avewright/proofwiki-math huggingface dataset:
Theorem: Accumulation Point of Sequence is not necessarily Limit Point

Let $T = \struct {S, \tau}$ be a topological space.
Let $\sequence {a_n}$ be a sequence in $T$.
Let $q \in S$ be an accumulation point of $\sequence {a_n}$.
Then it is not necessarily the case that $q$ is also a limit of $\sequence {a_n}$.

Proof:
Proof by Counterexample:
Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
Let $\sequence {a_n}$ be the sequence defined as:
{{begin-eqn}}
{{eqn | l = \sequence {a_n}
      | r = \begin {cases} 1 & : \text {$n$ odd} \\ n / 2 & : \text {$n$ even} \end {cases}
      | c = 
}}
{{eqn | r = \sequence {1, 1, 1, 2, 1, 3, 1, 4, \dotsc}
      | c = 
}}
{{end-eqn}}
Then $\sequence {a_n}$ has exactly one accumulation point, that is $1$.
However, $1$ is not a limit of $\sequence {a_n}$, as $\sequence {a_n}$ has no limit.
Hence the result.
{{qed}}
Theorem: Limit Point of Underlying Set of Sequence of Reciprocals and Reciprocals + 1

Let $\sequence {a_n}$ denote the sequence defined as:
{{begin-eqn}}
{{eqn | l = a_n
      | r = \begin {cases} \dfrac 2 {n + 1} & : \text {$n$ odd} \\ \\ 1 + \dfrac 2 n & : \text {$n$ even} \end {cases}
      | c = 
}}
{{eqn | r = \sequence {\dfrac 1 1, 1 + \dfrac 1 1, \dfrac 1 2, 1 + \dfrac 1 2, \dfrac 1 3, 1 + \dfrac 1 3, \dotsb}
      | c = 
}}
{{end-eqn}}
Let $\struct {\R, \tau}$ denote the real number line under the usual (Euclidean) topology.
Let $S$ denote the set of terms of $\sequence {a_n}$ considered as a subset of $\struct {\R, \tau_d}$.
Then $0$ is a limit point of $S$.

Proof:
Let $\epsilon \in \R_{>0}$ be a (strictly) positive real number.
Then the open interval $\openint {-\epsilon} \epsilon$ contains $0$ and all elements $a_m$ of $S$ such that $0 < \dfrac 1 m < \epsilon$.
Hence $\dfrac 1 m \in \openint {-\epsilon} \epsilon$.
Hence the result by definition of limit point of $S$.
{{qed}}
Theorem: Omega-Accumulation Point of Underlying Set of Sequence of Reciprocals and Reciprocals + 1

Let $\sequence {a_n}$ denote the sequence defined as:
{{begin-eqn}}
{{eqn | l = a_n
      | r = \begin {cases} \dfrac 2 {n + 1} & : \text {$n$ odd} \\ \\ 1 + \dfrac 2 n & : \text {$n$ even} \end {cases}
      | c = 
}}
{{eqn | r = \sequence {\dfrac 1 1, 1 + \dfrac 1 1, \dfrac 1 2, 1 + \dfrac 1 2, \dfrac 1 3, 1 + \dfrac 1 3, \dotsb}
      | c = 
}}
{{end-eqn}}
Let $\struct {\R, \tau}$ denote the real number line under the usual (Euclidean) topology.
Let $S$ denote the set of terms of $\sequence {a_n}$ considered as a subset of $\struct {\R, \tau_d}$.
Then $0$ is an $\omega$-accumulation point of $S$.

