hcju/proofwikiqa huggingface dataset:
=== Sufficient Condition === From Integer as Sum of Three Squares, every positive integer not of the form $4^n \paren {8 m + 7}$ can be expressed as the sum of three squares. Hence every positive integer $r$ such that $r \equiv 3 \pmod 8$ can likewise be expressed as the sum of three squares. From Square Modulo 8, the squares modulo $8$ are $0, 1$ and $4$. Thus for $r$ to be the sum of three squares, each of those squares needs to be congruent modulo $8$ to $1$. Thus each square is odd, and $r$ can be expressed in the form $8 n + 3$ as the sum of $3$ odd squares. {{qed|lemma}} === Necessary Condition === Suppose $r$ is the sum of $3$ odd squares. From Odd Square Modulo 8, each of these squares are equal to $1 \pmod 8$. Therefore: :$r \equiv 1 + 1 + 1 \pmod 8 \equiv 3 \pmod 8$ {{qed}} Category:Square Numbers Category:Number Theory eva4l6nkb4hyly5286vwlk3paew019b
Proof by induction: Let $\map P n$ be the proposition: :$V_n = \alpha^n + \beta^n$ === Basis for the Induction === We have: {{begin-eqn}} {{eqn | l = \alpha^0 + \beta^0       | r = 1 + 1       | c = Zeroth Power of Real Number equals One }} {{eqn | r = 2 }} {{eqn | r = V_0       | c = From definition }} {{eqn | l = \alpha^1 + \beta^1       | r = \frac {m + \Delta} 2 + \frac {m - \Delta} 2 }} {{eqn | r = m }} {{eqn | r = V_1       | c = From definition }} {{end-eqn}} Therefore $\map P 0$ and $\map P 1$ are true. This is the basis for the induction. === Induction Hypothesis === This is our induction hypothesis: :For some $k \in \N$, both $\map P k$ and $\map P {k + 1}$ are true. That is: :$V_k = \alpha^k + \beta^k$ :$V_{k + 1} = \alpha^{k + 1} + \beta^{k + 1}$ Now we need to show true for $n = k + 2$: :$\map P {k + 2}$ is true. That is: :$V_{k + 2} = \alpha^{k + 2} + \beta^{k + 2}$ === Induction Step === This is our induction step: First we notice that: {{begin-eqn}} {{eqn | l = \alpha^2       | r = \paren {\frac {m + \Delta} 2}^2 }} {{eqn | r = \frac 1 4 \paren {m^2 + 2 m \Delta + m^2 + 4}       | c = Square of Sum }} {{eqn | r = \frac 1 4 \paren {2 m^2 + 2 m \Delta + 4} }} {{eqn | r = \frac {m^2 + m \Delta} 2 + 1 }} {{eqn | r = m \alpha + 1 }} {{end-eqn}} Similarly: {{begin-eqn}} {{eqn | l = \beta^2       | r = \paren {\frac {m - \Delta} 2}^2 }} {{eqn | r = \frac 1 4 \paren {m^2 - 2 m \Delta + m^2 + 4}       | c = Square of Sum }} {{eqn | r = \frac 1 4 \paren {2 m^2 - 2 m \Delta + 4} }} {{eqn | r = \frac {m^2 - m \Delta} 2 + 1 }} {{eqn | r = m \beta + 1 }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = V_{k + 2}       | r = m V_{k + 1} + V_k       | c = Definition of the recursive sequence }} {{eqn | r = m \paren {\alpha^{k + 1} + \beta^{k + 1} } + \alpha^k + \beta^k       | c = Induction Hypothesis }} {{eqn | r = \paren {m \alpha + 1} \alpha^k + \paren {m \beta + 1} \beta^k }} {{eqn | r = \alpha^2 \alpha^k + \beta^2 \beta^k       | c = From above }} {{eqn | r = \alpha^{k + 2} + \beta^{k + 2} }} {{end-eqn}} This show that $\map P {k + 2}$ is true. By principle of mathematical induction, $\map P n$ is true for all $n \in \N$. {{qed}} Category:Binet Form tfa33g8gnzww7erm5y0s11gtermulp9
