avewright/proofwiki-math huggingface dataset:
Theorem: Addition of Cuts is Associative

Let $\alpha$, $\beta$ and $\gamma$ be cuts.
Let the operation of $\alpha + \beta$ be the sum of $\alpha$ and $\beta$.
Then:
:$\paren {\alpha + \beta} + \gamma = \alpha + \paren {\beta + \gamma}$

Proof:
$\paren {\alpha + \beta} + \gamma$ is the set of all rational numbers of the form $\paren {p + q} + r$ such that $p \in \alpha$, $q \in \beta$ and $r \in \gamma$.
Similarly, $\alpha + \paren {\beta + \gamma}$ is the set of all rational numbers of the form $p + \paren {q + r}$ such that $p \in \alpha$, $q \in \beta$ and $r \in \gamma$.
From Rational Addition is Associative we have that:
:$\paren {p + q} + r = p + \paren {q + r}$
The result follows.
{{qed}}
Theorem: Identity Element for Addition of Cuts

Let $\alpha$ be a cut.
Let $0^*$ be the rational cut associated with the (rational) number $0$:
:$0^* = \set {r \in \Q: r < 0}$
Then:
:$\alpha + 0^* = \alpha$
where $+$ denotes the operation of addition of cuts.

Proof:
Let $r \in \alpha + 0^*$.
$\alpha + 0^*$ is the set of all rational numbers of the form $p + q$ such that $p \in \alpha$, $q \in 0^*$, that is, $q < 0$.
It follows that:
:$p + q < p$
and so:
:$p + q \in \alpha$
that is:
:$r \in \alpha$
Let $r \in \alpha$.
Let $s \in \Q$ be a rational number such that $s > r$ and $s \in \alpha$.
This is possible because $r$ is not the greatest element of $\alpha$ by definition of a cut.
Let $q = r - s$.
Then $q < 0$ and so $q \in 0^*$.
Thus we have:
:$r = s + q$
and so:
:$r \in \alpha + 0^*$
We have that:
:$r \in \alpha \implies r \in \alpha + 0^*$
and:
:$r \in \alpha + 0^* \implies r \in \alpha$
and so by definition of set equality:
:$\alpha + 0^* = \alpha$
{{qed}}
Theorem: Existence of Upper and Lower Numbers of Cut whose Difference equal Given Rational

Let $\alpha$ be a cut.
Let $r \in \Q_{>0}$ be a (strictly) positive rational number.
Then there exist rational numbers $p$ and $q$ such that:
:$p \in \alpha, q \notin \alpha$
:$q - p = r$
such that $q$ is not the smallest upper number of $\alpha$.

Proof:
Let $s \in \alpha$ be a rational number.
For $n = 0, 1, 2, \ldots$ let $s_n = s + n r$.
Then there exists a unique integer $m$ such that:
:$s_m \in \alpha$
and:
:$s_{m + 1} \notin \alpha$
If $s_{m + 1}$ is not the smallest upper number of $\alpha$, take:
:$p = s_m$
:$q = s_{m + 1}$
If $s_{m + 1}$ ''is'' the smallest upper number of $\alpha$, take:
:$p = s_m + \dfrac r 2$
:$q = s_{m + 1} + \dfrac r 2$
The result follows.
{{qed}}
Theorem: Existence of Unique Inverse Element for Addition of Cuts

Let $\alpha$ be a cut.
Let $0^*$ be the rational cut associated with the (rational) number $0$:
:$0^* = \set {r \in \Q: r < 0}$
Then there exists a unique cut $\beta$ such that:
:$\alpha + \beta = 0^*$
where $+$ denotes the operation of addition of cuts.

Proof:
=== Proof of Uniqueness ===
Suppose $\alpha + \beta_1 = \alpha + \beta_2 = 0^*$.
We have:
{{begin-eqn}}
{{eqn | l = \beta_2
      | r = 0^* + \beta_2
      | c = Identity Element for Addition of Cuts
}}
{{eqn | r = \paren {\alpha + \beta_1} + \beta_2
      | c = Definition of $\beta_1$
}}
{{eqn | r = \paren {\alpha + \beta_2} + \beta_1
      | c = Addition of Cuts is Commutative and Addition of Cuts is Associative
}}
{{eqn | r = 0^* + \beta_1
      | c = Definition of $\beta_2$
}}
{{eqn | r = \beta_1
      | c = Identity Element for Addition of Cuts
}}
{{end-eqn}}
So if such a $\beta$ exists, it is indeed unique.
{{qed|lemma}}
Theorem: Ordering on Cuts is Compatible with Addition of Cuts

