hcju/proofwikiqa huggingface dataset:
=== Necessary Condition === Let $x \in S$ be a limit point of $H$. By definition of a limit point of $H$: :$\forall U \in \tau : x \in U$ satisfies $H \cap U \setminus \set x \ne \O$ By definition of a local basis of $T$: :$\BB_x \subseteq \tau$ The result follows. {{qed|lemma}} === Sufficient Condition === Let $x$ satisfy: :$\forall U \in \BB_x : H \cap U \setminus \set x \ne \O$ Let $V$ be any open neighborhood of $x$. By definition of a local basis of $T$: :$\exists U \in \BB : x \in U \subseteq V$ Then: :$H \cap U \setminus \set x \ne \O$ From the contrapositive statement of Subsets of Disjoint Sets are Disjoint: :$H \cap V \setminus \set x \ne \O$ Thus $x$ is a limit point of $H$ by definition. {{qed}} Category:Limit Points ed1ztna0nbks5iayet5eu6ds6u85jx0
=== Necessary Condition === Let $x \in S$ be an adherent point of $H$. By definition of an adherent point of $H$: :$\forall U \in \tau : x \in U$ satisfies $H \cap U \ne \O$ By definition of a local basis of $T$: :$\BB_x \subseteq \tau$ The result follows. {{qed|lemma}} === Sufficient Condition === Let $x$ satisfy: :$\forall U \in \BB_x : H \cap U \ne \O$ Let $V$ be any open neighborhood of $x$. By definition of a local basis of $T$: :$\exists U \in \BB : x \in U \subseteq V$ Then: :$H \cap U \ne \O$ From the contrapositive statement of Subsets of Disjoint Sets are Disjoint: :$H \cap V \ne \O$ Thus $x$ is an adherent point of $H$ by definition. {{qed}} Category:Adherent Points e266iczrqukz5c6eywtzap4udjt4qzq
Let $\overline {\paren{H^-}}$ denote the complement of $H^-$. Let $x \in \overline {\paren{H^-}}$. By definition of the closure of $H$: :$x$ is not a limit point of $H$. So: :$\exists \epsilon \in \R_{> 0} : \paren{\map {B_\epsilon} x \setminus \set x} \cap H = \O$ From Intersection with Set Difference is Set Difference with Intersection: :$\paren{\map {B_\epsilon} x \cap H} \setminus \set x = \O$ From Set Difference with Superset is Empty Set: :$\map {B_\epsilon} x \cap H \subseteq \set x$ By definition of the closure of $H$: :$x$ is not an isolated point of $H$. So: :$\map {B_\epsilon} x \cap H \ne \set x$ Thus: :$\map {B_\epsilon} x \cap H = \O$ From Empty Intersection iff Subset of Complement: :$\map {B_\epsilon} x \subseteq \overline {\paren{H^-}}$ It follows that $\overline {\paren{H^-}}$ is open in $M$. Thus $H^- $ is closed in $M$ by definition. {{qed}} Category:Set Closures gln0tdiitfl0gxjq2wpzk430zbd3zdy
Let $L \in X$. Then: :$\forall \epsilon_a \in \R_{> 0} : \exists N \in \N : \forall n \in \N : n > N \implies \norm {x_n - L}_a < \epsilon_a$ By equivalence of norms: :$\exists M \in \R_{> 0} : \norm {x_n - L}_b \le M \norm {x_n - L}_a < M \epsilon_a$ Let $\epsilon_b := M \epsilon_a$ Then: :$\forall \epsilon_b \in \R_{> 0}: \exists N \in \N : \forall n \in \N : n > N \implies \norm {x_n - L}_b < \epsilon_b$ Therefore, $\sequence{x_n}_{n \mathop \in \N}$ converges to $L$ also in $M_b$. {{qed}}
