hcju/proofwikiqa huggingface dataset:
Let $\N$ be defined as the minimal infinite successor set $\omega$. By definition of the ordering on minimal infinite successor set: :$m \le n \iff \begin {cases} m = n & \text {or} \\ m \in n & \end {cases}$ Thus it is sufficient to prove that exactly one of the following is the case: :$(1): \quad m \in n$ :$(2): \quad m = n$ :$(3): \quad n \in m$ The proof operates by induction: For each $n \in \omega$, let $\map S n$ be the set of all $m \in \omega$ which are comparable with $n$. Let $S$ be the set of all $n \in \N$ for which $\map S n = \omega$. Thus the proof is equivalent to demonstrating that $S = \omega$. === Basis for the Induction === First consider $\map S 0$. Clearly $0 \in \map S 0$ as $0 = 0$. Let $m \in \map S 0$. Then $m \notin 0$ as $0 = \O$. So either: :$(1): \quad 0 = m$, in which case $0 \in m^+$ by definition of successor set or: :$(2): \quad 0 \in m$, in which case, because $m \in m^+$ by definition of successor set, again $0 \in m^+$. In all cases, $m \in \map S 0 \implies m^+ \in \map S 0$. So $\map S 0 = \omega$ by induction. This is the basis for the induction. === Induction Hypothesis === The induction hypothesis is that: :$\map S k = \omega$ for some $k \in \omega$. It is now necessary to show that it follows that: :$\map S {k^+} = \omega$ === Induction Step === This is the induction step: Consider the set $\map S {k^+}$, given that $\map S k = \omega$. From the basis for the induction, we have that $k^+ \in \map S 0$. That is, $k^+$ is comparable with $0$. So $0$ is comparable with $k^+$ and so $0 \in \map S {k^+}$. Suppose $m \in \map S {k^+}$. Then either: :$(1): \quad k^+ \in m$, in which case $k^+ \in m^+$ or: :$(2): \quad k^+ = m$, in which case the same applies: $k^+ \in m^+$ or: :$(3): \quad m \in k^+$. In case $(3)$, either: :$(3 \text a): \quad m = k$, in which case $m^+ = k^+$ or: :$(3 \text b): \quad m \in k$. Case $(3 \text b)$ is treated as follows. We have that $m^+ \in \map S k$ by the induction hypothesis. Therefore, either: :$(4 \text a): \quad k \in m^+$ or: :$(4 \text b): \quad k = m^+$ or: :$(4 \text c): \quad m^+ \in k$. $(4 \text a)$ is not compatible with $m \in k$, because either: :$k \in m$ or :$k = m$ and so $k \subseteq m$ which contradicts Finite Ordinal is not Subset of one of its Elements. Both $(4 \text b)$ and $(4 \text c)$ imply that $m^+ \in n^+$. Hence $\map S {k^+} = \omega$. It follows by the Principle of Finite Induction that $S = \omega$. {{qed|lemma}} It then follows from Finite Ordinal is not Subset of one of its Elements that it is not possible for more than one of: :$(1): \quad m \in n$ :$(2): \quad m = n$ :$(3): \quad n \in m$ to be true at the same time. {{qed}}
Let $\N$ be defined as the von Neumann construction $\omega$. By definition of the ordering on von Neumann construction: :$m \le n \iff m \subseteq n$ From Von Neumann Construction of Natural Numbers is Minimally Inductive, $\omega$ is minimally inductive class under the successor mapping. Then from Minimally Inductive Class under Progressing Mapping induces Nest, $\omega$ is a nest in which: :$\forall m, n \in \omega: \map g x \subseteq y \lor y \subseteq x$ The result follows from Class under Progressing Mapping such that Elements are Sandwiched is Nest: :$\forall m, n \in \omega: m \subseteq n \lor n \subseteq m$ {{qed}}
