hcju/proofwikiqa huggingface dataset:
From Sequence in Indiscrete Space converges to Every Point, $\sequence {s_n}$ converges to every point of $S$. As $S$ is uncountable, the result follows. {{qed}}
{{AimForCont}} $T$ is a $T_{2 \frac 1 2}$ (completely Hausdorff) space. Then from Completely Hausdorff Space is Hausdorff Space, $T$ is a $T_2$ (Hausdorff) space. This contradicts the result Partition Topology is not Hausdorff. Hence the result, by Proof by Contradiction. {{qed}}
{{AimForCont}} $T$ is a $T_2$ (Hausdorff) space. Then from $T_2$ Space is $T_1$ Space, $T$ is a $T_1$ (Fréchet) space. This contradicts the result Partition Topology is not $T_1$. Hence the result, by Proof by Contradiction. {{qed}}
{{AimForCont}} $T$ is a $T_1$ (Fréchet) space. Then from $T_1$ Space is $T_0$ Space, $T$ is a $T_0$ (Kolmogorov) space. This contradicts the result Partition Topology is not $T_0$. Hence the result, by Proof by Contradiction. {{qed}}
From Odd-Even Topology is Second-Countable, $T$ is second-countable. The result follows from Second-Countable Space is Lindelöf. {{qed}}
From Odd-Even Topology is Second-Countable, $T$ is second-countable. The result follows from Second-Countable Space is Separable. {{qed}}
From Odd-Even Topology is Second-Countable, $T$ is second-countable. The result follows from Second-Countable Space is First-Countable. {{qed}}
Let $H = S \setminus \left\{{p}\right\}$. Let $V \subseteq H$ be any subset of $H$. As $p \notin V$, $V$ is a closed set of $T$. Thus, by definition of closed set $S \setminus V$ is open in $T$. By definition of subspace topology, $\left({S \setminus V}\right) \cap H$ is open in $T_H$. From Intersection with Set Difference is Set Difference with Intersection: :$\left({S \setminus V}\right) \cap H = \left({S \cap H}\right) \setminus V$ and so from Intersection with Subset is Subset: :$\left({S \setminus V}\right) \cap H = H \setminus V$ Thus $H \setminus V$ is open in $H$. But $H \setminus V$ is itself a subset of $H$, and so a closed set of $T$. Thus, by the same argument, $V$ is open in $H$. As $V$ is arbitrary, it follows that all subsets of $H$ are open in $H$. The result follows by definition of discrete space. {{qed}} Category:Particular Point Topology Category:Discrete Topology s5i7qez4qokagfqn03i9g1ch6cgr5nu
By definition, $\alpha$ is an accumulation point of $\sigma$ {{iff}}: :$\forall U \in \tau_0: \alpha \in U \implies \set {n \in \N: x_n \in U}$ is infinite. Both $\set 0$ and $\set {0, 1}$ contain $0$, which occurs an infinite number of times in $\sigma$. Hence, by definition, $0$ is an accumulation point of $\sigma$. {{qed}}
By definition, $\alpha$ is a limit point of $\sigma$ {{iff}}: :$\forall U \in \tau_0: \alpha \in U \implies \set {n \in \N: x_n \notin U}$ is finite. The only open set of $T$ containing $1$ is $\set {0, 1}$. It contains all but a finite number (that is: $0$) elements of $\sigma$. Hence, by definition, $1$ is a limit point of $\sigma$. {{qed}}
{{AimForCont}} $T$ is metacompact. From Metacompact Space is Countably Metacompact it follows that $T$ is countably metacompact. But we have that Infinite Particular Point Space is not Countably Metacompact. Hence the result by Proof by Contradiction. {{qed}} Category:Metacompact Spaces Category:Particular Point Topology 3axb4hycpe6yv0nrpnjxef7u3awkegb
