hcju/proofwikiqa huggingface dataset:
The proof proceeds by mathematical induction. For all $n \in \N_{> 0}$, let $\map P n$ be the proposition: :$\displaystyle \sum_{i \mathop = 1}^n x_i + \sum_{i \mathop = 1}^n y_i = \sum_{i \mathop = 1}^n \paren {x_i + y_i}$ === Basis for the Induction === $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^1 x_i + \sum_{i \mathop = 1}^1 y_i       | r = x_1 + y_1 }} {{eqn | r = \sum_{i \mathop = 1}^1 \paren {x_i + y_i} }} {{end-eqn}} This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$\displaystyle \sum_{i \mathop = 1}^k x_i + \sum_{i \mathop = 1}^k y_i = \sum_{i \mathop = 1}^k \paren {x_i + y_i}$ from which it is to be shown that: :$\displaystyle \sum_{i \mathop = 1}^{k + 1} x_i + \sum_{i \mathop = 1}^{k + 1} y_i = \sum_{i \mathop = 1}^{k + 1} \paren {x_i + y_i}$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^{k + 1} x_i + \sum_{i \mathop = 1}^{k + 1} y_i       | r = \paren {\sum_{i \mathop = 1}^k x_i + x_{k + 1} } + \paren {\sum_{i \mathop = 1}^k y_i + y_{k + 1} }       | c = {{Defof|Summation}} }} {{eqn | r = \left({\sum_{i \mathop = 1}^k x_i + \sum_{i \mathop = 1}^k y_i}\right) + \paren {x_{k + 1} + y_{k + 1} }       | c = Commutative Law of Addition and Associative }} {{eqn | r = \sum_{i \mathop = 1}^k \paren {x_i + y_i} + \paren {x_{k + 1} + y_{k + 1} }       | c = Induction Hypothesis }} {{eqn | r = \sum_{i \mathop = 1}^{k + 1} \paren {x_i + y_i}       | c = {{Defof|Summation}} }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \N_{> 0}: \sum_{i \mathop = 1}^n x_i + \sum_{i \mathop = 1}^n y_i = \sum_{i \mathop = 1}^n \paren {x_i + y_i}$ {{qed}} Category:Numbers Category:Proofs by Induction ibolccjs5zc8erhwzq2a6bct6mfwewf
For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition: :$\displaystyle \lambda \sum_{i \mathop = 1}^n x_i = \sum_{i \mathop = 1}^n \lambda x_i$ === Basis for the Induction === $P \left({1}\right)$ is the case: {{begin-eqn}} {{eqn | l = \lambda \sum_{i \mathop = 1}^1 x_i       | r = \lambda x_1 }} {{eqn | r = \sum_{i \mathop = 1}^1 \lambda x_i }} {{end-eqn}} This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is the induction hypothesis: :$\displaystyle \lambda \sum_{i \mathop = 1}^k x_i = \sum_{i \mathop = 1}^k \lambda x_i$ from which it is to be shown that: :$\displaystyle \lambda \sum_{i \mathop = 1}^{k + 1} x_i = \sum_{i \mathop = 1}^{k + 1} \lambda x_i$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \lambda \sum_{i \mathop = 1}^{k + 1} x_i       | r = \lambda \left({\sum_{i \mathop = 1}^k x_i + x_{k + 1} }\right)       | c = Definition of Summation }} {{eqn | r = \lambda \sum_{i \mathop = 1}^k x_i + \lambda x_{k + 1}       | c = Multiplication of Numbers Distributes over Addition }} {{eqn | r = \sum_{i \mathop = 1}^k \lambda x_i + \lambda x_{k + 1}       | c = Induction Hypothesis }} {{eqn | r = \sum_{i \mathop = 1}^{k + 1} \lambda x_i       | c = Definition of Summation }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \N_{> 0}: \lambda \sum_{i \mathop = 1}^n x_i = \sum_{i \mathop = 1}^n \lambda x_i$ {{qed}} Category:Numbers Category:Proofs by Induction s6gnp5lzhthp0s5hmwo7dcinjytg7f9