Let $\alpha$, $\beta$ and $\gamma$ be cuts.
Let the operation of $\alpha + \beta$ be the sum of $\alpha$ and $\beta$.
Let $\beta < \gamma$ denotes the strict ordering on cuts defined as:
:$\beta < \gamma \iff \exists p \in \Q: p \in \beta, p \notin \gamma$
Then:
:$\beta < \gamma \implies \alpha + \beta < \alpha + \gamma$

Proof:
By definition of the strict ordering on cuts and addition of cuts:
:$\alpha + \beta \le \alpha + \gamma$
{{explain|Details needed for the above}}
Suppose $\alpha + \beta = \alpha + \gamma$.
Then:
{{begin-eqn}}
{{eqn | l = \beta
      | r = 0^* + \beta
      | c = Identity Element for Addition of Cuts
}}
{{eqn | r = \paren {-\alpha} + \paren {\alpha + \beta}
      | c = {{Defof|Negative of Cut}}
}}
{{eqn | r = \paren {-\alpha} + \paren {\alpha + \gamma}
      | c = 
}}
{{eqn | r = 0^* + \gamma
      | c = 
}}
{{eqn | r = \gamma
      | c = 
}}
{{end-eqn}}
Thus:
:$\beta = \gamma$
and the result follows.
{{qed}}
Theorem: Ordering on Cuts is Compatible with Addition of Cuts/Corollary

Let $0^*$ denote the rational cut associated with the (rational) number $0$.
If:
:$\alpha > 0^*$ and $\gamma > 0^*$
then:
:$\alpha + \gamma > 0^*$

Proof:
From Ordering on Cuts is Compatible with Addition of Cuts
:$0^* + 0^* < 0^* + \alpha$
:$\alpha + 0^* < \alpha + \gamma$
The result follows from Ordering on Cuts is Transitive.
{{qed}}
Theorem: Existence of Unique Difference between Cuts

Let $\alpha$ and $\beta$ be cuts.
Then there exists exactly one cut $\gamma$ such that:
:$\alpha + \gamma = \beta$

Proof:
From Ordering on Cuts is Compatible with Addition of Cuts:
:$\gamma_1 \ne \gamma_2 \implies \alpha + \gamma_1 \ne \alpha + \gamma_2$
That demonstrates uniqueness.
Let:
:$\gamma = \beta + \paren {-\alpha}$
where $-\alpha$ is the negative of $\alpha$.
Then by Identity Element for Addition of Cuts:
{{begin-eqn}}
{{eqn | l = \alpha + \gamma
      | r = \alpha + \paren {\beta + \paren {-\alpha} }
      | c = Definition of $\gamma$
}}
{{eqn | r = \alpha + \paren {\paren {-\alpha} + \beta}
      | c = Addition of Cuts is Commutative
}}
{{eqn | r = \paren {\alpha + \paren {-\alpha} } + \beta
      | c = Addition of Cuts is Associative
}}
{{eqn | r = 0^* + \beta
      | c = Existence of Unique Inverse Element for Addition of Cuts
}}
{{eqn | r = \beta
      | c = Identity Element for Addition of Cuts
}}
{{end-eqn}}
The result follows.
{{qed}}
Theorem: Set of Cuts under Addition forms Abelian Group

Let $\CC$ denote the set of cuts.
Let $\struct {\CC, +}$ denote the algebraic structure formed from $\CC$ and the operation $+$ of addition of cuts.
Then $\struct {\CC, +}$ forms an abelian group.

Proof:
In the below, $\alpha$, $\beta$ and $\gamma$ denote arbitrary cuts.
Taking the abelian group axioms in turn:
Theorem: Product of Cuts is Cut

Let $\alpha$ and $\beta$ be cuts.
Let $\alpha \beta$ denote the product of cuts.
Then $\alpha \beta$ is also a cut.
Thus the operation of multiplication on the set of cuts is closed.

Proof:
{{ProofWanted}}
Category:Cuts
Category:Multiplication
m4f4p5wp4vizxgz4w7o28tg6sd15z9z
Theorem: Factorisation of z^n+a

Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $a \in \C$ be a complex number.
Then:
:$z^n + a = \ds \prod_{k \mathop = 0}^{n - 1} \paren {z - \alpha_k b}$
where:
:$\alpha_k$ are the complex $n$th roots of negative unity
:$b$ is any complex number such that $b^n = a$.

