avewright/proofwiki-math huggingface dataset:
Theorem: Monotone Function is of Bounded Variation

Let $a, b$ be real numbers with $a < b$.
Let $f : \closedint a b \to \R$ be a monotone function.
Then $f$ is of bounded variation.

Proof:
We use the notation from the definition of bounded variation.
Let $P = \set {x_0, x_1, \ldots, x_n}$ 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 $i \le n$.
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 ; \closedint a b}
	| r = \sum_{i \mathop = 1}^n \size {\map f {x_i} - \map f {x_{i - 1} } }
}}
{{eqn	| r = \sum_{i \mathop = 1}^n \paren {\map f {x_i} - \map f {x_{i - 1} } }
}}
{{eqn	| r = \map f {x_n} - \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 $i \le n$.
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 ; \closedint a b}
	| r = \sum_{i \mathop = 1}^n \size {\map f {x_i} - \map f {x_{i - 1} } }
}}
{{eqn	| r = -\sum_{i \mathop = 1}^n \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}}
Theorem: Differentiable Function with Bounded Derivative is of Bounded Variation

Let $a, b$ be real numbers with $a < b$.
Let $f : \closedint a b \to \R$ be a continuous function.
Let $f$ be differentiable on $\openint a b$, with bounded derivative.
Then $f$ is of bounded variation.

Proof:
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 ; \closedint a b}
      | 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}}
Theorem: Function of Bounded Variation is Bounded

Let $a, b$ be real numbers with $a < b$.
Let $f : \closedint a b \to \R$ be a function of bounded variation.
Then $f$ is bounded.

Proof:
We use the notation from the definition of bounded variation.
Since $f$ is of bounded variation, there exists $M \ge 0$ such that:
:$\map {V_f} {P ; \closedint a b} \le M$
for all finite subdivisions $P$ of $\closedint a b$.
Let $x$ be a real number with:
:$a < x < b$
Then $\set {a, x, b}$ is a finite subdivision of $\closedint a b$.
We have: 
:$\map {V_f} {\set {a, x, b} ; \closedint a b} = \size {\map f x - \map f a} + \size {\map f b - \map f x}$
Since $x \in \openint a b$ was arbitrary, we therefore have: 
:$\size {\map f x - \map f a} + \size {\map f b - \map f x} \le M$
for all $x \in \openint a b$.
We have:
{{begin-eqn}}
{{eqn	| l = \size {\map f x - \map f a} + \size {\map f b - \map f x}
	| o = \ge
	| r = \size {\map f x - \map f a}
}}
{{eqn	| o = \ge
	| r = \size {\size {\map f x} - \size {\map f a} }
	| c = Reverse Triangle Inequality: Real and Complex Fields
}}
{{eqn	| o = \ge
	| r = \size {\map f x} - \size {\map f a}
}}
{{end-eqn}}
So for all $x \in \openint a b$, we have: 
:$\size {\map f x} \le \size {\map f a} + M$
Since $M \ge 0$, this inequality is also satisfied for $x = a$. 
We therefore have: 
:$\size {\map f x} \le \map \max {\size {\map f a} + M, \size {\map f b} }$
for all $x \in \closedint a b$.
So $f$ is is bounded.
{{qed}}
Theorem: Total Variation is Non-Negative

Let $a, b$ be real numbers with $a < b$.
Let $f : \closedint a b \to \R$ be a function of bounded variation.
Let $\map {V_f} {\closedint a b}$ be the total variation of $f$ on $\closedint a b$.
Then: 
:$\map {V_f} {\closedint a b} \ge 0$
with equality {{iff}} $f$ is constant.

Proof:
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 $\closedint a b$.
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: 
:$\ds \map {V_f} {P ; \closedint a b} = \sum_{i \mathop = 1}^n \size {\map f {x_i} - \map f {x_{i - 1} } } \ge 0$
So, by the definition of supremum, we have: 
:$\ds \map {V_f} {\closedint a b} = \sup_P \paren {\map {V_f} {P ; \closedint a b} } \ge 0$
Now consider $f$ a constant function.
We show that $\map {V_f} {\closedint a b} = 0$.
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 ; \closedint a b}
	| r = \sum_{i \mathop = 1}^n \size {\map f {x_i} - \map f {x_{i - 1} } }
}}
{{eqn	| r = \sum_{i \mathop = 1}^n 0
}}
{{eqn	| r = 0
}}
{{end-eqn}}
We then have: 
{{begin-eqn}}
{{eqn	| l = \map {V_f} {\closedint a b}
	| r = \sup_P \paren {\map {V_f} {P ; \closedint a b} }
	| c = {{Defof|Total Variation of Real Function on Closed Bounded Interval}}
}}
{{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:
:$\map {V_f} {\closedint a b} = 0$
It remains to show that if $\map {V_f} {\closedint a b} = 0$, then $f$ is constant.
It suffices to show that if $f$ is not constant then $\map {V_f} {\closedint a b} > 0$.
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} ; \closedint a 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: 
:$\map {V_f} {\closedint a b} \ge \map {V_f} {\set {a, x, b} ; \closedint a b} > 0$
Hence the claim.
{{qed}}
Theorem: Norm Equivalence is Equivalence