Note that if $f$ is a bijection, the result Bijection is Open iff Closed applies. It is to be shown that if $f$ is not a bijection, this is not necessarily the case. This is achieved by Proof by Counterexample: Let $\struct {\R^2, d}$ be the real number plane with the usual (Euclidean) topology. Let $\rho: \R^2 \to \R$ be the first projection on $\R^2$ defined as: :$\forall \tuple{x, y} \in \R^2: \map \rho {x, y} = x$ From Projection on Real Euclidean Plane is Open Mapping, $\rho$ is an open mapping. From [[Projection on Real Euclidean Plane is not Closed Mapping], $\rho$ is not a closed mapping. The result is apparent. {{qed}}
{{begin-eqn}} {{eqn | n = 1       | l = \int x^m \cos a x \rd x       | r = \frac {x^m \sin a x} a + \frac {m x^{m - 1} \cos a x} {a^2} - \frac {m \paren {m - 1} } {a^2} \int x^{m - 2} \cos a x \rd x       | c = Primitive of Power of x by Cosine of a x }} {{eqn | n = 2       | ll= \leadsto       | l = \int x^{m - 2} \cos a x \rd x       | r = \frac {x^{m - 2} \sin a x} a + \frac {\paren {m - 2} x^{m - 3} \cos a x} {a^2} - \frac {\paren {m - 2} \paren {m - 3} } {a^2} \int x^{m - 4} \cos a x \rd x       | c = setting $m$ equal to $m - 2$ }} {{eqn | ll= \leadsto       | l = \int x^m \cos a x \rd x       | r = \frac {x^m \sin a x} a + \frac {m x^{m - 1} \cos a x} {a^2} - \frac {m \paren {m - 1} } {a^2} \paren {\frac {x^{m-2} \sin a x} a + \frac {\paren {m-2} x^{m - 3} \cos a x} {a^2} - \frac {\paren {m - 2} \paren {m - 3} } {a^2} \int x^{m - 4} \cos a x \rd x}       | c = substituting $(2)$ into $(1)$ above }} {{eqn | r = \frac {x^m \sin a x} a + \frac {m x^{m - 1} \cos a x} {a^2} - \frac {\paren m \paren {m - 1} x^{m - 2} \sin a x} {a^3} - \frac {\paren m \paren {m - 1} \paren {m - 2} x^{m - 3} \cos a x} {a^4} + \frac {\paren m \paren {m - 1} \paren {m - 2} \paren {m - 3} } {a^4} \int x^{m - 4} \cos a x \rd x       | c = Real Multiplication Distributes over Addition }} {{eqn | r = \sum_{k \mathop = 1}^{m + 1} \paren {m^{\underline {k - 1} } \frac {x^{m + 1 - k} } {a^k} \map \sin {x + \frac {\pi} 2 \paren { k - 1 } } }       | c = {{Defof|Falling Factorial}} }} {{end-eqn}} {{qed}} Category:Primitives involving Cosine Function 9jbue7rvy7yr9my9w6go3hxdfzxwvq8
A counterexample is demonstrated in Homeomorphic Non-Comparable Particular Point Topologies. {{qed}}
Let $\FF$ be a filter on $S$. Let $\BB$ be a filter basis of $\FF$. === $(1)$ implies $(2)$ === Let $x \in S$ be a limit Point of $\BB$ by definition $1$. Then by definition $\FF$ converges on $x$. By definition of convergent filter: :$\forall N_x \subseteq S: N_x \in \FF$ where $N_x$ is a neighborhood of $x$. That is, every neighborhood of $x$ is an element of $\FF$. By definition of filter basis: :$\forall U \in \FF: \exists V \in \BB: V \subseteq U$ But we have shown that $N_x \in \FF$. Therefore: :$\exists V \in \BB: V \subseteq N_x$ That is, every neighborhood of $x$ contains a set of $\BB$. Thus $x \in S$ is a limit Point of $\BB$ by definition $2$. {{qed|lemma}} === $(2)$ implies $(1)$ === Let $x \in S$ be a limit Point of $\BB$ by definition $2$. Then by definition every neighborhood of $x$ contains a set of $\BB$. By definition