avewright/proofwiki-math huggingface dataset:
Theorem: Sum of Functions of Bounded Variation is of Bounded Variation

Let $a, b$ be real numbers with $a < b$.
Let $f, g : \closedint a b \to \R$ be functions of bounded variation.
Let $\map {V_f} {\closedint a b}$ and $\map {V_g} {\closedint a b}$ be the total variations of $f$ and $g$ respectively, on $\closedint a b$. 
Then $f + g$ is of bounded variation on $\closedint a b$ with:
:$\map {V_{f + g} } {\closedint a b} \le \map {V_f} {\closedint a b} + \map {V_g} {\closedint a b}$
where $\map {V_{f + g} } {\closedint a b}$ denotes the total variation of $f + g$ on $\closedint a b$.

Proof:
For each finite subdivision $P$ of $\closedint a b$, write: 
:$P = \set {x_0, x_1, \ldots, x_n }$
with:
:$a = x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n = b$
Then:
{{begin-eqn}}
{{eqn	| l = \map {V_{f + g} } {P ; \closedint a b}
	| 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 ; \closedint a b} + \map {V_g} {P ; \closedint a b}
}}
{{end-eqn}}
Note that since $f$ and $g$ are of bounded variation, there exists $M, K \in \R$ such that: 
:$\map {V_f} {P ; \closedint a b} \le M$
:$\map {V_g} {P ; \closedint a b} \le K$
for all finite subdivisions $P$ of $\closedint a b$. 
We therefore have: 
:$\map {V_{f + g} } {P ; \closedint a b} \le M + K$
for all finite subdivisions $P$.
So $f + g$ is of bounded variation.
Note then that: 
{{begin-eqn}}
{{eqn	| l = \map {V_{f + g} } {\closedint a b}
	| r = \sup_P \paren {\map {V_{f + g} } {P ; \closedint a b} }
	| c = {{Defof|Total Variation of Real Function on Closed Bounded Interval}}
}}
{{eqn	| o = \le
	| r = \sup_P \paren {\map {V_f} {P ; \closedint a b} } + \sup_P \paren {\map {V_g} {P ; \closedint a b} }
}}
{{eqn	| r = \map {V_f} {\closedint a b} + \map {V_g} {\closedint a b}
	| c = {{Defof|Total Variation of Real Function on Closed Bounded Interval}}
}}
{{end-eqn}}
{{qed}}
Theorem: Open Sets in Vector Spaces with Equivalent Norms Coincide

Let $M_a = \struct {X, \norm {\, \cdot \, }_a}$ and $M_b = \struct {X, \norm {\, \cdot \,}_b}$ be normed vector spaces.
Let $U \subseteq X$ be an open set in $M_a$.
Suppose, $\norm {\, \cdot \, }_a$ and $\norm {\, \cdot \,}_b$ are equivalent norms, i.e. $\norm {\, \cdot \, }_a \sim \norm {\, \cdot \,}_b$.
Then $U$ is also open in $M_b$.

Proof:
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}}
Theorem: Local Basis Test for Isolated Point

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

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 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}}
Theorem: Local Basis Test for Limit Point

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

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 local basis of $T$:
:$\BB_x \subseteq \tau$
The result follows.
{{qed|lemma}}
Theorem: Local Basis Test for Adherent Point

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

Proof:
=== Necessary Condition ===
Let $x \in S$ be an adherent point of $H$.
By definition of an adherent point:
:$\forall U \in \tau : x \in U$ satisfies $H \cap U \ne \O$
By definition of a local basis of $T$:
:$\BB_x \subseteq \tau$
The result follows.
{{qed|lemma}}
Theorem: Closure of Subset of Metric Space is Closed

Let $M = \struct {A, d}$ be a metric space.
Let $H \subseteq A$ be a subset of $A$.
Let $H^-$ denote the closure of $H$.
Then $H^-$ is a closed set of $M$.

Proof:
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
oed2aumqq13pxtthqj1rh9of2ogix8f
Theorem: Convergent Sequences in Vector Spaces with Equivalent Norms Coincide

Let $M_a = \struct {X, \norm {\, \cdot \, }_a}$ and $M_b = \struct {X, \norm {\, \cdot \,}_b}$ be normed vector spaces.
Let $\sequence {x_n}_{n \mathop \in \N}$ be an convergent sequence in $M_a$.
Suppose, $\norm {\, \cdot \, }_a$ and $\norm {\, \cdot \,}_b$ are equivalent norms, i.e. $\norm {\, \cdot \, }_a \sim \norm {\, \cdot \,}_b$.
Then $\sequence {x_n}_{n \mathop \in \N}$ is also convergent in $M_b$.

