nyuuzyou/wikis huggingface dataset:
Degree of Product of Polynomials over Ring/Corollary 2

== Theorem ==
Let $\struct {D, +, \circ}$ be an integral domain whose zero is $0_D$.
Let $D \sqbrk X$ be the ring of polynomials over $D$ in the indeterminate $X$.
For $f \in D \sqbrk X$ let $\map \deg f$ denote the degree of $f$.
Then:
:$\forall f, g \in D \sqbrk X: \map \deg {f g} = \map \deg f + \map \deg g$
== Proof ==
An integral domain is a commutative and unitary ring with no proper zero divisors.
The result follows from Degree of Product of Polynomials over Ring: Corollary 1.
== Sources ==
* : Chapter $6$: Polynomials and Euclidean Rings: $\S 25$. Polynomials: Theorem $48$
* : $\S 3.2$: Polynomial rings: Lemma $3.7 \ \text{(iii)}$ 
Category:Degree of Product of Polynomials over Ring
Degree of Simple Algebraic Field Extension equals Degree of Algebraic Number

== Theorem ==
Let $F$ be a field.
Let $\theta \in \C$ be algebraic over $F$ of degree $n$.
Let $\map F \theta$ be the simple field extension of $F$ by $\theta$.
Then $\map F \theta$ is a finite extension of $F$ whose degree is:
:$\index {\map F \theta} F = n$
== Proof ==
Considered as a vector space over $F$, $\map F \theta$ is generated by the set $B$, where:
:$B := \set {1, \theta, \theta^2, \ldots, \theta^{n - 1} }$
But $S$ is linearly independent over $F$, because otherwise:
:$c_0 1 + c_1 \theta + c_2 \theta^2 + \dotsb + c_{n - 1} \theta^{n - 1} = 0$
with all the $c$s non-zero.
That would mean $\theta$ was the root of a polynomial whose degree was less than that of the minimal polynomial $\map m x$ of $\theta$, whose degree equals $n$.
Therefore $B$ is a basis of $\map F \theta$ over $F$.
Thus $\map F \theta$ is of dimension $\size B = n$.
Hence the $\index {\map F \theta} F = n$.
== Sources ==
* : Chapter $8$: Field Extensions: $\S 38$. Simple Algebraic Extensions: Theorem $73$
Category:Field Extensions
Degree of Sum of Polynomials

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring with unity whose zero is $0_R$.
Let $R \sqbrk X$ be the ring of polynomials over $R$ in the indeterminate $X$.
For $f \in R \sqbrk X$ let $\map \deg f$ be the degree of $f$.
Then:
:$\forall f, g \in R \sqbrk X: \map \deg {f + g} \le \max \set {\map \deg f, \map \deg g}$
== Proof ==
First we associate to $f = a_0 + a_1 X + \cdots + a_n X^n \in R \sqbrk X$ a formal vector $x_f = \tuple {a_0, a_1, \ldots, a_n, 0_R, \ldots} \in R^\infty$.
Let $x_f^i \in R$ denote the element at the $i$th position. 
Then:
:$\map \deg f = \sup \set {i \in \N : x_f^i \ne 0_R}$
The sum $+$ in the polynomial ring $R \sqbrk X$ gives rise to the following identity in $R^\infty$:
:$x_{f + g}^i = x_f^i + x_g^i$
Next, let $f, g \in R \sqbrk X$, and let $d = \max \set {\map \deg f, \map \deg g}$.
Then $x_f^i = 0_R = x_g^i$ for all $i > d$, so we have:
:$\map \deg {f + g} = \sup \set {i \in \N : x_{f + g}^i \ne 0_R} = \sup \set {i \in \N : x_f^i + x_g^i \ne 0_R} \le d$
== Sources ==
* : $\S 3.2$: Polynomial rings: Lemma $3.7 \ \text{(i)}$ 
Category:Polynomial Theory
Degree of Vertex/Examples/Impossible Order 4 Graph