Proof:
From $z^n + a = 0$ we have that:
:$z^n = - a$
Let $b = a^{1 / n}$, hence $b^n = a$, with $a, b \in \C$.
From Roots of Complex Number:
{{begin-eqn}}
{{eqn | l = z^{1 / n}
      | r = \set {a^{1 / n} e^{i \paren {\theta + 2 k \pi} / n}: k \in \set {0, 1, 2, \ldots, n - 1}, \theta = \arg -a}
      | c = $z^n = -a$ so we need $\theta = \arg -a$
}}
{{eqn | r = \set {a^{1 / n} e^{i \paren {2 k + 1} \pi / n}: k \in \set {0, 1, 2, \ldots, n - 1} }
      | c = $\theta = \pi$
}}
{{eqn | r = \set {b e^{i \paren {2 k + 1} \pi / n}: k \in \set {0, 1, 2, \ldots, n - 1} }
      | c = $b = a^{1 / n}$
}}
{{end-eqn}}
and so each of $\alpha_k b = b e^{i \paren {2 k + 1} \pi / n}$ is a unique root of $z^n + a$ with $n \in \Z_{>0}$.
From the corollary to the Polynomial Factor Theorem:
If $\zeta_1, \zeta_2, \ldots, \zeta_n \in \C$ such that all are different, and $\map P {\zeta_1} = \map P {\zeta_2} = \cdots = \map P {\zeta_n} = 0$, then:
:$\ds \map P z = k \prod_{j \mathop = 1}^n \paren {z - \zeta_j}$
where $k \in \C$.
$z^n + a$ is a monic polynomial, hence $k = 1$ in the above product.
Choose $\zeta_j = \alpha_{j - 1} b$ and we have the desired result.
{{qed}}
Category:Algebra
Category:Complex Roots
cmohxq4yoonr3xlwmz7weg2978u2u8u
Theorem: Multiplication of Cuts is Commutative

Let $\alpha$ and $\beta$ be cuts.
Let the $\alpha \beta$ be the product of $\alpha$ and $\beta$.
Then:
:$\alpha \beta = \beta \alpha$

Proof:
By definition, we have that:
:$\alpha \beta := \begin {cases}
\size \alpha \, \size \beta & : \alpha \ge 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha < 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha \ge 0^*, \beta < 0^* \\
\size \alpha \, \size \beta & : \alpha < 0^*, \beta < 0^*
\end {cases}$
where:
:$\size \alpha$ denotes the absolute value of $\alpha$
:$0^*$ denotes the rational cut associated with the (rational) number $0$
:$\ge$ denotes the ordering on cuts.
Let $\alpha \ge 0^*$ and $\beta \ge 0^*$.
:$\gamma := \alpha \beta$
where $\gamma$ is the set of all rational numbers $r$ such that:
:$\exists p \in \alpha, q \in \beta: r = p q$
$\alpha \beta$ is the set of all rational numbers $r$ either of the form $r < 0$ of the form $r = p q$ such that $p \in \alpha$ and $q \in \beta$.
Similarly, $\beta \alpha$ is the set of all rational numbers $r$ either of the form $r < 0$ of the form $r = q p$ such that $p \in \alpha$ and $q \in \beta$.
From Rational Multiplication is Commutative we have that:
:$p q = q p$
Thus we have that:
:$\size \alpha \, \size \beta = \size \beta \, \size \alpha$
and the result follows.
{{qed}}
Theorem: Multiplication of Cuts is Associative

Let $\alpha$, $\beta$ and $\gamma$ be cuts.
Let $\alpha \beta$ denote the product of $\alpha$ and $\beta$.
Then:
:$\paren {\alpha \beta} \gamma = \alpha \paren {\beta \gamma}$

