hcju/proofwikiqa huggingface dataset:
We use the notation from the definition of bounded variation. Let $P = \set {x_0, x_1, \ldots, x_{\size P - 1} }$ be a finite subdivision of $\closedint a b$. As $f$ is monotone, it is either increasing or decreasing. First consider the case of $f$ increasing, then:  :$\map f {x_i} \ge \map f {x_{i - 1} }$ for all $i \in \N$ with $i \le \size P - 1$. By the definition of the absolute value, we have:  :$\size {\map f {x_i} - \map f {x_{i - 1} } } = \map f {x_i} - \map f {x_{i - 1} }$ Therefore: {{begin-eqn}} {{eqn	| l = \map {V_f} P 	| r = \sum_{i \mathop = 1}^{\size P - 1} \size {\map f {x_i} - \map f {x_{i - 1} } } }} {{eqn	| r = \sum_{i \mathop = 1}^{\size P - 1} \paren {\map f {x_i} - \map f {x_{i - 1} } } }} {{eqn	| r = \map f {x_{\size P - 1} } - \map f {x_0} 	| c = Telescoping Series: Example 2 }} {{eqn	| r = \map f b - \map f a }} {{end-eqn}} for all finite subdivisions $P$. Note that $\map f b - \map f a$ is independent of the subdivision $P$. Therefore if $f$ is increasing, it is of bounded variation. Now consider the case of $f$ decreasing. We instead have: :$\map f {x_i} \le \map f {x_{i - 1} }$ for all $i \in \N$ with $i \le \size P - 1$. Hence, by the definition of the absolute value, we instead have: :$\size {\map f {x_i} - \map f {x_{i - 1} } } = -\paren {\map f {x_i} - \map f {x_{i - 1} } }$ Therefore: {{begin-eqn}} {{eqn	| l = \map {V_f} P 	| r = \sum_{i \mathop = 1}^{\size P - 1} \size {\map f {x_i} - \map f {x_{i - 1} } } }} {{eqn	| r = -\sum_{i \mathop = 1}^{\size P - 1} \paren {\map f {x_i} - \map f {x_{i - 1} } } }} {{eqn	| r = -\paren {\map f b - \map f a} 	| c = as in the previous calculation }} {{eqn	| r = \map f a - \map f b }} {{end-eqn}} for all finite subdivisions $P$. Note again that $\map f a - \map f b$ is independent of the subdivision $P$.  Therefore if $f$ is decreasing, it is of bounded variation. Hence, if $f$ is either increasing or decreasing, it is of bounded variation. Hence if $f$ is monotone, it is of bounded variation. {{qed}}
For each finite subdivision $P$ of $\closedint a b$, write:  :$P = \set {x_0, x_1, \ldots, x_n }$ with: :$a = x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n = b$ Since the derivative of $f$ is bounded, there exists some $M \in \R$ such that:  :$\size {\map {f'} x} \le M$ for all $x \in \openint a b$. By the Mean Value Theorem, for each $i \in \N$ with $i \le n$, there exists $\xi_i \in \openint {x_{i - 1} } {x_i}$ such that:  :$\map {f'} {\xi_i} = \dfrac {\map f {x_i} - \map f {x_{i - 1} } } {x_i - x_{i - 1} }$ Note that, from the boundedness of $f'$:  :$\size {\map {f'} {\xi_i} } \le M$ We also have from the fact that $x_i > x_{i - 1}$:  :$\size {x_i - x_{i - 1} } = x_i - x_{i - 1}$ So, for each $i$:  {{begin-eqn}} {{eqn	| l = \size {\map f {x_i} - \map f {x_{i - 1} } } 	| r = \size {\map {f'} {\xi_i} } \paren {x_i - x_{i - 1} } }} {{eqn	| o = \le 	| r = M \paren {x_i - x_{i - 1} } }} {{end-eqn}} We therefore have:  {{begin-eqn}} {{eqn	| l = \map {V_f} P 	| r = \sum_{i \mathop = 1}^n \size {\map f {x_i} - \map f {x_{i - 1} } } 	| c = using the notation from the definition of bounded variation }} {{eqn	| o = \le 	| r = M \sum_{i \mathop = 1}^n \paren {x_i - x_{i - 1} } }} {{eqn	| r = M \paren {x_n - x_0} 	| c = Telescoping Series: Example 2 }} {{eqn	| r = M \paren {b - a} }} {{end-eqn}} for all finite subdivisions $P$. So $f$ is of bounded variation.  {{qed}}
