hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | l = \paren {R \cup S} \setminus T       | r = \paren {R \cup S} \cap \overline T       | c = Set Difference as Intersection with Complement }} {{eqn | r = \paren {R \cap \overline T} \cup \paren {S \cap \overline T}       | c = Intersection Distributes over Union }} {{eqn | r = \paren {R \setminus T} \cup \paren {S \setminus T}       | c = Set Difference as Intersection with Complement }} {{end-eqn}} {{qed}} Category:Set Difference Category:Set Union Category:Distributive Operations 5iyvn5911vuketj7ydftdxfh554w9sg
Consider $R, S, T \subseteq \mathbb U$, where $\mathbb U$ is considered as the universe. {{begin-eqn}} {{eqn | l = R \setminus \paren {S \setminus T}       | r = R \cap \overline {\paren {S \cap \overline T} }       | c = Set Difference as Intersection with Complement }} {{eqn | r = R \cap \paren {\overline S \cup \overline {\paren {\overline T} } }       | c = De Morgan's Laws }} {{eqn | r = R \cap \paren {\overline S \cup T}       | c = Complement of Complement }} {{eqn | r = \paren {R \cap \overline S} \cup \paren {R \cap T}       | c = Intersection Distributes over Union }} {{eqn | r = \paren {R \setminus S} \cup \paren {R \cap T}       | c = Set Difference as Intersection with Complement }} {{end-eqn}} {{qed}}
Consider $R, S, T \subseteq \mathbb U$, where $\mathbb U$ is considered as the universe. {{begin-eqn}} {{eqn | l = R \setminus S       | r = R \cap \overline S       | c = Set Difference as Intersection with Complement }} {{eqn | r = \left({R \cap \overline S}\right) \cap \mathbb U       | c = Intersection with Universe }} {{eqn | r = \left({R \cap \overline S}\right) \cap \left({\overline T \cup T}\right)       | c = Union with Complement }} {{eqn | r = \left({R \cap \overline S \cap \overline T}\right) \cup \left({T \cap R \cap \overline S}\right)       | c = Intersection Distributes over Union }} {{eqn | o = \subseteq       | r = \left({R \cap \overline T}\right) \cup \left({T \cap \overline S}\right)       | c = Union of Intersections }} {{eqn | r = \left({R \setminus T}\right) \cup \left({T \setminus S}\right)       | c = Set Difference as Intersection with Complement }} {{end-eqn}} {{qed}} Category:Set Difference Category:Set Union cg8p0rvsy0xxxyzekbtn94m4ovyc2dd
By definition of ordered sum, we have that: * If $a, b \in S$, then $a \preceq b \iff a \preceq_1 b$. * Otherwise, if $a, b \in T$, then $a \preceq b \iff a \preceq_2 b$. * If neither of these is the case, then $a \in S, b \in T \iff a \preceq b$. First we show that $\preceq$ is connected. We note that as $\left({S, \preceq_1}\right)$ and $\left({T, \preceq_2}\right)$ are both tosets, then $\preceq_1$ and $\preceq_2$ are both connected. Let $a, b \in S \cup T$. There are three cases: : $(1): \quad a, b \in S$, in which case $a \preceq_1 b$ or $b \preceq_1 a$ : $(2): \quad a, b \notin S$, in which case $a \preceq_2 b$ or $b \preceq_2 a$ : $(3): \quad$ Otherwise, in which case either $a \in S, b \notin S$ or $a \notin S, b \in S$. :::By definition of $\preceq$ it follows that $a \preceq b$ or $b \preceq a$, but clearly not both. So in all cases $a \preceq b$ or $b \preceq a$, and $\preceq$ is connected. Now we check in turn each of the criteria for an ordering: === Reflexivity === We have by definition of set union: :$\forall a \in S \cup T: a \in S \lor a \in T$ and so either $a \preceq_1 a$ or $a \preceq_2 a$, showing $a \preceq