Let $S = \set {p, q, r, s}$. Let $\circ : S \times S \to S$ be defined by: :$p \circ q = q \circ p = r$ :$p \circ r = r \circ p = r$ :$q \circ r = r \circ q = r$ :$p \circ s = s \circ p = s$ :$q \circ s = s \circ q = s$ :$r \circ s = s \circ r = s$ :$p \circ p = p$ :$q \circ q = q$ :$r \circ r = r$ :$s \circ s = s$ Then a manual check shows that $\circ$ is: :closed :associative :commutative :idempotent So $\struct {S, \circ}$ is a semilattice with ordering :${\preceq} = \set {\tuple {p, p}, \tuple {q, q}, \tuple {r, r}, \tuple {s, s}, \tuple {p, r}, \tuple {q,r}, \tuple {p, s}, \tuple {q,s}, \tuple {r, s} }$ Let $T = \set {p, q, s}$. Then: :${\preceq \restriction_T} = \set {\tuple {p, p}, \tuple {q, q}, \tuple {s, s}, \tuple {p, s}, \tuple {q, s} }$ Then a manual check shows that all joins exist in $\struct{T, \preceq \restriction_T}$: :$p \vee q = q \vee p = s$ :$p \vee s = s \vee p = s$ :$q \vee s = s \vee q = s$ :$p \vee p = p$ :$q \vee q = q$ :$s \vee s = s$ So $\struct {T, \preceq \restriction_T}$ is a join semilattice with respect to $\preceq \restriction_T$. Now: :$p \vee q = s \ne r = p * q$ So $\vee$ is not the restiction of $\circ$ to $T$. That is, $\struct {T, \vee}$ is not a subsemilattice of $\struct {S, \circ}$ by definition. {{qed}}
{{begin-eqn}} {{eqn | l = H       | o = \subseteq       | r = F }} {{eqn | ll= \leadsto       | l = H^-       | o = \subseteq       | r = F^-       | c = Topological Closure of Subset is Subset of Topological Closure }} {{eqn | r = F       | c = Set is Closed iff Equals Topological Closure }} {{end-eqn}} {{qed}}
Let $x \notin F$. Then $x$ is in the open set $S \setminus F$. From Set Difference with Subset is Superset of Set Difference: :$S \setminus F \subseteq S \setminus H$ From Subsets of Disjoint Sets are Disjoint: :$S \setminus F \cap H = \O$ From Set is Open iff Neighborhood of all its Points: :$S \setminus F$ is an neighborhood of $x$ By definition of the closure of $H$: :$x \notin H^-$ We have shown that: :$S \setminus F \subseteq S \setminus H^-$ From Set Difference with Subset is Superset of Set Difference: :$H^- \subseteq F$ {{qed}}
The proof proceeds by induction. For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition: :There are $m^n$ partial derivatives of $u$ of order $n$. $\map P 0$ is the degenerate case where $f$ is not partially differentiated at all: :$\map f {x_1, x_2, \ldots, x_m}$ and it is apparent that there is only $1 = m^0$ such. Thus $\map P 0$ is seen to hold. === Basis for the Induction === $\map P 1$ is the case: :There are $m^1 = m$ partial derivatives of $u$ of order $1$. These can be instanced as: {{begin-eqn}} {{eqn | l = \map {f_1} {x_1, x_2, \ldots, x_m}       | r = \dfrac {\partial u} {\partial {x_1} } }} {{eqn | l = \map {f_2} {x_1, x_2, \ldots, x_m}       | r = \dfrac {\partial u} {\partial {x_2} } }} {{eqn | o = \vdots }} {{eqn | l = \map {f_m} {x_1, x_2, \ldots, x_m}       | r = \dfrac {\partial u} {\partial {x_m} } }} {{end-eqn}} and it is apparent that there are $m$ such. Thus $\map P 1$ is seen to hold. 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: :There are $m^k$ partial derivatives of $u$ of order $k$. from which it is to be shown that: :There are $m^{k + 1}$ partial derivatives of $u$ of order $k + 1$. === Induction Step === This is the induction step: Let $g$ be one of the partial derivatives of $u$ of order $k$. Then by the basis for the induction, there are $m$ partial derivatives of $g$. By the induction hypothesis, there are $m^k$ partial derivatives of $u$ of order $k$. Thus by the Product Rule for Counting, there are $m \times m^n$ partial derivatives of $u$ of order $k + 1$. That is, a total of $m^{k + 1}$ partial derivatives of $u$ of order $k + 1$. So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{\ge 0}:$ there are $m^n$ partial derivatives of $u$ of order $n$. {{qed}}