{{AimForCont}} $T$ is countably paracompact. From Countably Paracompact Space is Countably Metacompact it follows that $T$ is countably metacompact. But we have that Infinite Particular Point Space is not Countably Metacompact. Hence the result by Proof by Contradiction. {{qed}} Category:Countably Paracompact Spaces Category:Particular Point Topology t7f79a3os4f9qbju0y94gcsq21vefni
{{AimForCont}} $T^*_p$ is not a $T_2$ (Hausdorf) space. From $T_2$ Space is $T_1$ Space, $T^*_p$ is a $T_1$ (Fréchet) space. But this contradicts Closed Extension Topology is not $T_1$ Hence by Proof by Contradiction $T^*_p$ can not be a $T_2$ (Hausdorf) space. {{qed}}
By Underlying Set of Topological Space is Closed, $S$ is closed in $T$. By Closed Sets of Closed Extension Topology, $S$ is closed in $T^*_p$. {{Defof|Closed Extension Space}} gives: :$p \notin S$ :Every open set in $T^*_p$ is either $\O$ or it contains $p$. Thus no open set containing $S$ is disjoint from $\set p$. This shows that $T^*_p$ is not a $T_3$ space. {{qed}}
=== Sufficient Condition === Suppose $T^*_p$ is $T_4$. Then for any two disjoint closed sets $A, B \subseteq S$ there exist disjoint open sets $U, V \in \tau^*_p$ containing $A$ and $B$ respectively. However, for any non-empty set $U \in \tau^*_p$, $p \in U$. Hence no non-empty open sets in $T^*_p$ are disjoint. Therefore $T^*_p$ is a $T_4$ space vacuously: there do not exist two disjoint closed sets. {{qed|lemma}} === Necessary Condition === By Closed Sets of Closed Extension Topology, the closed sets of $T^*_p$ (apart from $S^*_p$) are the closed sets of $T$. Therefore $T^*_p$ does not contain two disjoint closed sets {{iff}} $T$ does not contain two disjoint closed sets. Hence $T^*_p$ is a $T_4$ space vacuously {{iff}} $T$ is a $T_4$ space vacuously. {{qed}}
=== Sufficient Condition === Let $T^*_p$ be a $T_5$ space. Then for any two separated sets $A, B \subseteq S$ there exist disjoint open sets $U, V \in \tau^*_p$ containing $A$ and $B$ respectively. However, for any non-empty set $U \in \tau^*_p$, $p \in U$. Hence no non-empty open sets in $T^*_p$ are separated. Therefore $T^*_p$ is a $T_5$ space vacuously: there do not exist two separated sets. {{qed|lemma}} === Necessary Condition === Let $A, B \in \tau^*_p$. By 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 $U$. Thus for all $A, B \ne \O$ it cannot be the case that $A^- \cap B = A \cap B^- = \O$. Therefore $T^*_p$ does not contain two separated sets {{iff}} $T$ does not contain two separated sets. Hence $T^*_p$ is a $T_5$ space vacuously {{iff}} $T$ is a $T_5$ space vacuously. {{qed}} {{proofread}}
Follows from: :Either-Or Topology is $T_5$ :$T_5$ Space is $T_4$ Space. {{qed}}
From Real Number Line is Second-Countable we have that $\struct {\R, \tau_d}$ is a second-countable space. The result follows from Second-Countable Space is First-Countable. {{qed}}
Let $\UU$ be an open cover of $T^*$. At least one $V \in \UU$ contains $p$. Because $p \notin S$, $V$ is not an open set of $T$. Therefore, by definition of the Alexandroff extension, $V$ must be the complement relative to $S^*$ of a closed, compact subset $\relcomp {S^*} V$ of $T$. Because $\relcomp {S^*} V$ is compact, it can be covered by a finite number of sets in $\UU$. So: :$T^*$ can be covered by $V$ and: :that finite number of sets in $\UU$ are a cover for $\relcomp {S^*} V$. That is, $\UU$ has a finite subcover. Hence the result by definition of compact topological space. {{qed}}
