p$ is in $U^-$. {{qed}}
Trivially, by definition, every open set in $T^*_p$ contains $p$. So: :$\forall U_1, U_2 \in \tau^*_p: p \in U_1 \cap U_2$ for $U_1, U_2 \ne \O$. {{qed}}
From Closure of Open Set of Closed Extension Space we have that: :$\forall U \in \tau^*_p: U \ne \O \implies U^- = S$ where $U^-$ is the closure of $U$. The result then follows by definition of irreducible space. {{qed}}
Let $n \ge 1$ be the degree of $p$. We argue that $p$ has at most $n$ roots in $F$. Let $A$ be the set of roots of $p$. Let $a \in A$. By the Polynomial Factor Theorem: :$p \left({x}\right) = q_1 \left({x}\right) \cdot \left({x - a}\right)$ :where $\deg q_1 = n - 1$. Let $a' \in A$ such that $a' \ne a$. Then since: :$p \left({a'}\right) = 0$ but: :$\left({a' - a}\right) \ne 0$ it follows that: :$q_1 \left({a'}\right) = 0$ Again by the Polynomial Factor Theorem: :$q_1 \left({x}\right) = q_2 \left({x}\right) \cdot \left({x - a'}\right)$ Therefore: :$p \left({x}\right) = q_2 \left({x}\right) \cdot \left({x - a'}\right) \cdot \left({x - a}\right)$ where $\deg q_2 = n-2$. We can repeat this procedure as long as there are still distinct roots in $A$. After the $i$th iteration we obtain a polynomial $q_i$ of degree $n-i$. Now the degree of $q_i$ decreases each time we factor out a root of $p$, so $q_n$ is necessarily a constant term. Therefore $q_n$ can share no roots with $p$. So this procedure must stop by at most the $n$th step. That is, we can pick at most $n$ distinct elements from the set $A$, and: :$\left\vert{A}\right\vert \le n$ {{qed}} Category:Polynomial Theory 6s0xud8pptm377rznzyfrdvezyq5w0x
Since $S \left[{x}\right]$ contains a copy of $S$ as constant polynomials, we have an injection $S \to S \left[{x}\right]$. We define an injection from $S \left[{x}\right]$ to the set $\mathcal F$ of finite sequences over $S$ as follows: Each polynomial in $f \in S \left[{x}\right]$ is of the form: :$f = a_0 + a_1 x + a_2 x^2 + \dotsb + a_n x^n$ where $a_n$ is non-zero and each $a_i$ is in $S$. We send each polynomial $f$ to the sequence of its coefficients $\left({a_0, \dotsc, a_n}\right)$. By the definition of equality of polynomials, this is injective. Now the set of finite sequences over $S$ is a countable union of sets of cardinality $\kappa$. From Cardinality of Infinite Union of Infinite Sets, $\mathcal F$ has cardinality $\kappa$. Therefore there is a bijection $\mathcal F \leftrightarrow S$. Composing this with the injection $S \left[{x}\right] \to \mathcal F$, we have an injection $S \left[{x}\right] \to S$. So by the Cantor-Bernstein-Schröder Theorem there is a bijection $S \left[{x}\right] \leftrightarrow S$. Hence, we have: :$\left\vert{S \left[{x}\right]}\right\vert = \left\vert{S}\right\vert = \kappa$ {{qed}} Category:Polynomial Theory p5prg956jy5h9i1jlhr1rm5qmxo5gqb
We have by definition that $S \in \tau_{\bar p}$, and as $p \notin \varnothing$ we have that $\varnothing \in \tau_{\bar p}$. Now let $U_1, U_2 \in \tau_{\bar p}$. By definition $p \notin U_1$ and $p \notin U_2$, and so $p \notin U_1 \cap U_2$ by definition of set intersection. So $U_1 \cap U_2 \in \tau_{\bar p}$. Now let $\mathcal U \subseteq \tau_{\bar p}$. We have that $\forall U \in \mathcal U: p \notin U$. Hence from Subset of Union $p \notin \bigcup \mathcal U$. So all the properties are fulfilled for $\tau_{\bar p}$ to be a topology on $S$. {{qed}}