== Examples of Degrees of Vertices ==
There exists no simple graph whose vertices have degrees $1, 3, 3, 3$.
== Proof ==
 there exists a simple graph $G = \struct {V, E}$ with $4$ vertices $v_1, v_2, v_3, v_4$ whose degrees are $1, 3, 3, 3$.
, let $\map \deg {v_1} = 1$.
Then each of $v_2$, $v_3$ and $v_4$ are of degree $3$.
So each of $v_2$, $v_3$ and $v_4$ are joined to all the other vertices.
That includes $v_1$.
So that means $v_1$ is joined to each of $v_2$, $v_3$ and $v_4$
But that means $v_1$ is of degree $3$.
This contradicts the supposition that $\map \deg {v_1} = 1$.
== Sources ==
* : Chapter $2$: Elementary Concepts of Graph Theory: $\S 2.1$: The Degree of a Vertex: Problem $4$
Category:Degrees of Vertices
Degree of Vertex/Examples/Party Puzzle

== Example of Degrees of Vertices ==
You arrive at a small party with your partner, which $3$ other couples are also attending.
Several handshakes took place.
Nobody shook hands with themselves or their partners.
Nobody shook hands with anyone else more than once.
After all the handshaking had taken place, you asked each person, including your partner, how many times they had shaken hands.
Every person replied with a different answer.
:$(1): \quad$ How many times did you shake hands?
:$(2): \quad$ How many times did your partner shake hands?
== Solution ==
Both you and your partner shook hands $3$ times.
== Proof ==
The situation can be modelled by a simple graph $G$ of order $8$.
The number of edges of $G$ corresponds to the number of handshakes.
As nobody shook hands with either themselves or their partner, there were $6$ people available for each person to shake hands with.
Each of the $7$ people other than you shook hands between $0$ and $6$ times.
So by the Pigeonhole Principle, you yourself shook hands the same number of times as someone else.
The total number of handshakes those $7$ people made was $0 + 1 + 2 + 3 + 4 + 5 + 6 = 21$.
By the corollary to the Handshake Lemma, the total number of handshakes is even.
Hence you yourself must have shaken hands an odd number of times in order make that total number even.
It is immediate that the person who shook hands $6$ must be the partner of the person who shook hands $0$ times, otherwise they would have shaken hands with each other.
The guest who shook hands $5$ times shakes hands with everybody except:
:his or her partner
:the person who shook hands $0$ times
leaving his or her partner to be the person who shook hands once.
The guest who shook hands $4$ times shakes hands with everybody except:
:his or her partner
:the person who shook hands $0$ times
:the person who shook hands once
leaving his or her partner to be the person who shook hands twice.
The situation now is that the final pair of guests both shook hands $3$ times.
One of these is you.
The following diagram illustrates the solution.
Note that the vertices corresponding to the other guests are labelled according to their degree.
:
== Sources ==
* : Chapter $2$: Elementary Concepts of Graph Theory: $\S 2.1$: The Degree of a Vertex: Problem $13$
Category:Degrees of Vertices
Degrees of Freedom (Statistics)/Examples/Arbitrary Example 1

== Example of Degrees of Freedom in context of Statistics ==
Consider a $2 \times 2$ contingency table with fixed marginal totals.
In this context there is only $1$ degree of freedom.
This is because once a total has been assigned to any one of the $4$ category cells, the remaining values are determined by the constraint that they must add up to the fixed marginal totals.
Take as an example the contingency table below:
:$\begin{array}{r|cc|c}
& \text {Column 1} & \text {Column 2} & \text {Row totals} \\
\hline
\text {Row 1} & a & b & 12 \\
\text {Row 2} & c & d & 13 \\
\hline
\text {Column totals} & 15 & 10
\end{array}$
If we arbitrarily assign $a = 10$, it follows that:
Similarly, if we set $a = 5$, then:
== Sources ==
* 
* 
Category:Examples of Degrees of Freedom (Statistics)
Degrees of Vertices determine Order and Size of Graph

== Theorem ==
Let $G = \struct {V, E}$ be a simple graph.
Let the degrees of each of the vertices of $G$ be given.
Then it is possible to determine both the order $\card V$ and size $\card E$ of $G$.
== Proof ==
Suppose we are given the degrees of each of the vertices of $G$.
By definition, $\card V$ is simply the count of the vertices.
So if we have been given the degrees of each of the vertices, we must know how many vertices there are.
Then from the Handshake Lemma:
:$\ds \sum_{v \mathop \in V} \map \deg v = 2 \card E$
That is, the sum of the degrees of all the vertices is equal to twice $\card E$.
So to find the size of $G$, add up the degrees of all the vertices and divide by $2$.
== Sources ==
* : Chapter $2$: Elementary Concepts of Graph Theory: $\S 2.1$: The Degree of a Vertex: Problem $5$
Category:Degrees of Vertices
Delambre's Analogies