Proof by induction. For all $n \in \N$, let $\map P n$ be the proposition: :$0 \times n = 0 = n \times 0$ === Basis for the Induction === By definition, we have: :$0 \times 0 = 0 = 0 \times 0$ Thus $\map P 0$ is seen to be true. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis $\map P k$: :$0 \times k = 0 = k \times 0$ Then we need to show that $\map P {k + 1}$ follows directly from $\map P k$: :$0 \times \paren {k + 1} = 0 = \paren {k + 1} \times 0$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = 0 \times \paren {k + 1}       | r = \paren {0 \times k} + 0       | c = {{Defof|Natural Number Multiplication}} }} {{eqn | r = 0 + 0       | c = Induction Hypothesis }} {{eqn | r = 0       | c = {{Defof|Natural Number Addition}} }} {{end-eqn}} By definition: :$\paren {k + 1} \times 0 = 0$ So $\map P k \implies \map P {k + 1}$, and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \N: 0 \times n = 0 = n \times 0$ {{qed}}
Checking the three defining properties of an equivalence relation in turn: === Reflexivity === For any Hilbert space $H$, we have the identity map $\operatorname{id}: H \to H$. Identity Mapping is Linear yields $\operatorname{id}$ to be a linear map. Also, Identity Mapping is Bijection ensures that $\operatorname{id}$ is a surjection. Lastly, for any $g, h \in H$, we have: :$\left\langle{g, h}\right\rangle = \left\langle{\operatorname{id} \left({g}\right), \operatorname{id} \left({h}\right)}\right\rangle$ Therefore, the three defining properties for a Hilbert space isomorphism are satisfied by $\operatorname{id}$. Hence, any Hilbert space $H$ is isomorphic to itself. It follows that Hilbert space isomorphism is reflexive. {{qed|lemma}} === Symmetry === Let $H, K$ be isomorphic Hilbert spaces, and let $U: H \to K$ be a Hilbert space isomorphism. From Hilbert Space Isomorphism is Bijection and the definition of bijection, it follows that $U$ has a two-sided inverse, $U^{-1}: K \to H$. Inverse of Linear Map is Linear yields $U^{-1}$ to be a linear map as well. Bijection iff Inverse is Bijection yields that $U^{-1}$ is also a surjection. For $k, l \in K$, the computation $\left\langle{k, l}\right\rangle_K = \left\langle{U U^{-1}k, U U^{-1}l}\right\rangle_K = \left\langle{U^{-1}k, U^{-1}l}\right\rangle_H$ shows that $U^{-1}$ preserves the inner product (where in the last step, the fact that $U$ is an isomorphism is used). Hence, $U^{-1}$ is an isomorphism, so $K$ and $H$ are also isomorphic. It follows that Hilbert space isomorphism is symmetric. {{qed|lemma}} === Transitivity === Let $H, K, L$ be Hilbert spaces, and let $U: H \to K, V: K \to L$ be Hilbert space isomorphisms. Consider the composition $VU : H \to L$. From Composition of Linear Maps is Linear, $VU$ is a linear map. Composite of Surjections is Surjection yields $VU$ to be a surjection. Lastly, for $g, h \in H$, we have: :$\left\langle{g, h}\right\rangle_H = \left\langle{Ug, Uh}\right\rangle_K = \left\langle{VUg, VUh}\right\rangle_L$ Hence $VU$ is a Hilbert space isomorphism, and $H$ and $L$ are isomorphic. It follows that Hilbert space isomorphism is transitive. {{qed|lemma}} Having verified these three conditions, it follows that Hilbert space isomorphism is an equivalence relation,  {{qed}}