a$. So $\preceq$ is reflexive. {{qed|lemma}} === Transitivity === Suppose $a \preceq b \preceq c$. We proceed on a case-by-case basis. There are eight of these, (indirectly) from Cardinality of Power Set of Finite Set. * Suppose $a, b, c \in S$. Then $a \preceq c$ by transitivity of $\preceq_1$. * Suppose $a, b \in S, c \notin S$. Then $a \preceq c$ by definition of $\preceq$. * Suppose $a \in S, b, c \notin S$. Then $a \preceq c$ by definition of $\preceq$. * Suppose $a, b, c \notin S$. Then $a \preceq c$ by transitivity of $\preceq_2$. There are four other cases to consider: * $a, c \in S, b \notin S$ * $b, c \in S, a \notin S$ * $c \in S, a, b \notin S$ * $b \in S, a, c \notin S$. None of these can happen as otherwise one of $a \npreceq b$ or $b \npreceq c$ would be the case. So we have shown that $\preceq$ is transitive. {{qed|lemma}} === Antisymmetry === Suppose $a \preceq b$ and $b \preceq a$. * It can not be the case that $a \in S$ and $b \notin S$ because then $b \npreceq a$. * It can not be the case that $a \notin S$ and $b \in S$ because then $a \npreceq b$. So either $a, b \in S$ or $a, b \notin S$. * If $a, b \in S$ then $a = b$ by antisymmetry of $\preceq_1$. * If $a, b \notin S$ then $a = b$ by antisymmetry of $\preceq_2$. Thus in all cases it can be seen that $a \preceq b \preceq a \implies a = b$. So $\preceq$ is antisymmetric. {{qed|lemma}} So we have shown that: * $\preceq$ is connected * $\preceq$ is reflexive, transitive and antisymmetric. Thus by definition, $\preceq$ is a total ordering and so $\left({S \cup T, \preceq}\right)$ is a toset. {{qed}}
By definition of ordered product, we have that: :$b_1 \prec_2 b_2 \implies \tuple {a_1, b_1} \prec \tuple {a_2, b_2}$ :$b_1 = b_2, a_1 \prec_1 a_2 \implies \tuple {a_1, b_1} \prec \tuple {a_2, b_2}$ :$b_1 = b_2, a_1 = a_2 \implies \tuple {a_1, b_1} = \tuple {a_2, b_2}$ We note that as $\struct {S_1, \preceq_1}$ and $\struct {S_2, \preceq_2}$ are both tosets, then $\preceq_1$ and $\preceq_2$ are both connected. Thus it is clear that $\preceq$ is connected. Now we check in turn each of the criteria for an ordering: === Reflexivity === We have by definition of set union: :$\forall \tuple {a, b} \in S_1 \times S_2: a = a \land b = b$ and so $\tuple {a, b} = \tuple {a, b}$. Thus $\tuple {a, b} \preceq \tuple {a, b}$ and so $\preceq$ is reflexive. {{qed|lemma}} === Transitivity === Suppose $\tuple {a_1, b_1} \preceq \tuple {a_2, b_2} \preceq \tuple {a_3, b_3}$. Suppose $\tuple {a_3, b_3} \prec \tuple {a_1, b_1}$. This would happen because: :$b_3 \prec_2 b_1$, which can't happen because $\preceq_2$ is transitive; :$b_3 = b_1$ and $a_3 \prec_1 a_1$, which can't happen because $\preceq_1$ is transitive. So we have shown that $\tuple {a_1, b_1} \preceq \tuple {a_3, b_3}$ and so $\preceq$ is transitive. {{qed|lemma}} === Antisymmetry === Suppose $\tuple {a_1, b_1} \preceq \tuple {a_2, b_2} \preceq \tuple {a_1, b_1}$. Suppose $\tuple {a_1, b_1} \ne \tuple {a_2, b_2}$. Then one of two cases holds: :$b_1 \ne b_2$, which can't happen because then either $\tuple {a_1, b_1} \prec \tuple {a_2, b_2}$ or $\tuple {a_2, b_2} \prec \tuple {a_1, b_1}$; :$b_1 = b_2$, and $a_1 \ne a_2$, which can't happen because then either $\tuple {a_1, b_1} \prec \tuple {a_2, b_2}$ or $\tuple {a_2, b_2} \prec \tuple {a_1, b_1}$. Thus in all cases it can be seen that $\tuple {a_1, b_1} \preceq \tuple {a_2, b_2} \preceq \tuple {a_1, b_1} \implies \tuple {a_1, b_1} = \tuple {a_2, b_2}$. So $\preceq$ is antisymmetric. {{qed|lemma}} So we have shown that: :$\preceq$ is connected :$\preceq$ is reflexive, transitive and antisymmetric. Thus by definition, $\preceq$ is a total ordering and so $\struct {S_1 \times S_2, \preceq}$ is a toset. {{qed}}