{{begin-eqn}} {{eqn | l = \sup_{x \mathop \in S} \set {\map f x} - \inf_{x \mathop \in S} \set {\map f x}       | r = \sup_{x \mathop \in S} \set {\map f x} + \sup_{x \mathop \in S} \set {-\map f x}       | c = Negative of Infimum is Supremum of Negatives }} {{eqn | r = \sup_{x, y \mathop \in S} \set {\map f x + \paren {-\map f y} }       | c = Supremum of Sum equals Sum of Suprema }} {{eqn | r = \sup_{x, y \mathop \in S} \set {\map f x - \map f y} }} {{eqn | r = \sup_{x, y \mathop \in S} \set {\size {\map f x - \map f y} }       | c = Supremum of Absolute Value of Difference equals Supremum of Difference }} {{end-eqn}} {{qed}} Category:Real Analysis kcdheyhqqm4yvk2a0xwcnftr9zctwmt
Let $\mathbb B = \set {0, 1}$.  Let $p$ be a proposition. Let the following two sets be defined: :$A = \set {x \in \mathbb B: x = 0 \lor p}$ :$B = \set {x \in \mathbb B: x = 1 \lor p}$ where $\lor$ denotes the disjunction operator. We have that: :$0 \in A$ and: :$1 \in B$ so both $A$ and $B$ are non-empty Then the set: :$X = \set {A, B}$ is a set of non-empty sets: By the axiom of choice, there exists a choice function: :$f: X \to \mathbb B$ since $\displaystyle \bigcup X = \mathbb B$. There are four cases: :$(1): \quad \map f A = \map f B = 0$ This means that $0 \in B$. But for that to happen, $\paren {0 = 1} \vee p$ must be true. So by Disjunctive Syllogism, $p$ is true. :$(2): \quad \map f A = \map f B = 1$ This means that $1 \in A$. Arguing similarly to case $(1)$, it follows that  $p$ is true in this case also. :$(3): \quad \map f A = 1 \ne \map f B = 0$ This means that $A \ne B$ (or otherwise $f$ would pick the same element). But if $p$ is true, that means: :$A = B = \mathbb B$ which is a contradiction. Therefore in this case: :$\neg p$ :$(4): \quad \map f A = 0 \ne \map f B = 1$ Using the same reasoning as in case $(3)$, it is seen that in this case: :$\neg p$ So by Proof by Cases: :$\paren {p \vee \neg p}$ That is the Law of Excluded Middle. {{qed}} {{Namedfor|Radu Diaconescu|name2 = Noah D. Goodman|name3 = John R. Myhill|cat = Diaconescu|cat2 = Goodman|cat3 = Myhill}}
=== Sufficient Condition === Let $x \in \partial A$. Let $U \in \BB$. By definition of basis, $U$ is an open set of $T$. Thus from Characterization of Boundary by Open Sets: :if $x \in U$ ::then $A \cap U \ne \O$ and $\relcomp S A \cap U \ne \O$. {{qed|lemma}} === Necessary Condition === Let $x$ be such that for every $U \in \BB$: :if $x \in U$ :then $A \cap U \ne \O$ and $\relcomp S A \cap U \ne \O$. By Characterization of Boundary by Open Sets, to prove that $x \in \partial A$ it is enough to prove that: :for every open set $U$ of $T$: ::if $x \in U$ then $A \cap U \ne \O$ and  $\relcomp S A \cap U \ne \O$. Let $U$ be an open set of $T$. Let $x \in U$. By definition of (analytic) basis, there exists $V \in \BB$ such that: :$x \in V \subseteq U$ By assumption: :$A \cap V \ne \O$ and: :$\relcomp S A \cap V \ne \O$ From the corollary to Set Intersection Preserves Subsets: :$A \cap V \subseteq A \cap U$ and: :$\relcomp S A \cap V \subseteq \relcomp S A \cap U$ So: :$A \cap U \ne \O$ and $\relcomp S A \cap U \ne \O$ and hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = \Int A \cup \partial A \cup \Int {A'}       | r = \Int A \cup \Int {A'} \cup \partial A       | c = Union is Associative, Union is Commutative }} {{eqn | r = \Int A \cup \Int {A'} \cup \paren {\map \cl A \cap \map \cl {A'} }       | c = Boundary is