Proof:
Let $\epsilon \in \R_{>0}$ be a (strictly) positive real number.
Then the open interval $\openint {-\epsilon} \epsilon$ contains $0$ and all elements $a_m$ of $S$ such that $0 < \dfrac 1 m < \epsilon$.
We have that:
:$\forall n \in \N: n \ge m$
have the property that $0 < \dfrac 1 n < \epsilon$.
Hence there are a countably infinite number of terms of $\sequence {a_n}$ such that $a_n \in \openint {-\epsilon} \epsilon$.
Hence the result by definition of $\omega$-accumulation point of $S$.
{{qed}}
Theorem: Accumulation Point of Sequence of Reciprocals and Reciprocals + 1

Let $\struct {\R, \tau}$ denote the real number line under the usual (Euclidean) topology.
Let $\sequence {a_n}$ denote the sequence in $\struct {\R, \tau}$ defined as:
{{begin-eqn}}
{{eqn | l = a_n
      | r = \begin {cases} \dfrac 2 {n + 1} & : \text {$n$ odd} \\ 1 + \dfrac 2 n & : \text {$n$ even} \end {cases}
      | c = 
}}
{{eqn | r = \sequence {\dfrac 1 1, 1 + \dfrac 1 1, \dfrac 1 2, 1 + \dfrac 1 2, \dfrac 1 3, 1 + \dfrac 1 3, \dotsb}
      | c = 
}}
{{end-eqn}}
Then $0$ is an accumulation point of $\sequence {a_n}$.

Proof:
Let $\epsilon \in \R_{>0}$ be a (strictly) positive real number.
Then the open interval $\openint {-\epsilon} \epsilon$ contains $0$ and all elements $a_m$ of $S$ such that $0 < \dfrac 1 m < \epsilon$.
We have that:
:$\forall n \in \N: n \ge m$
have the property that $0 < \dfrac 1 n < \epsilon$.
Hence there are a countably infinite number of terms of $\sequence {a_n}$ such that $a_n \in \openint {-\epsilon} \epsilon$.
{{explain|Why? The step from $0 < \dfrac 1 n < \epsilon$ to $0 < a_n < \epsilon$ is non-trivial.}}
Hence the result by definition of accumulation point of $\sequence {a_n}$.
{{qed}}
Theorem: Zero is not a Limit Point of Sequence of Reciprocals and Reciprocals + 1

Let $\struct {\R, \tau}$ denote the real number line under the usual (Euclidean) topology.
Let $\sequence {a_n}$ denote the sequence in $\struct {\R, \tau}$ defined as:
{{begin-eqn}}
{{eqn | l = a_n
      | r = \begin {cases} \dfrac 2 {n + 1} & : \text {$n$ odd} \\ \\ 1 + \dfrac 2 n & : \text {$n$ even} \end {cases}
      | c = 
}}
{{eqn | r = \sequence {\dfrac 1 1, 1 + \dfrac 1 1, \dfrac 1 2, 1 + \dfrac 1 2, \dfrac 1 3, 1 + \dfrac 1 3, \dotsb}
      | c = 
}}
{{end-eqn}}
Then $0$ is not a limit of $\sequence {a_n}$.

Proof:
The open interval $\openint {-1} 1$ contains $0$, and also contains all terms of $\sequence {a_n}$ with odd indices greater than $1$.
However, all terms of $\sequence {a_n}$ with even indices are outside $\openint {-1} 1$.
Hence $0$ cannot be a limit of $\sequence {a_n}$.
{{qed}}
Theorem: Lagrange's Theorem (Number Theory)

Let $\struct {\Z_p, +_p, \times_p}$ be the ring of integers modulo $p$ for some prime $p$.
Let $f$ be a polynomial in one variable of degree $n$ over $\Z_p$.
Then $f$ has at most $n$ roots in $\Z_p$.

Proof:
Proof by induction on $n$:
Theorem: 1 plus Power of 2 is not Perfect Power except 9

The only solution to:
:$1 + 2^n = a^b$
is:
:$\tuple {n, a, b} = \tuple {3, 3, 2}$
for positive integers $n, a, b$ with $b > 1$.