Let $m \in n$. Since $n$ is an ordinal, it is transitive. Therefore, it follows directly that $m \subseteq n$. Now let $m \subseteq n$. We have by hypothesis that $m \ne n$. From Natural Numbers are Comparable: Proof using Minimal Infinite Successor Set it follows that either $m \in n$ or $n \in m$. Suppose $n \in m$. Then $\exists x \in m: m \subseteq x$ which contradicts Finite Ordinal is not Subset of one of its Elements. The only option left is that $m \in n$. Hence the result. {{qed}}
Proof by induction: For all $k \in \N$, let $\map P k$ be the proposition: :$m < n \iff m + k < n + k$ $\map P 0$ is true, as this just says $m + 0 = m < n = n + 0$. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P j$ is true, where $j \ge 0$, then it logically follows that $\map P {j^+}$ is true. So this is our induction hypothesis: :$m < n \iff m + j < n + j$ Then we need to show: :$m < n \iff m + j^+ < n + j^+$ === Induction Step === This is our induction step: Let $m < n$. Then: {{begin-eqn}} {{eqn | l = m       | o = <       | r = n       | c =  }} {{eqn | ll= \leadstoandfrom       | l = m       | o = \subsetneq       | r = n       | c = Element of Finite Ordinal iff Subset }} {{eqn | ll= \leadstoandfrom       | l = m^+       | o = \subset       | r = n       | c = {{Defof|Successor Set}} }} {{eqn | ll= \leadstoandfrom       | l = m^+       | o = \subsetneq       | r = n^+       | c = {{Defof|Successor Set}} }} {{eqn | n = 1       | ll= \leadstoandfrom       | l = m^+       | o = <       | r = n^+       | c = Element of Finite Ordinal iff Subset }} {{end-eqn}} This gives: {{begin-eqn}} {{eqn | l = m + j       | o = <       | r = n + j       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \paren {m + j}^+       | o = <       | r = \paren {n + j}^+       | c = from $(1)$ above }} {{eqn | ll= \leadstoandfrom       | l = m + j^+       | o = <       | r = n + j^+       | c = {{Defof|Addition in Minimal Infinite Successor Set}} }} {{end-eqn}} So $\map P j \implies \map P {j^+}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall m, n, k \in \N: m < n \iff m + k < n + k$ {{qed}}
As $U$ is an isomorphism, it is necessarily surjective. Suppose now that $g, h \in H$ are such that $Ug = Uh$. Then as $U$ is a linear map, it follows that $U \left({g - h}\right) = \mathbf{0}_K$, the zero vector of $K$. From property $(3)$ of an isomorphism, conclude that: :$0 = \left\langle{\mathbf{0}_K, \mathbf{0}_K}\right\rangle_K = \left\langle{U \left({g - h}\right), U \left({g - h}\right)}\right\rangle_K = \left\langle{g - h, g - h}\right\rangle_H$ Property $(5)$ of an inner product ensures us that $g - h = \mathbf{0}_H$, i.e. $g = h$. Hence $U$ is injective. As $U$ is both injective and surjective, it is a bijection. {{qed}} Category:Hilbert Spaces avhhbg8lwg75uz4yiuep5vqe4ari45v
{{begin-eqn}} {{eqn | l = \phi \left({\log_a y}\right)       | r = \phi \left({\frac {\ln y}{\ln a} }\right)       | c = Change of Base of Logarithm }} {{eqn | r = \frac 1 {\ln a} \left({\phi \left({\ln y}\right)}\right)       | c = definition of linear operator (as $\frac 1 {\ln a}$ is constant) }} {{end-eqn}} {{qed}} Category:Linear Algebra Category:Logarithms 17whhp0hyp3jtcb4rjwxzv7tncm83vy