of filter basis, $\FF := \set {V \subseteq X: \exists U \in \BB: U \subseteq V}$ is a filter on $S$ {{iff}}: :$\forall V_1, V_2 \in \BB: \exists U \in \BB: U \subseteq V_1 \cap V_2$ But then letting $V_1 = V_2 = V$, we have: :$\forall V \in \BB: \exists U \in \BB: U \subseteq V$ and so $V \in \FF$. Thus every neighborhood of $x$ is an element of $\FF$. Thus, by definition, $\FF$ converges on $x$. Thus $x \in S$ is a limit Point of $\BB$ by definition $1$. {{qed}} Category:Limit Points of Filter Bases b686t3dvy3zu8xc8nkijwc3361g4f7g
We prove this using First Principle of Mathematical Induction. Let $\map P n$ be the proposition: :For every integer $m$ with $N < m \le N + s_{n + 1}$: ::$m$ can be expressed as a sum of distinct elements in $\set {s_1, s_2, \dots, s_n}$. === Basis for the induction === From our assumption above, $\map P k$ is true. This is the basis for the induction. === Induction Hypothesis === Suppose for some $n > k$, $\map P {n - 1}$ is true. That is, for every integer $m$ with $N < m \le N + s_n$: :$m$ can be expressed as a sum of distinct elements in $\set {s_1, s_2, \dots, s_{n - 1}}$. This is the induction hypothesis. We need to show that $\map P n$ is true. That is, for every integer $m$ with $N < m \le N + s_{n + 1}$: :$m$ can be expressed as a sum of distinct elements in $\set {s_1, s_2, \dots, s_n}$. === Induction Step === This is the induction step: By the induction hypothesis, for every integer $m$ with $N < m \le N + s_n$: :$m$ can be expressed as a sum of distinct elements in $\set {s_1, s_2, \dots, s_{n - 1}}$. Hence we only need to consider $N + s_n < m \le N + s_{n + 1}$. For $m$ in this range: :$N < m - s_n \le N + s_{n + 1} - s_n \le N + s_n$ By the induction hypothesis, $m - s_n$ can be expressed as a sum of distinct elements in $\set {s_1, s_2, \dots, s_{n - 1}}$. Hence $m$ can be expressed as a sum of distinct elements in $\set {s_1, s_2, \dots, s_{n - 1}, s_n}$. Hence $\map P n$ is true. By the First Principle of Mathematical Induction, $\map P n$ is true for all $n \ge k$. {{qed|lemma}} Since $1 \le s_n < s_{n + 1}$ for every $n \in N$: :$s_n \ge n$ for every $n \in \N$ Let $K > N$. Then: {{begin-eqn}} {{eqn | l = N + s_{K + k + 1}       | o = \ge       | r = N + K + k + 1 }} {{eqn | o = >       | r = K }} {{eqn | o = >       | r = N }} {{end-eqn}} By the result above, since $K + k \ge k$: :$K$ can be expressed as the sum of distinct elements in $\set {s_1, s_2, \dots, s_{K + k}}$. Thus every number greater than $N$ can be expressed as the sum of distinct elements in $S$. {{qed}} {{Namedfor|Hans-Egon Richert}} Category:Number Theory hu4tfm247b0ro2ijpsr8eu5hdkbi57h