Proof:
It suffices to prove the result for prime values of $b$.
For $n = 0$, it is clear that $1 + 2^0 = 2$ is not a perfect power.
For $n > 0$, $1 + 2^n$ is odd.
Hence for the equation to hold $a$ must be odd as well.
Writing $a = 2 m + 1$ we have:
{{begin-eqn}}
{{eqn | l = 1 + 2^n
      | r = \paren {2 m + 1}^b
}}
{{eqn | r = \sum_{i \mathop = 0}^b \binom b i \paren {2 m}^i \paren 1^{b - i}
      | c = Binomial Theorem
}}
{{eqn | r = \sum_{i \mathop = 0}^b \binom b i \paren {2 m}^i
}}
{{eqn | r = 1 + \sum_{i \mathop = 1}^b \binom b i \paren {2 m}^i
      | c = Binomial Coefficient with Zero
}}
{{eqn | l = 2^n
      | r = \sum_{i \mathop = 1}^b \binom b i \paren {2 m}^i
}}
{{eqn | r = 2 m \sum_{i \mathop = 1}^b \binom b i \paren {2 m}^{i - 1}
      | c = $m \ne 0$
}}
{{end-eqn}}
Since all factors of $2^n$ are powers of $2$:
:$\ds \sum_{i \mathop = 1}^b \binom b i \paren {2 m}^{i - 1}$ is a power of $2$.
But since each summand is non-negative:
:$\ds \sum_{i \mathop = 1}^b \binom b i \paren {2 m}^{i - 1} \ge 2$
we must have $\ds \sum_{i \mathop = 1}^b \binom b i \paren {2 m}^{i - 1}$ is even.
We have:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^b \binom b i \paren {2 m}^{i - 1}
      | r = \binom b 1 + \sum_{i \mathop = 2}^b \binom b i \paren {2 m}^{i - 1}
}}
{{eqn | r = b + 2 m \sum_{i \mathop = 2}^b \binom b i \paren {2 m}^{i - 2}
}}
{{eqn | o = \equiv
      | r = b
      | rr = \pmod 2
}}
{{end-eqn}}
Therefore we must have $b = 2$, the only even prime.
In that case:
{{begin-eqn}}
{{eqn | l = 2^n
      | r = \paren {2 m + 1}^2 - 1
}}
{{eqn | r = 4 m^2 + 4 m + 1 - 1
}}
{{eqn | r = 4 m \paren {m + 1}
}}
{{end-eqn}}
So $m$ and $m + 1$ are powers of $2$.
The only $m$ satisfying this is $1$, giving the solution:
{{begin-eqn}}
{{eqn | l = a
      | r = 2 m + 1
      | c = 
}}
{{eqn | r = 3
      | c = 
}}
{{eqn | l = 2^n
      | r = 3^2 - 1
      | c = 
}}
{{eqn | r = 8
      | c = 
}}
{{eqn | ll= \leadsto
      | l = n
      | r = 3
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Fermat Number is not Perfect Power

There exist no Fermat numbers which are perfect powers.

Proof:
Each Fermat number is in the form of $1 + 2^n$ for some $n \in \Z$.
This $n$ must also be a power of $2$.
From 1 plus Power of 2 is not Perfect Power except 9 we have:
:$1 + 2^n = a^b$
has only one solution $\tuple {n, a, b} = \tuple {3, 3, 2}$.
But $3$ is not a power of $2$.
Hence no Fermat numbers are perfect powers.
{{qed}}
Category:Fermat Numbers
blocil9c4mkehf14is8l2emcyx5njxl
Theorem: Uncountable Closed Ordinal Space is Countably Compact

Let $\Omega$ denote the first uncountable ordinal.
Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$.
Then $\closedint 0 \Omega$ is a countably compact space.

Proof:
We have:
:Closed Ordinal Space is Compact
:Compact Space is Countably Compact
{{qed}}
Theorem: Uncountable Open Ordinal Space is Countably Compact

Let $\Omega$ denote the first uncountable ordinal.
Let $\hointr 0 \Omega$ denote the open ordinal space on $\Omega$.
Then $\hointr 0 \Omega$ is a countably compact space.

