hcju/proofwikiqa huggingface dataset:
Let $D$ be an everywhere dense subset of $S$ which is countable, as is guaranteed as $T$ is separable. Consider the mapping $\Phi: S \to 2^{\powerset D}$ defined as: :$\forall x \in S: \map {\map \Phi x} A = 1 \iff A = D \cap U_x$ for some neighborhood $U_x$ of $x$ {{explain|It is not clear in Steen & Seeabch what is meant by $\Phi: S \to 2^{\powerset D}$ -- presumably $2^{\powerset D}$ is the ordinal which is the power set of the power set of $D$. It is also not clear what the notation $\map {\map \Phi x} A$ means -- in fact is may be the case that a transcription error has been committed.  Hence the proof cannot be attempted until these points have been cleared up.}} It is seen that if $T$ is a Hausdorff space, then $\Phi$ is an injection. It follows that: :$\card S \le \card {2^{\powerset D} } = 2^{2^{\aleph_0} }$ {{explain|the chain of reasoning leading to the above}}
Let $x, y$ both be cluster points of an ultrafilter $\FF$ on a set $S$. Then by definition: :$\forall U \in \FF: \set {x, y} \subseteq U$ But then we can create a finer filter $\FF \cup \set {\set x}$ on $S$. So $\FF$ is not an ultrafilter. {{qed}}
The relevant justifications are listed as follows: * Perfectly Normal implies Perfectly $T_4$ implies $T_4$ by definition. * Perfectly Normal Space is Completely Normal Space. * Completely Normal implies $T_5$ by definition. * Completely Normal Space is Normal Space. * $T_5$ space is $T_4$ space. * Normal implies $T_4$ by definition. * Normal Space is Tychonoff (Completely Regular) Space. * Completely Regular (Tychonoff) implies $T_{3 \frac 1 2}$ by definition. * Completely Regular (Tychonoff) Space is Regular Space. * $T_{3 \frac 1 2}$ Space is $T_3$ Space. * Regular implies $T_3$ by definition. * Completely Regular (Tychonoff) Space is Urysohn Space. * Urysohn Space is Completely Hausdorff Space. * Regular Space is Completely Hausdorff Space. * Regular Space is Semiregular Space. * Completely Hausdorff Space is $T_2$ (Hausdorff) Space. * Semiregular implies $T_2$ (Hausdorff) by definition. * $T_2$ (Hausdorff) Space is $T_1$ (Fréchet) Space. * $T_1$ (Fréchet) Space is $T_0$ (Kolmogorov) Space. {{qed}}
The relevant justifications are listed as follows: * Sequentially Compact Space is Countably Compact. * Compact Space is Countably Compact. * Compact Space is $\sigma$-Compact. * $\sigma$-Compact Space is Lindelöf. * Countably Compact Space is Weakly Countably Compact. * Countably Compact Space is Pseudocompact. {{qed}}
The relevant justifications are listed as follows: * Compact Space is Strongly Locally Compact. * Strongly Locally Compact Space is Weakly Locally Compact. * Locally Compact Space is Weakly Locally Compact. * Compact Space is Weakly $\sigma$-Locally Compact. * A weakly $\sigma$-locally compact is both weakly locally compact and $\sigma$-compact by definition. * $\sigma$-Compact Space is Lindelöf. {{qed}}
=== $(1) \implies (2)$ === By hypothesis, there exist $r_0, \ldots, r_{n-1} \in R$ such that: :$x^n + r_{n-1} x^{n-1} + \cdots + r_1 x + r_0 = 0$ So the powers $x^k$, $k \ge n$ can be written as an $R$-linear combination of: :$\left\{ {1, \ldots, x^{n-1} }\right\}$ Therefore this set generates $R \left[{x}\right]$. {{qed|lemma}} === $(2) \implies (3)$ === $B = R \left[{x}\right]$ trivially satisfies the required conditions. {{qed|lemma}} === $(3) \implies (4)$ === By $(3)$ we have an $R$-module $B$ such that $R \subseteq B$, $B$ is finitely generated over $R$. Also, $x \in