By definition, an open set $S \subseteq A$ is one where every point inside it is an element of an open ball contained entirely within that set. Let $x \in X$. An open ball of $x$ in $M$ is by definition a subset of $X$. Hence the result. {{qed}}
By definition, a subset $S \subseteq X$ is open if: :$\forall x \in X : \exists \epsilon \in \R_{>0} : \map {B_\epsilon} x \subseteq S$ Let $S = X$. {{AimForCont}} $X$ is not open. By De Morgan's laws: :$\exists x \in X : \forall \epsilon \in \R_{>0} : \map {B_\epsilon} x \cap \paren {X \setminus X} \ne \O$ Note that: :$X \setminus X = \O$. By Intersection with Empty Set: :$\map {B_\epsilon} x \cap \O = \O$ Hence: :$\exists x \in X : \forall \epsilon \in \R_{>0} : \O \ne \O$. Since Empty Set is Unique, we have a contradiction. {{qed}}
From Normed Vector Space is Open in Itself, $X$ is open in $M$. But: :$\varnothing = \relcomp X X$ where $\complement_X$ denotes the set complement relative to $X$. The result follows by definition of closed set. {{qed}}
By definition, an open set $S \subseteq X$ is one where every point inside it is an element of an open ball contained entirely within that set. That is, there are no points in $S$ which have an open ball some of whose elements are not in $S$. As there are no elements in $\O$, the result follows vacuously. {{qed}}
=== Sufficient Condition === {{:Sequence of Functions is Uniformly Cauchy iff Uniformly Convergent/Sufficient Condition}} === Necessary Condition === {{:Sequence of Functions is Uniformly Cauchy iff Uniformly Convergent/Necessary Condition}}
Fix some $\varepsilon \in \R_{> 0}$. As $\sequence {f_n}$ is uniformly Cauchy on $S$, there exists $N \in \N$ such that:  :$\size {\map {f_n} x - \map {f_m} x} < \dfrac {\varepsilon} 2$ for all $n, m > N$ and $x \in S$. Fix some $x \in S$.  Then:  :$\size {\map {f_n} x - \map {f_m} x} < \varepsilon$ for all $n, m > N$.  As $\varepsilon$ was arbitrary, the sequence $\sequence {\map {f_n} x}$ is therefore Cauchy. By Sequence is Cauchy iff Convergent, it follows that $\sequence {\map {f_n} x}$ is convergent. Define a function $f : S \to \R$ by:  :$\displaystyle \map f x = \lim_{n \to \infty} \map {f_n} x$ for all $x \in S$.  We aim to show that $f_n \to f$ uniformly. For all $x \in S$, we have: :$\displaystyle \lim_{m \to \infty} \size {\map {f_n} x - \map {f_m} x} = \size {\map {f_n} x - \map f x}$ We established that for all $\varepsilon \in \R_{> 0}$ we can find $N \in \N$ such that:  :$\size {\map {f_n} x - \map {f_m} x} < \dfrac \varepsilon 2$ for $x \in S$ and $n, m > N$.  We therefore have:  :$\size {\map {f_n} x - \map f x} \le \dfrac \varepsilon 2 < \varepsilon$ for all $x \in S$ and $n > N$. So $\sequence {f_n}$ converges uniformly to $f$ on $S$. {{qed|lemma}}
Fix some $\epsilon \in \R_{> 0}$. Since $f_n \to f$ uniformly, there exists some $N \in \N$ such that:  :$\size {\map {f_n} x - \map f x} < \dfrac \epsilon 2$ for all $x \in S$ and $n > N$.  Then if $x \in S$ and $n, m > N$, we have: {{begin-eqn}} {{eqn	| l = \size {\map {f_n} x - \map {f_m} x} 	| r = \size {\map {f_n} x - \map f x - \paren {\map {f_m} x - \map f x} } }} {{eqn	| r = \size {\map {f_n} x - \map f x} + \size {\map {f_m} x - \map f x} 	| c = Triangle Inequality for Real Numbers }} {{eqn	| o = < 	| r = \epsilon }} {{end-eqn}} Since $\epsilon$ was arbitrary, $\sequence {f_n}$ is uniformly Cauchy on $S$. {{qed}}