Proof:
By definition, we have that:
:$\alpha \beta := \begin {cases}
\size \alpha \, \size \beta & : \alpha \ge 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha < 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha \ge 0^*, \beta < 0^* \\
\size \alpha \, \size \beta & : \alpha < 0^*, \beta < 0^*
\end {cases}$
where:
:$\size \alpha$ denotes the absolute value of $\alpha$
:$0^*$ denotes the rational cut associated with the (rational) number $0$
:$\ge$ denotes the ordering on cuts.
Let $\alpha \ge 0^*$, $\beta \ge 0^*$ and $\gamma \ge 0^*$.
$\paren {\alpha \beta} \gamma$ is the set of all rational numbers $s$ of the form $s = \paren {p q} r$ such that $s < 0$ or $p \in \alpha$, $q \in \beta$ and $r \in \gamma$.
Similarly, $\alpha \paren {\beta \gamma}$ is the set of all rational numbers $s$ of the form $s = p \paren {q r}$ such that $s < 0$ or $p \in \alpha$, $q \in \beta$ and $r \in \gamma$.
From Rational Multiplication is Associative we have that:
::$\paren {p q} r = p \paren {q r}$
Thus we have that:
:$\size \alpha \paren {\size \beta \, \size \gamma} = \paren {\size \alpha \, \size \beta} \size \gamma$
and the result follows.
{{qed}}
Theorem: Multiplication of Cuts Distributes over Addition

Let $\alpha$, $\beta$ and $\gamma$ be cuts.
Let:
:$\alpha + \beta$ denote the sum of $\alpha$ and $\beta$.
:$\alpha \beta$ denote the product of $\alpha$ and $\beta$.
Then:
:$\alpha \paren {\beta + \gamma} = \alpha \beta + \alpha \gamma$

Proof:
By definition, we have that:
:$\alpha \beta := \begin {cases}
\size \alpha \, \size \beta & : \alpha \ge 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha < 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha \ge 0^*, \beta < 0^* \\
\size \alpha \, \size \beta & : \alpha < 0^*, \beta < 0^*
\end {cases}$
where:
:$\size \alpha$ denotes the absolute value of $\alpha$
:$0^*$ denotes the rational cut associated with the (rational) number $0$
:$\ge$ denotes the ordering on cuts.
Let $\alpha \ge 0^*$, $\beta \ge 0^*$ and $\gamma \ge 0^*$.
$\alpha \paren {\beta + \gamma}$ is the set of all rational numbers $s$ of the form:
:$s = p \paren {q + r}$
such that:
:$s < 0$
or:
:$p \in \alpha$, $q \in \beta$ and $r \in \gamma$.
$\alpha \beta + \alpha \gamma$ is the set of all rational numbers $s$ of the form:
:$s = p q + p r$
such that:
:$s < 0$
or:
:$p \in \alpha$, $q \in \beta$ and $r \in \gamma$.
From Rational Multiplication Distributes over Addition:
$p \paren {q + r} = p q + p r$
and the result follows.
{{qed}}
Theorem: Product of Cut with Zero Cut equals Zero Cut

Let $\alpha$ be a cut.
Let $0^*$ denote the rational cut associated with the (rational) number $0$.
Then:
:$\alpha 0^* = 0^*$
where $\alpha 0^*$ denotes the product of $\alpha$ and $0^*$.

Proof:
By definition, we have that:
:$\alpha \beta := \begin {cases}
\size \alpha \, \size \beta & : \alpha \ge 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha < 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha \ge 0^*, \beta < 0^* \\
\size \alpha \, \size \beta & : \alpha < 0^*, \beta < 0^*
\end {cases}$
where:
:$\size \alpha$ denotes the absolute value of $\alpha$
:$\size \alpha \, \size \beta$ is defined as in Multiplication of Positive Cuts
:$\ge$ denotes the ordering on cuts.
From Absolute Value of Cut is Zero iff Cut is Zero:
:$\size {0^*} = 0^*$
Thus:
:$\alpha 0^* := \begin {cases}
\size \alpha \, 0^* & : \alpha \ge 0^* \\
-\paren {\size \alpha \, 0^*} & : \alpha < 0^*
\end {cases}$
From Absolute Value of Cut is Greater Than or Equal To Zero Cut, we have that:
:$\size \alpha = \beta$
where $\beta \ge 0^*$.
By definition of Multiplication of Positive Cuts:
$\beta 0^*$ is the set of all rational numbers $r$ such that either:
:$r < 0$
or
:$\exists p \in \beta, q \in 0^*: r = p q$
where $p \ge 0$ and $q \ge 0$
By definition of $0^*$, it is not possible for $q \ge 0$.
Thus:
:$\beta 0^* = \set {r \in \Q: r < 0}$
Hence by definition of a cut:
:$\beta 0^* = 0^*$
By Identity Element for Addition of Cuts and definition of negative of cut:
:$-0^* = 0^*$
and the result follows.
{{qed}}
Theorem: Product of Cuts is Zero Cut iff Either Factor equals Zero Cut

Let $\alpha$ and $\beta$ be cuts.
Let $0^*$ denote the rational cut associated with the (rational) number $0$.
Then:
:$\alpha \beta = 0^*$
{{iff}}:
:$\alpha = 0^*$ or $\beta = 0^*$
where $\alpha \beta$ denotes the product of $\alpha$ and $\beta$.