B$, so $x B \subseteq B$ as required. {{qed|lemma}} === $(4) \implies (5)$ === By $(4)$ we have an $R \left[{x}\right]$-module $B$ that is finitely generated over $R$. Let $y$ lie in the annihilator $\operatorname{Ann}_{R \left[{x}\right]} \left({B}\right)$ We have that $1 \in B$. Then in particular $y \cdot 1 = 0$, and $y = 0$. Therefore, $B$ is faithful over $R \left[{x}\right]$. {{qed|lemma}} === $(5) \implies (1)$ === Let $B$ be as in $(5)$, say generated by $m_1, \ldots, m_n \in B$. Then there are $r_{i j} \in R$, $i, j = 1,\ldots, n$ such that: :$\displaystyle x \cdot m_i = \sum_{j \mathop = 1}^n r_{i j} m_j$ Let $b_{i j} = x \delta_{i j} - r_{i j}$ where $\delta_{i j}$ is the Kronecker delta. Then: :$\displaystyle \sum_{j \mathop = 1}^n b_{i j} m_j = 0, \quad i = 1, \ldots, n$ So, let $M = \left({b_{i j} }\right)_{1 \le i, j \le n}$. Then by Cramer's Rule: :$\left({\det M}\right) m_i = 0$, $i = 1, \ldots, n$ Since $\det M \in R \left[{x}\right]$, also $\det M \in \operatorname{Ann}_{R \left[{x}\right]} \left({B}\right)$. So $\det M = 0$ by hypothesis. But $\det M = 0$ is a monic polynomial in $x$ with coefficients in $R$. Thus $x$ is integral over $R$. {{qed}} Category:Algebraic Number Theory Category:Commutative Algebra 4kaiy1bq732m22rz278dj5jzit8839p
Let $b_1,\ldots, b_n$ generate $B$ over $A$. Let $c_1,\ldots, c_m$ generate $C$ over $B$. Then for any $x \in C$ there are $\beta_k \in B$, $k=1,\ldots, m$ such that: :$\displaystyle x = \sum_{k = 1}^m \beta_k c_k$ For any $i \in \{1,\ldots,m\}$ there exists $\alpha_{ij} \in A$, $j = 1,\ldots, n$ such that :$\displaystyle \beta_i = \sum_{j = 1}^n \alpha_{ij}b_j$ So :$\displaystyle x = \sum_{k = 1}^m \sum_{j = 1}^n \alpha_{kj} b_j c_k$ Therefore, $\{b_jc_k : j = 1,\ldots,n;\ k = 1,\ldots, m\}$ generates $C$ over $A$. {{qed}} Category:Commutative Algebra Category:Algebraic Number Theory sz9wj70oearyvyybtmztbg84dxkwuw5
For each $r \in R$, we have $r$ is a zero of $a - r = 0$. {{MissingLinks|zero}} Hence $R \subseteq C$, and in particular, $C$ is non-empty. By Subring Test it is sufficient to prove that if $x, y \in C$ then $x + \paren {-y}, x \circ y \in C$. Let $x, y \in C$. By Equivalent Definitions of Integral Dependence: $(1) \implies (2)$, $R \sqbrk x$ and $R \sqbrk y$ are finitely generated over $R$. Because $R \subseteq R \sqbrk x$, $R \sqbrk x \sqbrk y$ is also finitely generated over $R \sqbrk x$. Therefore, we have finitely generated extensions: :$R \hookrightarrow R \sqbrk x \hookrightarrow R \sqbrk x \sqbrk y = R \sqbrk {x, y}$ Hence, by Transitivity of Finite Generation, $R \sqbrk {x, y}$ is finitely generated over $R$. So by Equivalent Definition of Integral Dependence: $(2) \implies (1)$, every element of $R \sqbrk {x, y}$ is integral over $R$. In particular, $x + \paren {-y}$ and $x \circ y$ are integral over $R$. {{qed}} Category:Commutative Algebra Category:Algebraic Number Theory a4tpxtnv2uundr184ymd3rw6xox1tln
By the definition of localization, there exist unique homomorphisms: :$g : A_S \to \tilde A_S$ :$h : \tilde A_S \to A_S$ such that: :$h \circ \iota = \tilde \iota$ and: :$g \circ \tilde \iota = \iota$ Therefore: :$\tilde \iota = h \circ \iota = h \circ \left({g \circ \tilde \iota}\right) = \left({h \circ g}\right) \circ \tilde \iota$ The identity mapping also satisfies this equality (that is, $\tilde \iota = \left({h \circ g}\right) \circ \tilde \iota$). Therefore by the uniqueness of $h$ and $g$ we have: :$h \circ g = I_{\tilde A_S}$ where $I_{\tilde A_S}$ denotes the identity mapping on $\tilde A_S$. Similarly we have that: :$g \circ h = I_{A_S}$ where $I_{A_S}$ denotes the identity mapping on $A_S$. Therefore by Bijection iff Left and Right Inverse $g = \phi$ is a bijective homomorphisms, that is, an isomorphism. {{qed}} Category:Commutative Algebra q003p2eryv948l1wioz91exvyyozyl6