We use the notation from the definition of bounded variation. Note that by the definition of absolute value, we have:  :$\size x \ge 0$ for all $x \in \R$. Let $P$ be a finite subdivision of $\closedint a b$. Then:  :$\displaystyle \map {V_f} P = \sum_{i \mathop = 1}^{\size P - 1} \size {\map f {x_i} - \map f {x_{i - 1} } } \ge 0$ So, by the definition of supremum, we have:  :$\displaystyle V_f = \sup_P \paren {\map {V_f} P} \ge 0$ Let $f$ be constant. Then:  :$\size {\map f {x_i} - \map f {x_{i - 1} } } = 0$ for all $x_{i - 1}, x_i \in \closedint a b$. So, for any finite subdivision $P$ we have:  {{begin-eqn}} {{eqn	| l = \map {V_f} P 	| r = \sum_{i \mathop = 1}^{\size P - 1} \size {\map f {x_i} - \map f {x_{i - 1} } } }} {{eqn	| r = \sum_{i \mathop = 1}^{\size P - 1} 0 }} {{eqn	| r = 0 }} {{end-eqn}} We then have:  {{begin-eqn}} {{eqn	| l = V_f 	| r = \sup_P \paren {\map {V_f} P} 	| c = {{Defof|Total Variation}} }} {{eqn	| r = \sup \set {0} }} {{eqn	| r = 0 	| c = {{Defof|Supremum of Subset of Real Numbers}} }} {{end-eqn}} So if $f$ is constant, then $f$ is of bounded variation with: :$V_f = 0$ It remains to show that if $V_f = 0$, then $f$ is constant. It suffices to show that if $f$ is not constant then $V_f > 0$. Since $f$ is non-constant, we can pick $x \in \openint a b$ such that either:  :$\map f x \ne \map f a$ or: :$\map f x \ne \map f b$ So that either: :$\size {\map f x - \map f a} > 0$ or: :$\size {\map f b - \map f x} > 0$ Note that we then have:  {{begin-eqn}} {{eqn	| l = \map {V_f} {\set {a, x, b} } 	| r = \size {\map f x - \map f a} + \size {\map f b - \map f x} }} {{eqn	| o = > 	| r = 0 }} {{end-eqn}} We must then have:  :$V_f \ge \map {V_f} {\set {a, x, b} } > 0$ Hence the claim. {{qed}}
=== Reflexivity === Let $\norm {\, \cdot \,}$ be a norm on $X$. Then for all $x \in X$ we have that: :$\norm x = 1 \cdot \norm {x}$. Therefore: :$1 \cdot \norm x \le \norm x \le 1 \cdot \norm x$ Hence: :$\norm {\, \cdot \,} \sim \norm {\, \cdot \,}$. {{qed|lemma}} === Symmetry === Suppose, $\norm {\, \cdot \,}_a \sim \norm {\, \cdot \,}_b$. Then: :$\exists m, M \in \R_{> 0} : m \le M : \forall x \in X : m \norm x_b \le \norm x_a \le M \norm x_b$ Consider two inequalities, obtained by division by $M$ and $m$: :$\dfrac m M \norm x_b \le \dfrac 1 M \norm x_a \le \norm x_b$ :$\norm x_b \le \dfrac 1 m \norm x_a \le \dfrac M m \norm x_b$ Notice, that:  :$\dfrac 1 M \norm x_a \le \norm x_b \le \dfrac 1 m \norm x_a$ We have that $m \le M$ implies $\dfrac 1 m \ge \dfrac 1 M$. Define $C := \dfrac 1 m$ and $c := \dfrac 1 M$. Hence, $c \le C$ and: :$\norm {\, \cdot \,}_b \sim \norm {\, \cdot \,}_a$ {{qed|lemma}} === Transitivity === Suppose, $\norm {\, \cdot \,}_a \sim \norm {\, \cdot \,}_b$ and $\norm {\, \cdot \,}_b \sim \norm {\, \cdot \,}_c$.   