As $f: S \to T$ is a mapping, we have that: :$\forall x \in S: \tuple {x, y_1} \in f \land \tuple {x, y_2} \in f \implies y_1 = y_2$ From the definition of a subset, $x \in X \implies x \in S$, and so: :$\forall x \in X: \tuple {x, y_1} \in f \restriction_X \land \tuple {x, y_2} \in f \restriction_X \implies y_1 = y_2$ Also from the definition of a mapping: :$\forall x \in S: \exists y \in T: \tuple {x, y} \in f$ Again from the definition of a subset, $x \in X \implies x \in S$, and so: :$\forall x \in X: \exists y \in T: \tuple {x, y} \in f \restriction_X$ So $f \restriction_X: X \to T$ is a mapping. The fact that the domain of $f \restriction_X$ is $X$ follows from the definition of domain. The fact that the preimage of $f \restriction_X$ is also $X$ follows from Preimage of Mapping equals Domain. {{qed}}
Let $y \in f \sqbrk X$. Then by definition of the image of a subset: :$\exists x \in X: \map f x = y$ or equivalently: :$\exists x \in X: \tuple {x, y} \in f$ But then by definition of restriction of $f$ to $X$: :$f {\restriction_X} = \set {\tuple {x, y} \in f: x \in X}$ Thus by definition of the image set of $f {\restriction_X}$: :$\Img {f {\restriction_X} } = f {\restriction_X} \sqbrk X = \set {y \in T: \exists x \in X: \tuple {x, y} \in f {\restriction_X} }$ Hence it can be seen that: :$y \in \Img {f {\restriction_X} }$ So: :$f \sqbrk X \subseteq \Img {f {\restriction_X} }$ {{qed|lemma}} Now suppose that $y \in \Img {f {\restriction_X} }$. Then by definition of the image set of $f {\restriction_X}$: :$\exists x \in S: \tuple {x, y} \in f {\restriction_X}$ By definition of restriction of $f$ to $X$: :$x \in X$ Thus by definition of image of a subset: :$y \in f \sqbrk X$ So: :$\Img {f {\restriction_X} } \subseteq f \sqbrk X$ {{qed|lemma}} We have: :$f \sqbrk X \subseteq \Img {f {\restriction_X} }$ :$\Img {f {\restriction_X} } \subseteq f \sqbrk X$ The result follows by definition of set equality. {{qed}}
Let $\displaystyle B_i: i = 1, 2, \ldots, \binom n x$ be events such that: :$(1): \quad B_i$ is the $i$th possible way to see $x$ successes in $n$ Bernoulli trials :$(2): \quad \forall i \ne j: B_i \cap B_j = \varnothing$ We can see that: :$\forall i: \Pr \left({B_i}\right) = p^x \left({1-p}\right)^{n-x}$ This is true since there will be $x$ successes, each with probability $p$ of occurring, and $n-x$ failures each with probability $1-p$ of occurring. Furthermore we can assume independent trials and thus the result follows. See Bernoulli Process as Binomial Distribution for further analysis of this. Now our task becomes finding: :$\displaystyle \Pr \left({X = x}\right) = \Pr \left({\bigcup_{i \mathop = 1}^{\binom n x} B_i}\right)$ which is the probability of one of the $\binom n x$ outcomes occurring. Then by the Inclusion-Exclusion Principle considered as an extension of the Addition Law of Probability we have that for any countable union of events: :$\displaystyle \Pr \left({\bigcup_{i \mathop = 1}^n A_i}\right) = \sum_{i \mathop = 1}^n \Pr \left({A_i}\right) - \sum_{i \mathop \ne j: i, j \mathop = 1}^n \Pr \left({A_i \cap A_j}\right) - \Pr \left({\bigcap_{i \mathop = 1}^n A_i}\right)$ Fortunately in this case the above reduces to: :$\displaystyle \Pr \left({\bigcup_{i \mathop = 1}^n A_i}\right) = \sum_{i \mathop = 1}^n \Pr \left({A_i}\right)$ since the events are pairwise disjoint and $\Pr \left({\varnothing}\right) = 0$. Thus: {{begin-eqn}} {{eqn | l=\Pr \left({X = x}\right)       | r=\Pr \left({\bigcup_{i \mathop = 1}^{\binom n x} B_i}\right)       | c= }} {{eqn | r=\sum_{i \mathop = 1}^{\binom n x} \Pr \left({B_i}\right)       | c= }} {{eqn | r=\sum_{i \mathop = 1}^{\binom n x} p^x \left({1-p}\right)^{n-x}       | c= }} {{eqn | r=\binom n x p^x \left({1-p}\right)^{n-x}       | c= }} {{end-eqn}} {{Qed}} <!-- Our last task is to show that the PMF sums to unity. {{begin-eqn}} {{eqn | l=\sum_{x \mathop = 0}^{n}\Pr\left({X = x}\right)       | r=\sum_{x \mathop = 0}^{n}\binom n x p^x \left({1-p}\right)^{n-x}       | c= }} {{eqn | r=\left({p + \left({1 - p}\right)}\right)^n       | c= }} {{eqn | c=       | r= 1^n = 1 }} {{end-eqn}} --> Category:Binomial Distribution fgd8k86tv6aaif3715p8cwc6ypxoyh4
Let $S$ be the set of all chains of $\mathcal P$. $S \ne \varnothing$ since the empty set is an element of $S$.  From Subset Relation is Ordering, we have that $\left({S, \subseteq}\right)$ is partially ordered by inclusion. {{link wanted|Determine the correct link to 'partial ordering'}} Let $C$ be a totally ordered subset of $\left({S, \subseteq}\right)$. Then $\bigcup C$ is a chain in $C$ by Set of Chains is Chain Complete for Inclusion. This shows that $S$, ordered by set inclusion, is an inductive ordered subset. By applying Zorn's Lemma, the result follows. {{qed}}
Let $\preceq$ be an ordering on the set $\mathcal P$. Let $X$ be a chain in $\left({\mathcal P, \preceq}\right)$. By definition, a maximal chain in $\mathcal P$ that includes $X$ is a chain $Y$ in $\mathcal P$ such that $X \subseteq Y$ and there is no chain $Z$ in $\mathcal P$ with $X \subseteq Z$ and $Y \subsetneq Z$. Let us define $\mathcal C$ as: :$\mathcal C = \left\{{Y : Y}\right.$ is a chain in $\mathcal P$ and $\left.{X \subseteq Y}\right\}$ Then: * a maximal chain in $\mathcal P$ that includes $X$ and * a maximal element of $\mathcal C$ under the partial order induced on $\mathcal C$ by inclusion are one and the same. It thus suffices to show that $\mathcal C$ contains such a maximal element. According to Zorn's Lemma, $\mathcal C$ contains a maximal element if each chain in $\mathcal C$ has an upper bound in $\mathcal C$. Let $W$ be a chain in $\mathcal C$, and let $Z = \bigcup W$. Let $a, b \in Z$. Then $\exists A, B \in W: a \in A, b \in B$. Since $W$ is a chain in $\mathcal C$, one of $A$ and $B$ includes the other. Suppose WLOG $A \subseteq B$. Then $a, b \in B$. Since $B$ is a chain in $\mathcal P$, either $a \preceq b$ or $b \preceq a$. Therefore $Z$ is a chain in $\mathcal P$. By definition of $Z$, we have that $\forall A \in W: A \subseteq Z$. Thus $Z$ is an upper bound for $W$ under inclusion. Finally we have that: :$\forall A \in W: X \subseteq A$ and so $X \subseteq Z$. Thus $Z \in \mathcal C$ {{qed}}