By definition: :$\tau^*_{\bar p} = \set {U: U \in \tau} \cup \set {S^*_p}$ We have that $S^*_p \in \tau^*_{\bar p}$ by definition. We also have that $\O \in \tau$ so $\O \in \tau^*_{\bar p}$. Now let $U_1, U_2 \in \tau^*_{\bar p}$. Then $U_1, U_2 \in \tau \implies U_1 \cap U_2 \in \tau \implies U_1 \cap U_2 \in \tau^*_{\bar p}$. Finally consider $\UU \subseteq \tau^*_{\bar p}$. Assuming $S^*_p \notin \UU$ we have that $\UU \subseteq \tau$. So: :$\displaystyle \bigcup \UU \in \tau$ and so $\displaystyle \bigcup \UU \in \tau^*_p$. If $S^*_p \in \UU$ then $\displaystyle \bigcup \UU = S^*_p \in \tau^*_p$. So $\tau^*_p$ is a topology on $S \cup \set p$. {{qed}}
Directly apparent from the definitions of excluded point topology, discrete topological space and open extension space. {{qed}}
From Discrete Space is Paracompact, $T_X$ is paracompact. We have that the Sierpiński space $T_Y$ is a finite topological space. From Finite Topological Space is Compact, $T_Y$ is a compact space. {{finish|Steen and Seebach in Part $\text I$ chapter $3$ Compactness: Invariance Properties offer "If $X$ is compact, then in general $X \times Y$ has the compactness properties of $Y$." Whether this holds for $Y$ paracompact needs to be investigated.}}
Define the mappings: :$z_1, \ldots, z_n: I \to \R^m$ by: :$z_j = x^{\paren {j - 1} }$, $j = 1, \ldots, n$ Then: {{begin-eqn}} {{eqn | l = z_1'       | r = z_2 }} {{eqn | o = \vdots }} {{eqn | l = z_{n - 1}'       | r = z_n }} {{eqn | l = z_n'       | r = \map F {t, z_1, \ldots, z_n} }} {{end-eqn}} This is a system of $m n$ first order ODEs. By construction: :$\map {z_1} t = \map x t$ for all $t \in I$ and $\map {z_1} {t_0} = x_0$. Therefore we can take: :$y = \begin {pmatrix} z_1 \\ \vdots \\ z_{n - 1} \\ z_n \end {pmatrix}, \quad \tilde F: \begin {pmatrix} z_1 \\ \vdots \\ z_n \end{pmatrix} \mapsto \begin {pmatrix} z_2 \\ \vdots \\ z_n \\ \map F {t, z_1, \ldots, z_n} \end {pmatrix}$ {{qed}}
Let $T_X = \struct {X, \tau}$ be a countable discrete space. Let $T_Y = \struct {\set {0, 1}, \tau_0}$ be a Sierpiński space. Let $T_A = T_X \times T_Y$ be the product space of $T_X$ and $T_Y$ whose product topology is $\tau_A$ Then $T_A$ is paracompact from Product of Countable Discrete Space with Sierpiński Space is Paracompact. Let $T_B = \struct {S, \tau_p}$ be a countable particular point space. Then $T_B$ is not countably metacompact from Infinite Particular Point Space is not Countably Metacompact Let $\phi: T_A \to T_B$ be a mapping defined as: :$\forall \tuple {x, y} \in T_A: \map \phi {x, y} = \begin {cases} p & : y = 0 \\ x & : y = 1 \end {cases}$ Then $\phi$ is open and continuous as follows: Let $U \in \tau_A$ such that $U \ne \O$. By construction of $T_A$, $U$ is of the form $\tuple {U_A, U_B}$ where $U_A \in \tau$ and $U_B \in \tau_0$. By definition of $\tau_0$, $U_B \in \set {\set 0, \set {0, 1} }$. It follows that $U$ contains elements of the form $\tuple {x, 0}$. Hence by definition of $\phi$ we have that: :$\exists \tuple {x, y} \in U: \map \phi {x, y} = p$ and so: :$p \in \phi \sqbrk U$  By definition of $T_B$ it follows that $\phi \sqbrk U$ is an open set of $T_B$. Thus $\phi$ maps from open sets to open sets and so $\phi$ is an open mapping. {{qed|lemma}} Let $V \in T_B$ be an open set such that $V \ne \O$. Then by definition: :$p \in V$ Let $F = \phi^{-1} \sqbrk V$. As $p \in V$ we