From Underlying Set of Topological Space is Clopen, $S$ is clopen in $T$. From Closed Set Equals its Closure, $S = S^-$. So every point of $S$ is contained in an open set (that is, $S$) whose closure (that is, $S$ again) is compact (as  $T = \struct {S, \tau}$ itself is compact). That is precisely the definition of a strongly locally compact space. {{qed}}
Let $T = \struct {S, \tau}$ be a compact space. We have that: :Compact Space is Sigma-Compact :Compact Space is Weakly Locally Compact Hence by definition $T$ is weakly $\sigma$-locally compact space. {{qed}}
The relevant justifications are listed as follows: * Compact Space is Countably Compact. * Compact Space is Paracompact. * Fully Normal Space is Paracompact. * Countably Compact Space is Countably Paracompact. * Paracompact Space is Countably Paracompact. * Fully Normal Space is Fully $T_4$ by definition. * Fully $T_4$ Space is $T_4$. * Paracompact Space is Metacompact. * Metacompact Space is Countably Metacompact. * Countably Paracompact Space is Countably Metacompact. {{qed}}
From $T_2$ Space is $T_1$ Space and $T_1$ Space is $T_0$ Space, we note that in a $T_2$ (Hausdorff) space: :A fully $T_4$ space is a fully normal space :A $T_4$ space is a normal space :A $T_{3 \frac 1 2}$ space is a Tychonoff space :A $T_3$ space is a regular space all by definition. From Hausdorff Space is Fully $T_4$ iff Paracompact we have that: :Fully $T_4$ $\iff$ Paracompact. From that same result and Fully $T_4$ Space is $T_4$ Space: :Paracompact $\implies$ $T_4$ From Normal Space is Tychonoff Space it follows that: :$T_4$ $\implies$ $T_{3 \frac 1 2}$ From First-Countable Space is Sequentially Compact iff Countably Compact and Sequentially Compact Hausdorff Space is Regular it follows that: :First-Countable and Countably Compact $\implies$ $T_3$ The further justifications are listed as follows: :$\sigma$-Locally Compact Hausdorff Space is $T_4$ :Weakly $\sigma$-Locally Compact implies Weakly Locally Compact by definition. :$T_{3 \frac 1 2}$ Space is $T_3$ Space {{ProofWanted|Some results still to be written.}} {{qed}}
Let $x \in V \left({T}\right)$, so $f \left({x}\right) = 0$ for all $f \in T$. By definition, $J$ is the set of linear combinations of elements of $T$ over $k$. So any $g \in J$ is of the form :$g = k_1 t_1 + \cdots + k_r t_r$ with $k_i \in k$ and $t_i \in T$. Therefore: {{begin-eqn}} {{eqn|l=g \left({x}\right)      |r=k_1 t_1 \left({x}\right) + \cdots + k_r t_r \left({x}\right) }} {{eqn|l=      |r=0      |c=because $f \left({x}\right) = 0$ for all $f \in T$ }} {{end-eqn}} Therefore $x \in V \left({J}\right)$. Conversely, if $x \in V \left({J}\right)$, then $f \left({x}\right) = 0$ for all $f \in J$. But $T \subseteq J$, so in particular $f \left({x}\right) = 0$ for all $f \in T$. So $x \in V \left({T}\right)$. {{qed}} Category:Algebraic Geometry n4qyz7rgoqcle2fezabo58byr75biyn
The justifications are listed as follows: :Second-Countable $T_3$ Space is $T_5$ :Second-Countable Space is Lindelöf :Lindelöf $T_3$ Space is Paracompact :$T_3$ Space is Fully $T_4$ iff Paracompact :Fully $T_4$ Space is $T_4$ Space :$T_5$ Space is $T_4$ Space {{finish}}