Intersection of Closure with Closure of Complement }} {{eqn | r = \paren {\Int A \cup \Int {A'} \cup \map \cl A} \cap \paren {\Int A \cup \Int {A'} \cup \map \cl {A'} }       | c = Intersection Distributes over Union }} {{eqn | r = \paren {\Int A \cup \paren {\map \cl A}' \cup \map \cl A} \cap \paren {\Int A \cup \Int {A'} \cup \map \cl {A'} }       | c = Complement of Interior equals Closure of Complement }} {{eqn | r = \paren {\Int A \cup \paren {\paren {\map \cl A}' \cup \map \cl A} } \cap \paren {\Int A \cup \Int {A'} \cup \map \cl {A'} }       | c = Union is Associative }} {{eqn | r = \paren {\Int A \cup S} \cap \paren {\Int A \cup \Int {A'} \cup \map \cl {A'} }       | c = Union with Relative Complement }} {{eqn | r = \paren {\Int A \cup S} \cap \paren {\Int A \cup \Int {A'} \cup \paren {\Int A}'}       | c = Complement of Interior equals Closure of Complement }} {{eqn | r = \paren {\Int A \cup S} \cap \paren {\Int A \cup \paren {\Int A' \cup \Int {A'} } }       | c = Union is Associative, Union is Commutative }} {{eqn | r = \paren {\Int A \cup S} \cap \paren {S \cup \Int {A'} }       | c = Union with Relative Complement }} {{eqn | r = S \cap \paren {S \cup \Int {A'} }       | c = Union with Superset is Superset }} {{eqn | r = S \cap S       | c = Union with Superset is Superset }} {{eqn | r = S       | c = Intersection is Idempotent }} {{end-eqn}} {{qed}}
=== Sufficient Condition === Let $x \in A'$. Then $x$ is an accumulation point of $A$ by Definition:Set Derivative. Then by definition of accumulation point: :$(1): \quad x \in \left({A \setminus \left\{ {x}\right\} }\right)^-$ where $A^-$ denotes the closure of $A$. Let $U$ be an open set of $T$. Let $x \in U$. Then by $(1)$ and Condition for Point being in Closure: :$\left({A \setminus \left\{ {x}\right\}}\right) \cap U \ne \varnothing$ Then there exists $y$ being a point such that: :$(2): \quad y \in A \setminus \left\{ {x}\right\}$ and $y \in U$. Then $y \in A$ and $y \notin \left\{ {x}\right\}$ by definition of set difference. Hence by $(2)$ and the definitions of set intersection and singleton: :$y \in A \cap U$ and $x \ne y$ {{qed|lemma}} === Necessary Condition === Let $x$ be such that: :$(3): \quad$ for every open set $U$ of $T$ if $x \in U$, then there exists a point $y$ of $T$ such that $y \in A \cap U$ and $x \ne y$. It will be proved that: :for every open set $G$ of $T$: ::if $x \in G$ ::then $\left({A \setminus \left\{ {x}\right\} }\right) \cap U \ne \varnothing$. Let $G$ be an open set of $T$ such that $x \in G$. Then by $(3)$ there exists a point $y$ of $T$ such that: :$(4): \quad  y \in A \cap G$ and $x \ne y$ Then by the definitions of set intersection and singleton: :$y \in A$ and $y \notin \left\{ {x}\right\}$ By definition of set difference: :$y \in A \setminus \left\{ {x}\right\}$ By $(4)$ and the definition of set intersection: :$y \in G$ Hence: : $\left({A \setminus \left\{ {x}\right\} }\right) \cap U \ne \varnothing$ {{qed|lemma}} Then by Condition for Point being in Closure: :$x \in \left({A \setminus \left\{ {x}\right\} }\right)^-$ Then, by definition, $x$ is an accumulation point of $A$. Hence by definition of set derivative: :$x \in A'$ {{qed}}