have that $F$ is of the form: :$F = A \times B$ where: :$A \subseteq X$ and: :$B \in \set {\set 0, \set {0, 1} }$ Because $T_X$ is a countable discrete space, $A$ is open in $T_X$. By definition of $T_Y$, $B$ is open in $T_Y$. Hence by definition of product space, $A \times B = F$ is open in $T_A$. Thus the preimage of $V$ under $\phi$ is an open set. Hence $\phi$ is a continuous mapping. {{qed|lemma}} So we have constructed a mapping $\phi$ which is both open and continuous from a paracompact space to a space which is not countably metacompact. {{qed}}
Let $\map y t$ be a fixed point of the map $T$. That is: :$\displaystyle \map y t = x_0 + \int_{t_0}^t \map f {s, \map y s} \rd s$ Then: :$\displaystyle \map y {t_0} = x_0 + \int_{t_0}^{t_0} \map f {s, \map y s} \rd s = x_0$ By the fundamental theorem of calculus we have that $y$ is differentiable, and for $t \in \closedint a b$: {{begin-eqn}} {{eqn | l = \map {y'} t        | r = \frac {\d} {\d t} \int_{t_0}^t \map f {s, \map y s} \rd s       | c = Derivative of Constant }} {{eqn | r = \map f {t, \map y t}       | c = Fundamental Theorem of Calculus }} {{end-eqn}} This shows that $y$ is a solution to the ODE as claimed. {{Qed}} Category:Ordinary Differential Equations 483130d2qqb3grax2ihsd466suob03y
Let $S$ be a set. Let $\powerset S$ be the power set of $S$. By the Axiom of Choice, there is a choice function $c$ defined on $\powerset S \setminus \set \O$. We will use $c$ and the Principle of Transfinite Induction to define a bijection between $S$ and some ordinal. Intuitively, we start by pairing $\map c S$ with $0$, and then keep extending the bijection by pairing $\map c {S \setminus X}$ with $\alpha$, where $X$ is the set of elements already dealt with. === Basis for the Induction === $\alpha = 0$ Let $s_0 = \map c S$. === Inductive Step === Suppose $s_\beta$ has been defined for all $\beta < \alpha$. If $S \setminus \set {s_\beta: \beta < \alpha}$ is empty, we stop. Otherwise, define: :$s_\alpha := \map c {S \setminus \set {s_\beta: \beta < \alpha} }$ The process eventually stops, else we have defined bijections between subsets of $S$ and arbitrarily large ordinals. {{handwaving}} Now, we can impose a well-ordering on $S$ by embedding it via $s_\alpha \to \alpha$ into the ordinal $\beta = \displaystyle {\bigcup_{s_\alpha \mathop \in S} \alpha}$ and using the well-ordering of $\beta$. {{qed}}
=== Necessary Condition === Suppose the Axiom of Choice holds. Then the Well-Ordering Theorem holds by the Well-Ordering Theorem itself. {{qed|lemma}} === Sufficient Condition === Assume the Well-Ordering Theorem holds. Let $\mathcal F$ be an arbitrary collection of sets. By assumption all sets can be well-ordered. Hence the set $\bigcup \mathcal F$ of all elements of sets contained in $\mathcal F$ is well-ordered by some ordering $<$. By definition, in a well-ordered set, every subset has a unique least element. Also, note that each set in $\mathcal F$ is a subset of $\bigcup \mathcal F$. Thus, we may define the choice function $c$: :$\forall X \in \mathcal F: c: \mathcal F \to \bigcup \mathcal F$ by letting $c \left({X}\right)$ be the least element of $X$ under $<$. {{qed}} Category:Set Theory Category:Axiom of Choice 85sp3pc1j6a9vxdjqwosz1qf6ftobab
The result follows from the definition of the Tychonoff topology and Equivalence of Definitions of Initial Topology. {{qed}}