Proof:
By definition, we have that:
:$\alpha \beta := \begin {cases}
\size \alpha \, \size \beta & : \alpha \ge 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha < 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha \ge 0^*, \beta < 0^* \\
\size \alpha \, \size \beta & : \alpha < 0^*, \beta < 0^*
\end {cases}$
where:
:$\size \alpha$ denotes the absolute value of $\alpha$
:$\size \alpha \, \size \beta$ is defined as in Multiplication of Positive Cuts
:$\ge$ denotes the ordering on cuts.
First we note that from Absolute Value of Cut is Zero iff Cut is Zero:
:$\size {0^*} = 0^*$
Theorem: Cut Associated with 1 is Identity for Multiplication of Cuts

Let $\alpha$ be a cut.
Let $1^*$ denote the rational cut (rational) number $1$.
Then:
:$\alpha 1^* = \alpha$
where $\alpha 1^*$ denote the product of $\alpha$ and $1^*$.

Proof:
By definition, we have that:
:<nowiki>$\alpha \beta := \begin {cases}
\size \alpha \, \size \beta & : \alpha \ge 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha < 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha \ge 0^*, \beta < 0^* \\
\size \alpha \, \size \beta & : \alpha < 0^*, \beta < 0^*
\end {cases}$</nowiki>
where:
:$\size \alpha$ denotes the absolute value of $\alpha$
:$\size \alpha \, \size \beta$ is defined as in Multiplication of Positive Cuts
:$0^*$ denotes the rational cut associated with the (rational) number $0$.
:$\ge$ denotes the ordering on cuts.
We have that:
:$0 < \dfrac 1 2 < 1$
and so:
:$\dfrac 1 2 \notin 0^*$
but:
:$\dfrac 1 2 \in 1^*$
Thus by definition of the strict ordering of cuts:
:$1^* > 0^*$
Thus by definition of absolute value of $\alpha$:
:$\size {1^*} = 1^*$
Thus:
:$\alpha 1^* := \begin {cases}
\size \alpha \, 1^* & : \alpha \ge 0^* \\
-\paren {\size \alpha \, 1^*} & : \alpha < 0^*
\end {cases}$
Let $\alpha > 0^*$.
Let $r \in \alpha 1^*$.
Then by definition of Multiplication of Positive Cuts, either:
:$r < 0$
or
:$\exists p \in \beta, q \in 1^*: r = p q$
where $p \ge 0$ and $q \ge 0$.
Because $0 \le q < 1$ it follows that $p q < p$.
Thus if $r > 0$ it follows that $r < p$.
That is:
:$r \in \alpha$
If $r < 0$ then $r \in \alpha$ because $0^* < \alpha$.
Thus in either case $r \in \alpha$.
That is:
:$\alpha 1^* \le \alpha$
{{finish|tedious}}
Theorem: Multiplication of Positive Cuts preserves Ordering

Let $0^*$ denote the rational cut associated with the (rational) number $0$.
Let $\alpha$, $\beta$ and $\gamma$ be cuts such that:
:$0^* < \alpha < \beta$
:$0^* < \gamma$
where $<$ denotes the strict ordering on cuts.
Then
:$\alpha \gamma < \beta \gamma$
where $\alpha \gamma$ denotes the product of $\alpha$ and $\gamma$.

Proof:
{{ProofWanted|bored with this}}
Theorem: Sum of Rational Cuts is Rational Cut

Let $p \in\ Q$ and $q \in \Q$ be rational numbers.
Let $p^*$ and $q^*$ denote the rational cuts associated with $p$ and $q$.
Then:
:$p^* + q^* = \paren {p + q}^*$
Thus the operation of addition on the set of rational cuts is closed.