Let $\left({\mathcal C, \subseteq}\right)$ be the set of all chains in $P$ ordered by inclusion.  By Set of Chains is Chain Complete for Inclusion, $\mathcal C$ is a chain complete ordered set. Now define $f: \mathcal C \to \mathcal C$ as follows: : If $C$ is a maximal chain then $f \left({C}\right) = C$. : Otherwise $f$ chooses arbitrarily, using the axiom of choice, some chain $D$ which strictly contains $C$. By construction, $f \left({C}\right) \supseteq C$. By the above, $\mathcal C$ is chain complete. Therefore the Bourbaki-Witt Fixed Point Theorem applies and $f$ has a fixed point $F \left({M}\right) = M$. But by the above construction, the only fixed points of $f$ are maximal chains. Therefore $M$ is a maximal chain. {{qed}}
First we note that, from the Law of Identity: : $P \models P$ Suppose $U \models P$. So we have: * $U \models P$ * $P \models P$ So by definition of model: : $U \cup P \models P$ It also follows that if $U \cup P \models P$, then: * $U \models P$ * $P \models P$ Thus we have shown that $U \models P$ iff $U \cup P \models P$. {{qed}} Category:Propositional Logic 18u0q1g3rrl1c1ypn3yirbp7dytwi45
=== Sufficient Condition === Suppose that $U \models P$. Let $\mathcal M$ be a model such that: :$\mathcal M \models U \cup \left\{{\neg P}\right\}$ {{explain|This is not what Logical Consequence with Union says, but perhaps is indicative of what it should say}} Then from Logical Consequence with Union, we have that: :$\mathcal M \models U$ and by definition of semantic consequence: :$\mathcal M \models P$ So we have that $\mathcal M \models P$ and $\mathcal M \models \neg P$ By definition of contradiction, it follows that there can be no such model. So $U \models P$ is a sufficient condition for $U \cup \left\{{\neg P}\right\}$ to have no models. {{qed|lemma}} === Necessary Condition === Suppose that $U \cup \left\{{\neg P}\right\}$ has no models. There are the following possibilities: * $\neg P$ has no models, in which case it is itself a contradiction. Hence from Tautology and Contradiction, $P$ is a tautology. Hence $P$ is true under every model. Hence whatever $U$ may be, every model of $U$ is also a model of $P$: :$U \models P$ * $U$ has no models, in which case every model of $U$ is also a model of $P$ vacuously, and so: :$U \models P$ * There exists at least one model of $U$, but each one does not model $\neg P$. Let $\mathcal M$ be such a model. We have that: :$\mathcal M \not \models \neg P$ so by definition of negation: :$\mathcal M \models P$ Thus $\mathcal M \models U$ implies that $\mathcal M \models P$. Hence by definition: :$U \models P$ In all cases $U \models P$. So $U \models P$ is a necessary condition for $U \cup \left\{{\neg P}\right\}$ to have no models. {{qed}}
Let $\mathbb S = \set S$. Then from the definition of set union: :$\displaystyle \bigcup \mathbb S = \set {x: \exists X \in \mathbb S: x \in X}$ from which it follows directly that: :$\displaystyle \bigcup \mathbb S = \set {x: x \in S}$ as $S$ is the only set in $\mathbb S$. That is: :$\displaystyle \bigcup \mathbb S = S$ {{qed}}
Let $\mathbb S = \O$. Then from the definition: :$\displaystyle \bigcup \mathbb S = \set {x: \exists X \in \mathbb S: x \in X}$ from which it follows directly: :$\displaystyle \bigcup \mathbb S = \O$ as there are no sets in $\mathbb S$. {{qed}}
Let $\mathbb S = \O$. Then from the definition: :$\displaystyle \bigcap \mathbb S = \set {x: \forall X \in \mathbb S: x \in X}$ Consider any $x \in \mathbb U$. Then as $\mathbb S = \O$, it follows that: :$\forall X \in \mathbb S: x \in X$ from the definition of vacuous truth. It follows directly that: :$\displaystyle \bigcap \mathbb S = \set {x: x \in \mathbb U}$ That is: :$\displaystyle \bigcap \mathbb S = \mathbb U$ {{qed}}