Then: :$\exists m_{a b}, M_{a b} \in \R_{> 0} : m_{a b} \le M_{a b} : \forall x \in X : m_{a b} \norm x_b \le \norm x_a \le M_{a b} \norm x_b$  :$\exists m_{b c}, M_{b c} \in \R_{> 0} : m_{b c} \le M_{b c} : \forall x \in X : m_{b c} \norm x_c \le \norm x_b \le M_{b c} \norm x_c$ Generate $2$ more inequalities by multiplying the second inequality by $m_{a b}$ and $M_{a b}$: :$m_{a b} m_{b c} \norm x_c \le m_{a b} \norm x_b \le m_{a b} M_{b c} \norm x_c$ :$M_{a b} m_{b c} \norm x_c \le M_{a b} \norm x_b \le M_{a b} M_{b c} \norm x_c$ From above it follows that: {{begin-eqn}} {{eqn | l = m_{a b} m_{b c} \norm x_c       | o = \le        | r = m_{a b} \norm x_b  }} {{eqn | o = \le        | r = \norm x_a  }} {{eqn | o = \le        | r = M_{a b} \norm x_b  }} {{eqn | o = \le        | r = M_{a b} M_{b c} \norm x_c }} {{end-eqn}} Define $k := m_{a b} m_{b c}$ and $K := M_{a b} M_{b c}$. Then $k \le K$ and: :$k \norm x_c \le \norm x_a \le K \norm x_c$ Therefore: :$\norm {\, \cdot \,}_a \sim \norm {\, \cdot \,}_c$ {{qed}}
From Definite Integral of Constant, if $\map f x = 0$ for all $x \in \closedint a b$, then:  :$\displaystyle \int_a^b \map f x \rd x = 0$ We want to show that if: :$\displaystyle \int_a^b \map f x \rd x = 0$ then: :$\map f x = 0$ for all $x \in \closedint a b$. Since $\map f x \ge 0$, by Relative Sizes of Definite Integrals: :$\displaystyle \int_a^b \map f x \rd x \ge 0$ It therefore suffices to show that if:  :$\map f x > 0$ for some $x \in \closedint a b$ then: :$\displaystyle \int_a^b \map f x \rd x > 0$ We split this into three cases:  :$x = a$ :$a < x < b$ :$x = b$ since continuity on endpoints is defined slightly differently. Consider the case:  :$a < x < b$ By continuity at $x$ we have that there exists $\delta > 0$ such that:  :for all $y \in \openint {x - \delta} {x + \delta}$ we have $\size {\map f y - \map f x} < \dfrac {\map f x} 2$ In particular for $y \in \openint {x - \delta} {x + \delta}$ we have:  :$0 < \dfrac {\map f x} 2 < \map f y$ Pick $\delta$ sufficiently small so that $\openint {x - \delta} {x + \delta} \subseteq \closedint a b$ We then have: {{begin-eqn}} {{eqn	| l = \int_a^b \map f y \rd y 	| r = \int_a^{x - \delta} \map f y \rd y + \int_{x - \delta}^{x + \delta} \map f y \rd y + \int_{x + \delta}^b \map f y \rd y 	| c = Sum of Integrals on Adjacent Intervals for Continuous Functions }} {{eqn	| o = \ge 	| r = \int_{x - \delta}^{x + \delta} \map f y \rd y 	| c = Relative Sizes of Definite Integrals gives $\displaystyle \int_a^{x - \delta} \map f y \rd y + \int_{x + \delta}^b \map f y \rd y \ge 0$ }} {{eqn	| o = > 	| r = \delta \map f x 	| c = Relative Sizes of Definite Integrals, Definite Integral of Constant }} {{eqn	| o = > 	| r = 0 }} {{end-eqn}} In the case $x = a$, by the definition of right continuity, there exists $\delta > 0$ such that:  :for all $x \in \openint a {a + \delta}$ we have $\size {\map f x - \map f a} < \dfrac {\map f a} 2$ That is, there exists some $x \in \openint a b$ such that:  :$\map f x > \dfrac {\map f a} 2 > 0$ So the former proof applies in this case.  The case $x = b$ follows similarly. In the case $x = b$, by the definition of left continuity, there exists $\delta > 0$ such that:  :for all $x \in \openint {b - \delta} b$ we have $\size {\map f x - \map f b} < \dfrac {\map f b} 2$ That is, there exists some $x \in \openint a b$ such that:  :$\map f x > \dfrac {\map f b} 2 > 0$ So, again, the proof for the case $a < x < b$ applies. We have covered all three cases, so we are done. {{qed}}