Let $\delta = \epsilon - \norm {x - y}$. From the definition of open ball, this is strictly positive, since $y \in \map {B_\epsilon} x$. If $z \in \map {B_\delta} y$, then $\norm {y - z} < \delta$. So: : $\norm {x - z} \le \norm {x - y} + \norm {y - z} < \norm {x - y} + \delta = \epsilon$ Thus $z \in \map {B_\epsilon} x$. So $\map {B_\delta} y \subseteq \map {B_\epsilon} x$. {{qed}}
Let $\epsilon \in \R_{> 0}$.  Since $f_n \to f$ uniformly, there exists some $N \in \N$ such that:  :$\size {\map {f_n} x - \map f x} < \dfrac \epsilon 3$ for all $x \in S$ and $n \ge N$. Since $f_N$ is continuous at $x$, there exists some $\delta > 0$ such that:  :for all $y$ with $\size {x - y} < \delta$, we have $\size {\map {f_N} x - \map {f_N} y} < \dfrac \epsilon 3$ Then for $y$ with $\size {x - y} < \delta$ we have:  {{begin-eqn}} {{eqn	| l = \size {\map f x - \map f y} 	| r = \size {\map f x - \map {f_N} x + \map {f_N} x - \map {f_N} y + \map {f_N} y - \map f y} }} {{eqn	| r = \size {\paren {\map f x - \map {f_N} x} + \paren {\map {f_N} x - \map {f_N} y} + \paren {\map {f_N} y - \map f y} } }} {{eqn	| o = \le 	| r = \size {\map f x - \map {f_N} x} + \size {\map {f_N} x - \map {f_N} y} + \size {\map {f_N} y - \map f y} 	| c = Triangle Inequality for Real Numbers }} {{eqn	| o = < 	| r = 3 \times \frac \epsilon 3 }} {{eqn	| r = \epsilon }} {{end-eqn}} Since $\epsilon$ was arbitrary, $f$ is continuous at $x$. {{qed}}
Let $y \in \map {B_\epsilon} x$. From Open Ball of Point Inside Open Ball in Normed Vector Space, there exists $\delta \in \R_{>0}$ such that $\map {B_\delta} y \subseteq \map {B_\epsilon} x$ The result follows from the definition of open set. {{qed}}
Let $\displaystyle x \in \bigcap_{i \mathop = 1}^n U_i$. For each $i \in \closedint 1 n$, we have $x \in U_i$. Thus: : $\exists \epsilon_i > 0: \map {B_{\epsilon_i}} x \subseteq U_i$ where $\map {B_{\epsilon_i}} x$ is the open $\epsilon_i$-ball of $x$. Let $\displaystyle \epsilon = \min_{i \mathop = 1}^n \set {\epsilon_i}$. So: : $\epsilon > 0$. Let $y \in \map {B_\epsilon} x$. Then $\norm {x - y} < \epsilon$ Hence: :$\forall i \in \closedint 1 n : \norm {x - y} < \epsilon_i$ In other words: : $\map {B_\epsilon} x \subseteq \map {B_{\epsilon_i}} x \subseteq U_i$ for all $i \in \closedint 1 n$. So: : $\displaystyle \map {B_\epsilon} x \subseteq \bigcap_{i \mathop = 1}^n U_i$ The result follows. {{qed}}
Let $I$ be any indexing set. Let $U_i$ be open in $M$ for all $i \in I$. Let $\displaystyle x \in \bigcup_{i \mathop \in I} U_i$. Then $x \in U_k$ for some $k \in I$. Since $U_k$ is open in $M$: : $\displaystyle \exists \epsilon > 0: \map {B_\epsilon} x \subseteq U_k \subseteq \bigcup_{i \mathop \in I} U_i$ where $\map {B_\epsilon} x$ is the open $\epsilon$-ball of $x$ in $M$. The result follows. {{qed}}