=== Necessary Condition === Let $T = \struct {S, \tau}$ be a $T_1$ space. By definition, $T$ is a $T_1$ space {{iff}} all points of $S$ are closed in $T$. We have that $S$ is open in $T$ by definition of a topology. Thus by definition of the Alexandroff extension, $S$ is open in $T^*$. So as $S = S^* \setminus \set p$ is open in $T^*$, it follows that $\set p$ is closed in $T^*$. That is, $p$ is a closed point of $T^*$. Now let $x \in S^*$ such that $x \ne p$. As $T$ is a $T_1$ space, $\set x$ is closed in $T$. From Finite Topological Space is Compact, $\set x$ is a compact subset of $T$. Thus by definition of the Alexandroff extension, $S^* \setminus \set x$ is open in $T^*$. Thus by definition $\set x$ is closed in $T^*$. That is, $x$ is a closed point of $T^*$. Thus it has been shown that all points in $T^*$ are closed points of $T^*$. Thus by definition $T^*$ is a $T_1$ space. {{qed|lemma}} === Sufficient Condition === Let $T^*$ be a $T_1$ space. By definition of $T$ is a subspace of $T^*$. By $T_1$ Property is Hereditary, it follows that $T$ is a $T_1$ space. {{qed}}
=== Necessary Condition === Let $T = \struct {S, \tau}$ be a locally compact Hausdorff space. Let $x, y \in S$. Then as $T$ is Hausdorff, there exist two disjoint open sets $U, V \in \tau$ containing $x$ and $y$ respectively. But by definition of the Alexandroff extension on $S$, $U$ and $V$ are also open sets of $T^*$. That covers the case where both $x \ne p$ and $y \ne p$. {{WLOG}}, let $y = p$. As $T$ is a locally compact Hausdorff space, $x$ has a compact neighborhood $N_x$. By definition of neighborhood, there exists an open set $U$ of $T$ containing $x$. By definition of the Alexandroff extension, $U$ is an open set of $T^*$. By Compact Subspace of Hausdorff Space is Closed, $N_x$ is closed in $T$. Let $V = S^* \setminus N_x$. By definition of $V$, $x \notin V$ and $p \in V$. By definition of the Alexandroff extension, $V$ is an open set of $T^*$. By Intersection with Complement is Empty iff Subset: :$U \cup V = \O$ So we have two open sets $U, V$ of $T^*$ such that $x \in U, p \in V$ and $U \cup V = \O$. Thus by definition $T^*$ is a $T_2$ space. {{qed|lemma}} === Sufficient Condition === Let $T^*$ be a $T_2$ space. By definition of $T$ is a subspace of $T^*$. By $T_2$ Property is Hereditary, it follows that $T$ is a $T_2$ space. It remains to be shown that $T$ is also (weakly) locally compact. Let $x \in S$. As $T^*$ is a $T_2$ space, we can find. :an open set $U$ of $T^*$ containing $x$ and: :an open set $V$ of $T^*$ containing $p$ such that $U \cap V = \O$. Then by definition of the Alexandroff extension, $S^* \setminus V$ is a compact (closed) neighborhood of $x$. Thus by definition $S$ is a locally compact Hausdorff space. {{qed}}
We have: :Alexandroff Extension is Compact :Compact Hausdorff Space is T4. {{qed}}
From Condition for Alexandroff Extension to be $T_2$ Space, $T^*$ is a $T_2$ space {{iff}} $\struct {\Q, \tau_d}$ is a locally compact Hausdorff Space. But from Rational Number Space is not Locally Compact Hausdorff Space, $\struct {\Q, \tau_d}$ is not a locally compact Hausdorff Space. Hence the result. {{qed}}
From Condition for Alexandroff Extension to be $T_1$ Space, $T^*$ is a $T_1$ space {{iff}} $\struct {\Q, \tau_d}$ is also a $T_1$ space. From Rational Numbers form Metric Space, $\struct {\Q, d}$ is a metric space. From Metric Space fulfils all Separation Axioms, $\struct {\Q, \tau_d}$ is a $T_1$ space. Hence the result. {{qed}}