From Continuous Real Function is Darboux Integrable, $f$ is Darboux integrable on $\closedint a b$. Let $F : \closedint a b \to \R$ be a real function defined by:  :$\displaystyle \map F x = \int_a^x \map f x \rd x$ We are assured that this function is well-defined, since $f$ is integrable on $\closedint a b$. From Fundamental Theorem of Calculus: First Part, we have:  :$F$ is continuous on $\closedint a b$ :$F$ is differentiable on $\openint a b$ with derivative $f$ Note that:  {{begin-eqn}} {{eqn	| l = \map {F'} x 	| r = \map f x }} {{eqn	| o = \ge 	| r = 0 }} {{end-eqn}} for all $x \in \openint a b$.  We therefore have, by Real Function with Positive Derivative is Increasing:  :$F$ is increasing on $\closedint a b$. However, by hypothesis: {{begin-eqn}} {{eqn	| l = \map F b 	| r = \int_a^b \map f x \rd x }} {{eqn	| r = 0 }} {{eqn	| r = \int_a^a \map f x \rd x 	| c = Definite Integral on Zero Interval }} {{eqn	| r = \map F a }} {{end-eqn}} So, it must be the case that:  :$\map F x = 0$ for all $x \in \closedint a b$. We therefore have, from Derivative of Constant:  :$\map {F'} x = \map f x = 0$ for all $x \in \closedint a b$ as required. {{qed}}
Sum down each column, then sum across: :$\begin{array}{r|cccccccccc} \displaystyle \paren {\map \zeta k - 1} &  &  &  &  &  &  &  \\ \hline \paren {\map \zeta 2 - 1} & \paren {\dfrac {1} {2}}^2 & \paren {\dfrac {1} {3}}^2 & \paren {\dfrac {1} {4}}^2 & \paren {\dfrac {1} {5}}^2 & \cdots & \paren {\dfrac {1} {n}}^2 & \cdots \\ \paren {\map \zeta 3 - 1}  & \paren {\dfrac {1} {2}}^3 & \paren {\dfrac {1} {3}}^3 & \paren {\dfrac {1} {4}}^3 & \paren {\dfrac {1} {5}}^3 & \cdots & \paren {\dfrac {1} {n}}^3 & \cdots \\ \paren {\map \zeta 4 - 1}  & \paren {\dfrac {1} {2}}^4 & \paren {\dfrac {1} {3}}^4 & \paren {\dfrac {1} {4}}^4 & \paren {\dfrac {1} {5}}^4 & \cdots & \paren {\dfrac {1} {n}}^4 & \cdots \\ \paren {\map \zeta 5 - 1}  & \paren {\dfrac {1} {2}}^5 & \paren {\dfrac {1} {3}}^5 & \paren {\dfrac {1} {4}}^5 & \paren {\dfrac {1} {5}}^5 & \cdots & \paren {\dfrac {1} {n}}^5 & \cdots \\ \cdots  & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ \hline \sum_{k \mathop = 2}^\infty \paren {\map \zeta k - 1}  & \paren {\dfrac 1 1} \paren {\dfrac 1 2} & \paren {\dfrac 1 2} \paren {\dfrac 1 3} & \paren {\dfrac 1 3} \paren {\dfrac 1 4} & \paren {\dfrac 1 4} \paren {\dfrac {1} {5}} & \cdots & \paren {\dfrac {1} {n-1}} \paren {\dfrac {1} {n}} & \cdots \\ \end{array}$ First, summing down the array: {{begin-eqn}} {{eqn | l = \paren {\dfrac {1} {n} }^2 \paren {1 + \paren {\dfrac 1 n } + \paren {\dfrac 1 n }^2 + \cdots }       | r = \paren {\dfrac {1} {n} }^2 \paren {\dfrac 1 {1 - \dfrac 1 {n} } } }} {{eqn | r = \paren {\dfrac {1} {n} }^2 \paren {\dfrac n {n - 1} }       | rr = = \paren {\dfrac 1 {n} } \paren {\dfrac 1 {n-1} }       | c = Sum of Geometric Sequence }} {{end-eqn}} Next, summing across, we note: {{begin-eqn}} {{eqn | l = \paren {\dfrac 1 {n} } \paren {\dfrac 1 {n-1} }       | r = \paren {\dfrac 1 {n-1} } - \paren {\dfrac 1 {n} } }} {{end-eqn}} Therefore, {{begin-eqn}} {{eqn | l = \paren {\dfrac 1 1 } \paren {\dfrac 1 2 } + \paren {\dfrac 1 2 } \paren {\dfrac 1 3 } + \paren {\dfrac 1 3 } \paren {\dfrac 1 4 } + \cdots       | r = \paren {\dfrac 1 1 - \dfrac 1 2 } + \paren {\dfrac 1 2 - \dfrac 1 3 } + \paren {\dfrac 1 3 - \dfrac 1 4 } + \cdots }} {{eqn | r = \paren {\dfrac 1 1 - \paren {\dfrac 1 2 - \dfrac 1 2 } - \paren {\dfrac 1 3 - \dfrac 1 3 } - \paren {\dfrac 1 4 - \dfrac 1 4 } - \cdots }       | c = regroup terms }} {{eqn | r = \paren {\dfrac 1 1 - \paren {0 } - \paren {0 } - \paren {0 } - \cdots }       | rr = = 1       | c =  }} {{end-eqn}} Hence, {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 2}^\infty \paren {\displaystyle \paren {\map \zeta k - 1} }       | r = 1       | c =  }} {{end-eqn}} {{qed}}
