hcju/proofwikiqa huggingface dataset:
We prove the contrapositive: :$T$ is not a Baire space {{iff}} there exists a non-empty open set of $T$ which is meager in $T$. We have by definition of Baire space: :$T$ is a Baire space {{iff}} the interior of the union of any countable set of closed sets of $T$ which are nowhere dense is empty. Therefore: :$T$ is not a Baire space {{iff}} the interior of the union of some countable set of closed sets of $T$ which are nowhere dense is non-empty. We have by definition of meager: :$A$ is meager in $T$ {{iff}} it is a countable union of subsets of $S$ which are nowhere dense in $T$. Here $I$ denotes a countable indexing set. === Sufficient Condition === Suppose $T$ is not a Baire space. Then there is a countable set of closed sets of $T$ which are nowhere dense with the interior of its union non-empty. Let $\FF = \set {F_i: i \in I}$ be such a set. Then $\paren {\bigcup \FF}^\circ$ is a non-empty open set. Consider the set $\FF' = \set {F_i \cap \paren {\bigcup \FF}^\circ: i \in I}$. From Intersection is Subset: :$F_i \cap \paren {\bigcup \FF}^\circ \subseteq F_i$ for any $i \in I$ Since $F_i$ are nowhere dense, so is $F_i \cap \paren {\bigcup \FF}^\circ$. We have: {{begin-eqn}} {{eqn | l = \bigcup \FF'       | r = \bigcup_{i \mathop \in I} \paren {F_i \cap \paren {\bigcup \FF}^\circ} }} {{eqn | r = \paren {\bigcup \FF}^\circ \cap \bigcup_{i \mathop \in I} F_i       | c = Intersection Distributes over Union }} {{eqn | r = \paren {\bigcup \FF}^\circ \cap \bigcup \FF }} {{eqn | r = \paren {\bigcup \FF}^\circ       | c = Set Interior is Largest Open Set; Intersection with Subset is Subset }} {{end-eqn}} Therefore $\bigcup \FF'$ is meager in $T$. Hence we have shown the existence of a non-empty open set of $T$ which is meager in $T$. {{qed|lemma}} === Necessary Condition === Suppose there exists a non-empty open set of $T$ which is meager in $T$. Let $\displaystyle A = \bigcup_{i \in I} A_i$ be such a set. Then $A$ is a non-empty open set, and $A_i$ are nowhere dense for every $i \in I$. Thus for every $i$, $\paren {A_i^-}^\circ = \O$. By Closure of Topological Closure equals Closure: :$\paren {\paren {A_i^-}^-}^\circ = \paren {A_i^-}^\circ = \O$. By Set is Closed iff Equals Topological Closure, $A_i^-$ are closed sets of $T$ which are nowhere dense. We also have: {{begin-eqn}} {{eqn | l = A       | r = A^\circ       | c = Interior of Open Set }} {{eqn | r = \paren {\bigcup_{i \in I} A_i}^\circ }} {{eqn | o = \subseteq       | r = \paren {\bigcup_{i \in I} A_i^-}^\circ       | c = Set is Subset of its Topological Closure; Set Union Preserves Subsets; Interior of Subset }} {{end-eqn}} Hence there is a countable set of closed sets of $T$ which are nowhere dense with the interior of its union non-empty. Therefore $T$ is not a Baire space. {{qed}}
Let $S \left({\MM}\right)$ denote the set containing all complete types over $M$ of every number of free variables. Let $\kappa = \left\vert{S \left({\MM}\right)}\right\vert$. Use a bijection between $\kappa$ and $S \left({\MM}\right)$ to write the elements of $S \left({\MM}\right)$ as $p_\alpha$ for $\alpha < \kappa$. For each $\alpha < \kappa$, let $N_\alpha$ be an elementary extension of $\MM$ which has an ordered tuple $\bar a_\alpha$ realizing $p_\alpha$. Such extensions and tuples exist by Type is Realized in some Elementary Extension. We will construct the extension claimed by the theorem as the union over a chain of elementary extensions defined using Transfinite Induction. At each step, we use the Elementary Amalgamation Theorem to add on $N_\alpha$. Base case $\alpha = 0$: Let $\BB_0 = \MM$. Note that $\BB_0$ is an elementary extension of $\MM$ by choice of $\NN_0$. Limit ordinals $\alpha \le \kappa$: Let $\displaystyle \BB_\alpha = \bigcup_{\beta \mathop < \alpha} \BB_\beta$ $\BB_\alpha$ is an elementary extension of $\MM$ by Union of Elementary Chain is Elementary Extension. Successor ordinals $\alpha + 1$ for $\alpha < \kappa$: We have that $\BB_\alpha$ and $\NN_\alpha$ are both elementary extensions of $\MM$. By the Elementary Amalgamation Theorem, there is: :an elementary extension $\BB_{\alpha + 1}$ of $\BB_\alpha$ and: :an elementary embedding $g_\alpha: \NN_\alpha \to \BB_{\alpha + 1}$ which is the identity on $M$ viewed as a subset of $\NN_\alpha$. Note that $\map {g_\alpha} {\bar a_\alpha}$ realizes $p_\alpha$ in $\BB_{\alpha + 1}$ since $g_\alpha$ is elementary. Thus $\BB_{\alpha + 1}$ is an elementary extension of $\MM$ which contains $\map {g_\alpha} {\bar a_\alpha}$ realizing $p_\alpha$. Now, let $\displaystyle \BB = \bigcup_{\alpha < \kappa} \BB_\alpha$. $\BB$ is an elementary extension of each $\BB_\alpha$ by Union of Elementary Chain is an Elementary Extension. In particular, $\BB$ is an elementary extension of $\BB_0 = \MM$ We have that: :each $p_\alpha$ is realized in $\BB_{\alpha + 1}$ by the corresponding $\map {g_\alpha} {\bar a_\alpha}$ and: :$\BB$ is an elementary extension of each $\BB_{\alpha + 1}$ Thus we have that $\map {g_\alpha} {\bar a_\alpha}$ realizes $p_\alpha$ in $\BB$. {{qed}} Category:Model Theory nydyqgv6evfgf48xvhhom5vczk5retb
From Baire Space iff Open Sets are Non-Meager, all open sets of $T$ are non-meager in $T$. But $T$ itself is an open set of $T$ by definition of topological space. Hence the result. {{qed}} Category:Baire Spaces Category:Non-Meager Spaces kzh9g5nfshlvd9lkp0529aeaoy5y7a1
The rational numbers $\Q$ form a metric space. We have that the Rationals are Everywhere Dense in Topological Space of Reals. We also have that the Rational Numbers are Countably Infinite. The result follows from the definition of separable space. {{qed}}
Follows from: :Real Number Line is Second-Countable :Second-Countable Space is Separable {{qed}}
Let $S' = S \times \set {a, b}$. Let $F' \subseteq S'$ such that $F'$ is closed in $T \times D$. Then $F' = F \times \set {a, b}$ or $F' = F \times \O$ by definition of the double pointed topology. If $F' = F \times \O$ then $F' = \O$ from Cartesian Product is Empty iff Factor is Empty, and the result is trivial. So suppose $F' = F \times \set {a, b}$. From Open and Closed Sets in Multiple Pointed Topology it follows that $F$ is closed in $T$. Let $y' = \tuple {y, q} \in \relcomp {S'} {F'}$. Then $y \notin F$. Suppose that $T$ is a $T_3$ space. Then by definition: :For any closed set $F$ of $T$ and any point $y \in S$ such that $y \notin F$ there exist disjoint open sets $U, V \in \tau$ such that $F \subseteq U$, $y \in V$. Then $y' \in V \times \set {a, b}$ and $F' \subseteq U \times \set {a, b}$ and: :$U \times \set {a, b} \cap V \times \set {a, b} = \O$ demonstrating that $T \times D$ is a $T_3$ space. Now suppose that $T \times D$ is a $T_3$ space. Then $\exists U', V' \in S': y' \in V'$ and $F' \subseteq U'$ such that $U' \cap V' = \O$. As $D$ is the indiscrete topology it follows that: :$U' = U \times \set {a, b}$ :$V' = V \times \set {a, b}$ for some $U, V \subseteq T$. From Open and Closed Sets in Multiple Pointed Topology it follows that $U$ and $V$ are open in $T$. As $U' \cap V' = \O$ it follows that $U \cap V = \O$. It follows that $F$ and $y$ fulfil the conditions that make $T$ a $T_3$ space. Hence the result.