From Alexandroff Extension of Rational Number Space is $T_1$ Space, $T^*$ is a $T_1$ space. From Alexandroff Extension of Rational Number Space is not Hausdorff, $T^*$ is not a $T_2$ (Hausdorff) space. From Completely Hausdorff Space is Hausdorff Space, $T^*$ is not a $T_{2 \frac 1 2}$ (completely Hausdorff) space. {{ProofWanted|Chain of dependencies which need to be verified when I'm less tired.}}
By definition, $p$ is a dispersion point of $T^*$ {{iff}}: :$\Q^*$ is a connected set in $T^*$ :$\Q^* \setminus \set p$ is totally disconnected in $T^*$. From Alexandroff Extension of Rational Number Space is Connected, $\Q^*$ is a connected set in $T^*$. It remains to be shown that $\Q^* \setminus \set p$ is totally disconnected in $T^*$. By definition, $\Q^* \setminus \set p$ is the rational number space $\struct {\Q, \tau_d}$. The result follows from Rational Numbers are Totally Disconnected. {{qed}}
By definition $T^*$ is a connected space of $T^*$ {{iff}} it admits no separation. {{AimForCont}} $T^*$ does admit a separation. That is, there exist open sets $A, B \in \tau^*$ such that $A, B \ne \O$, $A \cup B = \Q^*$ and $A \cap B = \O$. That is, both $A$ and $B = \relcomp {\Q^*} A$ are open in $T^*$. {{WLOG}}, Let $p \in A$. Then $B$ is a closed and compact subset of $\Q$. From Compact Set of Rational Numbers is Nowhere Dense, $B$ is nowhere dense. From Set is Closed iff Equals Topological Closure, $B$ equals its closure. Thus by definition of nowhere dense, the interior of $B$ is empty. We have that $B$ is open in $T^*$ and $p \notin B$. Thus it follows by definition of Alexandroff extension that $B$ is open in $\struct {\Q, \tau_d}$. But from Interior of Open Set, $B$ equals its interior. This contradicts the deduction above that the interior of $B$ is empty. Hence by Proof by Contradiction $B$ cannot be open in $T^*$. That is, $A$ and $B$ cannot form a separation of $T^*$. Thus $T^*$ is a connected space by definition. {{qed}}
From Particular Point of Alexandroff Extension of Rational Number Space is Dispersion Point, $p$ is a dispersion point of $T^*$. The result follows from Set with Dispersion Point is Biconnected. {{qed}}
The strategy here is to demonstrate that every sequence in $T^*$ is either contained in a compact subspace of $T^*$, or must contain a subsequence which converges to $p$. {{ProofWanted}}
We have that Hilbert Sequence Space is Metric Space. It remains to be shown that it is complete. Recall that from Real Number Line is Complete Metric Space, $\R$ is a complete metric space. Let $x^1, x^2, x^3, \ldots$ be a Cauchy sequence $\ell^2$. Then for each $i \in \N_{>0}$, we have that $\sequence { {x_i}^j}_{j \mathop \in \N_{>0} }$ is a Cauchy sequence in the complete metric space $\R$. Thus $\sequence { {x_i}^j}_{j \mathop \in \N_{>0} }$ converges to a point in $\R$, say $x_i$. Then if $x = \sequence {x_i}$, the points $x - x_j$ eventually belong to $H$. So $x = \paren {x - x_j} + x^j$ must also belong to $H$. Thus $\map {d_2} {x, x^j} = 0$. Hence the result. {{qed}}
Consider the set $H$ of all points of $\ell^2$ which have finitely many rational coordinates and all the rest zero. Then $H$ forms a countable subset of $A$ which is (everywhere) dense. {{finish|Demonstrate that it is (everywhere) dense.}} The result follows by definition of separable space. {{qed}}
From Hilbert Sequence Space is Separable, $\ell^2$ is a separable space. We also have that Hilbert Sequence Space is Metric Space. The result follows from Separable Metric Space is Second-Countable. {{qed}}
