nyuuzyou/wikis huggingface dataset:
Infinite Limit Theorem

== Theorem ==
Let $f$ be a real function of $x$ of the form 
:$\map f x = \dfrac {\map g x} {\map h x}$
Further, let $g$ and $h$ be continuous on some open interval $\mathbb I$, where $c$ is a constant in $\mathbb I$.
If:
:$(1): \quad \map g c \ne 0$
:$(2): \quad \map h c = 0$
:$(3): \quad \forall x \in \mathbb I: x \ne c \implies \map h x \ne 0$
then the limit of $f$ as $x \to c$ will not exist, and:
:$\ds \lim_{x \mathop \to c^+} \map f x = +\infty$ or $-\infty$
:$\ds \lim_{x \mathop \to c^-} \map f x = +\infty$ or $-\infty$
== Proof ==
To prove the claim, it will suffice to show that for each $N \in \R_{>0}$, one can find an $\epsilon > 0$ such that:
:$\size {x - c} < \epsilon \implies \size {\map f x} \ge N$
So fix $N \in \R_{>0}$.
First, by continuity of $g$, find an $\epsilon_1 > 0$ such that:
:$\size {x - c} < \epsilon_1 \implies \size {\map g x - \map g c} < \size {\dfrac {\map g c} 2}$
which by corollary $1$ to the Reverse Triangle Inequality implies that:
:$\size {x - c} < \epsilon_1 \implies \size {\map g c} - \size {\map g x} < \size {\dfrac {\map g c} 2}$
and by rearranging, this becomes:
:$\size {x - c} < \epsilon_1 \implies \size {\map g x} > \size {\dfrac {\map g c} 2}$
Next, by continuity of $h$, find an $\epsilon_2 > 0$ such that $\epsilon_1 > \epsilon_2$ and:
:$\size {x - c} < \epsilon_2 \implies \size {\map h x} < \dfrac {\size {\map g c} } {2 N}$
From Ordering of Reciprocals, this means:
:$\size {x - c} < \epsilon_2 \implies \dfrac 1 {\size {\map h x} } > \dfrac {2 N} {\size {\map g c} }$
These together imply that, whenever $0 < \size {x - c} < \epsilon_2$, we have:
Thus $\epsilon_2$ is a valid choice for the sought $\epsilon$.
Hence the result.
== Intuition ==
Consider the graph of the reciprocal function at the origin.
Immediately to the right of the origin, as positive numbers become smaller, the function value becomes larger.
Similarly, immediately to the left of the origin, as the absolute value of negative numbers becomes smaller, the absolute value of the function value becomes larger, that is, more and more negative.
== Warning ==
This theorem is absolutely not saying that $\dfrac c 0 = \infty$ when dealing with real numbers. Division by Zero is undefined.
== Sources ==
* : $\S 1.5$
Category:Real Analysis
Infinite Measure is Measure

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Then the infinite measure $\mu$ on $\struct {X, \Sigma}$ is a measure.
== Proof ==
Let us verify the measure axioms $(1)$, $(2)$ and $(3')$ for $\mu$.
=== Proof of $(3')$ ===
We have by definition $\map \mu \O = 0$.
=== Proof of $(1)$ ===
Let $S \in \Sigma$.
Then:
:$\map \mu S = + \infty > 0$ for $S \ne \O$
:$\map \mu S = 0 \ge 0$ for $S = \O$
So $\map \mu S \ge 0$ for all $S \in \Sigma$.
=== Proof of $(2)$ ===
It is to be shown that (for a sequence $\sequence {S_n}_{n \in \N}$ of pairwise disjoint sets):
:$\ds \sum_{n \mathop = 1}^\infty \map \mu {S_n} = \map \mu {\bigcup_{n \mathop = 1}^\infty S_n}$
Suppose $S_n = \O$ for all $n \in \N$.
Then by definition of $\mu$:
:$\ds \map \mu {S_n} = \map \mu {\bigcup_{n \mathop = 1}^\infty S_n} = \map \mu \O = 0$
Thus, the desired equation becomes:
:$\ds \sum_{n \mathop = 1}^\infty 0 = 0$
which trivially holds.
Suppose $S_n \ne \O$ for some $n \in \N$.
Then $S_n \subseteq \ds \bigcup_{n \mathop = 1}^\infty S_n \ne \O$.
:$\ds \map \mu {S_n} = \map \mu {\bigcup_{n \mathop = 1}^\infty S_n} = + \infty$
:$\ds \sum_{n \mathop = 1}^\infty \map \mu {S_n} = \map \mu {S_n} = + \infty = \map \mu {\bigcup_{n \mathop = 1}^\infty S_n}$
Thus $\ds \sum_{n \mathop = 1}^\infty \map \mu {S_n} = \map \mu {\bigcup_{n \mathop = 1}^\infty S_n}$.
The axioms are fulfilled, and it follows that $\mu$ is a measure.
Category:Measures
Infinite Number of Chen Primes