Proof:
Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$.
From Uncountable Closed Ordinal Space is Countably Compact, $\closedint 0 \Omega$ is a countably compact space.
So every sequence in $\hointr 0 \Omega$ has an accumulation point in $\closedint 0 \Omega$.
{{LinkWanted|sequence in $\hointr 0 \Omega$ has an accumulation point in $\closedint 0 \Omega$}}
But $\Omega$ cannot be an accumulation point of any sequence in $\closedint 0 \Omega$.
{{LinkWanted|$\Omega$ cannot be an accumulation point of any sequence in $\closedint 0 \Omega$}}
So every sequence in $\hointr 0 \Omega$ has an accumulation point in $\hointr 0 \Omega$.
This means that $\hointr 0 \Omega$ is countably compact.
{{qed}}
Theorem: Uncountable Open Ordinal Space is not Metacompact

Let $\Omega$ denote the first uncountable ordinal.
Let $\hointr 0 \Omega$ denote the open ordinal space on $\Omega$.
Then $\hointr 0 \Omega$ is not a metacompact space.

Proof:
{{AimForCont}} $\hointr 0 \Omega$ is a metacompact space.
From Uncountable Open Ordinal Space is Countably Compact we have that $\hointr 0 \Omega$ is a countably compact space.
From Metacompact Countably Compact Space is Compact it follows that $\hointr 0 \Omega$ is a compact space.
But from Open Ordinal Space is not Compact Space this contradicts the fact that $\hointr 0 \Omega$ is not a compact space.
Hence the result by Proof by Contradiction.
{{qed}}
Theorem: Uncountable Open Ordinal Space is not Paracompact

Let $\Omega$ denote the first uncountable ordinal.
Let $\hointr 0 \Omega$ denote the open ordinal space on $\Omega$.
Then $\hointr 0 \Omega$ is not a paracompact space.

Proof:
{{AimForCont}} $\hointr 0 \Omega$ is a paracompact space.
From Paracompact Space is Metacompact, it follows that $\hointr 0 \Omega$ is a metacompact space.
But from Uncountable Open Ordinal Space is not Metacompact this contradicts the fact that $\hointr 0 \Omega$ is not a metacompact space.
Hence the result by Proof by Contradiction.
{{qed}}
Theorem: Uncountable Open Ordinal Space is not Lindelöf

Let $\Omega$ denote the first uncountable ordinal.
Let $\hointr 0 \Omega$ denote the open ordinal space on $\Omega$.
Then $\hointr 0 \Omega$ is not a Lindelöf space.

Proof:
{{AimForCont}} $\hointr 0 \Omega$ is a Lindelöf space.
From Ordinal Space is Completely Normal, $\hointr 0 \Omega$ is a completely normal.
From Sequence of Implications of Separation Axioms, $\hointr 0 \Omega$ is a $T_3$ space.
From $T_3$ Lindelöf Space is Paracompact, it follows that $\hointr 0 \Omega$ is a paracompact space.
But this contradicts the fact that from Uncountable Open Ordinal Space is not Paracompact, $\hointr 0 \Omega$ is not a paracompact space.
Hence the result by Proof by Contradiction.
{{qed}}
Theorem: Uncountable Open Ordinal Space is not Sigma-Compact

Let $\Omega$ denote the first uncountable ordinal.
Let $\hointr 0 \Omega$ denote the open ordinal space on $\Omega$.
Then $\hointr 0 \Omega$ is not a $\sigma$-compact space.

Proof:
{{AimForCont}} $\hointr 0 \Omega$ is a $\sigma$-compact space.
From Sigma-Compact Space is Lindelöf Space, $\hointr 0 \Omega$ is a Lindelöf space.
But this contradicts the fact that from Uncountable Open Ordinal Space is not Lindelöf, $\hointr 0 \Omega$ is not a Lindelöf space.
Hence the result by Proof by Contradiction.
{{qed}}
Theorem: Uncountable Closed Ordinal Space is Lindelöf

Let $\Omega$ denote the first uncountable ordinal.
Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$.
Then $\closedint 0 \Omega$ is a Lindelöf space.