Let $\struct {S_1, \tau_1}$ and $\struct {S_2, \tau_2}$ be $\sigma$-compact. Then $\displaystyle S_1 = \bigcup_{n \mathop = 1}^\infty C_n$ and $\displaystyle S_2 = \bigcup_{n \mathop = 1}^\infty D_n$ where all the $C_n$ and $D_n$ are compact sets. From Tychonoff's Theorem, the set $\KK = \set {C_n \times D_m: n, m \in \N}$ is a countable set of compact sets of $S_1 \times S_2$ with the product topology. Consider $\tuple {a, b} \in S_1 \times S_2$. Then $a \in S_1$ and $b \in S_2$. Using the $\sigma$-compactness of $S_1$ and $S_2$, $\exists n, m$ such that $a \in C_n$ and $b \in D_m$. So $\tuple {a, b} \in C_n \times D_m \in \KK$. Thus, $\bigcup \KK = S_1 \times S_2$ and then $S_1 \times S_2$ is $\sigma$-compact. Using induction on the number of factors of the product we get the result. {{finish|Complete the inductive proof.}}
Pick any $\tuple {x_1, \dots, x_n} \in S$. Since each of $\struct {S_i, \tau_i}$ is weakly locally compact, $x_i$ has a compact neighborhood. That is, there is a compact set $N_i$ such that: :$\exists U_i \in \tau_i: x_i \in U_i \subseteq N_i$ By Cartesian Product of Subsets/Family of Subsets, we have: :$\displaystyle \tuple {x_1, \dots, x_n} \in \prod_{i \mathop = 1}^n U_i \subseteq \prod_{i \mathop = 1}^n N_i$ By definition of product space: :$\displaystyle \prod_{i \mathop = 1}^n U_i \in \tau$ By Topological Product of Compact Spaces/Finite Product: :$\displaystyle \prod_{i \mathop = 1}^n N_i$ is compact Therefore $\displaystyle \prod_{i \mathop = 1}^n N_i$ is a compact neighborhood of $\tuple {x_1, \dots, x_n}$. As $\tuple {x_1, \dots, x_n}$ is arbitrary, $\struct {S, \tau}$ is weakly locally compact. {{qed}}
Let $x = \family {x_\alpha} \in S$. Let $\BB_\alpha = \set {N^\alpha_i : i \in \N}$ be a countable local basis for $x_\alpha$ in $\struct {S_\alpha, \tau_\alpha}$ for each $\alpha \in I$. Let $\pr_\alpha$ denote the projection of $S$ onto $S_\alpha$. Let $\LL_\alpha = \set {\map {\pr_\alpha^{-1} } {N^\alpha_i}: N^\alpha_i \in B_\alpha}$. Let $\KK_J = \displaystyle \set {\bigcap_{\alpha \mathop \in J} L_\alpha : \forall \alpha \in J: L_\alpha \in \LL_\alpha}$ for $J \subset I$, $\size J < \infty$. From Product Space Local Basis Induced from Factor Spaces Local Bases: :$\displaystyle \BB_x = \bigcup_{J \mathop \subset I \mathop , \size J \mathop < \infty} \KK_J$ forms a local basis of $x$ in the product space $\struct {S_\alpha, \tau_\alpha}$ Now, each of the $\LL_\alpha$'s is countable.  Since the $\KK_J$'s can be identified with a finite product of countable sets, they are each countable From Countable Union of Countable Sets is Countable, $\BB_x$ forms a countable local basis of $x$ in the product space $\struct {S_\alpha, \tau_\alpha}$. The result follows. {{qed}}
Let $\BB_\alpha = \set {N^\alpha_i : i \in \N}$ be a countable basis for $\tau_\alpha$ for each $\alpha \in I$. Let $\pr_\alpha$ denote the projection of $S$ onto $S_\alpha$. Let $\LL_\alpha = \set {\map {\pr_\alpha^{-1} } {N^\alpha_i}: N^\alpha_i \in B_\alpha}$. Let $\KK_J = \displaystyle \set {\bigcap_{\alpha \mathop \in J} L_\alpha : \forall \alpha \in J: L_\alpha \in \LL_\alpha}$ for $J \subset I$, $\size J < \infty$. From Product Space Basis Induced from Factor Space Bases: :$\displaystyle \BB = \bigcup_{J \mathop \subset I \mathop , \size J \mathop < \infty} \KK_J$ forms a basis of the product space. Now, each of the $\LL_\alpha$'s is countable.  Since the $\KK_J$'s can be identified with a finite product of countable sets, they are each countable From Countable Union of Countable Sets is Countable, $\BB$ forms a countable basis of $S$. {{qed}}