=== Sufficient Condition === Let $x \in A'$. By Characterization of Derivative by Open Sets: For every open set $U$ of $T$: :if $x \in U$ :then there exists a point $y$ of $T$ such that $y \in A \cap U$ and $x \ne y$ As the elements of $\BB$ are all open sets, it follows that: For every open set $U \in \BB$: :if $x \in U$ :then there exists a point $y$ of $T$ such that $y \in A \cap U$ and $x \ne y$ {{qed|lemma}} === Necessary Condition === Let $x$ be such that: $(1): \quad$ for every subset $U \in \BB$, there exists a point $y$ of $T$ such that $y \in A \cap U$ and $x \ne y$. By Characterization of Derivative by Open Sets, to prove that $x \in A'$ it is enough to prove: :for every open set $U$ of $T$: ::if $x \in U$ ::then there exists a point $y$ of $T$ such that $y \in A \cap U$ and $x \ne y$. Let $U$ be an open set of $T$. Let that $x \in U$. Then by definition of local basis: :there exists $V \in \BB$ such that: ::$x \in V \subseteq U$ By $(1)$: :there exists a point $y$ of $T$ such that $y \in A \cap V$ and $x \ne y$. By the corollary to Set Intersection Preserves Subsets: :$A \cap V \subseteq A \cap U$ and so: :$y \in A \cap V \implies y \in A \cap U$ and so $y$ fulfils the conditions of the hypothesis. {{qed}}
Let $x \in A'$. By Condition for Point being in Closure it is enough to prove that: :for every open set $G$ of $T$: :: if $x \in G$ :: then $A \cap G \ne \varnothing$. Let $G$ be an open set of $T$. Let $x \in G$. Then by Characterization of Derivative by Open Sets: : there exists a point $\exists y \in S: y \in A \cap G \land x \ne y$.   Hence: :$A \cap G \ne \varnothing$ Hence the result. {{qed}}
=== Closure Subset of Union === It is to be proved that: :$A^- \subseteq A \cup A'$ Let $x \in A^-$. In the case where $x \in A$ then $x \in A \cup A'$ by definition of set union. Let: :$(1): \quad x \notin A$ From Characterization of Derivative by Open Sets, to prove $x \in A'$ it is enough to show that: :for every open set $U$ of $T$: ::if $x \in U$ ::then there exists a point $y$ of $T$ such that $y \in A \cap U$ and $x \ne y$. Let $U$ be an open set of $T$. Let $x \in U$. Then by Condition for Point being in Closure: :$A \cap U \ne \varnothing$ Then by definition of empty set: :$\exists y \in S: y \in A \cap U$ By definition of set intersection: :$y \in A$ and $y \in U$ But as $x \notin A$ it follows by definition of set intersection that: :$x \notin A \cap U$ So by $(1)$: :$x \ne y$ Thus $y$ fulfils the conditions of the hypothesis, and so: :$x \in A'$ Hence by definition of set union: :$x \in A \cup A'$ Thus in all cases: :$x \in A^- \implies x \in A \cup A'$ and so: :$A^- \subseteq A \cup A'$ {{qed|lemma}} === Union Subset of Closure === It is to be proved that: :$A \cup A' \subseteq A^-$ By Set is Subset of its Topological Closure: :$A \subseteq A^-$ By Derivative is Included in Closure: :$A' \subseteq A^-$      Hence by Union of Subsets is Subset: :$A \cup A' \subseteq A^-$ {{qed|lemma}} Hence by definition of set equality: :$A^- = A \cup A'$
Let $A \subseteq B$. Let $x \in A'$. By Characterization of Derivative by Open Sets it is enough to prove that: :for every open set $G$ of $T$: ::if $x \in G$ ::then there exists $y$ such that $y \in B \cap G$ and $x \ne y$. Let $G$ be an open set of $T$. Let $x \in G$. Then by Characterization of Derivative by Open Sets: :there exists a point $y$ of $T$ such that $y \in A \cap G$ and $x \ne y$. By the corollary to Set Intersection Preserves Subsets: :$A \cap G \subseteq B \cap G$ Hence: :$y \in B \cap G$ and $x \ne y$. The conditions of the hypothesis are thus fulfilled, and: :$x \in B'$ Thus by definition of subset: :$A' \subseteq B'$ {{qed}}