Proof:
We have:
:Closed Ordinal Space is Compact
:Compact Space is Lindelöf
{{qed}}
Theorem: Uncountable Closed Ordinal Space is Sigma-Compact

Let $\Omega$ denote the first uncountable ordinal.
Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$.
Then $\closedint 0 \Omega$ is a $\sigma$-compact space.

Proof:
We have:
:Closed Ordinal Space is Compact
:Compact Space is $\sigma$-Compact Space
{{qed}}
Theorem: Uncountable Open Ordinal Space is Sequentially Compact Space

Let $\Omega$ denote the first uncountable ordinal.
Let $\hointr 0 \Omega$ denote the open ordinal space on $\Omega$.
Then $\hointr 0 \Omega$ is a sequentially compact space.

Proof:
We have that:
:Uncountable Open Ordinal Space is First-Countable
:Uncountable Open Ordinal Space is Countably Compact
The result follows from First-Countable Space is Sequentially Compact iff Countably Compact.
{{qed}}
Theorem: Recurrence Relation for Euler Numbers

Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Then:
{{begin-eqn}}
{{eqn | l = E_{2 n}
      | r = -\sum_{k \mathop = 0}^{n - 1} \dbinom {2 n} {2 k} E_{2 k}
      | c = 
}}
{{eqn | r = -\paren {\binom {2 n} 0 E_0 + \binom {2 n} 2 E_2 + \binom {2 n} 4 E_4 + \cdots + \binom {2 n} {2 n - 2} E_{2 n - 2} }
      | c = 
}}
{{end-eqn}}
where $E_n$ denotes the $n$th Euler number.

Proof:
{{begin-eqn}}
{{eqn | q = \forall n \in \Z_{>0}
      | l = \sum_{k \mathop = 0}^n \binom {2 n} {2 k} E_{2 k}
      | r = 0
      | c = Sum of Euler Numbers by Binomial Coefficients Vanishes
}}
{{eqn | ll= \leadsto
      | l = \sum_{k \mathop = 0}^{n - 1} \dbinom {2 n} {2 k} E_{2 k} + \dbinom {2 n} {2 n} E_{2 n}
      | r = 0
      | c = separating out case where $k = n$
}}
{{eqn | ll= \leadsto
      | l = E_{2 n}
      | r = -\sum_{k \mathop = 0}^{n - 1} \dbinom {2 n} {2 k} E_{2 k}
      | c = Binomial Coefficient with Self: $\dbinom {2 n} {2 n} = 1$
}}
{{end-eqn}}
{{qed}}
Theorem: Integer to Power of Multiple of Order/Corollary

Then $\map \phi n$ is a multiple of $c$, where $\map \phi n$ is the Euler phi function of $n$.

Proof:
From Euler's Theorem (Number Theory):
:$a^{\map \phi n} \equiv 1 \pmod n$
Applying Integer to Power of Multiple of Order we see that $\map \phi n$ is a multiple of $c$.
{{qed}}
Theorem: Divisor of Fermat Number/Euler's Result

Then $m$ is in the form:
:$k \, 2^{n + 1} + 1$
where $k \in \Z_{>0}$ is an integer.