It suffices to show that if $f$ is discontinuous at $c$: :there exists a real sequence $\sequence {x_n}$ in $A$ and $x \in \R$ such that $\sequence {x_n}$ converges to $c$ but $\sequence {\map f {x_n} }$ does not converge to $\map f c$. As $f$ is discontinuous, there exists some $\varepsilon > 0$ such that for all $\delta > 0$: :there exists $x \in A$ with $\size {x - c} < \delta$ we have $\size {\map f x - \map f c} \ge \varepsilon$. Using this property, we can construct a sequence $\sequence {x_n}$ as follows: :for each $n \in \N$, pick $x_n \in A$ such that $\size {x_n - c} \le \dfrac 1 n$ and $\size {\map f {x_n} - \map f c} \ge \varepsilon$ Note that since:  :$\displaystyle \lim_{n \to \infty} \frac 1 n = 0$ We have by the squeeze theorem for sequences of real numbers:  :$\displaystyle \lim_{n \to \infty} \size {x_n - c} = 0$ so $\sequence {x_n}$ converges to $c$. However $\sequence {\map f {x_n} }$ cannot converge to $\map f c$ since: :$\size {\map f {x_n} - \map f c} \ge \varepsilon > 0$ for all $n \in \N$. Therefore, our $\sequence {x_n}$ satisfies our original demand. {{qed|lemma}}
Let $c \in \R$.  Let $\sequence {x_n}$ be a sequence in $A$ that converges to $c$. Let $\varepsilon \in \R_{> 0}$.  Since $f$ is continuous at $c$, there exists $\delta > 0$ such that:  :for all $x \in A$ with $\size {x - c} < \delta$, we have $\size {\map f x - \map f c} < \varepsilon$. Additionally, since $\sequence {x_n}$ converges to $c$, there exists $N \in \N$ such that:  :for all $n > N$ we have $\size {x_n - c} < \delta$. Therefore, since $x_n \in A$ for all $n \in \N$:  :for all $n > N$, we have $\size {\map f {x_n} - \map f c} < \varepsilon$. Since $\varepsilon$ was arbitrary, we have:  :$\sequence {\map f {x_n} }$ converges to $\map f c$. {{qed}}
Let $H^i$ be the set of isolated points of $H$ in $M$. From Isolated Point in Metric Space iff Isolated Point in Topological Space: :$H^i$ equals the set of isolated points of $H$ in the topological space $T$. Let $H'$ be the set of limit points of $H$ in $M$. From Limit Point in Metric Space iff Limit Point in Topological Space: :$H'$ equals the set of limit points of $H$ in the topological space $T$. Let $H^-$ denote the closure of $H$ in the metric space $M$. By definition of the closure of $H$ in the metric space $M$ :$H^- = H' \cup H^i$ Let $\map \cl H$ denote the closure of the closure of $H$ in the topological space $T$. By definition of the closure of $H$ in the topological space $T$ :$\map \cl H = H' \cup H^i$ Thus: :$H^- = \map \cl H$ {{qed}} Category:Set Closures k5csayawx16jp5g37kf7ulvg56ln4bc
Proof by Counterexample: Let $T = \struct {\R, \tau}$ denote the right order topology on $\R$. Let $H \subseteq \R$ be a finite subset of $\R$. Let $\Omega$ denote the set of $\omega$-accumulation points of $H$. From Finite Set of Right Order Topology with Omega-Accumulation Points is not Closed, $H \cup \Omega$ is not a closed set of $T$. {{qed}}
Proof by Counterexample: Let $T = \struct {\R, \tau}$ denote the right order topology on $\R$. Let $H \subseteq \R$ be a finite subset of $\R$. Let $\CC$ denote the set of condensation points of $H$. From Finite Set of Right Order Topology with Condensation Points is not Closed, $H \cup \CC$ is not a closed set of $T$. {{qed}}