Proof:
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}}
Theorem: Open Mapping is not necessarily Closed Mapping

Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $f: T_1 \to T_2$ be a mapping which is not a bijection.
Let $f$ be an open mapping.
Then it is not necessarily the case that $f$ is also a closed mapping.

Proof:
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}}
Theorem: Primitive of Power of x by Cosine of a x/Corollary

:$\ds \int x^m \cos a x \rd x = \sum_{k \mathop = 1}^{m + 1} \paren {m^{\underline {k - 1} } \frac {x^{m + 1 - k} } {a^k} \map {\sin} {x + \dfrac {\pi} 2 \paren {k - 1} } }$
where $m^{\underline {k - 1} }$ denotes the $k - 1$th falling factorial of $m$.

Proof:
{{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
icxngujkdijjl6c655jdloxtwttrx1f
Theorem: Homeomorphism may Exist between Non-Comparable Topologies

Let $S$ be a set.
Let $T_1 = \struct {S, \tau_1}$ and $T_2 = \struct {S, \tau_2}$ be topological spaces defined on the underlying set $S$.
Let $\tau_1$ and $\tau_2$ be non-comparable.
Then it may possibly be the case that $T_1$ and $T_2$ are homeomorphic.

Proof:
A counterexample is demonstrated in Homeomorphic Non-Comparable Particular Point Topologies.
{{qed}}
Theorem: Equivalence of Definitions of Limit Point of Filter Basis

{{TFAE|def = Limit Point of Filter Basis}}
Let $T = \struct {S, \tau}$ be a topological space.
Let $\FF$ be a filter on the underlying set $S$ of $T$.
Let $\BB$ be a filter basis of $\FF$.

Proof:
Let $\FF$ be a filter on $S$.
Let $\BB$ be a filter basis of $\FF$.
Theorem: Richert's Theorem

Let $S = \set {s_1, s_2, \dots}$ be an infinite set of (strictly) positive integers, with the property:
:$s_n < s_{n + 1}$ for every $n \in \N$
Suppose there exists some integers $N, k$ such that every integer $n$ with $N < n \le N + s_{k + 1}$:
:$n$ can be expressed as a sum of distinct elements in $\set {s_1, s_2, \dots, s_k}$
:$s_{i + 1} \le 2 s_i$ for every $i \ge k$
Then for any $n > N$, $n$ can be expressed as a sum of distinct elements in $S$.

Proof:
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 $\set {s_1, s_2, \dots, s_n}$.
Theorem: Equivalence of Definitions of Filter Basis

{{TFAE|def = Filter Basis}}
Let $S$ be a set.
Let $\FF$ be a filter on $S$.

Proof:
=== $(1)$ implies $(2)$ ===
Let $\BB$ be a filter basis of $\FF$ by definition $1$.
{{Recall|Filter Basis|index = 1}}
{{:Definition:Filter Basis/Definition 1}}
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}}
Theorem: Number as Sum of Distinct Primes

For $n \ne 1, 4, 6$, $n$ can be expressed as the sum of distinct primes.

Proof:
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
Theorem: Multiple of Function of Bounded Variation is of Bounded Variation

Let $a, b, k$ be real numbers with $a < b$.
Let $f : \closedint a b \to \R$ be a functions of bounded variation.
Let the total variation of $f$ on $\closedint a b$ be $\map {V_f} {\closedint a b}$.
Then $k f$ is of bounded variation with: 
:$\map {V_{k f} } {\closedint a b} = \size k \map {V_f} {\closedint a b}$
where $\map {V_{k f} } {\closedint a b}$ is the total variation of $k f$ on $\closedint a b$.