Proof:
It is sufficient to prove the result for prime divisors.
The general argument for all divisors follows from the argument:
:$\paren {a \, 2^c + 1} \paren {b \, 2^c + 1} = a b \, 2^{2 c} + \paren {a + b} \, 2^c + 1 = \paren {a b \, 2^c + a + b} \, 2^c + 1$
So the product of two factors of the form preserves that form.
Let $n \in \N$.
Let $p$ be a prime divisor of $F_n = 2^{2^n} + 1$.
Then we have:
{{begin-eqn}}
{{eqn | l = 2^{2^n}
      | o = \equiv
      | r = 1
      | rr= \pmod p
}}
{{eqn | ll= \leadsto
      | l = 2^{2^{n + 1} }
      | o = \equiv
      | r = 1
      | rr= \pmod p
      | c = Congruence of Powers: squaring both sides
}}
{{end-eqn}}
From Integer to Power of Multiple of Order, the order of $2$ modulo $p$ divides $2^{n + 1}$ but not $2^n$.
Therefore it must be $2^{n + 1}$.
Hence:
{{begin-eqn}}
{{eqn | q = \exists k \in \Z
      | l = \map \phi p
      | r = k \, 2^{n + 1}
      | c = {{Corollary|Integer to Power of Multiple of Order}}
}}
{{eqn | ll= \leadsto
      | l = p - 1
      | r = k \, 2^{n + 1}
      | c = Euler Phi Function of Prime
}}
{{eqn | ll= \leadsto
      | l = p
      | r = k \, 2^{n + 1} + 1
}}
{{end-eqn}}
{{qed}}
Theorem: Divisor of Fermat Number/Refinement by Lucas

Let $n \ge 2$.
Then $m$ is in the form:
:$k \, 2^{n + 2} + 1$

Proof:
It is sufficient to prove the result for prime divisors.
The general argument for all divisors follows from the argument:
:$\paren {a \, 2^c + 1} \paren {b \, 2^c + 1} = a b \, 2^{2 c} + \paren {a + b} \, 2^c + 1 = \paren {a b \, 2^c + a + b} \, 2^c + 1$
So the product of two factors of the form preserves that form.
Let $p$ be a prime divisor of $F_n = 2^{2^n} + 1$.
From Euler's Result:
:$\exists q \in \Z: p = q \, 2^{n + 1} + 1$
Since $n \ge 2$, $q \, 2^{n + 1}$ is divisible by $2^{2 + 1} = 8$.
Hence:
:$p \equiv 1 \pmod 8$
By Second Supplement to Law of Quadratic Reciprocity:
:$\paren {\dfrac 2 p} = 1$
so $2$ is a quadratic residue modulo $p$.
Hence:
:$\exists x \in \Z: x^2 = 2 \pmod p$
We have shown $2^{2^n} \equiv -1 \pmod p$ and $2^{2^{n + 1} } \equiv 1 \pmod p$.
By Congruence of Powers:
:$x^{2^{n + 1} } \equiv 2^{2^n} \equiv -1 \pmod p$
:$x^{2^{n + 2} } \equiv 2^{2^{n + 1}} \equiv 1 \pmod p$
From Integer to Power of Multiple of Order, the order of $x$ modulo $p$ divides $2^{n + 2}$ but not $2^{n + 1}$.
Therefore it must be $2^{n + 2}$.
Hence:
{{begin-eqn}}
{{eqn | q = \exists k \in \Z
      | l = \map \phi p
      | r = k \, 2^{n + 2}
      | c = {{Corollary|Integer to Power of Multiple of Order}}
}}
{{eqn | l = p - 1
      | r = k \, 2^{n + 2}
      | c = Euler Phi Function of Prime
}}
{{eqn | l = p
      | r = k \, 2^{n + 2} + 1
}}
{{end-eqn}}
{{qed}}
Theorem: Set Closure is Smallest Closed Set/Normed Vector Space

Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Let $S$ be a subset of $X$:
:$S \subseteq X$
Let $S^-$ be the closure of $S$.
Then $S^-$ is the smallest closed set which contains $S$.

Proof:
Let $F$ be a closed set in $X$.
Suppose $S \subseteq F$.
Theorem: Bounded Real Function may not be of Bounded Variation