== Theorem ==
Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
=== Sine by Sine ===
=== Sine by Cosine ===
=== Cosine by Sine ===
=== Cosine by Cosine ===
== Also known as ==
== Also see ==
* Napier's Analogies
== Historical Note ==
== Sources ==
* : Chapter $\text I$. Spherical Trigonometry: $16$. Delambre's and Napier's analogies.
* 
* 
* 
* 
Category:Spherical Trigonometry
Category:Delambre's Analogies
Delambre's Analogies/Also known as

== Delambre's Analogies: Also known as ==
Delambre's Analogies are also known as Gauss's Formulas, or Gauss's Formulae.
However, there are so many results and theorems named for that prefers to settle for .
The names of the individual formulas are not standard, but needs some way to distinguish between them. Any advice on this matter is welcome.
== Sources ==
* 
* 
Category:Delambre's Analogies
Delambre's Analogies/Cosine by Cosine

== Delambre's Analogies ==
Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
Then:
:$\cos \dfrac c 2 \cos \dfrac {A + B} 2 = \sin \dfrac C 2 \cos \dfrac {a + b} 2$
== Proof ==
In the below, we have:
:$s = \dfrac {a + b + c} 2$
Thus:
== Also known as ==
== Also see ==
* Napier's Analogies
== Sources ==
* : Chapter $\text I$. Spherical Trigonometry: $16$. Delambre's and Napier's analogies.
* 
* 
Category:Delambre's Analogies
Delambre's Analogies/Cosine by Sine

== Delambre's Analogies ==
Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
Then:
:$\cos \dfrac c 2 \sin \dfrac {A + B} 2 = \cos \dfrac C 2 \cos \dfrac {a - b} 2$
== Proof ==
In the below, we have:
:$s = \dfrac {a + b + c} 2$
Thus:
== Also known as ==
== Also see ==
* Napier's Analogies
== Sources ==
* : Chapter $\text I$. Spherical Trigonometry: $16$. Delambre's and Napier's analogies.
* 
* 
Category:Delambre's Analogies
Delambre's Analogies/Sine by Cosine

== Delambre's Analogies ==
Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
Then:
:$\sin \dfrac c 2 \cos \dfrac {A - B} 2 = \sin \dfrac C 2 \sin \dfrac {a + b} 2$
== Proof ==
In the below, we have:
:$s = \dfrac {a + b + c} 2$
Thus:
== Also known as ==
== Also see ==
* Napier's Analogies
== Sources ==
* : Chapter $\text I$. Spherical Trigonometry: $16$. Delambre's and Napier's analogies.
* 
* 
Category:Delambre's Analogies
Delambre's Analogies/Sine by Sine

== Delambre's Analogies ==
Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
Then:
:$\sin \dfrac c 2 \sin \dfrac {A - B} 2 = \cos \dfrac C 2 \sin \dfrac {a - b} 2$
== Proof ==
In the below, we have:
:$s = \dfrac {a + b + c} 2$
Thus:
== Also known as ==
== Also see ==
* Napier's Analogies
== Sources ==
* : Chapter $\text I$. Spherical Trigonometry: $16$. Delambre's and Napier's analogies.
* 
* 
Category:Delambre's Analogies
Deletable Prime/Sequence

== Sequence of Deletable Primes ==
The sequence of deletable primes begins:
:$2$, $3$, $5$, $7$, $13$, $17$, $23$, $29$, $31$, $37$, $43$, $47$, $53$, $59$, $67$, $71$, $73$, $79$, $83$, $97$, $103$, $107$, $\ldots$
== Sources ==
* 
Category:Deletable Primes
Deleted Integer Topology is First-Countable

== Corollary to Deleted Integer Topology is Second-Countable ==
Let $S = \R_{\ge 0} \setminus \Z$.
Let $\tau$ be the deleted integer topology on $S$.
The topological space $T = \struct {S, \tau}$ is first-countable.
== Proof ==
From Deleted Integer Topology is Second-Countable, $T$ is second-countable.
The result follows from Second-Countable Space is First-Countable.
== Sources ==
* : Part $\text {II}$: Counterexamples: $7$. Deleted Integer Topology: $5$
Category:First-Countable Spaces
Category:Deleted Integer Topology
Deleted Integer Topology is Lindelöf