From Hilbert Sequence Space is Second-Countable, $\ell^2$ is a second-countable space. The result follows from Second-Countable Space is Lindelöf. {{qed}}
From Metric Space fulfils all Separation Axioms, $\ell^2$ is a Hausdorff space. Let $x = \sequence {x_i} \in A$ be a point of $\ell^2$. From Point in Hilbert Sequence Space has no Compact Neighborhood, $x$ has no compact neighborhood. Hence the result by definition of locally compact Hausdorff space. {{qed}}
From Metric Space fulfils all Separation Axioms, $\ell^2$ is a Hausdorff space. The result follows from Compact Subspace of Hausdorff Space is Closed. {{qed}}
Let $x \in H^\circ$, where $H^\circ$ denotes the interior of $H$. By definition, $H^\circ$ is an open set of $\ell^2$ containing $x$. Again by definition, $H$ is a neighborhood of $x$. But from Point in Hilbert Sequence Space has no Compact Neighborhood, $x$ has no compact neighborhood in $\ell^2$. Thus $H$ cannot be compact in $\ell^2$. {{qed}}
From Hilbert Sequence Space is Metric Space, $\ell^2$ is a metric space. Let $x = \sequence {x_i} \in A$ be a point of $\ell^2$. Consider the closed $\epsilon$-ball of $x$ in $\ell^2$: :$\map { {B_\epsilon}^-} x := \set {y \in A: \map {d_2} {x, y} \le \epsilon}$ for some $\epsilon \in \R_{>0}$. Consider the point: :$\sequence {y_n} = \tuple {x_1, x_2, \ldots, x_{n - 1}, x_n + \epsilon, x_{n + 1}, \ldots}$ We have that $\sequence {y_n} \in \map { {B_\epsilon}^-} x$. But for $m \ne n$ we have that: :$\map {d_2} {y_m, y_n} = \epsilon \sqrt 2$ and so $\sequence {y_n}$ has no convergent subsequence. Thus $\map { {B_\epsilon}^-} x$ is not compact. So $x$ has no compact neighborhood. {{qed}}
By Compact Subset of Hilbert Sequence Space is Closed, $H$ is a closed set of $\ell^2$. From Set is Closed iff Equals Topological Closure: :$H^- = H$ where $H^-$ denotes the closure of $H$. From Subset of Hilbert Sequence Space with Non-Empty Interior is not Compact: :$H^\circ = \O$ where $H^\circ$ denotes the interior of $H$. Thus: :$\left({H^-}\right)^\circ = \O$ and the result follows by definition of nowhere dense. {{qed}}
By Compact Subset of Hilbert Sequence Space is Nowhere Dense, a compact subset of $\ell^2$ is nowhere dense in $\ell^2$. We have that Hilbert Sequence Space is Complete Metric Space. Hence $\ell^2$ is non-meager. {{explain|Find a result that shows a complete metric space is non-meager. S&S seem to imply that it follows.}} It follows that $\ell^2$ is not $\sigma$-compact. {{explain|How?}} {{qed}}
Let $x = \sequence {x_i}$ and $y = \sequence {y_i}$. Consider the mapping $f: \closedint 0 1 \to \ell^2$ defined as: :$\forall t \in \closedint 0 1: \map f t = t x + \paren {1 - t} y = \sequence {t x_i + \paren {1 - t} y_i}$ {{begin-eqn}} {{eqn | l = \sum_{i \mathop \ge 0} \paren {t x_i + \paren {1 - t} y_i}^2       | r = \sum_{i \mathop \ge 0} \paren {t^2 x_i^2 + \paren {1 - t}^2 y_i^2 + 2 t \paren {1 - t} x_i y_i}       | c =  }} {{eqn | r = t^2 \sum_{i \mathop \ge 0} x_i^2 + \paren {1 - t}^2 \sum_{i \mathop \ge 0} y_i^2 + 2 t \paren {1 - t} \sum_{i \mathop \ge 0} x_i y_i       | c =  }} {{end-eqn}} which is convergent. {{explain|Find the appropriate rules to prove the above statement.}} Then $f$ is an injective path joining $x$ to $y$. Hence the result. {{qed}}