By Limit of Uniformly Convergent Sequence of Integrable Functions is Integrable, $f$ is integrable. We have: {{begin-eqn}} {{eqn	| l = \size {\int_a^b \map f x \rd x - \int_a^b \map {f_n} x \rd x} 	| r = \size {\int_a^b \paren {\map f x - \map {f_n} x} \rd x} }} {{eqn	| o = \le 	| r = \int_a^b \size {\map f x - \map {f_n} x} \rd x 	| c = Triangle Inequality for Integrals }} {{eqn	| o = \le 	| r = \paren {b - a} \sup_{x \in \closedint a b} \size {\map f x - \map {f_n} x} 	| c = Upper and Lower Bounds of Integral }} {{end-eqn}} Let $\varepsilon \in \R_{> 0}$. Since $f_n \to f$ uniformly, we can find $N \in \N$ such that for $n > N$ we have: :$\displaystyle \sup_{x \in \closedint a b} \size {\map f x - \map {f_n} x} < \frac \varepsilon {b - a}$ by the definition of uniform convergence. So, for $n > N$ we have:  :$\displaystyle \paren {b - a} \sup_{x \in \closedint a b} \size {\map f x - \map {f_n} x} < \varepsilon$ and hence: :$\displaystyle \size {\int_a^b \map f x \rd x - \int_a^b \map {f_n} x \rd x} < \varepsilon$  Since $\varepsilon$ was arbitrary it follows that:  :$\displaystyle \lim_{n \to \infty} \int_a^b \map {f_n} x \rd x = \int_a^b \map f x \rd x$ as required. {{qed}} Category:Integral Calculus Category:Uniform Convergence hpgnx1cl23avi8v6oqmw2otmsa8ocqx
From Union with Superset is Superset: :$A \cup B = A \iff B \subseteq A$ Let $B$ occur. Let $\omega$ be the outcome of $\EE$. Let $\omega \in B$. That is, by definition of occurrence of event, $B$ occurs. Then by definition of subset: :$\omega \in A$ Thus by definition of occurrence of event, $A$ occurs. Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = T_1 \cup T_2       | r = \overline {\overline {T_1 \cup T_2} }       | c = Complement of Complement }} {{eqn | r = \overline {\overline T_1 \cap \overline T_2}       | c = De Morgan's Laws: Complement of Union }} {{end-eqn}} {{qed}}
Let $I$ be an indexing set (either finite or infinite). Let $\displaystyle \bigcap_{i \mathop \in I} V_i$ be the intersection of a indexed family of closed sets of $M$ indexed by $I$. By definition of closed set, each of $X \setminus V_i$ are by definition open in $M$. From De Morgan's laws: Difference with Intersection: :$\displaystyle X \setminus \bigcap_{i \mathop \in I} V_i = \bigcup_{i \mathop \in I} \paren {X \setminus V_i}$ We have that $\displaystyle \bigcup_{i \mathop \in I} \paren {X \setminus V_i}$ is the union of a indexed family of open sets of $M$ indexed by $I$. By Union of Open Sets of Normed Vector Space is Open, $\displaystyle \bigcup_{i \mathop \in I} \paren {X \setminus V_i} = X \setminus \bigcap_{i \mathop \in I} V_i$ is likewise open in $M$. Then by definition of closed set, $\displaystyle \bigcap_{i \mathop \in I} V_i$ is closed in $M$. {{qed}}
From Empty Set is Open in Normed Vector Space, $\O$ is open in $M$. But: :$X = \relcomp X \O$ where $\complement_X$ denotes the set complement relative to $X$. The result follows by definition of closed set. {{qed}}
Let $\sequence {s_n}$ be a sequence of real functions $S \to \R$ with:  :$\displaystyle \map {s_n} x = \sum_{k \mathop = 1}^n \map {f_k} x$ for each $n \in \N$ and $x \in S$. By the definition of uniform convergence of an infinite series: :$\displaystyle \sum_{n \mathop = 1}^\infty f_n$ is uniformly convergent {{iff}} $\sequence {s_n}$ is uniformly convergent. By Sequence of Functions is Uniformly Cauchy iff Uniformly Convergent: :$\sequence {s_n}$ is uniformly convergent {{iff}} it is uniformly Cauchy. That is for all $\varepsilon \in \R_{> 0}$ there exists $N \in \N$ such that for all $m, n > N$ we have: :$\size {\map {s_n} x - \map {s_m} x} < \varepsilon$ for all $x \in S$. Note that if $n = m$:  :$\size {\map {s_n} x - \map {s_m} x} = 0 < \varepsilon$ {{WLOG}}, we can therefore take $n > m$. If $n > m$, then: {{begin-eqn}} {{eqn	| l = \size {\map {s_n} x - \map {s_m} x} 	| r = \size {\sum_{k \mathop = 1}^n \map {f_k} x - \sum_{k \mathop = 1}^m \map {f_k} x} }} {{eqn	| r = \size {\sum_{k \mathop = m + 1}^n \map {f_k} x} }} {{end-eqn}} So $\sequence {s_n}$ is uniformly Cauchy {{iff}} for all $\varepsilon \in \R_{> 0}$ there exists $N \in \N$ such that: :$\displaystyle \size {\sum_{k \mathop = m + 1}^n \map {f_k} x} < \varepsilon$ for all $n > m > N$ and $x \in S$. {{qed}}