== Corollary to Deleted Integer Topology is Second-Countable ==
Let $S = \R_{\ge 0} \setminus \Z$.
Let $\tau$ be the deleted integer topology on $S$.
The topological space $T = \struct {S, \tau}$ is Lindelöf.
== Proof ==
From Deleted Integer Topology is Second-Countable, $T$ is second-countable.
The result follows from Second-Countable Space is Lindelöf.
== Sources ==
* : Part $\text {II}$: Counterexamples: $7$. Deleted Integer Topology: $5$
Category:Lindelöf Spaces
Category:Deleted Integer Topology
Deleted Integer Topology is Second-Countable

== Theorem ==
Let $S = \R_{\ge 0} \setminus \Z$.
Let $\tau$ be the deleted integer topology on $S$.
Then the topological space $T = \struct {S, \tau}$ is second-countable.
=== Corollary 1 ===
=== Corollary 2 ===
=== Corollary 3 ===
== Proof ==
Let $\Z_{>0}$ be understood as the set of strictly positive integers:
:$\Z_{>0} = \set {x \in \Z: x > 0} = \set {1, 2, 3, \ldots}$
From Basis for Partition Topology, the set:
:$\BB = \set {\openint {n - 1} n: n \in Z_{>0} }$
is a basis for $T$.
There is an obvious one-to-one correspondence $\phi: \Z_{> 0} \leftrightarrow \BB$ between $\Z_{> 0}$ and $\BB$:
:$\forall x \in \Z_{>0}: \map \phi x = \openint {x - 1} x$
But $\Z_{>0} \subseteq \Z$ and Integers are Countably Infinite.
So from Subset of Countably Infinite Set is Countable, $\Z_{>0}$ is countable.
Thus $\BB$ is also countable by definition of countability.
So we have that $T$ has a countable basis, and so is second-countable by definition.
== Sources ==
* : Part $\text {II}$: Counterexamples: $7$. Deleted Integer Topology: $5$
Category:Second-Countable Spaces
Category:Deleted Integer Topology
Deleted Integer Topology is Separable

== Corollary to Deleted Integer Topology is Second-Countable ==
Let $S = \R_{\ge 0} \setminus \Z$.
Let $\tau$ be the deleted integer topology on $S$.
The topological space $T = \struct {S, \tau}$ is separable.
== Proof ==
From Deleted Integer Topology is Second-Countable, $T$ is second-countable.
The result follows from Second-Countable Space is Separable.
== Sources ==
* : Part $\text {II}$: Counterexamples: $7$. Deleted Integer Topology: $5$
Category:Separable Spaces
Category:Deleted Integer Topology
Deleted Integer Topology is Weakly Countably Compact

== Theorem ==
Let $S = \R_{\ge 0} \setminus \Z$.
Let $\tau$ be the deleted integer topology on $S$.
Then the topological space $T = \struct {S, \tau}$ is weakly countably compact.
== Proof ==
Let $A \subseteq S$ such that $A$ is infinite.
Let $x \in A$.
Let $n \in \Z$ such that $n < x < n + 1$.
By the definition of the deleted integer topology, $\openint n {n + 1}$ is open in $T$.
We have that $\openint n {n + 1}$ is infinite.
Take some $y \in \openint n {n + 1}$ such that $x \ne y$.
Now we claim that $y$ is a limit point of $A$.
Let $N_y$ be an open neighbourhood of $y$.
To be open in $T$, $N_y$ needs to be the union of open intervals of the form $\openint k {k + 1}$ for some $k \in \Z_{\ge 0}$.
Therefore to contain $y$ and be open at the same time, $\openint n {n + 1} \subseteq N_y$.
Thus $x \in N_y$ and $y \ne x \in A \cap N_y \ne \O$.
This is the definition of a limit point.
So, by definition, $T$ is weakly countably compact.
== Sources ==
* : Part $\text {II}$: Counterexamples: $7$. Deleted Integer Topology: $5$
Category:Weakly Countably Compact Spaces
Category:Deleted Integer Topology
Deleted Integer Topology is not Countably Compact