=== Derivative of Union subset of Union of Derivatives === It will be shown that: :$\left({A \cup B}\right)' \subseteq A' \cup B\,'$ Let $x \in \left({A \cup B}\right)'$. By definition of set derivative: :$x$ is an accumulation point of $A \cup B$.      Then by definition of accumulation point of set: :$(1): \quad x \in \left({\left({A \cup B}\right) \setminus \left\{ {x}\right\} }\right)^-$ where $A^-$ denotes the closure of $A$. By Union Distributes over Difference: :$\left({A \cup B}\right) \setminus \left\{ {x}\right\} = \left({A \setminus \left\{ {x}\right\} }\right) \cup \left({B \setminus \left\{ {x}\right\} }\right)$ Then by Closure of Finite Union equals Union of Closures: :$\left({\left({A \cup B}\right) \setminus \left\{ {x}\right\} }\right)^- = \left({A \setminus \left\{ {x}\right\} }\right)^- \cup \left({B \setminus \left\{ {x}\right\} }\right)^-$ Then by $(1)$ and definition of set union: :$x \in \left({A \setminus \left\{ {x}\right\} }\right)^-$ or $x \in \left({B \setminus \left\{ {x}\right\} }\right)^-$ Then by definition of accumulation point of set: :$x$ is an accumulation point of $A$ or $x$ is an accumulation point of $B$. Then by definition of set derivative: :$x \in A'$ or $x \in B\,'$ Hence by definition of set union: :$x \in A' \cup B\,'$ === Union of Derivatives subset of Derivative of Union === By Set is Subset of Union: :$A \subseteq A \cup B$ and $B \subseteq A \cup B$ Then by Derivative of Subset is Subset of Derivative: :$A' \subseteq \left({A \cup B}\right)'$ and $B\,' \subseteq \left({A \cup B}\right)'$ Hence by Union of Subsets is Subset: :$A' \cup B\,' \subseteq  \left({A \cup B}\right)'$ {{qed}}
Let: : $(1): \quad x \in A''$. Aiming for a contradiction, suppose that $x \notin A'$. Then by Characterization of Derivative by Open Sets there exists an open subset $G$ of $T$ such that: : $(2): \quad x \in G$ and : $(3): \quad \lnot \exists y: y \in A \cap G \land x \ne y$. By definition of $T_1$ space: :$\left\{{x}\right\}$ is closed. Then by Open Set minus Closed Set is Open: :$(4): \quad G \setminus \{x\}$ is open. By $(1)$, $(2)$, and Characterization of Derivative by Open Sets there exists a point $y$ of $T$ such that: :$(5): \quad y \in A' \cap G$ and :$(6): \quad x \ne y$. Then by definition of intersection: : $y \in A'$. Then by $(6)$ and definition of set difference: :$(7): \quad y \in A' \setminus \{x\}$. By definition of intersection and $(5)$: : $y \in G$. By $(6)$ and definition of singleton: :$y \notin \left\{{x}\right\}$ Then by definition of set difference: :$(8): \quad y \in G \setminus \{x\}$ We will prove: :$(9): \quad G \cap \left({A \setminus \{x\}}\right) = \varnothing$ :Aiming for a contradiction suppose that: :: $G \cap \left({A \setminus \left\{{x}\right\}}\right) \ne \varnothing$. :Then by definition of the empty set there exists $g$ such that: ::$g \in G \cap \left({A \setminus \left\{{x}\right\}}\right)$ :Hence by definition of intersection: :: $g \in G$ and :: $g \in A \setminus \left\{{x}\right\}$. :Then