{{begin-eqn}} {{eqn | l = T_1 \cap T_2       | r = \overline {\overline {T_1 \cap T_2} }       | c = Complement of Complement }} {{eqn | r = \overline {\overline T_1 \cup \overline T_2}       | c = De Morgan's Laws: Complement of Intersection }} {{end-eqn}} {{qed}}
Let $\gcd \set {a, b, c} = d_1$. From definition: :$d_1 \divides a$, $d_1 \divides b$ and $d_1 \divides c$. By Common Divisor Divides GCD: :$d_1 \divides \gcd \set {a, b}$ and $d_1 \divides \gcd \set {a, c}$. By Common Divisor Divides GCD again: :$d_1 \divides \gcd \set {\gcd \set {a, b}, \gcd \set {a, c} }$. On the other hand, let $\gcd \set {\gcd \set {a, b}, \gcd \set {a, c} } = d_2$. From definition: :$d_2 \divides \gcd \set {a, b}$ and $d_2 \divides \gcd \set {a, c}$. From definition again: :$d_2 \divides a$, $d_2 \divides b$ and $d_2 \divides c$. Hence $d_2 \divides \gcd \set {a, b, c}$. Since $\gcd \set {\gcd \set {a, b}, \gcd \set {a, c} }$ and $\gcd \set {a, b, c}$ divide each other, by Absolute Value of Integer is not less than Divisors they must be equal. {{qed}} Category:Greatest Common Divisor Category:LCM of 3 Integers in terms of GCDs of Pairs of those Integers 1breqfkqm4awy95mtgjcqzl0krs2783
From Intersection with Subset is Subset: :$A \cap B = A \iff A \subseteq B$ Let $A$ occur. Let $\omega$ be the outcome of $\EE$. Let $\omega \in A$. That is, by definition of occurrence of event, $A$ occurs. Then by definition of subset: :$\omega \in B$ Thus by definition of occurrence of event, $B$ occurs. Hence the result. {{qed}}
By definition: :$A \subseteq \Omega$ and: :$\overline A = \relcomp \Omega A$ From Union with Relative Complement: :$A \cup \overline A = \Omega$ We then have from Kolmogorov axiom $(2)$ that: :$\map \Pr \Omega = 1$ The result follows by definition of certainty. {{qed}} {{LEM|Union with Relative Complement}} Category:Complementary Events Category:Set Union mbw4g8v1z5vt8du09qkcxemct1o1k0x
Let $A^-$ denote the closure of $A$. It is required to be shown that $x \in A^-$ {{iff}}, for every open neighborhood $U$ of $x$, the intersection $A \cap U$ is non-empty. For a subset $H \subseteq S$, let $H^{\complement}$ denote the relative complement of $H$ in $S$. We have that: {{begin-eqn}} {{eqn | l = A \cap U       | r = \O }} {{eqn | ll= \leadstoandfrom       | l = A       | o = \subseteq       | r = U^{\complement}       | c = Note that $U^{\complement}$ is closed }} {{eqn | ll= \leadstoandfrom       | l = A^-       | o = \subseteq       | r = U^{\complement}       | c = Set Closure is Smallest Closed Set in Topological Space }} {{eqn | ll= \leadstoandfrom       | l = U       | o = \subseteq       | r = \paren {A^-}^{\complement}       | c = Relative Complement inverts Subsets and Relative Complement of Relative Complement }} {{eqn | ll= \leadstoandfrom       | l = A^- \cap U       | o = =       | r = \O       | c =  }} {{end-eqn}} Thus: : $x \in U \iff x \notin A^-$ The result follows from the Rule of Transposition.