For each $\alpha \in I$, let $D_\alpha$ be a countable everywhere dense subset of $\struct {S_\alpha, \tau_\alpha}$. Let $D = \displaystyle \prod_{\alpha \mathop \in I} D_\alpha$. From Countable Union of Countable Sets is Countable, $D$ is a countable set of $S$. From Natural Basis of Tychonoff Topology, the set $\BB$ of cartesian products of the form $\displaystyle \prod_{\alpha \mathop \in I} U_\alpha$ where: :for all $\alpha \in I : U_\alpha \in \tau_\alpha$ :for all but finitely many indices $\alpha : U_\alpha = X_\alpha$ is a basis for $\tau$. From Analytic Basis Characterization of Denseness, to show that $D$ is an everywhere dense subset in $\struct {S, \tau}$ it is sufficient to show that: :$\forall U \in \BB : U \cap D \ne \O$ Let $U = \displaystyle \prod_{\alpha \mathop \in I} U_\alpha \in \BB$. Then: :$\forall \alpha \in I : U_\alpha \cap D_\alpha \ne \O$ For all $\alpha \in I$, let $x_\alpha \in U_\alpha \cap D_\alpha$. Then: {{begin-eqn}} {{eqn | l = \family{x_\alpha}_{\alpha \in I}       | o = \in        | r = \prod_{\alpha \mathop \in I} \paren {U_\alpha \cap D_\alpha} }} {{eqn | r = \paren {\prod_{\alpha \mathop \in I} U_\alpha} \bigcap \paren {\prod_{\alpha \mathop \in I}  D_\alpha}       | c = General Case of Cartesian Product of Intersections }} {{eqn | r = U \cap D }} {{end-eqn}} Thus: :$U \cap D \ne \O$ The result follows. {{qed}}
=== Proof for Compactness === {{:Compactness is Preserved under Continuous Surjection}} === Proof for $\sigma$-Compactness === {{:Sigma-Compactness is Preserved under Continuous Surjection}} === Proof for Countable Compactness === {{:Countable Compactness is Preserved under Continuous Surjection}} === Proof for Sequential Compactness === {{:Sequential Compactness is Preserved under Continuous Surjection}} === Proof for Lindelöf Property === {{:Lindelöf Property is Preserved under Continuous Surjection}}
Let $U \in \tau$. It follows from the definition of Tychonoff topology that $U$ can be expressed as: :$\displaystyle U = \bigcup_{j \mathop \in J} \bigcap_{k \mathop = 1}^{n_j} \map {\pr_{i_{k,j} }^{-1}} { U_{k,j} }$ where $J$ is an arbitrary index set, $n_j \in \N$, $i_{k,j} \in \set {1,2}$, and $U_{k,j} \in \tau_{i_{k,j} }$. For all $i \in \set {1, 2}$, define $V_{i, k, j} \in \tau_i$ by: :$V_{i, k, j} = \begin {cases} U_{k, j} & : i = i_{k, j} \\ S_i & : i \ne i_{k, j} \end {cases}$ By definition of projection: :$\map {\pr_{i_{k, j} }^{-1} } {U_{k, j} } = \displaystyle \prod_{i \mathop = 1}^2 V_{i, k, j}$ {{WLOG}}, we consider $\pr_1$. We have: {{begin-eqn}} {{eqn | l = \map {\pr_1} U       | r = \bigcup_{j \mathop \in J} \map {\pr_1} {\bigcap_{k \mathop = 1}^{n_j} \map {\pr_{i_{k, j} }^{-1} } {U_{k, j} } }       | c = Image of Union under Relation: Family of Sets }} {{eqn | r = \bigcup_{j \mathop \in J} \map {\pr_1} {\bigcap_{k \mathop = 1}^{n_j} \prod_{i \mathop = 1}^2 V_{i, k, j} } }} {{eqn | r = \bigcup_{j \mathop \in J} \map {\pr_1} {\prod_{i \mathop = 1}^2 \bigcap_{k \mathop = 1}^{n_j} V_{i, k, j} }       | c = Cartesian Product of Intersections }} {{eqn | r = \bigcup_{j \mathop \in J} \bigcap_{k \mathop = 1}^{n_j} V_{1, k, j}       | c = {{Defof|Projection (Mapping Theory)|Projection}} }} {{end-eqn}} As $\displaystyle \bigcup_{j \mathop \in J} \bigcap_{k \mathop = 1}^{n_j} V_{1, k, j} \in \tau_1$, it follows that $\pr_1$ is open. The proof for $\pr_2$ is the same, mutatis mutandis.