Let $S' = S \times \set {a, b}$. Let $H' \subseteq S'$ such that $H$ is closed in $T \times D$. Then $H' = H \times \set {a, b}$ or $H' = H \times \O$ by definition of the double pointed topology. If $H' = H \times \O$ then $H' = \O$ from Cartesian Product is Empty iff Factor is Empty, and the result is trivial. So suppose $H' = H \times \set {a, b}$. From Open and Closed Sets in Multiple Pointed Topology it follows that $H$ is closed in $T$. Suppose that $T$ is a $T_4$ space. Then by definition: :For any two disjoint closed sets $A, B \subseteq S$ there exist disjoint open sets $U, V \in \tau$ containing $A$ and $B$ respectively. Then $A \times \set {a, b} \subseteq U \times \set {a, b}$ and $B \times \set {a, b} \subseteq V \times \set {a, b}$ and: :$U \times \set {a, b} \cap V \times \set {a, b} = \O$ demonstrating that $T \times D$ is a $T_4$ space. Now suppose that $T \times D$ is a $T_4$ space. Then $\exists U', V' \in S': A' \subseteq U'$ and $B' \subseteq V'$ such that $U' \cap V' = \O$. As $D$ is the indiscrete topology it follows that: :$U' = U \times \set {a, b}$ :$V' = V \times \set {a, b}$ for some $U, V \subseteq T$. From Open and Closed Sets in Multiple Pointed Topology it follows that $U$ and $V$ are open in $T$. As $U' \cap V' = \O$ it follows that $U \cap V = \O$. It follows that $A$ and $B$ fulfil the conditions that make $T$ a $T_4$ space. Hence the result.
=== $T_0$, $T_1$, $T_2$ and $T_{2 \frac 1 2}$ Axioms === From: :Double Pointed Topology is not $T_0$ :$T_1$ Space is $T_0$ Space :$T_2$ Space is $T_1$ Space :Completely Hausdorff Space is $T_2$ Space we have that $T \times D$ is not $T_0$, $T_1$ or $T_2$ or $T_{2 \frac 1 2}$. {{qed|lemma}} From: :Separation Axioms on Double Pointed Topology: $T_3$ Axiom $T \times D$ is a $T_3$ space {{iff}} $T$ is. {{qed|lemma}} From: :Separation Axioms on Double Pointed Topology: $T_{3 \frac 1 2}$ Axiom $T \times D$ is a $T_{3 \frac 1 2}$ space {{iff}} $T$ is. {{qed|lemma}} From: :Separation Axioms on Double Pointed Topology: $T_4$ Axiom $T \times D$ is a $T_4$ space {{iff}} $T$ is. {{qed|lemma}} From: :Separation Axioms on Double Pointed Topology: $T_5$ Axiom $T \times D$ is a $T_5$ space {{iff}} $T$ is. {{qed}} {{expand|investigate whether we can directly use Separation Properties Preserved under Topological Product}}
By definition of multiple pointed topology, $T \times D$ is the product space of $T$ with $D$: :$T \times D = \left({S \times A, \tau}\right)$ where $\tau$ is the Tychonoff topology on $S \times A$. Consider the set $H \times A$. We have that: :$H \times A = \operatorname{pr}_1^{-1} \left({H}\right)$ where $\operatorname{pr}_1: T \times D \to T$ is the first projection from $T \times D$ to $T$. By definition of the Tychonoff topology, $\operatorname{pr}_1: T \times D \to T$ is a continuous mapping. === Open Sets === By definition of Tychonoff topology, if $H$ is open in $T$ then $H \times A$ is open in $T \times D$. As the Tychonoff Topology is Coarsest Topology such that Projections are Continuous, it follows that if $H$ is not open in $T$ then $H \times A$ is not open in $T \times D$. {{qed|lemma}} === Closed Sets === Let $H \subseteq S$ be closed in $T$. We have that $\operatorname{pr}_1: T \times D \to T$ is a continuous mapping. By Continuity