Let $X$ be a vector space.
Let $\norm {\, \cdot \,}_a$ and $\norm {\, \cdot \,}_b$ be equivalent norms on $X$.
Denote this relation by $\sim$: 
:$\norm {\, \cdot \,}_a \sim \norm {\, \cdot \,}_b$.
Then $\sim$ is an equivalence relation.

Proof:
=== 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}}
Theorem: Continuous Non-Negative Real Function with Zero Integral is Zero Function

Let $a, b$ be real numbers with $a < b$. 
Let $f : \closedint a b \to \R$ be a continuous function. 
Let: 
:$\map f x \ge 0$
for all $x \in \closedint a b$.
Let: 
:$\ds \int_a^b \map f x \rd x = 0$
Then $\map f x = 0$ for all $x \in \closedint a b$.

Proof:
From Definite Integral of Constant, if $\map f x = 0$ for all $x \in \closedint a b$, then: 
:$\ds \int_a^b \map f x \rd x = 0$
We want to show that if:
:$\ds \int_a^b \map f x \rd x = 0$
then:
:$\map f x = 0$ for all $x \in \closedint a b$.
Since $\map f x \ge 0$, by Relative Sizes of Definite Integrals:
:$\ds \int_a^b \map f x \rd x \ge 0$
It therefore suffices to show that if: 
:$\map f x > 0$ for some $x \in \closedint a b$
then:
:$\ds \int_a^b \map f x \rd x > 0$
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 $\ds \int_a^{x - \delta} \map f y \rd y + \int_{x + \delta}^b \map f y \rd y \ge 0$
}}
{{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}}
Theorem: Continuous Non-Negative Real Function with Zero Integral is Zero Function

Let $a, b$ be real numbers with $a < b$. 
Let $f : \closedint a b \to \R$ be a continuous function. 
Let: 
:$\map f x \ge 0$
for all $x \in \closedint a b$.
Let: 
:$\ds \int_a^b \map f x \rd x = 0$
Then $\map f x = 0$ for all $x \in \closedint a b$.

Proof:
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: 
:$\ds \map F x = \int_a^x \map f x \rd x$
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}}
Theorem: Sequential Continuity is Equivalent to Continuity in the Reals/Sufficient Condition

Let $A \subseteq \R$ be a subset of the real numbers.
Let $c \in A$.
Let $f: A \to \R$ be a real function.
Then if $f$ is continuous at $c$:
:for each sequence $\sequence {x_n}$ in $A$ that converges to $c$, the sequence $\sequence {\map f {x_n} }$ converges to $\map f c$.

Proof:
It suffices to show that if $f$ is discontinuous at $c$:
:there exists a real sequence $\sequence {x_n}$ in $A$ 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 $\size {x - c} < \delta$ such that $\size {\map f x - \map f c} \ge \varepsilon$.
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: 
:$\ds \lim_{n \mathop \to \infty} \frac 1 n = 0$
We have by the Squeeze Theorem for Real Sequences: 
:$\ds \lim_{n \mathop \to \infty} \size {x_n - c} = 0$
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.
Theorem: Sequential Continuity is Equivalent to Continuity in the Reals/Necessary Condition

Let $A \subseteq \R$ be a subset of the real numbers.
Let $c \in A$.
Let $f : A \to \R$ be a real function.
Then if $f$ is continuous at $c$:
:for each sequence $\sequence {x_n}$ in $A$ that converges to $c$, the sequence $\sequence {\map f {x_n} }$ converges to $\map f c$.