=== $(1)$ implies $(2)$ === Let $\BB$ be a filter basis of $\FF$ by definition $1$. Then by definition: :$\BB \subset \powerset S$ such that $\O \notin \BB$ and $\BB \ne \O$ and $\FF := \set {V \subseteq S: \exists U \in \BB: U \subseteq V}$ is a filter on $S$ {{iff}}: :$\forall V_1, V_2 \in \BB: \exists U \in \BB: U \subseteq V_1 \cap V_2$ Let $U \in \FF$. Then by definition of $\FF$: :$\exists V \in \BB: V \subseteq U$ Thus $\BB$ is a filter basis of $\FF$ by definition $2$. {{qed|lemma}} === $(2)$ implies $(1)$ === Let $\BB$ be a filter basis of $\FF$ by definition $2$. By definition, $\BB$ is a '''filter basis''' of $\FF$ {{iff}}: :$\forall U \in \FF: \exists V \in \BB: V \subseteq U$ Let $V_1, V_2 \in \BB$. Then: :$V_1, V_2 \in \FF$ By definition of filter: :$V_1 \cap V_2 \in \FF$ and so: :$\exists U \in \BB: U \subseteq V_1 \cap V_2$ Thus $\BB$ is a filter basis of $\FF$ by definition $1$. {{qed}} Category:Filter Bases qvl6ivhbdzhxpjlz7byc879nsga1kb2
Let $S = \set {s_n}_{n \mathop \in N}$ be the set of primes. Then $S = \set {2, 3, 5, 7, 11, 13, \dots}$. By Bertrand-Chebyshev Theorem: :$s_{n + 1} \le 2 s_n$ for all $n \in \N$. We observe that every integer $n$ where $6 < n \le 6 + s_6 = 19$ can be expressed as a sum of distinct elements in $\set {s_1, \dots, s_5} = \set {2, 3, 5, 7, 11}$. Hence the result by Richert's Theorem. {{qed|lemma}} Here is a demonstration of our claim: {{begin-eqn}} {{eqn | l = 1       | o = \text {is}       | r = \text {less than the smallest prime } 2 }} {{eqn | l = 2       | r = 2 }} {{eqn | l = 3       | r = 3 }} {{eqn | l = 4       | o = \ne       | r = 2 + 3 }} {{eqn | l = 5       | r = 5 }} {{eqn | l = 6       | o = \ne       | r = 2 + 3 \text { or } 2 + 5 }} {{eqn | l = 7       | r = 7 }} {{eqn | l = 8       | r = 3 + 5 }} {{eqn | l = 9       | r = 2 + 7 }} {{eqn | l = 10       | r = 2 + 3 + 5 }} {{eqn | l = 11       | r = 11 }} {{eqn | l = 12       | r = 2 + 3 + 7 }} {{eqn | l = 13       | r = 2 + 11 }} {{eqn | l = 14       | r = 3 + 11 }} {{eqn | l = 15       | r = 3 + 5 + 7 }} {{eqn | l = 16       | r = 5 + 11 }} {{eqn | l = 17       | r = 2 + 3 + 5 + 7 }} {{eqn | l = 18       | r = 2 + 5 + 11 }} {{eqn | l = 19       | r = 3 + 5 + 11 }} {{end-eqn}} {{qed}} Category:Prime Numbers moh4pi0nyd1rhipoc6kqau1oqjzdqar
By Multiple of Function of Bounded Variation is of Bounded Variation, we have that:  :$-g$ is of bounded variation. So, by Sum of Functions of Bounded Variation is of Bounded Variation, we have that:  :$f + \paren{-g} = f - g$ is of bounded variation with:  :$V_{f - g} \le V_f + V_{-g}$ where $V_{-g}$ is the total variation of $-g$.  We have, by Multiple of Function of Bounded Variation is of Bounded Variation: {{begin-eqn}} {{eqn	| l = V_{-g} 	| r = \size {-1} V_g }} {{eqn	| r = V_g }} {{end-eqn}} so:  :$V_{f - g} \le V_f + V_g$ {{qed}}
For each finite subdivision $P$ of $\closedint a b$, write:  :$P = \set {x_0, x_1, \ldots, x_n }$ with: :$a = x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n = b$ By Function of Bounded Variation is Bounded:  :$f$ and $g$ are bounded.  