by definition of set difference: ::$g \in A$  :Hence by definition of intersection: :: $g \in A \cap G$ :Then by $(3)$: :: $x = g$ :Hence this by definition of singleton contadicts with $g \notin \left\{{x}\right\}$ obtained by definition of set difference. :Thus $G \cap \left({A \setminus \left\{{x}\right\}}\right) = \varnothing$. Define $U = G \setminus \{x\}$ as an open set by $(4)$. By $(5)$ and definition of set difference: :$y \in A'$ Then by $(8)$ and Characterization of Derivative by Open Sets there exists a point $q$ of $T$ such that :$(10): \quad q \in A \cap U$ and :$(11): \quad y \ne q$. By $(10)$ and definition of intersection: : $q \in A$ By $(11)$ and definition of singleton: :$q \notin \left\{{y}\right\}$ Then by definition of set difference :$(12): \quad q \in A \setminus \left\{{y}\right\}$. By definition of intersection: : $q \in U$. Then by $(12)$ and by definition of set difference :$q \ne x$ and $q \in A$. Then by definition of set difference: : $q \in A \setminus \{x\}$ and  :$q \in G$. Hence this contradicts with $(9)$ by definition of intersection. Thus the result by Proof by Contradiction. {{qed}}
{{begin-eqn}} {{eqn | l = A'^-       | r = A' \cup A''       | c = Closure Equals Union with Derivative }} {{eqn | o = \subseteq       | r = A' \cup A'       | c = $A'' \subseteq A'$ by Derivative of Derivative is Subset of Derivative in T1 Space }} {{eqn | r = A'       | c = Union is Idempotent }} {{end-eqn}} So: :$A'^- \subseteq A'$ Then by  definition of closure: :$A' \subseteq A'^-$ Hence the result by definition of set equality. {{qed}}
Let $\ds x \in \bigcup_{A \mathop \in \FF} A'$. Then by definition of union there exists $A \in \FF$ such that: :$(1): \quad x \in A'$ By Set is Subset of Union: :$\ds A \subseteq \bigcup_{A \mathop \in \FF} A$ Then by Derivative of Subset is Subset of Derivative: :$\ds A' \subseteq \paren {\bigcup_{A \mathop \in \FF} A}'$ Hence by $(1)$ the result: :$\ds x \in \paren {\bigcup_{A \mathop \in \FF} A}'$ follows by definition of subset. {{qed}}
=== Sufficient Condition === Let $x \in H$ be an isolated point in $H$. Then by definition of isolated point: :$\exists U \in \tau: H \cap U = \left\{ {x}\right\}$ That is, by definition of Definition of uniqueness: :$\lnot \forall U \in \tau: \left({x \in U \implies \exists y \in S: \left({y \in H \cap U \land x \ne y}\right)}\right)$ Hence by Characterization of Derivative by Open Sets: :$x \notin A'$ where $A'$ denotes the derivative of $A$. Thus by definition of derivative: :$x$ is not an accumulation point of $H$. {{qed|lemma}} === Necessary Condition === Let $x \in H$ not be an accumulation point of $H$. Thus by definition of derivative: :$x \notin A'$ Hence: {{begin-eqn}} {{eqn | l = \lnot \forall U \in \tau       | o = :       | r = \left({x \in U \implies \exists y \in S: \left({y \in H \cap U \land x \ne y}\right)}\right)       | c = Characterization of Derivative by Open Sets }} {{eqn | l = \exists U \in \tau       | o = :       | r = \lnot \left({x \in U \implies \exists y \in S: \left({y \in H \cap U \land x \ne y}\right)}\right)       | c = Denial of Universality }} {{eqn | l = \exists U \in \tau       | o = :       | r = \left({x \in U \land \lnot \exists y \in S: \left({y \in H \cap U \land x \ne y}\right)}\right)       | c = Conjunction with Negative Equivalent to Negation of Implication }} {{eqn | l = \exists U \in \tau       | o = :       | r = \left({x \in U \land \forall y \in S: \lnot \left({y \in H \cap U \land x \ne y}\right)}\right)       | c = Denial of Existence }} {{eqn | l = \exists U \in \tau       | o = :       | r = \left({x \in U \land \forall y \in S: \left({y \in H \cap U \implies x = y}\right)}\right)       | c = Implication Equivalent to Negation of Conjunction with Negative }} {{eqn | l = \exists U \in \tau       | o = :       | r = H \cap U = \left\{ {x}\right\}       | c = Definition of Uniqueness, and $x \in H$ }} {{end-eqn}} Thus by definition of isolated point: :$x$ is an isolated point in $H$. {{qed}} Category:Isolated Points Category:Accumulation Points grmf8cl1o8om1tbe6spc1mrqkf15c74
$x$ is an isolated point in $H$ $\iff$ $x \in H$ and $x$ is not an accumulation point of $H$ by Point is Isolated iff not Accumulation Point $\iff$ $x \in H$ and $x \notin H'$ by definition of derivative $\iff$ $x \in H \setminus H'$ by definition of set difference. {{qed}}
$A$ is dense-in-itself $\iff$ every $x \in A$ is not isolated point in $A$ by definition of dense-in-itself $\iff$ every $x \in A$ is accumulation point of $A$ by Point is Isolated iff not Accumulation Point $\iff$ every $x \in A$ belongs to $A'$ by definition of derivative $\iff$ $A \subseteq A'$ by definition of subset. {{qed}} Category:Denseness Category:Set Derivatives q22ronc5mp9vnewyqchvq7sa5szs7g8
Let $A$ be dense-in-itself. Then by Dense-in-itself iff Subset of Derivative: :$(1): \quad A \subseteq A'$ where $A'$ denotes the derivative of $A$. By Derivative of Derivative is Subset of Derivative in $T_1$ Space: :$(2): \quad A'' \subseteq A'$ By Dense-in-itself iff Subset of Derivative it is sufficient to proof that: :$A^- \subseteq \left({A^-}\right)'$ Thus: {{begin-eqn}} {{eqn | l = \left({A^-}\right)'       | r = \left({A \cup A'}\right)'       | c = Closure Equals Union with Derivative }} {{eqn | r = A' \cup A''       | c = Derivative of Union is Union of Derivatives }} {{eqn | r = A'       | c = $(2)$ and Union with Superset is Superset }} {{eqn | r = A \cup A'       | c = $(1)$ and Union with Superset is Superset }} {{eqn | r = A^-       | c = Closure Equals Union with Derivative }} {{end-eqn}} {{qed}}
Let $f: S \to T$ be a mapping. Let: :$Y = \set {t \in T: \exists s \in S: f \left({s}\right) = t}$ Then by definition: :$Y$ is the image of $f$ by definition 1. But by definition of image of subset under mapping: :$Y = f \left[{S}\right]$ Thus $Y$ is the image of $f$ by definition 2. Hence the result. {{qed}} Category:Mapping Theory Category:Images 2qslu9e12e6ytzvvz4jzv7rgjlp6rux
By Dense-in-itself iff Subset of Derivative: :$\forall A \in \mathcal F: A \subseteq A'$ where $A'$ denotes the derivative of $A$. Then by Set Union Preserves Subsets: :$\displaystyle \bigcup \mathcal F \subseteq \bigcup_{A \in \mathcal F} A'$ By Union of Derivatives is Subset of Derivative of Union: :$\displaystyle \bigcup_{A \in \mathcal F} A' \subseteq \left({\bigcup \mathcal F}\right)'$ Then by Subset Relation is Transitive: :$\displaystyle \bigcup \mathcal F \subseteq \left({\bigcup \mathcal F}\right)'$ The result follows by Dense-in-itself iff Subset of Derivative. {{qed}}