=== Necessary Condition === Let $x \in H$ be an isolated point of $H$. By definition of an isolated point: :$\exists U \in \tau: U \cap H  = \set x$ By definition of a synthetic basis of $T$: :$\exists V \in \BB: x \in V \subseteq U$ From Set Intersection Preserves Subsets: :$V \cap H \subseteq U \cap H  = \set x$ From Singleton of Element is Subset: :$\set x \subseteq V \cap H$ From set equality: :$V \cap H  = \set x$ {{qed|lemma}} === Sufficient Condition === Let $U \in \BB : U \cap H = \set x$. By definition of synthetic basis of $T$: :$U \in \tau$ Then $x$ is an isolated point of $H$ by definition. {{qed}} Category:Isolated Points 5ljnuyr7azebz0poirvgt64csf3uyra
=== 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 synthetic basis of $T$: :$\BB \subseteq \tau$ The result follows. {{qed|lemma}} === Sufficient Condition === Let $x$ satisfy: :$\forall U \in \BB : x \in U$ satisfies $H \cap U \setminus \set x \ne \O$ Let $V$ be any open neighborhood of $x$. By definition of a synthetic 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 s2rci5ndexlisnsdafo9zdx1chhymx6
=== 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 synthetic basis of $T$: :$\BB \subseteq \tau$ The result follows. {{qed|lemma}} === Sufficient Condition === Let $x$ satisfy: :$\forall U \in \BB : x \in U$ satisfies $H \cap U \ne \O$ Let $V$ be any open neighborhood of $x$. By definition of a synthetic 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 c29nij74r4twhq1997ysbb0fpd0t097
By Corollary to Liouville's Constant is Transcendental, all numbers of the form: {{begin-eqn}} {{eqn | l = \sum_{n \mathop \ge 1} \frac {a_n} {10^{n!} }       | r = \frac {a_1} {10^1} + \frac {a_2} {10^2} + \frac {a_3} {10^6} + \frac {a_4} {10^{24} } + \cdots       | c =  }} {{end-eqn}} where :$a_1, a_2, a_3, \ldots \in \set {1, 2, \ldots, 9}$ are Liouville numbers. Therefore each sequence in $\set {1, 2, \ldots, 9}$ defines a unique Liouville number. By Set of Infinite Sequences is Uncountable, there are uncountable sequences in $\set {1, 2, \ldots, 9}$. As the set of Liouville numbers has an uncountable subset, it is also uncountable by Sufficient Conditions for Uncountability. {{qed}} Category:Transcendental Numbers e83747fu0fg9s0g9ssbgyh2by3acyt2
From Open Balls form Local Basis for Point of Metric Space, the set: :$\BB_x = \set {\map {B_\epsilon} x : \epsilon \in \R_{>0} }$ is a local basis of $x$. From Local Basis Test for Isolated Point: :$x$ is an isolated point of $H$ in $T$ {{iff}} $\exists \epsilon \in \R_{>0}: \map {B_\epsilon} x \cap H = \set x$ By definition of an isolated point in $M$: :$x$ is an isolated point of $H$ in $T$ {{iff}} $x$ is an isolated point of $H$ in $M$ {{qed}} Category:Isolated Points dutqp017g6dzcek07ew4iqp1wdenny4
From Open Balls form Local Basis for Point of Metric Space, the set: :$\BB_x = \set{\map {B_\epsilon} x : \epsilon \in \R_{>0}}$ is a local basis of $x$. From Local Basis Test for Limit Point: :$x$ is a limit point of $H$ in $T$ {{iff}} $\forall \epsilon \in \R_{>0}: H \cap \map {B_\epsilon} x \setminus \set x \ne \O$ By definition of a limit point in $M$: :$x$ is a limit point of $H$ in $T$ {{iff}} $x$ is an limit point of $H$ in $M$ {{qed}} Category:Limit Points onrvxa4fd0s35dgsj7tyyamqf3pozjl