So, there exists $A, B \in \R$ such that:  :$\size {\map f x} \le B$ :$\size {\map g x} \le A$ for all $x \in \closedint a b$. Then: {{begin-eqn}} {{eqn	| l = \map {V_{f \times g} } P 	| r = \sum_{i \mathop = 1}^n \size {\map {\paren {f \times g} } {x_i} - \map {\paren {f \times g} } {x_{i - 1} } } 	| c = using the notation from the definition of bounded variation }} {{eqn	| r = \sum_{i \mathop = 1}^n \size {\map f {x_i} \map g {x_i} - \map f {x_{i - 1} } \map g {x_{i - 1} } } }} {{eqn	| r = \sum_{i \mathop = 1}^n \size {\map f {x_i} \map g {x_i} - \map f {x_{i - 1} } \map g {x_i} + \map f {x_{i - 1} } \map g {x_i} - \map f {x_{i - 1} } \map g {x_{i - 1} } } }} {{eqn	| o = \le 	| r = \sum_{i \mathop = 1}^n \size {\map f {x_i} \map g {x_i} - \map f {x_{i - 1} } \map g {x_i} } + \sum_{i \mathop = 1}^n \size {\map f {x_{i - 1} } \map g {x_i} - \map f {x_{i - 1} } \map g {x_{i - 1} } } 	| c = Triangle Inequality }} {{eqn	| r = \sum_{i \mathop = 1}^n \size {\map g {x_i} } \size {\map f {x_i} - \map f {x_{i - 1} } } + \sum_{i \mathop = 1}^n \size {\map f {x_{i - 1} } } \size {\map g {x_i} - \map g {x_{i - 1} } } }} {{eqn	| o = \le 	| r = A \sum_{i \mathop = 1}^n \size {\map f {x_i} - \map f {x_{i - 1} } } + B \sum_{i \mathop = 1}^n \size {\map g {x_i} - \map g {x_{i - 1} } } 	| c = since $\size {\map g {x_i} } \le A$ and $\size {\map f {x_{i - 1} } } \le B$ }} {{eqn	| r = A \map {V_f} P + B \map {V_g} P }} {{end-eqn}} Since $f$ and $g$ are of bounded variation, there exists $M, K \in \R$ such that:  :$\map {V_f} P \le M$ :$\map {V_g} P \le K$ for all finite subdivisions $P$. We therefore have:  :$\map {V_{f \times g} } P \le A M + B K$ so $f \times g$ is of bounded variation. Further, we have:  {{begin-eqn}} {{eqn	| l = V_{f \times g} 	| r = \sup_P \paren {\map {V_{f \times g} } P} 	| c = {{Defof|Total Variation}} }} {{eqn	| o = \le 	| r = \sup_P \paren {A \map {V_f} P} + \sup_P \paren {B \map {V_g} P} }} {{eqn	| r = A \sup_P \paren {\map {V_f} P} + B \sup_P \paren {\map {V_g} P} 	| c = Multiple of Supremum }} {{eqn	| r = A V_f + B V_g }} {{end-eqn}} {{qed}}
We have that $\sequence {x_n}_{n \mathop \in \N}$ is a Cauchy sequence in $M_a$. Then: :$\forall \epsilon_a \in \R_{> 0} : \exists N \in \N : \forall n, m \in \N : n, m > N \implies \norm {x_n - x_m}_a < \epsilon_a$ By equivalence of norms: :$\exists M \in \R_{> 0} : \norm {x_n - x_m}_b \le M \norm {x_n - x_m}_a < M \epsilon_a$ Let $\epsilon_b := M \epsilon_a$ Then: :$\forall \epsilon_b \in \R_{> 0} : \exists N \in \N : \forall n \in \N : n, m > N \implies \norm {x_n - x_m}_b < \epsilon_b$ Therefore, $\sequence {x_n}_{n \mathop \in \N}$ is also a Cauchy sequence in $M_b$. {{qed}}
Proof by Counterexample: Take $a = 0$, $b = 1$.  Let $f : \closedint 0 1 \to \R$ have:  :$\map f x = \sqrt x$ for all $x \in \closedint 0 1$.  Note that $f$ is increasing, so by Monotone Function is of Bounded Variation: :$f$ is of  bounded variation. By Derivative of Power, $f$ is differentiable on $\openint 0 1$ with derivative: :$\map {f'} x = \dfrac 1 {2 \sqrt x}$ Note that this is unbounded as $x \to 0^+$. So $f$ does not have a bounded derivative, despite being of bounded variation. {{qed}}