Proof by induction: First we note that if $k = 0$ then $m \times k = 0 = n \times k$ whatever $m$ and $n$ are, and the proposition clearly does not hold. So, for all $k \in \N \setminus \set 0$, let $\map P k$ be the proposition: :$m < n \iff m \times k < n \times k$ From Identity Element of Natural Number Multiplication is One: :$m \times 1 = m < n = n \times 1$ and so $\map P 1$ is true. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P j$ is true, where $j \ge 1$, then it logically follows that $\map P {j^+}$ is true. So this is our induction hypothesis: :$m < n \iff m \times j < n \times j$ Then we need to show: :$m < n \iff m \times j^+ < n \times j^+$ === Induction Step === This is our induction step: Let $m < n$. Then: {{begin-eqn}} {{eqn | l = m       | o = <       | r = n       | c =  }} {{eqn | ll= \leadstoandfrom       | l = m \times j       | o = <       | r = n \times j       | c = Induction Hypothesis }} {{eqn | ll= \leadstoandfrom       | l = \paren {m \times j} + m       | o = <       | r = \paren {n \times j} + m       | c = Ordering on Natural Numbers is Compatible with Addition }} {{end-eqn}} Similarly: :$m < n \iff \paren {m \times j} + n < \paren {n \times j} + n$ But then from Ordering on Natural Numbers is Compatible with Addition, we also have: :$m < n \iff \paren {m \times j} + m < \paren {m \times j} + n$ So from Natural Number Ordering is Transitive, we have: :$m < n \iff \paren {m \times j} + m < \paren {n \times j} + n$ This gives: {{begin-eqn}} {{eqn | l = m       | o = <       | r = n       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \paren {m \times j} + m       | o = <       | r = \paren {n \times j} + n       | c = from above }} {{eqn | ll= \leadstoandfrom       | l = m \times j^+       | o = <       | r = n \times j^+       | c = {{Defof|Natural Number Multiplication}} }} {{end-eqn}} So $\map P j \implies \map P {j^+}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall m, n, k \in \N, k \ne 0: m \times n \iff m \times k < n \times k$ {{qed}}
Let $k < m, m < n$. By definition it follows that $k \in m, m \in n$. We have from Element of Finite Ordinal iff Subset that: :$k \in m \iff k \subseteq m$ :$m \in n \iff m \subseteq n$ It follows from Subset Relation is Transitive that $k \subseteq n$. Hence the result. {{qed}} Category:Natural Numbers 3h3h6rhf4et2e2b2hq753efl4aepldy
Proof by induction: For all finite ordinals $n$, let $P \left({n}\right)$ be the proposition: :$x \subsetneq n \implies \exists m \in \N: m < n: m \sim x$ $P \left({0}\right)$ is vacuously true, as there are no proper subsets of $0 = \varnothing$. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k^+}\right)$ is true. So this is our induction hypothesis: :$x \subsetneq k \implies \exists m \in \N: m < k: m \sim x$ Then we need to show: :$x \subsetneq k^+ \implies \exists m \in \N: m < k^+: m \sim x$ === Induction Step === This is our induction step: Let $x \subsetneq k^+$. Then either: :$(1) \quad x \subsetneq k$, in which case the induction hypothesis applies or: :$(2) \quad x = k$, in which case the result is trivially true or: :$(3) \quad k \in x$. In case $(3)$, we find a finite ordinal $j \in k$ such that $j \notin x$. Then we define a mapping $f$ on $x$ as: :$\forall i \in x: f \left({i}\right) = \begin{cases} i & : i \ne k \\ j & : i = k \end{cases}$ Clearly $f$ is injective and $f$ maps $x$ into $k$. {{Questionable|mapping should be defined somehow differently in order to show '''explicitly''' that x is equivalent to some element of k}} So the image of $x$ under $f$ is either equal to $k$  or by the induction hypothesis equivalent to some element of $k$. Consequently, $x$ is always equivalent to some element of $k$. So $P \left({k}\right) \implies P \left({k^+}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \N: x \subsetneq n \implies \exists m \in \N: m < n: m \sim x$ {{qed}}