== Theorem ==
Let $S = \R_{\ge 0} \setminus \Z$, and let $\tau$ be the deleted integer topology on $S$.
Then the topological space $T = \struct {S, \tau}$ is not countably compact.
== Proof ==
By definition, the deleted integer topology is a partition topology.
Let $\PP$ be the partition which is the basis for $T$, that is:
:$\PP = \set {\openint {n - 1} n: n \in \Z_{> 0} }$
Then $\PP$ is a countable open cover of $S$ which has no finite subcover.
Hence the result.
== Sources ==
* : Part $\text {II}$: Counterexamples: $7$. Deleted Integer Topology: $5$
Category:Countably Compact Spaces
Category:Deleted Integer Topology
Delta-Algebra is Sigma-Algebra

== Theorem ==
Every $\delta$-algebra is a $\sigma$-algebra.
== Proof ==
Let $\DD$ be a $\delta$-algebra whose unit is $\mathbb U$.
Let $A_1, A_2, \ldots$ be a countably infinite collection of elements of $\DD$.
Then:
Thus $\DD$ is also a $\sigma$-algebra.
== Also see ==
* Sigma-Algebra is Delta-Algebra
Category:Sigma-Algebras
Denial of Existence/Examples/x less than or equal to 3

== Examples of Denial of Existence ==
Let $S \subseteq \R$ be a subset of the real numbers.
Let $P$ be the statement:
:$\exists x \in S: x \le 3$
The negation of $P$ is the statement written in its simplest form as:
:$\forall x \in S: x > 3$
== Proof ==
== Examples ==
== Sources ==
* : Chapter $1$: Sets and Logic: Exercise $1 \ \text{(ii)}$
Category:Examples of De Morgan's Laws
Denial of Existence/Examples/x less than or equal to 3/Examples/0 to 3

== Example of Denial of Universality: $\forall x \in S: x \le 3$ ==
Let $P$ be the statement:
:$\exists x \in S: x \le 3$
and $\lnot P$ its negation:
:$\forall x \in S: x > 3$
Let $S = \closedint 0 3$ where $\closedint \cdot \cdot$ denotes a closed real interval.
Then we have that:
:$P$ is true
and consequently:
:$\lnot P$ is false
== Proof ==
The truth of $P$ can be demonstrated by citing $x \in S: x = 2$.
Thus $2$ is an instance of an $x \in S$ such that $x > 3$ that has been shown to exist.
Hence its negation is false.
== Sources ==
* : Chapter $1$: Sets and Logic: Exercise $1 \ \text {(ii)} \, \text {(b)}$
Category:Examples of De Morgan's Laws
Denial of Existence/Examples/x less than or equal to 3/Examples/2, 3, 4

== Example of Denial of Existence: $\forall x \in S: x \le 3$ ==
Let $P$ be the statement:
:$\exists x \in S: x \le 3$
and $\lnot P$ its negation:
:$\forall x \in S: x > 3$
Let $S = \set {2, 3, 4}$.
Then we have that:
:$P$ is true
and consequently:
:$\lnot P$ is false
== Proof ==
The falsehood of $\lnot P$ can be demonstrated by citing $x \in S: x = 2$.
Thus $4$ is a counterexample to the assertion that all $x \in S$ are such that $x > 3$.
Hence its negation $P$ is true.
== Sources ==
* : Chapter $1$: Sets and Logic: Exercise $1 \ \text {(ii)} \, \text {(a)}$
Category:Examples of De Morgan's Laws
Denial of Universality/Examples/x less than or equal to 3

== Example of Denial of Universality ==
Let $S \subseteq \R$ be a subset of the real numbers.
Let $P$ be the statement:
:$\forall x \in S: x \le 3$
The negation of $P$ is the statement written in its simplest form as:
:$\exists x \in S: x > 3$
== Proof ==
== Examples ==
== Sources ==
* : Chapter $1$: Sets and Logic: Exercise $1 \ \text{(i)}$
Category:Examples of De Morgan's Laws
Denial of Universality/Examples/x less than or equal to 3/Examples/0 to 3

== Example of Denial of Universality: $\forall x \in S: x \le 3$ ==
Let $P$ be the statement:
:$\forall x \in S: x \le 3$
and $\lnot P$ its negation:
:$\exists x \in S: x > 3$
Let $S = \closedint 0 3$ where $\closedint \cdot \cdot$ denotes a closed real interval.
Then we have that:
:$P$ is true
and consequently:
:$\lnot P$ is false
== Proof ==
By definition of closed real interval:
:$\forall x \in S: 0 \le x \le 3$
and so:
:$\forall x \in S: x \le 3$
Hence its negation is false.
== Sources ==
* : Chapter $1$: Sets and Logic: Exercise $1 \ \text {(i)} \, \text {(b)}$
Category:Examples of De Morgan's Laws
Denial of Universality/Examples/x less than or equal to 3/Examples/2, 3, 4