Let $T$ have the specified property. By definition of a normal space, for $T$ to be normal, it has to be both $T_4$ space and a $T_1$ space. From Urysohn's Lemma Converse, $T$ is a $T_4$ space. It remains to be shown that $T$ is not necessarily a $T_1$ space. This is done by Proof by Counterexample: Let $S$ be a set and let $\PP$ be a partition on $S$ which is specifically not the (trivial) partition of singletons. Let $T = \struct {S, \tau}$ be the partition space whose basis is $\PP$. From Partition Topology is $T_4$, we have that $T$ is a $T_4$ space. From Urysohn's Lemma, for all $A, B \subseteq S$ be closed sets of $T$ such that $A \cap B = \O$, there exists an Urysohn function for $A$ and $B$. From Partition Topology is not $T_1$, $T$ is not a $T_1$ space. The result follows. {{qed}}
Proof by Counterexample: Let $S$ be a set and let $\PP$ be a partition on $S$ which is specifically not the (trivial) partition of singletons. Let $T = \struct {S, \tau}$ be the partition space whose basis is $\PP$. From Partition Topology is $T_{3 \frac 1 2}$, we have that $T$ is a $T_{3 \frac 1 2}$ space. From Partition Topology is not Hausdorff, $T$ is not a $T_2$ (Hausdorff) space. The result follows. {{qed}}
Let $T = \struct {S, \tau}$ be a finite complement topology on an infinite set $S$. Let $D = \struct {\set {0, 1}, \vartheta}$ be the indiscrete topology on two points. Let $T \times D$ be the double pointed topology on $T$. From Double Pointed Finite Complement Topology is Compact, $T \times D$ is compact. From Double Pointed Finite Complement Topology fulfils no Separation Axioms, $T \times D$ is not a $T_0$ space, $T_1$ space, $T_2$ space, $T_3$ space, $T_4$ space or $T_5$ space. {{qed}}
By definition of everywhere dense, $A$ is everywhere dense {{iff}}: :$A^- = S$ where $A^-$ is the closure of $A$. That happens {{iff}}: {{begin-eqn}} {{eqn | l = \paren {\relcomp S A}^\circ       | r = \relcomp S {A^-}       | c = Complement of Interior equals Closure of Complement }} {{eqn | r = \relcomp S S }} {{eqn | r = \O       | c = Relative Complement with Self is Empty Set }} {{end-eqn}} Hence the result. {{qed}} Category:Denseness Category:Set Interiors c5ojyec4e826itgus8uvohb8qc6tea1
Let $A, B \in \TT$ both be greatest sets of $\TT$. Since $A$ is the greatest set: :$B \subseteq A$ Since $B$ is the greatest set: :$A \subseteq B$ Hence, by definition of set equality: :$A = B$ Therefore the greatest set of $\TT$ is unique. {{qed}} Category:Set Theory Category:Order Theory ks9f2fsa7hkz4doxp5oul9yulae66pn
Let $S = \set {0, 1}$ and $\TT = \set {\set 0, \set 1} \in \powerset S$. Then since $\set 0 \not \subseteq \set 1$: :$\set 1$ is not the greatest set of $\TT$. Similarly, since $\set 1 \not \subseteq \set 0$: :$\set 0$ is not the greatest set of $\TT$. Therefore $\TT$ has no greatest set. {{qed}} Category:Set Theory Category:Order Theory kar4bgf3c1wsygg84qvh6b2ranls27q
Let $A, B \in \TT$ both be smallest sets of $\TT$. Since $A$ is the smallest set: :$A \subseteq B$ Since $B$ is the smallest set: :$B \subseteq A$ Hence, by definition of set equality: :$A = B$ Therefore the smallest set of $\TT$ is unique. {{qed}} Category:Set Theory Category:Order Theory m04postjr6kfx509nqofattbnx1958e