Defined from Closed Sets, it follows that $H \times A$ is closed in $T \times D$. As the Tychonoff Topology is Coarsest Topology such that Projections are Continuous, it follows that if $H$ is not closed in $T$ then $H \times A$ is not closed in $T \times D$. {{qed|lemma}} === Cartesian Product with Empty Set === From Cartesian Product is Empty iff Factor is Empty, $H \times \varnothing = \varnothing$. From Open and Closed Sets in Topological Space, $\varnothing$ is both open and closed in $T \times D$. {{qed|lemma}} === Neither Open Nor Closed === Let $\varnothing \subset H_A \subset A$. That is, $H_A \ne \varnothing$ and $H_A \ne A$. So $H_A$ is neither open nor closed in the indiscrete topology on $A$ We have that: :$S \times H_A = \operatorname{pr}_2^{-1} \left({H_A}\right)$ where $\operatorname{pr}_2: T \times D \to D$ is the second projection from $T \times D$ to $D$. By definition of the Tychonoff topology, $\operatorname{pr}_2: T \times D \to D$ is a continuous mapping. We have that the Tychonoff Topology is Coarsest Topology such that Projections are Continuous. So if $H_A$ is neither open nor closed in $D$ then $S \times H_A$ is neither open nor closed in $T \times D$. Hence, by definition of the Tychonoff topology, nor is $H \times H_A$. {{qed|lemma}} {{qed}} Category:Multiple Pointed Topology Category:Open Sets Category:Closed Sets Category:Clopen Sets 93opzeu96qlq6kz77rctynhnws9au81
Let us check the two conditions for $\mathcal B$ to be a synthetic basis in turn. === $(B1)$: Covering === From open set axiom $(O3)$, we have $S_i \in \tau_i$ for all $i \in I$. Thus $S = \displaystyle \prod_{i \mathop \in I} S_i \in \mathcal B$. Hence by Set is Subset of Union: General Result, we have: :$S \subseteq \displaystyle \bigcup \mathcal B$ so $\mathcal B$ is a cover for $S$. {{qed|lemma}} === $(B2)$: Intersections are Unions === Let $A = \displaystyle \prod_{i \mathop \in I} U_i$ and $B = \displaystyle \prod_{i \mathop \in I} V_i$ be in $\mathcal B$. Then by Cartesian Product of Intersections, we have: :$A \cap B = \displaystyle \prod_{i \mathop \in I} \left({U_i \cap V_i}\right)$ By open set axiom $(O2)$, $U_i \cap V_i \in \tau_i$ for all $i \in I$. Hence $A \cap B \in \mathcal B$, and in particular, by Union of Singleton: :$A \cap B = \displaystyle \bigcup \left\{{A \cap B}\right\}$ {{qed|lemma}} Having verified both conditions, we conclude $\mathcal B$ is a synthetic basis on $S$. {{qed}}
Let $x \in S$ be an open point of $T$. That is: :$\set x \in \tau$ Recall that: :a topological space is non-meager if it is not meager and: :a topological space is meager {{iff}} it is a countable union of subsets of $S$ which are nowhere dense in $S$. {{AimForCont}} that $T$ is meager. Let: :$\displaystyle T = \bigcup \SS$ where $\SS$ is a countable set of subsets of $S$ which are nowhere dense in $S$. Then: :$\exists H \in \S S: x \in H$ and so: :$\set x \subseteq H$ We have that $H$ is nowhere dense in $T$. By definition, its closure $H^-$ contains no open set of $T$ which is non-empty. But from Set is Subset of its Topological Closure we have that: :$H \subseteq H^-$ So by Subset Relation is Transitive: :$\set x \subseteq H^-$ So $H$ is not nowhere dense. Therefore $T$ cannot be a countable union of subsets of $S$ which are nowhere dense in $S$. That is, $T$ is not meager. Hence the result by definition of non-meager. {{qed}}