From Fréchet Space (Functional Analysis) is Metric Space, $\struct {\R^\omega, d}$ is a metric space. It remains to be demonstrated that $\struct {\R^\omega, d}$ is complete. {{ProofWanted}}
Let $f: M \to \R^\omega$ be the mapping defined as: :$\forall x \in M: \map f x = \sequence {\map d {x, x_i} }$ where $\set {x_i}$ is a countable dense subset of $A$. It remains to be shown that $f$ is a homeomorphism. {{ProofWanted}}
As defined, $M$ is a subspace of the Hilbert sequence space $\ell^2$. We have that Hilbert Sequence Space is Metric Space. The result follows from Subspace of Metric Space is Metric Space. {{qed}}
Let $x = \sequence {x_i}$ be an arbitrary element of $M_1$. Let $f: M_1 \to M_2$ be the mapping defined as: :$\forall x \in M_1: \map f x = \tuple {x_1, 2 x_2, 3 x_3, \ldots}$ Then $f$ is seen to be a bijection. It remains to be shown that an open set in $M_1$ is mapped to an open set in $M_2$ by $f$. {{ProofWanted}}
We have that Hilbert Cube is Metric Space. The result follows from Metric Space is Completely Normal. {{qed}}
Consider the set $H$ of all points of $M$ which have finitely many rational coordinates and all the rest zero. Then $H$ forms a countable subset of $A$ which is (everywhere) dense. {{finish|Demonstrate that it is (everywhere) dense.}} The result follows by definition of separable space. {{qed}}
From Hilbert Cube is Separable, $M$ is a separable space. We also have that Hilbert Cube is Metric Space. The result follows from Separable Metric Space is Second-Countable. {{qed}}
Let $M'$ be the metric space defined as: :$M' = \displaystyle \prod_{k \mathop \in \N} \closedint 0 1$ under the Tychonoff topology. By definition, $\closedint 0 1$ is the closed unit interval under the usual (Euclidean) topology. From Hilbert Cube is Homeomorphic to Countable Infinite Product of Real Number Unit Intervals, $M$ is homeomorphic to $M'$. From Closed Real Interval is Compact in Metric Space, $\closedint 0 1$ is a compact space. But $M'$ is also a compact space by Tychonoff's Theorem. From Compactness is Preserved under Continuous Surjection, if $M'$ is compact then so is $M$. {{qed}} {{expand|Add the direct first-principles proof from S&S as Proof 2.}}
Let $x = \sequence {x_i}$ and $y = \sequence {y_i}$. Consider the mapping $f: \closedint 0 1 \to I^\omega$ defined as: :$\forall t \in \closedint 0 1: \map f t = t x + \paren {1 - t} y = \sequence {t x_i + \paren {1 - t} y_i}$ {{begin-eqn}} {{eqn | l = \sum_{i \mathop \ge 0} \paren {t x_i + \paren {1 - t} y_i}^2       | r = \sum_{i \mathop \ge 0} \paren {t^2 x_i^2 + \paren {1 - t}^2 y_i^2 + 2 t \paren {1 - t} x_i y_i}       | c =  }} {{eqn | r = t^2 \sum_{i \mathop \ge 0} x_i^2 + \paren {1 - t}^2 \sum_{i \mathop \ge 0} y_i^2 + 2 t \paren {1 - t} \sum_{i \mathop \ge 0} x_i y_i       | c =  }} {{end-eqn}} which is convergent. {{explain|Find the appropriate rules to prove the above statement.}} Then $f$ is an injective path joining $x$ to $y$. Hence the result. {{qed}}
By definition, an open ray of $S$ is defined as a subset of $S$ of one of the two forms: :An upward-pointing ray: :: $\left\{{x \in S: s \prec x}\right\}$ (the strict upper closure of $s$), denoted $s^\succ$ :A downward-pointing ray: :: $\left\{{x \in S: x \prec s}\right\}$ (the strict lower closure of $s$), denoted $s^\prec$ from a given point $s \in S$. Thus the set of open rays of $S$ can be expressed as: :the set of all subsets of $S$ of the form: ::$\left\{ {s^\succ: s \in S}\right\}$ and: :the set of all subsets of $S$ of the form: ::$\left\{ {s^\prec: s \in S}\right\}$ This is precisely the set ${\Uparrow} \left({S}\right) \cup {\Downarrow} \left({S}\right)$, where: :${\Uparrow} \left({S}\right) = \left\{{s^\succ: s \in S}\right\}$ :${\Downarrow} \left({S}\right) = \left\{{s^\prec: s \in S}\right\}$ Hence the result, by definition of the topology on $S$ generated by a sub-basis of $S$. {{qed}} Category:Order Topology pz2nj18hygmqx7gsgq00ky02n6kbyti