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

Proof:
Let $a = 0$, $b = 1$.
Define $f : \closedint 0 1 \to \R$ by: 
:$\map f x = \begin{cases} 1 & x \in \Q \\ 0 & x \not \in \Q \end{cases}$
For each finite subdivision $P$ of $\closedint 0 1$, write: 
:$P = \set {x_0, x_1, \ldots, x_n }$
with:
:$0 = x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n = 1$
For each such subdivision also write:
:$\ds \map {V_f} {P ; \closedint 0 1} = \sum_{i \mathop = 1}^n \size {\map f {x_i} - \map f {x_{i - 1} } }$
It suffices to find a sequence of finite subdivisions $\sequence {P_n}$ such that: 
:$\ds \lim_{n \mathop \to \infty} \map {V_f} {P_n ; \closedint 0 1} = \infty$
by the definition of bounded variation.
For each $n \in \N$, let $P_n$ be a finite subdivision of size $n + 2$.
Let $\sequence {x_n}_{0 \mathop \le i \mathop \le n + 2}$ be the sequence forming this subdivision.
Note that from Between two Rational Numbers exists Irrational Number:
:it is possible to select an irrational number strictly between two rational numbers.
Similarly from Between two Real Numbers exists Rational Number:
:it is possible to select a rational number strictly between two irrational numbers.
We can therefore define $x_i$ for $1 \le i \le n + 1$ as follows:
:if $i$ is odd, let $x_i$ be an irrational number between $x_{i - 1}$ and $b$
:if $i$ is even, let $x_i$ be a rational number between $x_{i - 1}$ and $b$.
Let $x_0 = 0$ and $x_{n + 2} = 1$ to complete the definition.
Then:
{{begin-eqn}}
{{eqn	| l = \map {V_f} {P_n ; \closedint 0 1}
	| r = \sum_{i \mathop = 1}^{n + 2} \size {\map f {x_i} - \map f {x_{i - 1} } }
}}
{{eqn	| o = \ge
	| r = \sum_{i \mathop = 1}^{n + 1} \size {\map f {x_i} - \map f {x_{i - 1} } }
}}
{{end-eqn}}
We omit the term $i = n + 2$ for convenience.
Note that:
:if $x_i$ is rational then $x_{i + 1}$ is irrational.
That is:
:if $\map f {x_i} = 1$ then $\map f {x_{i + 1} } = 0$.
Similarly:
:if $x_i$ is irrational then $x_{i + 1}$ is rational.
That is:
:if $\map f {x_i} = 0$ then $\map f {x_{i + 1} } = 1$.
So for all $1 \le i \le n + 1$, we have:
:$\size {\map f {x_i} - \map f {x_{i - 1} } } = 1$
Hence:
:$\ds \sum_{i \mathop = 1}^{n + 1} \size {\map f {x_i} - \map f {x_{i - 1} } } = n + 1$
So:
:$\ds \map {V_f} {P_n ; \closedint 0 1} \ge n + 1$
giving:
:$\ds \lim_{n \mathop \to \infty} \map {V_f} {P_n ; \closedint 0 1} = \infty$
as required.
{{qed}}
Theorem: Element is Loop iff Member of Closure of Empty Set

Let $M = \struct{S, \mathscr I}$ be a matroid.
Let $x \in S$.
Then:
:$x$ is a loop {{iff}} $x \in \map \sigma \O$
where $\map \sigma \O$ denotes the closure of the empty set.

Proof:
From Element is Loop iff Rank is Zero:
:$x$ is a loop {{iff}} $\map \rho {\set x} = 0$
where $\rho$ is the rank function of $M$.
Now:
{{begin-eqn}}
{{eqn | r = x \in \map \sigma \O
      | o =
}}
{{eqn | ll= \leadstoandfrom
      | r = x \sim \O
      | o =
      | c = {{Defof|Closure Operator (Matroid)|Closure Operator}}
}}
{{eqn | ll= \leadstoandfrom
      | r = \map \rho {\set x} = \map \rho \O
      | o =
      | c = {{Defof|Depends Relation (Matroid)|Depends Relation}}
}}
{{eqn | ll= \leadstoandfrom
      | r = \map \rho {\set x} = 0
      | o =
      | c = Rank of Empty Set is Zero
}}
{{end-eqn}}
The result follows.
{{qed}}
Theorem: Singleton is Dependent implies Rank is Zero/Corollary