== Example of Denial of Universality: $\forall x \in S: x \le 3$ ==
Let $P$ be the statement:
:$\forall x \in S: x \le 3$
and $\lnot P$ its negation:
:$\exists x \in S: x > 3$
Let $S = \set {2, 3, 4}$.
Then we have that:
:$P$ is false
and consequently:
:$\lnot P$ is true
== Proof ==
The falsehood of $P$ can be demonstrated by citing $x \in S: x = 4$.
Thus $4$ is a counterexample to the assertion that all $x \in S$ are such that $x \le 3$.
Hence its negation is true.
== Sources ==
* : Chapter $1$: Sets and Logic: Exercise $1 \ \text {(i)} \, \text {(a)}$
Category:Examples of De Morgan's Laws
Denominators of Simple Continued Fraction are Strictly Increasing

== Theorem ==
Let $N \in \N \cup \set \infty$ be an extended natural number.
Let $\sqbrk {a_0, a_1, a_2, \ldots}$ be a simple continued fraction in $\R$ of length $N$.
Let $q_0, q_1, q_2, \ldots$ be its denominators.
Then with the possible exception of $q_0 = q_1$, the sequence $\sequence {q_n}$ is strictly increasing.
== Proof ==
By definition of simple continued fraction, all partial denominators of $\sqbrk {a_0, a_1, a_2, \ldots}$ are strictly positive integers, with the possible exception of $a_0$.
So:
:$q_1 = a_1 \ge 1 = q_0$
:$q_2 = a_2 a_1 + 1 \ge a_1 + 1 > a_1 = q_1$
hence:
:$1 = q_0 \le q_1 < q_2$
Suppose $q_k > q_{k - 1} \ge 1$ for some $k \ge 2$.
Then:
:$q_{k + 1} = a_{k + 1} q_k + q_{k - 1} \ge q_k + q_{k - 1} \ge q_k + 1 > q_k$
So, by induction, $\sequence {q_n}$ is strictly increasing except when possibly $q_0 = q_1 = 1$.
== Also see ==
* Lower Bounds for Denominators of Simple Continued Fraction
Category:Simple Continued Fractions
Denominators of Simple Continued Fraction are Strictly Positive

== Theorem ==
Let $n \in \N \cup \set \infty$ be an extended natural number.
Let $\tuple {a_0, a_1, \ldots}$ be a simple continued fraction in $\R$ of length $n$.
Let $q_0, q_1, q_2, \ldots$ be its denominators.
Then for $0 \leq k \leq n$ we have $q_k > 0$.
== Proof ==
Follows from:
:$q_0 = 1$ by definition
:Denominators of Simple Continued Fraction are Strictly Increasing.
Category:Simple Continued Fractions
Dense-in-itself Subset of T1 Space is Infinite

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a topological space which is $T_1$ (Fréchet).
Let $H \subseteq T$ be dense-in-itself.
Then $H$ is infinite.
== Proof ==
Proof by Contradiction:
 $H$ is finite.
From Finite $T_1$ Space is Discrete, $H$ has the discrete topology.
From Discrete Space is not Dense-In-Itself it then follows that $H$ can not be dense-in-itself.
So for $H$ to be dense-in-itself, it must be infinite.
== Sources ==
* : Part $\text {II}$: Counterexamples: $23 \text { - } 24$. Fort Space: $8$
Category:Denseness
Category:T1 Spaces
Category:Infinite Sets
Dense-in-itself iff Subset of Derivative

== Theorem ==
Let $T$ be a topological space.
Let $A \subseteq T$.
Then:
:$A$ is dense-in-itself
:
:$A \subseteq A'$
where
:$A'$ denotes the derivative of $A$.
== Proof ==
:$\qquad A$ is dense-in-itself
:$\leadstoandfrom$ every $x \in A$ is not an isolated point in $A$ by definition of dense-in-itself
:$\leadstoandfrom$ every $x \in A$ is an accumulation point of $A$ by Point is Isolated iff not Accumulation Point
:$\leadstoandfrom$ every $x \in A$ is an element of $A'$ by definition of derivative
:$\leadstoandfrom A \subseteq A'$ by definition of subset.
Category:Denseness
Category:Set Derivatives