Let $S = \set {0, 1}$ and $\TT = \set {\set 0, \set 1} \in \powerset S$. Then since $\set 0 \not \subseteq \set 1$: :$\set 0$ is not the smallest set of $\TT$. Similarly, since $\set 1 \not \subseteq \set 0$: :$\set 1$ is not the smallest set of $\TT$. Therefore $\TT$ has no smallest set. {{qed}} Category:Set Theory Category:Order Theory 3ifxyjo0zvbdefzijhv3jh37ipfsiw5
Let $y \in S$. Since $f^\to$ is a surjection: :$\exists U \in \powerset S: \map {f^\to} U = \set y$ From definition of direct image mapping: :$\set y \ne \O \implies U \ne \O$ Let $x \in U$. Then from definition of direct image mapping: :$\map f x = y$ Since $y$ is arbitrary, $f$ is a surjection. {{qed}}
Let $A \subseteq \R$ be the set of all points on $\R$ defined as: :$A := \set {\dfrac 1 n : n \in \Z_{>0} }$ Let $M = \struct {A, \tau_d}$ be the integer reciprocal space under the usual (Euclidean) topology. Let $\sequence {x_n}$ be a sequence in $A$ that converges to the limit $l \in A$. From Integer Reciprocal Space contains Cauchy Sequence with no Limit Point, $\sequence {x_n}$ is a Cauchy sequence in $M$ which does not converge in $M$. {{qed}}
In the following, $H^-$ denotes the closure of the set $H$. {{begin-eqn}} {{eqn | l = \paren {\bigcup_{H \mathop \in \mathbb H} H}^\circ       | r = T \setminus \paren {T \setminus \bigcup_{H \mathop \in \mathbb H} H}^-       | c = Complement of Interior equals Closure of Complement }} {{eqn | r = T \setminus \paren {\paren {\bigcap_{H \mathop \in \mathbb H} \paren {T \setminus H} }^-}       | c = De Morgan's Laws: Difference with Union }} {{end-eqn}} At this point we note that: :$(1): \quad \displaystyle \paren {\bigcap_{H \mathop \in \mathbb H} \paren {T \setminus H} }^- \subseteq \bigcap_{H \mathop \in \mathbb H} \paren {T \setminus H}^-$ from Closure of Intersection is Subset of Intersection of Closures. Then we note that: :$\displaystyle T \setminus \paren {\bigcap_{H \mathop \in \mathbb H} \paren {T \setminus H}^-} \subseteq T \setminus \paren {\paren {\bigcap_{H \mathop \in \mathbb H} \paren {T \setminus H} }^-} $ from $(1)$ and Set Complement inverts Subsets. Then we continue: {{begin-eqn}} {{eqn | l = T \setminus \paren {\bigcap_{H \mathop \in \mathbb H} \paren {T \setminus H}^-}       | r = T \setminus \paren {\bigcap_{H \mathop \in \mathbb H} T \setminus H^\circ}       | c = Complement of Interior equals Closure of Complement }} {{eqn | r = T \setminus \paren {T \setminus \paren {\bigcup_{H \mathop \in \mathbb H} H^\circ} }       | c = De Morgan's Laws: Difference with Union }} {{eqn | r = \bigcup_{H \mathop \in \mathbb H} H^\circ       | c = Relative Complement of Relative Complement }} {{end-eqn}} {{qed}}
We show that the complement $X \setminus \map {B_\epsilon^-} x$ is open in $M$. Let $y \in X \setminus \map {B_\epsilon^-} x$. Then by definition of closed ball: :$\norm {x - y} > \epsilon$ Put: :$\delta := \norm {x - y} - \epsilon > 0$ Then: :$\norm {x - y} - \delta = \epsilon$ Let $z \in \map {B_\delta} y$. Then: {{begin-eqn}} {{eqn | l = \norm {x - z}       | o = \ge        | r = \norm {x - y} - \norm {y - z}       | c = Reverse Triangle Inequality }} {{eqn | o = >       | r = \norm {x - y} - \delta }} {{eqn | r = \epsilon }} {{end-eqn}} and so: :$z \notin \map {B_\epsilon^-} x$ Then: :$\map {B_\delta} y \subseteq X \setminus \map {B_\epsilon^-} x$ so $X \setminus \map {B_\epsilon^-} x$ is open in $M$. Hence, by definition of closed set: :$\map {B_\epsilon^-} x$ is closed in $M$. {{qed}}