As $T$ is a trivial topological space, by definition $S$ is a singleton: $S = \set s$, say. Then $\set s$ is an open set. That is, $s$ is an open point. The result follows from Space with Open Point is Non-Meager. {{qed}}
Let $V \subset S$ be a closed set. By Continuity Defined from Closed Sets, we have that $U_i = \paren {f \restriction_{C_i} }^{-1} \sqbrk V$ is also closed. From the definition of a restriction, we have that $U_i = C_i \cap f^{-1} \sqbrk V$. Therefore, we can compute: {{begin-eqn}} {{eqn | l = \paren {f \restriction_{C_i} }^{-1} \sqbrk V       | r = C \cap f^{-1} \sqbrk V       | c = {{Defof|Restriction of Mapping}} }} {{eqn | r = \paren {\bigcup_{i \mathop = 1}^n C_i} \cap f^{-1} \sqbrk V       | c = Definition of $C$ }} {{eqn | r = \bigcup_{i \mathop = 1}^n \paren {C_i \cap f^{-1} \sqbrk V}       | c = Intersection Distributes over Union }} {{eqn | r = \bigcup_{i \mathop = 1}^n U_i       | c = Definition of $U_i$ }} {{end-eqn}} That is, $U = \paren {f \restriction_{C_i} }^{-1} \sqbrk V$ is the intersection of finitely many closed sets. Therefore, $U$ is itself closed by definition of a topology. It follows by Continuity Defined from Closed Sets that $f \restriction_C$ is also continuous. {{qed}} {{finish|Counterexample for infinitely many closed sets.}}
The idea of the proof is: : to go to some elementary equivalent structure where a type is realized and: : to interpret a new relation symbol as the singleton containing some realization of the type. This lets us write sentences about the extension saying that: :there must exist an element satisfying the relation and: :every element satisfying the relation satisfies all of the formulas in the type. Then using bigness, these sentences carry over into $\mathcal M$. Let $p$ be a complete $n$-type over $A$, where $A$ is a subset of the universe of $\mathcal M$ with cardinality strictly less than $\kappa$. We are to show that $p$ is realized in $\mathcal M$. Let $\mathcal L_A$ be the language obtained from $\mathcal L$ by adding constant symbols for each element of $A$. Viewing $\mathcal M$ as an $\mathcal L_A$-structure in the natural way, let $\operatorname {Th}_A (\mathcal M)$ be the collection of $\mathcal L_A$-sentences satisfied by $\mathcal M$. By definition of type, $p \cup \operatorname {Th}_A \left({\mathcal M}\right)$ is satisfiable by some $\mathcal L_A$-structure $\mathcal N$. Since $\mathcal N$ satisfies $p$, some $n$-ordered tuple $\bar b$ in it realizes $p$. Now, let $\mathcal L_A^*$ be the language obtained from $\mathcal L_A$ by adding a new relation symbol $R$. We can extend $\mathcal N$ to be an $\mathcal L_A^*$-structure by interpreting the symbol $R$ as the singleton $R^\mathcal N = \left\{{\bar b}\right\}$. Since $\mathcal N \models \operatorname {Th}_A \left({\mathcal M}\right)$, we have that $\mathcal M$ and $\mathcal N$ are elementary equivalent as $\mathcal L_A$-structures. We have that $\mathcal M$ is $\kappa$-big by assumption. Hence there is some relation $R^\mathcal M$ on the universe of $\mathcal M$ such that: :$\left({\mathcal M, R^\mathcal M}\right)$ is elementary equivalent to $\left({\mathcal N, R^\mathcal