For a fixed $z_0 \in D$ and $z \in D$ we consider the function: :$\displaystyle \map F z = \int_\gamma \map f w \rd w$ where $\gamma$ is any (simple) contour starting at $z_0$ and ending at $z$. By Primitive of Function on Connected Domain, $F$ is a primitive of $f$. Since $F$ is analytic and $F' = f$, we conclude that $f$ is analytic as well. {{qed}}
=== Necessary Condition === Let $\RR$ be an antisymmetric relation. Let $\tuple {a, b} \in \RR \cap \RR^{-1}$. That means: :$\tuple {a, b} \in \RR$ and :$\tuple {a, b} \in \RR^{-1}$ which means, by definition of inverse relation: :$\tuple {b, a} \in \RR$ But as $\RR$ is antisymmetric, that means $a = b$. Thus: :$\tuple {a, b} = \tuple {a, a}$ and so: :$\tuple {a, b} \in \Delta_S$ where $\Delta_S$ is the diagonal relation. Thus from the definition of subset: :$\RR \cap \RR^{-1} \subseteq \Delta_S$ Hence, by definition, $\RR$ is coreflexive. {{qed|lemma}} === Sufficient Condition === Let $\RR \cap \RR^{-1}$ be coreflexive. Hence by definition of coreflexive: :$\RR \cap \RR^{-1} \subseteq \Delta_S$ Let $\tuple {a, b} \in \RR$ and $\tuple {b, a} \in \RR$. That is, by definition of inverse relation: :$\tuple {a, b} \in \RR$ and :$\tuple {a, b} \in \RR^{-1}$ That is: :$\tuple {a, b} \in \RR \cap \RR^{-1}$ But as $\RR \cap \RR^{-1} \subseteq \Delta_S$ it follows that $a = b$. So by definition $\RR$ is antisymmetric. {{Qed}}
By the definition of the min operation, either: :$\min \left({x, y}\right) = x$ or :$\min \left({x, y}\right) = y$ So $\min$ is closed on $S$. Then we have that the min operation is associative: : $\forall x, y, z \in S: \min \left({x, \min \left({y, z}\right)}\right) = \min \left({\min \left({x, y}\right), z}\right)$ Hence the result, by definition of semigroup. {{qed}}
Let $x, y \in S$.  We have that the max operation is commutative: : $\max \left({x, y}\right) = \max \left({y, x}\right)$  Hence the result, by definition of commutative semigroup. {{qed}}
=== Necessary Condition === Let $\mathcal R$ be a connected relation. By definition of relation: : $\mathcal R \subseteq S \times S$ : $\mathcal R^{-1} \subseteq S \times S$ : $\Delta_S \subseteq S \times S$ So from Union is Smallest Superset (and indeed, trivially): : $\mathcal R \cup \mathcal R^{-1} \cup \Delta_S \subseteq S \times S$ Let $\left({a, b}\right) \in S \times S$. Suppose $a = b$. Then $\left({a, b}\right) \in \Delta_S$ by definition of the diagonal relation. Then by Set is Subset of Union: :$\left({a, b}\right) \in \mathcal R \cup \mathcal R^{-1} \cup \Delta_S$ Suppose otherwise, that is, that $a \ne b$. As $\mathcal R$ is connected, either: :$\left({a, b}\right) \in \mathcal R$ or: :$\left({b, a}\right) \in \mathcal R$ From the definition of inverse relation, either: :$\left({a, b}\right) \in \mathcal R$ or: :$\left({a, b}\right) \in \mathcal R^{-1}$ That is: :$\left({a, b}\right) \in R \cup \mathcal R^{-1}$ Again by Set is Subset of Union: :$\left({a, b}\right) \in \mathcal R \cup \mathcal R^{-1} \cup \Delta_S$ So, by definition of subset: :$S \times S \subseteq \mathcal R \cup \mathcal R^{-1} \cup \Delta_S$ Hence, by definition of set equality: :$\mathcal R \cup \mathcal R^{-1} \cup \Delta_S = S \times S$ {{qed|lemma}} === Sufficient Condition === Let $\mathcal R \cup \mathcal R^{-1} \cup \Delta_S = S \times S$. Let $\left({a, b}\right) \in S \times S$. Suppose $a \ne b$. By definition of diagonal relation: :$\left({a, b}\right) \notin \Delta_S$ So, by definition of set union: :$\left({a, b}\right) \in \mathcal R$ or: :$\left({a, b}\right) \in \mathcal R^{-1}$ That is, by definition of inverse relation: :$\left({a, b}\right) \in \mathcal R$ or: :$\left({b, a}\right) \in \mathcal R$ So by definition $\mathcal R$ is connected. {{Qed}}