Let $\phi$ be a mapping which is surjective, continuous and open. Let $T_A$ be weakly locally compact. Take $b \in S_B$. Let $V$ be a neighbourhood of $b$. Since $\phi$ is surjective: :$\forall y \in S_B: \exists x \in S_A: x \in \map {\phi^{-1} } y$ From the weak local compactness of $T_A$ and the continuity of $\phi$, there exists a compact neighbourhood $K$ of $x$ such that $\phi \sqbrk K \subseteq V$. Since $K$ is a neighbourhood of $x$, then $x \in K^\circ$ and $y \in \phi \sqbrk {K^\circ} \subseteq \phi \sqbrk K$, where $K^\circ$ is the interior of $K$. $\phi$ is an open mapping and $K^\circ$ is an open set, so $\phi \sqbrk {K^\circ}$ is also open. Finally we get that $y \in \phi \sqbrk K \subseteq V$, where $\phi \sqbrk K$ is a compact neighbourhood. Thus $T_B$ is weakly locally compact. {{qed}}
First note that Projection from Product Topology is Continuous. Also note that Projection is Surjection. It follows from Compactness Properties Preserved under Continuous Surjection that: :Compact Space :$\sigma$-Compact Space :Countably Compact Space :Sequentially Compact Space :Lindelöf Space are all preserved under projections. Next note that Projection from Product Topology is Open. It follows from Local Compactness is Preserved under Open Continuous Surjection that local compactness is preserved under projections. It follows from Weak Local Compactness is Preserved under Open Continuous Surjection that weak local compactness is preserved under projections. The final result is that Paracompactness is Preserved under Projections. {{finish|Finish these results off.}}
First note that Projection from Product Topology is Continuous. It follows from Continuous Image of Separable Space is Separable that separability is preserved under projections. Next note that Projection from Product Topology is Open. It follows from Countability Axioms Preserved under Open Continuous Surjection that: :First-Countability :Second-Countability are preserved under projections. {{qed}}
=== Proof for First-Countability === {{:First-Countability is Preserved under Open Continuous Surjection}} === Proof for Second-Countability === {{:Second-Countability is Preserved under Open Continuous Surjection}}
Checking in turn each of the criteria for equivalence: === Reflexivity === We have that $\set a \subseteq S$ is a (degenerate) connected set of $S$. So $a$ is in the same connected set as itself and so $a \sim a$. So $\sim$ has been shown to be reflexive. {{qed|lemma}} === Symmetry === If $x \sim y$ then $x$ is in the same connected set as $y$ by definition. Trivially it follows that $y$ is in the same connected set as $x$ and so $y \sim x$. So $\sim$ has been shown to be symmetric. {{qed|lemma}} === Transitivity === Let $x \sim y$ and $y \sim z$. Then by definition: :$x$ is in the same connected set as $y$ :$y$ is in the same connected set as $z$ By Union of Connected Sets with Non-Empty Intersections is Connected it follows that $x$ is in the same connected set as $z$. Thus so $x \sim z$. So $\sim$ 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}}