Let $\map {B_1} 0 = \set {x \in X : \norm x < 1}$ be an open ball. Let $\map {B_1^-} 0 = \set {x \in X : \norm x \le 1}$ be a closed ball. Then: :$\displaystyle X = \Bbb S \bigcup \relcomp X {\Bbb S}$ where :$\displaystyle \relcomp X {\Bbb S} = \map {B_1} 0 \bigcup \paren {X \setminus \map {B_1^-} 0}$ is the relative complement of $\Bbb S$ in $X$. We have that Closed Ball is Closed in Normed Vector Space. By definition, $X \setminus \map {B_1^-} 0$ is open. Furthermore, Open Ball is Open Set in Normed Vector Space. By Union of Open Sets of Normed Vector Space is Open, $\relcomp X {\Bbb S}$ is open. By definition, $\Bbb S$ is closed. {{qed}}
Every open set of $T^*_p = \struct {S^*_p, \tau^*_p}$ except $\O$ contains the point $p$ by definition. So every open set $U \in \tau^*_p$ such that $x \in U$ contains $p$. So by definition of the limit point of a set, $x$ is a limit point of $U$. {{qed}}
Let $\sequence {a_i}$ be a convergent sequence in $T$ whose limit is $\alpha$. Then by definition every open set in $T$ containing $\alpha$ contains all but a finite number of terms of $\sequence {a_i}$. This includes the open set $\set {\alpha, p}$. Thus $\sequence {a_i}$ is such that, except for a finite number of terms, all are equal either to $\alpha$ or $p$. Thus $\sequence {a_i}$ is such that, except for a finite number of terms, all such that $a_i \ne p$, are equal to $\alpha$. The result follows. {{qed}}
Let $\beta$ be an accumulation point of $\sequence {a_i}$. Then by definition: :all open sets of $T$ which contain $\beta$ also contain an infinite number of terms of $\sequence {a_i}$. This condition applies to the open set $\set {\beta, p}$. So $\set {\beta, p}$ contains an infinite number of terms of $\sequence {a_i}$. The result follows. {{qed}}
Let $\epsilon$ be a positive real number. Since $f$ is absolutely continuous, there exists real $\delta > 0$ such that for all collections of disjoint closed real intervals $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq I$ with:  :$\displaystyle \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta$ we have:  :$\displaystyle \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \epsilon$ By the Reverse Triangle Inequality, we have:  :$\size {\map f {b_i} - \map f {a_i} } \ge \size {\size {\map f {b_i} } - \size {\map f {a_i} } }$ Therefore:  :$\displaystyle \sum_{i \mathop = 1}^n \size {\size {\map f {b_i} } - \size {\map f {a_i} } } \le \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \epsilon$ whenever: :$\displaystyle \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta$ Since $\epsilon$ was arbitrary: :$\size f$ is absolutely continuous. {{qed}} Category:Absolutely Continuous Functions fp6q7lzsazm506llf7tqahhf1e5yh6a