== Theorem ==
There exists an infinite number of Chen primes.
That is, there exists an infinite number of pairs:
:$\set {p, p + 2}$
where:
:$p$ is a prime
:$p + 2$ is either a prime or a semiprime.
== Proof ==
== Historical Note ==
== Sources ==
* : Chapter $7$: Patterns in Numbers: Fermat
Category:Prime Numbers
Infinite Number of Even Fermat Pseudoprimes

== Theorem ==
Despite their relative rarity, there exist an infinite number of even Fermat pseudoprimes.
== Proof ==
In the context of Wells, Fermat pseudoprime probably means Fermat pseudoprimes to the base $2$.
Consider the equation:
:$2^m - 2 \equiv 0 \pmod m$
Any $m$ satisfying the above equation is a Fermat pseudoprime.
We show that for each even $m$ satisfying the above equation, there exists a prime $p$ such that $m p$ also satisfies the equation.
Write $m = 2 n$.
Then $2^{2 n - 1} - 1 \equiv 0 \pmod n$.
By Zsigmondy's Theorem, there exists a prime $p$ such that:
:$p \divides 2^{2 n - 1} - 1$
:$p \nmid 2^k - 1$ for all $k < 2 n - 1$
By this we have $2^{2 n - 1} \equiv 1 \pmod p$.
By Fermat's Little Theorem $2^{p - 1} \equiv 1 \pmod p$.
Let $d = \gcd \set {2 n - 1, p - 1}$.
We have $d \le 2 n - 1$ and $2^d \equiv 1 \pmod p$.
This gives $d = 2 n - 1$ and thus $2 n - 1 \divides p - 1$.
Thus there exists $x \in \N$ such that:
:$p = \paren {2 n - 1} x + 1 > n$
This shows that $p \perp n$.
From $2 p n - 1 = 2 n - 1 + 2 n \paren {p - 1} = \paren {2 n - 1} \paren {1 + 2 n x}$ we have:
:$2^{m p - 1} = 2^{\paren {2 n - 1} \paren {1 + 2 n x}} \equiv 1 \pmod {n, p}$
By Chinese Remainder Theorem:
:$2^{m p - 1} \equiv 1 \pmod {n p}$
By Congruence by Product of Moduli:
:$2^{m p} \equiv 2 \pmod {m p}$
showing that $m p$ is also an even Fermat pseudoprime.
The first even Fermat pseudoprime is $161 \, 038$.
We can use the above method repeatedly to construct infinitely many even Fermat pseudoprimes, though it is impractical since these numbers grow exponentially.
== Sources ==
* 
* : $161,038$
Category:Fermat Pseudoprimes
Infinite Number of Integers which are Sum of 3 Sixth Powers in 2 Ways

== Theorem ==
There exist an infinite number of positive integers which can be expressed as the sum of $3$ sixth powers in $2$ different ways.
== Proof ==
There are many parametric solutions to $x^6 + y^6 + z^6 = u^6 + v^6 + w^6$. One is given by:
This set of solutions also satisfy:
== Sources ==
* 
* : $160,426,514$
* : $160,426,514$
Category:Sixth Powers
Category:160,426,514
Infinite Number of Primes of form 4n - 1