Proof:
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 $\map f c$.
Theorem: Metric Closure and Topological Closure of Subset are Equivalent

Let $M = \struct{A, d}$ be a metric space.
Let $T = \struct{A, \tau}$ be the topological space with the topology induced by $d$.
Let $H \subseteq A$.
Then:
:the metric closure of $H$ in $M$ equals the topological closure of $H$ in $T$

Proof:
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 $H$ 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
qpjcnpknk8v44yv2x84eumsptxjymvw
Theorem: Set together with Omega-Accumulation Points is not necessarily Closed

Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
Let $\Omega$ denote the set of $\omega$-accumulation points of $H$.
Then it is not necessarily the case that $H \cup \Omega$ is a closed set of $T$.

Proof:
;Proof by Counterexample
Let $T = \struct {\R, \tau}$ denote the right order topology on $\R$.
Let $H \subseteq \R$ be a non-empty finite subset of $\R$.
Let $\Omega$ denote the set of $\omega$-accumulation points of $H$.
Since $H$ is finite, it has no $\omega$-accumulation points, that is, $\Omega = \O$.
Therefore $H \cup \Omega = H$ is finite.
From Closed Sets of Right Order Space on Real Numbers, closed sets of $T$ are either infinite or is the empty set.
Since $H \cup \Omega$ is finite, it is not a closed set of $T$.
{{qed}}
Theorem: Set together with Condensation Points is not necessarily Closed

Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
Let $\CC$ denote the set of condensation points of $H$.
Then it is not necessarily the case that $H \cup \CC$ is a closed set of $T$.

Proof:
;Proof by Counterexample
Let $T = \struct {\R, \tau}$ denote the right order topology on $\R$.
Let $H \subseteq \R$ be a non-empty finite subset of $\R$.
Let $\CC$ denote the set of condensation points of $H$.
Since $H$ is finite, it has no condensation points, that is, $\CC = \O$.
Therefore $H \cup \CC = H$ is finite.
From Closed Sets of Right Order Space on Real Numbers, closed sets of $T$ are either infinite or is the empty set.
Since $H \cup \CC$ is finite, it is not a closed set of $T$.
{{qed}}
Theorem: Basis Test for Isolated Point

Let $T = \struct {S, \tau}$ be a topological space.
Let $\BB$ be a synthetic basis of $T$.
Let $H \subseteq S$.
Then $x \in H$ is an isolated point of $H$ {{iff}}:
:$\exists U \in \BB : U \cap H = \set x$

Proof:
=== 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}}
Theorem: Basis Test for Limit Point

Let $T = \struct {S, \tau}$ be a topological space.
Let $\BB$ be a synthetic basis of $T$.
Let $H \subseteq S$.
Then $x \in S$ is a limit point of $H$ {{iff}}:
:$\forall U \in \BB : x \in U$ satisfies $H \cap U \setminus \set x \ne \O$

Proof:
=== 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}}
Theorem: Basis Test for Adherent Point

Let $T = \struct {S, \tau}$ be a topological space.
Let $\BB$ be a synthetic basis of $T$.
Let $H \subseteq S$.
Then $x \in S$ is an adherent point of $H$ {{iff}}:
:$\forall U \in \BB : x \in U$ satisfies $H \cap U \ne \O$

Proof:
=== 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}}
Theorem: Set of Liouville Numbers is Uncountable

The set of Liouville numbers is uncountable.

Proof:
By {{Corollary|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
03hl9ym8txl83m27c3resma0em30lmc
Theorem: Isolated Point in Metric Space iff Isolated Point in Topological Space

Let $M = \struct {A, d}$ be a metric space.
Let $T = \struct {A, \tau}$ be the topological space with the topology induced by $d$.
Let $H \subseteq A$.
Let $x \in H$
Then:
:$x$ is an isolated point of $H$ in $M$ {{iff}} $x$ is an isolated point of $H$ in $T$

Proof:
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
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Theorem: Limit Point in Metric Space iff Limit Point in Topological Space

Let $M = \struct {A, d}$ be a metric space.
Let $T = \struct {A, \tau}$ be the topological space with the topology induced by $d$.
Let $H \subseteq A$.
Then:
:$x \in H$ is a limit point in $M$ {{iff}} $x$ is a limit point in $T$

Proof:
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
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Theorem: Boundary of Boundary is not necessarily Equal to Boundary

Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
Let $\partial H$ denote the boundary of $H$.
While it is true that:
:$\map \partial {\partial H} \subseteq \partial H$
it is not necessarily the case that:
:$\map \partial {\partial H} = \partial H$

Proof:
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}}