From Boundary of Boundary is Contained in Boundary, we have that: :$\map \partial {\partial H} \subseteq \partial H$ It remains to be proved that the equality does not always hold. Proof by Counterexample: Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space. Let $H \subseteq S$ such that $H \ne \O$ and $H \ne S$. From Boundary of Subset of Indiscrete Space: :$\partial H = S$ From Boundary of Boundary of Subset of Indiscrete Space: :$\map \partial {\partial H} = \O$ The result follows. {{qed}}
For each finite subdivision $P$ of $\closedint a b$, write:  :$P = \set {x_0, x_1, \ldots, x_n }$ with: :$a = x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n = b$ Then: {{begin-eqn}} {{eqn	| l = \map {V_{f + g} } P 	| r = \sum_{i \mathop = 1}^n \size {\map {\paren {f + g} } {x_i} - \map {\paren {f + g} } {x_{i - 1} } } 	| c = using the notation from the definition of bounded variation }} {{eqn	| r = \sum_{i \mathop = 1}^n \size {\paren {\map f {x_i} - \map f {x_{i - 1} } } + \paren {\map g {x_i} - \map g {x_{i - 1} } } } }} {{eqn	| o = \le 	| r = \sum_{i \mathop = 1}^n \size {\map f {x_i} - \map f {x_{i - 1} } } + \sum_{i \mathop = 1}^n \size {\map g {x_i} - \map g {x_{i - 1} } } 	| c = Triangle Inequality }} {{eqn	| r = \map {V_f} P + \map {V_g} P }} {{end-eqn}} Note that 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 $P$ of $\closedint a b$.  We therefore have:  :$\map {V_{f + g} } P \le M + K$ for all finite subdivisions $P$. So $f + g$ is of bounded variation. Note then that:  {{begin-eqn}} {{eqn	| l = V_{f + g} 	| r = \sup_P \paren {\map {V_{f + g} } P} 	| c = {{Defof|Total Variation}} }} {{eqn	| o = \le 	| r = \sup_P \paren {\map {V_f} P} + \sup_P \paren {\map {V_g} P} }} {{eqn	| r = V_f + V_g 	| c = {{Defof|Total Variation}} }} {{end-eqn}} {{qed}}
By definition of equivalent norms: :$\exists m,M \in \R_{> 0} : m \le M : \forall x \in X:  m \norm x_b \le \norm x_a \le M \norm x_b$ Since $U$ is open in $M_a$: :$\forall x \in U : \exists \epsilon_a \in \R_{> 0} : \map {B_{\epsilon_a}} x \subseteq U$ where $\map {B_{\epsilon_a}} x$ stands for an open ball, defined as: :$\map {B_{\epsilon_a}} x := \set {\forall y \in X : \norm {x - y}_a < \epsilon_a}$ Define an open ball $\map {B_{\epsilon_b}} x$ as: :$\map {B_{\epsilon_b}} x := \set {\forall y \in X : \norm {x - y}_b < \epsilon_b}$ Then we have that: :$m \norm {x - y}_b \le \norm {x - y}_a < \epsilon_a$ So far, $\epsilon_b$ was unspecified. Define $\epsilon_b := \dfrac {\epsilon_a} m$. So: :$\forall x \in U : y \in \map {B_{\epsilon_a}} x \implies y \in \map {B_{\epsilon_b}} x$ with $\epsilon_a = m \epsilon_b$. Hence: :$\forall x \in U : \exists \epsilon_b \in \R_{> 0} : \map {B_{\epsilon_b}} x \subseteq U$ {{qed}}
=== Necessary Condition === Let $x \in H$ be an isolated point of $H$. By definition of an isolated point: :$\exists U \in \tau: U \cap H  = \set x$ By definition of a local basis of $T$: :$\exists V \in \BB_x : x \in V \subseteq U$ From Set Intersection Preserves Subsets: :$V \cap H \subseteq U \cap H  = \set x$ From Singleton of Element is Subset: :$\set x \subseteq V \cap H$ From set equality: :$V \cap H  = \set x$ {{qed|lemma}} === Sufficient Condition === Let $U \in \BB_x : U \cap H = \set x$. By definition of local basis of $T$: :$U \in \tau$ Then $x$ is an isolated point of $H$ by definition. {{qed}} Category:Isolated Points 60b92i4by6anligwiu78ea676ya5pwh