Proof:
From Sum of Cuts is Cut, $p^* + q^*$ is a cut.
Let $r \in p^* + q^*$.
Then:
:$r = s + t$
where $s < p$ and $t < q$
Thus:
:$r < p + q$
and so:
:$r \in \paren {p + q}^*$
Hence:
:$p^* + q^* \subseteq \paren {p + q}^*$
{{qed|lemma}}
Let $r \in \paren {p + q}^*$.
Then:
:$r < p + q$
Let:
:$h = p + q - r$
:$s = p - \dfrac h 2$
:$t = q - \dfrac h 2$
Then:
:$s \in p^*$
:$t \in q^*$
Hence:
:$r = s + t$
and so:
:$r \in p^* + q^*$
So:
:$\paren {p + q}^* \subseteq p^* + q^*$
{{qed|lemma}}
Hence the result by definition of set equality.
{{Qed}}
Theorem: Product of Rational Cuts is Rational Cut

Let $p \in\ Q$ and $q \in \Q$ be rational numbers.
Let $p^*$ and $q^*$ denote the rational cuts associated with $p$ and $q$.
Then:
:$p^* q^* = \paren {p q}^*$
Thus the operation of multiplication on the set of rational cuts is closed.

Proof:
From Product of Cuts is Cut, $p^* q^*$ is a cut.
{{ProofWanted}}
Theorem: Ordering of Rational Cuts preserves Ordering of Associated Rational Numbers

Let $p \in\ Q$ and $q \in \Q$ be rational numbers.
Let $p^*$ and $q^*$ denote the rational cuts associated with $p$ and $q$.
Then:
:$p^* < q^* \iff p < q$
where $p^* < q^*$ denotes the strict ordering on cuts defined as:
:$\beta < \gamma \iff \exists p \in \Q: p \in \beta, p \notin \gamma$

Proof:
Let $p < q$.
Then $p \notin p^*$ but $q \in q^*$.
Thus $p^* < q^*$ by definition of the strict ordering on cuts .
Let $p^* < q^*$.
Then:
:$\exists r \in \Q: r \notin p^*, r \in q^*$
Hence:
:$p \le r < q$
and so:
:$p < q$
{{qed}}
Theorem: Exists Rational Cut Between two Cuts

Let $\alpha$ and $\beta$ be cuts.
Let $\alpha < \beta$, where $<$ denotes the strict ordering on cuts.
Then there exists a rational cut $r^*$ associated with the rational number $r$ such that:
:$\alpha < r^* < \beta$

Proof:
Let $\alpha < \beta$.
Then there exists a rational number $p$ such that $p \in \beta$ and $p \notin \alpha$.
Let $r \in \Q$ such that $r > p$ and $r \in \beta$.
Because $r \in \beta$ and $r \notin r^*$, we have that $r^* < \beta$.
Because $p \in r^*$ and $p \notin \alpha$, we have that $\alpha < r^*$.
{{qed}}
Theorem: Condition for Rational Cut to be Less than Given Cut

Let $\alpha$ be a cut.
Let $p^*$ be the rational cut associated with a rational number $p$.
Then:
:$p \in \alpha$
{{iff}}:
:$p^* < \alpha$
where $<$ denotes the strict ordering on cuts.

Proof:
Let $p$ be a rational number such that $p \in \alpha$.
Then by definition of rational cut:
:$p \notin p^*$
Thus:
:$p \in \alpha \implies p^* < \alpha$
{{qed|lemma}}
Let $p^* < \alpha$.
Then there exists a rational number $q$ such that $q \in \alpha$ and $q \notin p$.
Thus $q \ge p$.
But as $q \in \alpha$ it follows that $p \in \alpha$.
{{Qed}}
Theorem: Continuous Mapping to Product Space/General Result

Let $X$ be a topological space.
Let $\family {Y_i}_{i \mathop \in I}$ be an indexed family of topological spaces for some indexing set $I$.
Let $\ds Y = \prod_{i \mathop \in I} Y_i$ be the product space of $\family {Y_i}_{i \mathop \in I}$. 
For each $i \in I$, let $\pr_i: X \to X_i$ denote the $i$th projection on $X$:
:$\forall \family {x_j}_{j \mathop \in I} \in X: \map {\pr_i} {\family {x_j}_{j \mathop \in I} } = x_i$
Let $f$ be a mapping from $X$ to $Y$. 
Then $f$ is continuous {{iff}} $\pr_i \circ f$ is continuous for all $i \in I$.