Proof:
For each finite subdivision $P$ of $\closedint a b$, write: 
:$P = \set {x_0, x_1, \ldots, x_n }$
with:
:$a = x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n = b$
Then:
{{begin-eqn}}
{{eqn	| l = \map {V_{k f} } {P ; \closedint a b}
	| r = \sum_{i \mathop = 1}^n \size {k \map f {x_i} - k \map f {x_{i - 1} } }
	| c = using the notation from the definition of bounded variation
}}
{{eqn	| r = \size k \sum_{i \mathop = 1}^n \size {\map f {x_i} - \map f {x_{i - 1} } }
}}
{{eqn	| r = \size k \map {V_f} {P ; \closedint a b}
}}
{{end-eqn}}
Since $f$ is of bounded variation, there exists $M \in \R$ such that: 
:$\map {V_f} {P ; \closedint a b} \le M$
for all finite subdivisions $P$. 
So:
:$\map {V_{k f} } {P ; \closedint a b} \le \size k M$
So $k f$ is of bounded variation.
We then have: 
{{begin-eqn}}
{{eqn	| l = \map {V_{k f} } {\closedint a b}
	| r = \sup_P \paren {\map {V_{k f} } {P ; \closedint a b} }
	| c = {{Defof|Total Variation of Real Function}}
}}
{{eqn	| r = \sup_P \paren {\size k \map {V_f} {P ; \closedint a b} }
}}
{{eqn	| r = \size k \sup_P \paren {\map {V_f} {P ; \closedint a b} }
	| c = Multiple of Supremum
}}
{{eqn	| r = \size k \map {V_f} {\closedint a b}
}}
{{end-eqn}}
{{qed}}
Category:Bounded Variation
Category:Total Variation of Real Function
f2czygoevo1me56tv1q8srulcn7m2rg
Theorem: Difference of Functions of Bounded Variation is of Bounded Variation

Let $a, b$ be real numbers with $a < b$.
Let $f, g : \closedint a b \to \R$ be functions of bounded variation.
Let $\map {V_f} {\closedint a b}$ and $\map {V_g} {\closedint a b}$ be the total variations of $f$ and $g$ respectively, on $\closedint a b$. 
Then $f - g$ is of bounded variation with:
:$\map {V_{f - g} } {\closedint a b} \le \map {V_f} {\closedint a b} + \map {V_g} {\closedint a b}$
where $V_{f - g}$ denotes the total variation of $f - g$ on $\closedint a b$.

Proof:
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: 
:$\map {V_{f - g} } {\closedint a b} \le \map {V_f} {\closedint a b} + \map {V_{-g} } {\closedint a b}$
where $V_{-g}$ is the total variation of $-g$ on $\closedint a b$. 
We have, by Multiple of Function of Bounded Variation is of Bounded Variation:
{{begin-eqn}}
{{eqn	| l = \map {V_{-g} } {\closedint a b}
	| r = \size {-1} \map {V_g} {\closedint a b}
}}
{{eqn	| r = \map {V_g} {\closedint a b}
}}
{{end-eqn}}
so: 
:$\map {V_{f - g} } {\closedint a b} \le \map {V_f} {\closedint a b} + \map {V_g} {\closedint a b}$
{{qed}}
Theorem: Reciprocal of Function of Bounded Variation Bounded away from Zero is of Bounded Variation

Let $a, b$ be real numbers with $a < b$.
Let $f: \closedint a b \to \R$ be functions of bounded variation.
Let $f$ be bounded away from zero. 
That is, there exists $M \in \R$ such that: 
:$\size {\map f x} \ge M > 0$
for all $x \in \closedint a b$. 
Let $\map {V_f} {\closedint a b}$ be the total variation of $f$ on $\closedint a b$.
Then:
:$\dfrac 1 f$ is of bounded variation
with:
:$\map {V_{1 / f} } {\closedint a b} \le \dfrac {\map {V_f} {\closedint a b} } {M^2}$
where $\map {V_{1 / f} } {\closedint a b}$ denotes the total variation of $\dfrac 1 f$ on $\closedint a b$.