N}\right)$ as $\mathcal L_A^*$-structures. But $\left({\mathcal N, R^\mathcal N}\right)$ satisfies the $\mathcal L_A^*$-sentences: :$\exists \bar x R \left({\bar x}\right)$ and :$\forall \bar x \left({R \left({\bar x}\right) \to \phi \left({\bar x}\right)}\right)$ for each $\phi \left({\bar x}\right)$ in $p$. Hence $\left({\mathcal M, R^\mathcal M}\right)$ also satisfies these sentences, and so there must be some $\bar d \in R^\mathcal M$ realizing $p$. So $\mathcal M$ realizes $p$. {{qed}} Category:Model Theory 78f6cxjvoukbaly9ux17vxwtj90s35c
We show that the unit interval $\hointr 0 1$ is uncountable, from which uncountability of $\R$ follows immediately. {{AimForCont}} that $\hointr 0 1$ is countable. Clearly $\hointr 0 1$ is not finite because $\displaystyle \frac 1 n \in \hointr 0 1$ are distinct for all $n \in \N$. Therefore an injection $\hointr 0 1 \hookrightarrow \N$ enumerates $\hointr 0 1$ with a subset of the natural numbers. By relabeling, we can associate each $x \in \hointr 0 1$ to exactly one natural number to obtain a bijection. (By hypothesis, such a mapping can be constructed). Let $g$ be such a correspondence: {{begin-eqn}} {{eqn | l = 1       | o = \leftrightarrow       | r = 0.d_{11} d_{12} d_{13} \ldots }} {{eqn | l = 2       | o = \leftrightarrow       | r = 0.d_{21} d_{22} d_{23} \ldots }} {{eqn | o = \vdots }} {{eqn | l = n       | o = \leftrightarrow       | r = 0.d_{n1} d_{n2} d_{n3} \ldots }} {{eqn | o = \vdots }} {{end-eqn}} where juxtaposition of digits describes the decimal expansion of a number. By Existence of Base-$N$ Representation, any decimal expansion of a real number exists and is unique, or else has exactly two representative strings. In this case, if there are exactly two representations, one will have an infinite trail of $9$s, and one will terminate. {{MissingLinks|this is proven somewhere, find it}} Restrict $g$ such that there exists no infinite strings of $9$s in the decimal expansions in the question, i.e. to the set: :$S := \left\{ {f: \N \to \set {0, 1, 2, \ldots, 9} }\middle\vert\,{ \forall M \in \N: \exists m \ge M: \map f m \ne 9}\right\}$ of sequences in $\set {0, 1, 2, \ldots, 9}$ not ending in infinitely many $9$s. Define $g'$ as the restriction of $g$ to $S$: :$g' := g \restriction S$ That is, construct $g'$ such that there exist no infinite strings of $9$s in the decimal expansions in question. From Injection to Image is Bijection, $g'$ is one-to-one, and consequently is onto. For every $k \in \N$ define $f_k = d_{k k} + 1$ taken modulo $10$.  That is, $f: 0 \mapsto 1, 1 \mapsto 2, \dots, 8 \mapsto 9, 9 \mapsto 0$. Let $y$ be defined by the decimal expansion: :$y = 0.f_1 f_2 f_3 \ldots$ Now: :$y$ differs from $\map {g'} 1$ in the first digit of the decimal expansion :$y$ differs from $\map {g'} 2$ the second digit of the decimal expansion and generally the $n$th digit of the decimal expansion of $\map {g'} n$ and $y$ is different. So $y$ can be none of the numbers $\map {g'} n$ for $n \in \N$. But $g'$ is a surjection. From this contradiction it is deduced that $\hointr 0 1$ is not countable. {{proofread}} {{qed}}