Proof:
Suppose $f$ is continuous.
From Projection from Product Topology is Continuous $\pr_i$ is continuous for each $i \in I$.
By Composite of Continuous Mappings is Continuous it follows that $\pr_i \circ f$ is continuous for each $i \in I$.
Conversely, suppose that each $\pr_i \circ f$ is continuous. 
Let $U = \pr_{i_1}^{-1} \sqbrk {U_{i_1} } \cap \cdots \cap \pr_{i_n}^{-1} \sqbrk {U_{i_n} }$ be an open set in the natural basis of the product topology for $Y$.
Note that by the definition of natural basis, $U$ is '''constrained''' to be the intersection of a '''finite number''' of sets of $\mathbb S$.
Then by Preimage of Intersection under Mapping we find:
:$f^{-1} \sqbrk U = f^{-1} \sqbrk {\pr_{i_1}^{-1} \sqbrk {U_{i_1} } } \cap \cdots \cap f^{-1} \sqbrk {\pr_{i_n}^{-1} \sqbrk {U_{i_n} } }$
Since each $\pr_{i_j} \circ f$ is continuous, each:
:$\paren {\pr_{i_j} \circ f}^{-1} \sqbrk {U_{i_j} } = f^{-1} \sqbrk {\pr_{i_j}^{-1} \sqbrk {U_{i_j} } }$ 
is open.
Therefore $f^{-1} \sqbrk U$ is open.
By the definition of basis, each open set of $Y$ can be written as a union of sets in the natural basis of the product topology for $Y$.
From Preimage of Union under Mapping, the union of the preimages of open sets of $Y$ equals the preimages of the union of those sets.
Thus the preimage of an open set of $Y$ is open in $X$.
Hence $f$ is continuous.
{{qed}}
Category:Product Topology
Category:Continuous Mappings
s7zhu7twfgfc2fii1ujl9x6dkk26312
Theorem: Set of Cuts forms Ordered Field

Let $\CC$ denote the set of cuts.
Let $\struct {\CC, +, \times, \le}$ denote the ordered structure formed from $\CC$ and:
:the operation $+$ of addition of cuts
:the operation $\times$ of multiplication of cuts
:the ordering $\le$ of cuts.
Then $\struct {\CC, +, \times, \le}$ is an ordered field.

Proof:
First we show that $\struct {\CC, +, \times}$ is a field by demonstrating that it fulfils the field axioms:
{{:Axiom:Field  Axioms}}
It has been established from Set of Cuts under Addition forms Abelian Group that $\struct {\CC, +}$ forms an abelian group.
Thus $\text A 0$ through to $\text A 4$ are fulfilled.
It remains to verify that $\struct {\CC, +, \times}$ fulfils the remaining field axioms.
Theorem: Set of Rational Cuts forms Ordered Field

Let $\RR$ denote the set of rational cuts.
Let $\struct {\RR, +, \times, \le}$ denote the ordered structure formed from $\RR$ and:
:the operation $+$ of addition of cuts
:the operation $\times$ of multiplication of cuts
:the ordering $\le$ of cuts.
Then $\struct {\RR, + \times, \le}$ is an ordered field.

Proof:
We demonstrate that $\struct {\RR, +, \times}$ is a field by showing it is a subfield of the structure $\struct {\CC, +, \times}$, where $\CC$ denotes the set of all cuts.
We do this by establishing that all $4$ criteria of the Subfield Test are satisfied.
We note that $0^* \in \RR$, where $0^*$ is the rational cut associated with the (rational) number $0$:
:$0^* = \set {r \in \Q: r < 0}$
So $\RR \ne \O$.
Thus criterion $(1)$ is satisfied.
{{qed|lemma}}
{{finish}}
{{qed}}
Theorem: Box Topology contains Product Topology

Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of topological spaces where $I$ is an arbitrary index set.
Let $X$ be the cartesian product of $\family {X_i}_{i \mathop \in I}$:
:$\ds X := \prod_{i \mathop \in I} X_i$
Let $\tau$ be the product topology on $X$.
Let $\tau'$ be the box topology on $X$.
Then:
:$\tau \subseteq \tau'$

Proof:
From Natural Basis of Product Topology and Basis for Box Topology, it follows immediately that the natural basis $\BB$ for the product topology is contained in the basis $\BB'$ for the box topology.
From Corollary of Basis Condition for Coarser Topology, it follows that $\tau \subseteq \tau'$.
{{qed}}
Category:Product Topology
Category:Box Topology
mf81o8aj9z0zddrxcpagas1en4g537n