From the nature of the identity mapping, we have: :$I_S \circ I_S = I_S$ from which it follows by definition that $I_S$ is the inverse of itself. {{Qed}}
The mapping $q_{\RR_f}: S \to S / \RR_f$ follows from the definition of quotient mapping. The mapping $h$ is justified by Condition for Mapping from Quotient Set to be Well-Defined. {{qed}}
Let $k \in \N$. Let $j \ge k$. Then by as many applications as necessary of Subset Relation is Transitive, we have: :$S_k \subseteq S_j$ Now $s \in S$ means, by definition of set union, that: :$\exists S_k \subseteq S: s \in S_k$ Then from above: :$j \ge k \implies S_k \subseteq S_j$ it follows directly that: :$\forall s \in S: \exists k \in \N: \forall j \ge k: s \in S_j$ from the definition of subset. {{qed}}
Suppose $f: S \to T$ and $g: S \to T$ are both extensions of $f_i$ to $S$ for all $i \in \N$ such that $f \ne g$. We have been given the domain and codomain of $f$. Suppose there exists any $g: S \to T$ which is different from $f$. This must be because $f$ and $g$ do not agree throughout their entire domain. Thus: :$\exists x \in S: f \left({x}\right) \ne g \left({x}\right)$ Then by definition of set union: :$\exists S_k \subseteq S: x \in S_k$. From the definition, both $f$ and $g$ are extensions of $S_k$. But this means they must agree on $S_k$. That is: :$\forall x \in S_k: f \left({x}\right) = g \left({x}\right)$ This contradicts the supposition that $f \left({x}\right) \ne g \left({x}\right)$. Thus: :$\forall x \in S: f \left({x}\right) = g \left({x}\right)$ and $f$ has been shown to equal $g$. Hence the result. {{qed}}
Let $w = (n [n - 1])$, that is, the transposition of the $n^{th}$ and $n - 1^{th}$ letters that we consider $A_n$ acting on. Then the permutation $x^{-1} w$ is the $y$ of the theorem. {{qed}}
Taking the group axioms in turn: === $\text G 0$: Closure === From the Sum Rule for Continuous Functions, if $f$ and $g$ are continuous real functions then so is $f + g$. Thus closure is demonstrated. {{qed|lemma}} === $\text G 1$: Associativity === Pointwise Addition is Associative. {{qed|lemma}} === $\text G 2$: Identity === The constant function $f_0$ defined as: :$\forall x \in \R: \map {f_0} x = 0$ fulfils the role of the Identity: :$\forall x \in \R: \map {f_0} x + \map f x = 0 + \map f x = \map f x = \map f x + 0 = \map f x = \map {f_0} x$ Note that $f_0 \in C$ as the Constant Function is Uniformly Continuous, and hence continuous. {{qed|lemma}} === $\text G 3$: Inverses === From the Multiple Rule for Continuous Functions, if $\map f x$ is continuous then so is $\map g x$ where: :$\forall x \in \R: \map g x = -\map f x$. Then we note that: :$\forall x \in \R: \map f x + \paren {-\map f x} = 0 = \paren {-\map f x} + \map f x$ So every element has an inverse. {{qed|lemma}} All the group axioms are satisfied, hence the result. {{qed}}
Proof by induction: For all $n \in \N_{>0}$, let $\map P n$ be the proposition: :For any sets $S_1, S_2, \ldots, S_n \in \tau$, it follows that $\displaystyle \bigcap_{i \mathop = 1}^n S_i \in \tau$. Let $\mathbb S$ be any finite subset of $\tau$. From Intersection of Empty Set, we have that: :$\mathbb S = \O \implies \bigcap \mathbb S = S$ From the definition of a topology, we have that $S \in \tau$. Hence $\map P 0$ is true. From Intersection of Singleton, we have that: :$\mathbb S = S_1 \implies \bigcap \mathbb S = S_1$ Thus $\map P 1$ is trivially true. === Basis for the Induction === $\map P 2$ is the case $S_1 \cap S_2 \in \O$, which is our axiom: :The intersection of any two elements of $\tau$ is an element of $\tau$. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :For any sets $S_1, S_2, \ldots, S_k \in \tau$, it follows that $\displaystyle \bigcap_{i \mathop = 1}^k S_i \in \tau$. Then we need to show: :For any sets $S_1, S_2, \ldots, S_{k + 1} \in \tau$, it follows that $\displaystyle \bigcap_{i \mathop = 1}^{k + 1} S_i \in \tau$. === Induction Step === This is our induction step: Consider the set $\displaystyle \bigcap_{i \mathop = 1}^{k + 1} S_i$. This is $\displaystyle \paren {\bigcap_{i \mathop = 1}^k S_i} \cap S_{k + 1}$. But from the induction hypothesis, we know that: :$\displaystyle \paren {\bigcap_{i \mathop = 1}^k S_i} \in \tau$ So from the base case, it follows that: :$\displaystyle \bigcap_{i \mathop = 1}^{k + 1} S_i \in \tau$ So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore, for any sets $S_1, S_2, \ldots, S_n \in \tau$, it follows that $\displaystyle \bigcap_{i \mathop = 1}^n S_i \in \tau$. {{qed|lemma}} The converse follows directly from the above. In particular, the fact that $S$ is itself an element of $\tau$ follows from the definition of Intersection of Empty Set. {{qed}}
Axiom $(\text O 1)$ for a topology states that: :$\displaystyle \forall \AA \subseteq \tau: \bigcup \AA \in \tau$ By Empty Set is Subset of All Sets, we have that $\O \subseteq \tau$. Hence, by Union of Empty Set, we have: :$\displaystyle \O = \bigcup \O \in \tau$ {{qed}} Category:Topology Category:Empty Set pdpbbr3m5pjhiq9ruxgomaqd0n75bgg
It follows directly from the definition of being coarser that: :$\tau_1 \le \tau_2 \iff \tau_1 \subseteq \tau_2$ By Subset Relation is Ordering it thus follows that $\le$ is an ordering. From Topologies Not Always Comparable by Coarseness, it follows that such an ordering is not always total. {{qed}}
From the definition, if $V$ is a closed set of $X$, then $X \setminus V$ is an open set of $X$. Let $\mathbb V$ be any arbitrary set of closed sets of $X$. Then by De Morgan's Laws: Difference with Intersection, we have: :$\displaystyle X \setminus \bigcap \mathbb V = \bigcup_{V \mathop \in \mathbb V} \left({X \setminus V}\right)$ First, let $\tau$ be a topology on $X$. We have that: :Intersection of Closed Sets is Closed in Topological Space :Finite Union of Closed Sets is Closed in Topological Space :By Open and Closed Sets in Topological Space, $\O$ and $X$ are both closed in $X$. Thus the properties as listed above hold. {{qed|lemma}} Now, suppose the properties: :$(1): \quad$ Any intersection of arbitrarily many closed sets of $X$ under $\tau$ is a closed set of $X$ under $\tau$ :$(2): \quad$ The union of any finite number of closed sets of $X$ under $\tau$ is a closed set of $X$ under $\tau$ :$(3): \quad X$ and $\O$ are both closed sets of $X$ under $\tau$. all hold. That means $\displaystyle \bigcap \mathbb V$ is closed. So $\displaystyle X \setminus \bigcap \mathbb V = \bigcup_{V \mathop \in \mathbb V} \paren {X \setminus V}$ is open. Thus we have that the union of arbitrarily many open sets of $X$ under $\tau$ is an open set of $X$ under $\tau$. Similarly we deduce that the intersection of any finite number of open sets of $X$ under $\tau$ is an open set of $X$ under $\tau$. And of course by Open and Closed Sets in Topological Space, $\O$ and $X$ are both open in $X$. So $\tau$ is a topology on $X$. {{qed}}