=== Necessary Condition === Suppose every neighborhood of $x$ contains infinitely many points of $H$. Let $N$ be an open neighborhood of $x$. Then $H \cap N$ is infinite, and so is $H \cap \paren {N \setminus \set x}$. Since $N$ is arbitrary, by {{Defof|Limit Point of Set}}, $x$ is a limit point of $H$. {{qed|lemma}} === Sufficient Condition === We prove the contrapositive: Suppose there exists some neighborhood $V$ of $x$ containing only finitely many points of $H$. Then there exists some open set $N \subseteq V$ such that $x \in N$. By Set Intersection Preserves Subsets, $N \cap H \subseteq V \cap H$. By Subset of Finite Set is Finite, $N \cap H$ is finite. For each $y \in H \cap N$ and $y \ne x$, since $T$ is $T_1$: :$\exists U_y \in \tau: x \in U_y, y \notin U_y$ Consider the set $\displaystyle U = N \cap \bigcap_{y \mathop \in H \cap \paren {N \setminus \set x} } U_y$. This is a finite intersection of open sets. By General Intersection Property of Topological Space, $U \in \tau$. By {{Defof|Set Intersection}}, since $x \in N$ and $x \in U_y$ for all $y \in H \cap \paren {N \setminus \set x}$, we have $x \in U$. Therefore $U$ is an open neighborhood of $x$. Let $z \in H \cap \paren {U \setminus \set x}$. Then: {{begin-eqn}} {{eqn | o =        | r = z \in H \land z \in U \setminus \set x       | c = {{Defof|Set Intersection}} }} {{eqn | o = \leadsto       | r = z \in H \land z \notin \set x \land z \in U       | c = {{Defof|Set Complement}} }} {{eqn | o = \leadsto       | r = z \in H \land z \notin \set x \land z \in N \land \paren {\forall y} \paren {y \in H \cap \paren {N \setminus \set x} \implies z \in U_y}       | c = {{Defof|Set Intersection}} }} {{eqn | o = \leadsto       | r = z \in H \land z \in N \setminus \set x \land \paren {\forall y} \paren {y \in H \cap \paren {N \setminus \set x} \implies z \in U_y}       | c = {{Defof|Set Complement}} }} {{eqn | o = \leadsto       | r = z \in H \cap \paren {N \setminus \set x} \land \paren {\forall y} \paren {y \in H \cap \paren {N \setminus \set x} \implies z \in U_y}       | c = {{Defof|Set Intersection}} }} {{eqn | o = \leadsto       | r = z \in H \cap \paren {N \setminus \set x} \land z \in U_z       | c = Universal Instantiation }} {{end-eqn}} But from definition of $U_z$:  :$z \in H \cap \paren {N \setminus \set x} \implies z \notin U_z$ Which is a contradiction. This shows that $H \cap \paren {U \setminus \set x} = \O$. By {{Defof|Limit Point of Set}}, $x$ is not a limit point of $H$. Thus the contrapositive is proved. {{qed}} Category:Limit Points Category:T1 Spaces blp2niyph0ixy0brxggaf1uwx1ct2v8
Let $U \in \tau$ be an open set of $T$ such that $x \in T$. The fact that such a $U$ exists follows from the fact that $S$ is open in $T$ and $x \in S$. Then as $\left\{{x}\right\} \subseteq S$ it follows that $\left\{{x}\right\} \cap U = \left\{{x}\right\}$ from Intersection with Subset is Subset. So by definition, $x$ is an isolated point of $\left\{{x}\right\}$. From Topological Space is Discrete iff All Points are Isolated, unless $T$ is the discrete space on $S$, not all elements of $T$ are isolated points of $T$. {{qed}} Category:Isolated Points Category:Singletons 1xustjkquotedgz9fhc6j1rit9rvbxb
From Singleton Point is Isolated, $x$ is isolated in $\left\{{x}\right\}$. So by definition $\left\{{x}\right\}$ is not dense-in-itself. {{qed}} Category:Denseness Category:Singletons akms0y3s9xd4yoyqbiuid33kcvymq78