Proof:
For each finite subdivision $P$ of $\closedint a b$, write: 
:$P = \set {x_0, x_1, \ldots, x_n }$
with:
:$a = x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n = b$
Note that: 
:$\dfrac 1 {\map f x} \le \dfrac 1 M$
for all $x \in \closedint a b$.
We then have: 
{{begin-eqn}}
{{eqn	| l = \map {V_{1 / f} } {P ; \closedint a b}
	| r = \sum_{i \mathop = 1}^n \size {\frac 1 {\map f {x_i} } - \frac 1 {\map f {x_{i - 1} } } }
	| c = using the notation from the definition of bounded variation
}}
{{eqn	| r = \sum_{i \mathop = 1}^n \size {\frac {\map f {x_{i - 1} } - \map f {x_i} } {\map f {x_i} \map f {x_{i - 1} } } }
}}
{{eqn	| r = \sum_{i \mathop = 1}^n \frac {\size {\map f {x_i} - \map f {x_{i - 1} } } } {\size {\map f {x_i} \map f {x_{i - 1} } } }
}}
{{eqn	| o = \le
	| r = \frac 1 {M^2} \sum_{i \mathop = 1}^n \size {\map f {x_i} - \map f {x_{i - 1} } }
	| c = since $\size {\map f {x_i} \map f {x_{i - 1} } } \le M^2$ 
}}
{{eqn	| r = \frac {\map {V_f} {P ; \closedint a b} } {M^2}
}}
{{end-eqn}}
Since $f$ is of bounded variation, there exists $K \in \R$ such that: 
:$\map {V_f} {P ; \closedint a b} \le K$
for every finite subdivision $P$. 
Therefore: 
:$\map {V_{1 / f} } {P ; \closedint a b} \le \dfrac K {M^2}$
So $\dfrac 1 f$ is of bounded variation.
We also have: 
{{begin-eqn}}
{{eqn	| l = \map {V_{1 / f} } {\closedint a b}
	| r = \sup_P \paren {\map {V_{1 / f} } {P ; \closedint a b} }
	| c = {{Defof|Total Variation of Real Function on Closed Bounded Interval}}
}}
{{eqn	| o = \le
	| r = \sup_P \paren {\frac {\map {V_f} {P ; \closedint a b} } {M^2} }
}}
{{eqn	| r = \frac 1 {M^2} \sup_P \paren {\map {V_f} {P ; \closedint a b} }
	| c = Multiple of Supremum
}}
{{eqn	| r = \frac {\map {V_f} {\closedint a b} } {M^2}
	| c = {{Defof|Total Variation of Real Function on Closed Bounded Interval}}
}}
{{end-eqn}}
{{qed}}
Theorem: Cauchy Sequences in Vector Spaces with Equivalent Norms Coincide

Let $M_a = \struct {X, \norm {\, \cdot \, }_a}$ and $M_b = \struct {X, \norm {\, \cdot \,}_b}$ be normed vector spaces.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a Cauchy sequence in $M_a$.
Suppose, $\norm {\, \cdot \, }_a$ and $\norm {\, \cdot \,}_b$ are equivalent norms, that is:
:$\norm {\, \cdot \, }_a \sim \norm {\, \cdot \,}_b$
Then $\sequence {x_n}_{n \mathop \in \N}$ is also a Cauchy sequence in $M_b$.

Proof:
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}}
Theorem: Differentiable Function of Bounded Variation may not have Bounded Derivative

Let $a, b$ be real numbers with $a < b$.
Let $f : \closedint a b \to \R$ be a continuous function of bounded variation.
Let $f$ be differentiable on $\openint a b$.
Then $f'$ is not necessarily bounded.

Proof:
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}}
Theorem: Existence of Urysohn Function does not guarantee Normal Space

Let $T = \struct {S, \tau}$ be a regular space.
Let $T$ have the property that:
:For all closed sets $A, B \subseteq S$ of $T$ such that $A \cap B = \O$, there exists an Urysohn function for $A$ and $B$.
Then it is not necessarily the case that $T$ is a normal space.

Proof:
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.
;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 Space 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 Space is not $T_1$, $T$ is not a $T_1$ space.
The result follows.
{{qed}}
Theorem: T3.5 Space is not necessarily T2

Let $T = \struct {S, \tau}$ be a be a $T_{3 \frac 1 2}$ space.
Then it is not necessarily the case that $T$ is a $T_2$ (Hausdorff) space.

Proof:
;Proof by Counterexample
Let $S$ be a set.
Pet $\PP$ 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 Space is $T_{3 \frac 1 2}$, we have that $T$ is a $T_{3 \frac 1 2}$ space.
From Partition Space is not $T_2$, $T$ is not a $T_2$ space.
The result follows.
{{qed}}
Theorem: T0 Space is not necessarily T1

Let $T = \struct {S, \tau}$ be a be a $T_0$ space.
Then it is not necessarily the case that $T$ is a $T_1$ space.

Proof:
;Proof by Counterexample
Let $T = \struct {S, \tau_p}$ be a particular point space such that $S$ is not a singleton.
From Particular Point Space is $T_0$, we have that $T$ is a $T_0$ space.
From Non-Trivial Particular Point Space is not $T_1$, $T$ is not a $T_1$ space.
The result follows.
{{qed}}
Theorem: T0 Space is not necessarily T1

Let $T = \struct {S, \tau}$ be a be a $T_0$ space.
Then it is not necessarily the case that $T$ is a $T_1$ space.

Proof:
;Proof by Counterexample
Let $T$ be the overlapping interval space.
From Overlapping Interval Space fulfils no Separation Axioms but $T_0$, we have that $T$ satisfies none of the Tychonoff separation axioms except for the $T_0$ axiom.
That is:
:$T$ is a $T_0$ space
but:
:$T$ is not a $T_1$ space.
The result follows.
{{qed}}