First of all, note that Absolutely Convergent Generalized Sum Converges assures us that the left expression in above equality is defined. Suppose that there exists an $\epsilon > 0$ such that $\displaystyle \left\Vert{\sum \left\{{v_i: i \in I}\right\}}\right\Vert > \sum \left\{{\left\Vert{v_i}\right\Vert: i \in I}\right\} + \epsilon$. This supposition is seen to be equivalent to $\displaystyle \left\Vert{\sum \left\{{v_i: i \in I}\right\}}\right\Vert > \sum \left\{{\left\Vert{v_i}\right\Vert: i \in I}\right\}$. {{explain|Expand below sentence to be more clear}} Then, by definition of a generalized sum, there necessarily exists a finite subset $F$ of $I$ with: :$\displaystyle \left\Vert{\sum_{i \in F} v_i}\right\Vert > \left\Vert{\sum \left\{{v_i: i \in I}\right\}}\right\Vert - \epsilon > \sum \left\{{\left\Vert{v_i}\right\Vert: i \in I}\right\}$ However, using the standard triangle equality on this ''finite'' sum (i.e., axiom '''N3''' for a norm, repetitively), we also have: :$\displaystyle \left\Vert{\sum_{i \in F} v_i}\right\Vert \le \sum_{i \in F} \left\Vert{v_i}\right\Vert \le \sum \left\{{\left\Vert{v_i}\right\Vert: i \in I}\right\}$ Here the second inequality follows from Generalized Sum is Monotone. These two estimates constitute a contradiction, and therefore such an $\epsilon$ cannot exist. Hence $\displaystyle \left\Vert{\sum \left\{{v_i: i \in I}\right\}}\right\Vert \le \sum \left\{{\left\Vert{v_i}\right\Vert: i \in I}\right\}$. {{qed}} Category:Generalized Sums Category:Banach Spaces Category:Triangle Inequality abiohhjxt7y574z92vr2w74cszc9tgy
Denote, for $\lambda \in \R$, $I_{> \lambda} := \left\{ {i \in I: a_i > \lambda}\right\}$. Then as $\displaystyle \sum \left\{ {a_i: i \mathop \in I}\right\}$ converges, necessarily all of the sets $I_{> \frac 1 n}$ are finite. However, we have that $\displaystyle I_{>0} = \bigcup_{n \mathop = 1}^\infty I_{> \frac 1 n}$. From Countable Union of Countable Sets is Countable, it follows that $I_{>0}$ is countable. {{qed}} Category:Generalized Sums Category:Real Analysis 5tigvqux50ppkhvjgo0z8x8uoehpxf5
=== Inverse of Natural Logarithm implies Limit of Sequence === {{:Equivalence of Definitions of Real Exponential Function/Inverse of Natural Logarithm implies Limit of Sequence}} === Limit of Sequence implies Sum of Series === {{:Equivalence of Definitions of Real Exponential Function/Limit of Sequence implies Sum of Series}} === Limit of Sequence implies Extension of Rational Exponential === {{:Equivalence of Definitions of Real Exponential Function/Limit of Sequence implies Extension of Rational Exponential}} === Extension of Rational Exponential implies Solution of Differential Equation === {{:Equivalence of Definitions of Real Exponential Function/Extension of Rational Exponential implies Differential Equation}} === Sum of Series equivalent to Solution of Differential Equation === {{:Equivalence of Definitions of Real Exponential Function/Sum of Series equivalent to Differential Equation}} === Inverse of Natural Logarithm equivalent to Solution of Differential Equation === {{:Equivalence of Definitions of Real Exponential Function/Inverse of Natural Logarithm equivalent to Differential Equation}} <!-- == Proof 2 == {{:Equivalence of Definitions of Real Exponential Function/Proof 2}} -->