:$x$ is a loop {{iff}} $\map \rho {\set x} = 0$

Proof:
By definition of a loop:
:$x$ is a loop {{iff}} $\set x \notin \mathscr I$
From Singleton is Dependent implies Rank is Zero:
:if $\set x \notin \mathscr I$ then $\map \rho {\set x} = 0$
From Singleton is Independent implies Rank is One:
:if $\set x \in \mathscr I$ then $\map \rho {\set x} = 1$
It follows that:
:$\set x \notin \mathscr I$ {{iff}} $\map \rho {\set x} = 0$
{{qed}}
Theorem: Superset of Dependent Set is Dependent/Corollary

Let $A \subseteq S$.
Let $x \in A$.
If $x$ is a loop then $A$ is dependent.

Proof:
Let $x$ be a loop.
By definition of a loop:
:$\set x \notin \mathscr I$
By definition of a dependent subset:
:$\set x$ is a dependent subset
From Singleton of Element is Subset:
:$\set x \subseteq A$
From Superset of Dependent Set is Dependent:
:$A$ is a dependent subset
{{qed}}
Theorem: Closure of Subset contains Loop

Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $x$ be a loop of $M$.
Let $A \subseteq S$.
Then:
:$x \in \map \sigma A$
where $\map \sigma A$ denotes the closure of $A$.

Proof:
By definition of the closure of $A$:
:$x \in \map \sigma A$ {{iff}} $x \sim A$
where $\sim$ is the depends relation on $M$.
By definition of the depends relation:
:$x \sim A$ {{iff}} $\map \rho {A \cup \set x} = \map \rho A$
where $\rho$ is the rank function on $M$.
So it remains to show that:
:$\map \rho {A \cup \set x} = \map \rho A$
By definition of the rank function:
:$\map \rho {A \cup \set x} = \max \set {\size X : X \subseteq A \cup \set x \land X \in \mathscr I}$
From Max Operation Equals an Operand:
:$\exists X \in \mathscr I : X \subseteq A \cup \set x \land \size X = \map \rho {A \cup \set x}$
From the contrapositive statement of Set is Dependent if Contains Loop:
:$x \notin X$
Now:
{{begin-eqn}}
{{eqn | l = X
      | r = \paren{A \cup \set x} \cap X
      | c = Intersection with Subset is Subset
}}
{{eqn | r = \paren{A \cap X} \cup \paren {\set x \cap X}
      | c = Intersection Distributes over Union
}}
{{eqn | r = \paren{A \cap X} \cup \O
      | c = Intersection With Singleton is Disjoint if Not Element
}}
{{eqn | r = \paren{A \cap X}
      | c = Union with Empty Set
}}
{{eqn | ll = \leadsto
      | l = X
      | o = \subseteq
      | r = A
      | c = Intersection with Subset is Subset
}}
{{end-eqn}}
From Max Operation Yields Supremum of Operands:
:$\size X \le \max \set {\size Y : Y \subseteq A \land Y \in \mathscr I} = \map \rho A$
From Rank Function is Increasing:
:$\map \rho A \le \map \rho {A \cup \set x} = \size X$
Thus:
:$\map \rho A = \size X = \map \rho {A \cup \set x}$
{{qed}}
Theorem: Element is Loop iff Singleton is Circuit

Let $M = \struct{S, \mathscr I}$ be a matroid.
Let $x \in S$.
Then:
:$x$ is a loop {{iff}} $\set x$ is a circuit

Proof:
=== Necessary Condition ===
Let $x$ be a loop.
By definition of a loop:
:$\set x$ is a dependent subset of $S$
Let $A \subseteq \set x$ be a dependent subset.
From Power Set of Singleton:
:$\powerset {\set x} = \set{\O, \set x}$
By matroid axiom $(\text I 1)$:
:$\O$ is an independent subset
Then:
:$A = \set x$
It follows that:
:$\set x$ is a minimal dependent subset of $S$.
Then $\set x$ be a circuit by definition. 
{{qed|lemma}}