=== $(1)$ implies $(2)$ === Let $A \subseteq S$ be convex in $S$ by definition 1. Then by definition: :$\forall x, y \in A: \forall z \in S: x \preceq z \preceq y \implies z \in A$ Let $x, y, z \in A$. Let $x \prec z \prec y$. Then by definition: :$x \preceq z \preceq y$ and so by hypothesis $z \in A$. That is: :$\forall x, y \in A: \forall z \in S: x \prec z \prec y \implies z \in A$ Thus $A \subseteq S$ is convex in $S$ by definition 2. {{qed|lemma}} === $(2)$ implies $(1)$ === Let $A \subseteq S$ be convex in $S$ by definition 2. Then by definition: :$\forall x, y \in A: \forall z \in S: x \prec z \prec y \implies z \in A$ Let $x, y, z \in A$. Let $x \preceq z \preceq y$. First suppose $z \ne x$ and $z \ne y$. Then by definition of the relation $\prec$: :$x \prec z \prec y$ and so by hypothesis $z \in A$. Let $x = z$ or $x = y$ We have that $x \in A$ and $y \in A$. So $z \in A$. That is, whether $x \prec z \prec y$ or $z = x$ or $x = y$: :$\forall x, y \in A: \forall z \in S: x \preceq z \preceq y \implies z \in A$ Thus $A \subseteq S$ is convex in $S$ by definition 1. {{qed}} Category:Convex Sets j14llnw892z6x324zvjwpcjz7bdf9e0
There are a number of cases to investigate. === Open Interval === Let $I = \openint a b$ be an open interval: :$I = \set {x \in S: a \prec x \prec b}$ Let $s, t, x \in I$ such that $s \prec x \prec t$. Then by definition: :$a \prec s \prec x$ and: :$x \prec t \prec b$ and so: :$a \prec x \prec b$ and $x \in I$. Thus we have: :$\forall s, t, x, \in I: s \prec x \prec t \implies x \in I$ and $I$ is convex by definition. {{qed|lemma}} === Left Half-Open Interval === Let $I = \hointl a b$ be a left half-open interval: :$I = \set {x \in S: a \prec x \preceq b}$ Let $s, t, x \in I$ such that $s \prec x \prec t$. Then by definition: :$a \prec s \prec x$ and: :$x \prec t \preceq b$ and so: :$a \prec x \preceq b$ and $x \in I$. Thus we have: :$\forall s, t, x, \in I: s \prec x \prec t \implies x \in I$ and $I$ is convex by definition. {{qed|lemma}} === Right Half-Open Interval === Let $I = \hointr a b$ be a right half-open interval: :$I = \set {x \in S: a \preceq x \prec b}$ Let $s, t, x \in I$ such that $s \prec x \prec t$. Then by definition: :$a \preceq s \prec x$ and: :$x \prec t \prec b$ and so: :$a \preceq x \prec b$ and $x \in I$. Thus we have: :$\forall s, t, x, \in I: s \prec x \prec t \implies x \in I$ and $I$ is convex by definition. {{qed|lemma}} === Closed Interval === Let $I = \closedint a b$ be a closed interval: :$I = \set {x \in S: a \preceq x \preceq b}$ Let $s, t, x \in I$ such that $s \prec x \prec t$. Then by definition: :$a \preceq s \prec x$ and: :$x \prec t \preceq b$ and so: :$a \preceq x \preceq b$ and $x \in I$. Thus we have: :$\forall s, t, x, \in I: s \prec x \prec t \implies x \in I$ and $I$ is convex by definition. {{qed}}
Consider the open ray of $S$: :$R = \set {x \in S: a \prec x}$ for some $a \in S$. From Ray is Convex, $R$ is a convex set of $S$. But $R$ is not an interval of $S$. {{qed}}