=== Necessary Condition === Let every open neighborhood $U$ of $x$ satisfy: :$A \cap U \ne \O$ Let $N$ be any neighborhood of $x$. By definition of a neighborhood: :$\exists V \in \tau : x \in V \subseteq N$ From Set is Open iff Neighborhood of all its Points, $V$ is an open neighborhood of $x$. Thus: :$A \cap V \ne \O$ By the contrapositive statement of Subsets of Disjoint Sets are Disjoint: :$A \cap N \ne \O$ Since $N$ was arbitrary then every neighborhood $N$ of $x$ satisfies: :$A \cap N \ne \O$ {{qed|lemma}} === Sufficient Condition === Let every neighborhood $N$ of $x$ satisfy: :$A \cap N \ne \O$ By definition, every open neighborhood $U$ of $x$ is a neighborhood of $x$. So every open neighborhood $U$ of $x$ satisfies: :$A \cap U \ne \O$
Let $T = \struct {S, \tau}$ be a Fortissimo space. === Necessary Condition === By Finite Topological Space is Compact, all finite sets are compact in $T$. {{qed|lemma}} === Sufficient Condition === We prove the contrapositive. Let $A$ be an infinite subset of $S$. Let $C$ be a countably infinite subset of $A$ that does not contain $p$. For each $x \in A$, $C \setminus \set x$ is a countably infinite set. Hence $\relcomp S {C \setminus \set x} = \relcomp S C \cup \set x$ is open in $T$. Thus $\CC = \set {\relcomp S C \cup \set x: x \in A}$ is an open cover of $A$. However, each set in $\CC$ contains exactly $1$ element in $C$, so a finite subset of $\CC$ can only contain a finite number of elements in $C$. Therefore $\CC$ has no finite subcover. Hence $A$ is not compact. This shows that if a set in Fortissimo space is compact, it must be finite. {{qed}} Category:Compact Spaces Category:Fortissimo Space jclupkhtqcxchvfqyn4lf2jpzpo26qi
=== Definition $1$ is equivalent to Definition $3$ === This is demonstrated in Boundary is Intersection of Closure with Closure of Complement. {{qed|lemma}} === Definition $2$ is equivalent to Definition $3$ === Let $x \in S$. By definition of the closure: :$x \in H^-$ {{iff}} every neighborhood $N$ of $x$ satisfies $H \cap N \ne \O$ :$x \in \paren{\overline H}^-$ {{iff}} every neighborhood $N$ of $x$ satisfies $\overline H \cap N \ne \O$ Therefore $x \in H^- \cap \paren{\overline H}^-$ {{iff}} every neighborhood $N$ of $x$ satisfies: :$H \cap N \ne \O$ and :$\overline H \cap N \ne \O$ {{qed}} Category:Boundaries 0oiw3cdykopt9xx4g9gsbubbas49fah
Let $\sequence {s_n}$ be sequence of real functions $S \to \R$ with:  :$\displaystyle \map {s_n} x = \sum_{k \mathop = 1}^n \map {f_n} x$ for each $n \in \N$ and $x \in S$. By Combination Theorem for Continuous Functions: Sum Rule: :$s_n$ is continuous at $x$ for all $n \in \N$. Since additionally $s_n \to f$ uniformly, we have by Uniformly Convergent Sequence of Continuous Functions Converges to Continuous Function:  :$f$ is continuous at $x$. {{qed}}
Let $x \in S$. Then $f_n$ is continuous at $x$ for all $n \in \N$. Since: :$\displaystyle \sum_{n \mathop = 1}^\infty f_n$ converges uniformly to $f$, we have by Uniformly Convergent Series of Continuous Functions Converges to Continuous Function: :$f$ is continuous at $x$. As $x \in S$ was arbitrary, we have that:  :$f$ is continuous. {{qed}} Category:Uniform Convergence Category:Continuous Functions j1ahir88958io84stzghg2mres6r4bv