From Derivative of Exponential Function and Exponential of Zero, each definition of $\exp$ satisfies the following: :$ (1): \quad D_x \exp = \exp$ :$ (2): \quad \exp \left({ 0 }\right) = 1$ on $\R$. From  Exponential Function is Well-Defined, such a solution is  unique. Thus they all are all  equivalent. {{qed}}
Let $x \in \operatorname{Dom} \left({f}\right)$ such that $f \left({x}\right) = y$. Then by definition $x \in \operatorname{Dom} \left({g}\right)$ and $g \left({x}\right) = y$. Thus by definition of subset, $f \subseteq g$. We have that Subset Relation is Ordering. Hence the result. {{qed}}
From Subset Relation is Ordering, we know that $\subseteq$ is an ordering, hence also a preordering. Now let $F, G \in \mathcal F$. From Set Union Preserves Subsets, conclude that $F \cup G \subseteq I$ as $F, G \subseteq I$. From Cardinality of Set Union, $\card {F \cup G} \le \card F + \card G$, implying that $F \cup G$ is a finite set. Hence $F \cup G \in \mathcal F$. Furthermore, $F \subseteq F \cup G$ and $G \subseteq F \cup G$. It follows that $\struct {\mathcal F, \subseteq}$ is a directed set. {{qed}} Category:Preorder Theory Category:Set Theory 8i9hubqhfeanlj598gtqtiraigcss6g
We are given that $\left({S, \preceq}\right)$ is an ordered set. === Antireflexive === Follows immediately: :$\forall a \in S: a = a \implies a \not \prec a$ {{explain}} {{qed|lemma}} === Transitive === Suppose $a, b, c \in S$ such that $a \preceq b, b \preceq c$. $a \preceq b \land b \preceq c \implies a \preceq c$ from transitivity of $\preceq$. Now suppose $a \preceq b, b \preceq c, a = c$. Then $a \preceq b$ but because of the antisymmetry of $\preceq$, it is not then possible for $b \preceq c$. So a condition for transitivity to be violated will not arise in this circumstance. If either $a = b$ or $b = c$, then $a \not \prec b$ and $b \not \prec c$ by antireflexivity of $\prec$. So $\prec$ is shown to be transitive. {{qed|lemma}} === Asymmetric === A direct application of Antireflexive and Transitive Relation is Asymmetric. Hence the result. {{qed}} Let $a \preceq b$. Either $a = b$ or $a \ne b$ by the Law of Excluded Middle. If $a = b$ we are done. Suppose $a \ne b$. Then it follows that $a \prec b$. Thus $a \preceq b \implies (a = b) \lor (a \prec b)$. {{qed}} Now let $(a = b) \lor (a \prec b)$. If $a = b$ then $a \preceq b$ by the reflexivity of $\preceq$. If $a \prec b$ then $a \preceq b$ by definition. {{qed}}
Let $a$ and $b$ both be smallest elements of $S$. Then by definition: :$\forall y \in S: a \preceq y$ :$\forall y \in S: b \preceq y$ Thus it follows that: :$a \preceq b$ :$b \preceq a$ But as $\preceq$ is an ordering, it is antisymmetric. Hence by definition of antisymmetric, $a = b$. {{qed}}
Let $a$ and $b$ both be greatest elements of $S$. Then by definition: :$\forall y \in S: y \preceq a$ :$\forall y \in S: y \preceq b$ Thus it follows that: :$b \preceq a$ :$a \preceq b$ But as $\preceq$ is an ordering, it is antisymmetric. Hence, by definition of antisymmetric relation, $a = b$. {{qed}}