Let $\mathcal F$ be the equivalence class of $\mathcal A$ under the relation of compatibility. By Compatibility of Atlases is Equivalence Relation, this is indeed an equivalence relation. By definition we have $\mathcal A \in \mathcal F$. By Relation Partitions Set iff Equivalence, $\mathcal F$ is an element of the partition of equivalence classes. By definition, the elements of a partition are pairwise disjoint. Therefore if $\mathcal G \ne \mathcal F$ is an element of the partition, we must have: : $\mathcal A \notin \mathcal G$ Therefore $\mathcal A$ belongs to exactly one differentiable structure. {{qed}} {{explain|Clarification is needed as to why this result should be categorised in :Category:Manifolds.}} Category:Manifolds 1d6a163jw7q2sy421hs3o2njg2xs4ya
Let $T$ be the disjoint union $T = R \cup \paren {S \setminus \Img \phi}$. Define $\theta : T \to S$ as follows: :If $x \in R$, then $\map \theta x = \map \phi x$ :If $x \in \paren {S \setminus \Img \phi}$ then $\map \theta x = x$ We claim that $\theta$ is an isomorphism. ''Injectivity'': Let $\map \theta x = \map \theta y$. Since $\theta \sqbrk R = \phi \sqbrk R$, we have $\theta \sqbrk R \cap \paren {S \setminus \Img \phi} = \O$. Therefore either $x, y \in R$ or $x, y \in \paren {S \setminus \Img \phi}$. If $x, y \in R$ then $\map \phi x = \map \phi y$, so $\theta$ is injective because $\phi$ is. If $x, y \in \paren {S \setminus \Img \phi}$ then $\theta$ acts as the identity and $x = y$. ''Surjectivity'': Under $\theta$, $R$ surjects onto $\Img \phi = \phi \sqbrk R$ by definition. Moreover: :$\theta \sqbrk {S \setminus \Img \phi} = I \sqbrk {S \setminus \Img \phi} = \paren {S \setminus \Img \phi}$ where $I$ denotes the identity mapping. Therefore $\theta$ is surjective everywhere. Now we endow $T$ with addition and multiplication: for $x, y \in T$, let :$x + y = \map {\theta^{-1} } {\map \theta x + \map \theta y}$ :$x y = \map {\theta^{-1} } {\map \theta x \, \map \theta y}$ so trivially we have :$\map \theta {x + y} = \map \theta x + \map \theta y$ :$\map \theta {x y} = \map \theta x \, \map \theta y$ Therefore $T$ is the isomorphic image of a ring, and an isomorphism is precisely a map that preserves the ring axioms. Thus $T$ is a ring isomorphic to $S$ containing $R$ as a subring. {{qed}} {{proof wanted|I wonder whether this page might be no more than an application of the Embedding Theorem.}} Category:Ring Isomorphisms Category:Ring Monomorphisms orqom74s8aylmkd8ikavss1u965boik
Let $H$ be a clopen set in $T$. By definition, $H$ is open and so $H \in \tau$. But as $H$ is also closed, by definition $\relcomp S H \in \tau$ where $\complement_S$ denotes complement relative to $S$. Thus $H$ and $\relcomp S H$ are both open such that: :$H \cap \relcomp S H = \O$ from Intersection with Relative Complement is Empty :$H \cup \relcomp S H = S$ from Union with Relative Complement and so forming a partition of $T$. As $H$ and $\relcomp S H$ are both closed, it follows from Closed Set Equals its Closure that: :$H \cap \paren {\relcomp S H}^- = \O = H^- \cap \relcomp S H$ and so by definition $H$ and $\relcomp S H$ are separated. {{AimForCont}} that $H$ does not contain the components of all of its points. That is: :$\exists x \in H: \map {\operatorname {Comp}_x} T \nsubseteq H$ where $\map {\operatorname {Comp}_x} T$ is the component of $x$ in $T$. Then: :$\exists z \in \map {\operatorname {Comp}_x} T: z \in H \cup \relcomp S H$ This means that $H \cup \relcomp S H$ is connected. This contradicts the fact that $H$ and $\relcomp S H$ are separated. The result follows by Proof by Contradiction. {{qed}}