== Theorem ==
There are infinitely many prime numbers of the form $4 n - 1$.
== Proof ==
 the contrary.
That is, suppose there is a finite number of prime numbers of the form $4 n - 1$.
Let there be $k$ of them: $p_1, p_2, \ldots, p_k$.
Let $S = \set {p_1, p_2, \ldots, p_k}$.
Let $N$ be constructed as:
:$\ds N = 4 \prod_{i \mathop = 1}^k p_i - 1$
If $N$ is a prime number, then it is of the form $4 n - 1$.
But then we have that $N \notin S$, which means that $S$ is not the complete set of prime numbers of the form $4 n - 1$.
Therefore $N$ must be composite.
Suppose all the prime factors of $N$ are of the form $4 n + 1$.
Then from Product of Integers of form $4 n + 1$ it follows that $N$ itself is of the form $4 n + 1$.
Therefore $N$ must have at least one prime factor $p$ of the form $4 n - 1$.
Suppose $p \in S$.
We have that:
:$\ds p \divides 4 \prod_{i \mathop = 1}^k p_i$
and so:
:$\ds p \divides 4 \prod_{i \mathop = 1}^k p_i - p$
So:
:$N = q p + \left({p - 1}\right)$
where:
:$q = \dfrac {\ds 4 \prod_{i \mathop = 1}^k p_i - p} p$
So applying the Euclidean Algorithm:
So if $p \in S$, then it cannot be a divisor of $N$.
Therefore there must be a prime numbers of the form $4 n - 1$ which is not in $S$.
Therefore $S$ is not the complete set of prime numbers of the form $4 n - 1$.
Therefore the assumption that $S$ is finite was false.
Hence the result.
== Also presented as ==
Some sources present this result as:
:There are infinitely many prime numbers congruent to $3 \pmod 4$.
== Sources ==
* : Chapter $2$: Some Properties of $\Z$: Exercise $2.15$
* : $\S 2.2$: Divisibility and factorization in $\mathbf Z$: Exercise $8$
* : Chapter $\text {B}.16$: The Sequence of Primes: Theorem $3$
Category:Arithmetic
Category:Number Theory
Category:Prime Numbers
Infinite Ordinal can be expressed Uniquely as Sum of Limit Ordinal plus Finite Ordinal

== Theorem ==
Let $x$ be an ordinal.
Suppose $x$ satisfies $\omega \subseteq x$.
Then $x$ has a unique representation as $\paren {y + z}$ where $y$ is a limit ordinal and $z$ is a finite ordinal.
== Proof ==
Take $K_{II}$ to be the set of all limit ordinals.
Then set $y = \bigcup \set {w \in K_{II}: w \le x}$
The set $\set {w \in K_{II}: w \le x}$ is non-empty because $\omega \subseteq x$.
By Union of Ordinals is Least Upper Bound, $y \in K_{II}$ and $y \le x$.
By Ordinal Subtraction when Possible is Unique, there is a unique $z$ such that $x = \paren {y + z}$
Assume $\omega \le z$.
Then, again by Ordinal Subtraction when Possible is Unique:
:$z = \paren {\omega + w}$
and so:
But $y + \omega$ is a limit ordinal by Limit Ordinals Preserved Under Ordinal Addition:
:$\O < \omega \implies y < y + \omega$
This contradicts the fact that $y$ is the largest limit ordinal smaller than $x$.
Therefore, $z \in \omega$.
Thus we have proven that such a selection of $y$ and $z$ exists.
Suppose $z$ and $w$ both satisfy:
:$\paren {y + w} = \paren {y + z}$
By Ordinal Addition is Left Cancellable, we have $w = z$.
Thus $z$ is unique.
It remains to prove uniqueness for $y$.
Suppose that $x = \paren {y + u}$ and $x = \paren {w + z}$.
, assume further that $y \le w$.
Then:
 that $\exists m: n = m^+$.
This is clearly a contradiction.
Hence:
:$n = \O$
and so:
:$w = y$
The result follows.
== Sources ==
* : $\S 8.13$
Category:Ordinal Arithmetic
Infinite Particular Point Space is not Compact

== Theorem ==
Let $T = \struct {S, \tau_p}$ be an infinite particular point space.
Then $T$ is not compact.
== Proof ==
Consider the open cover of $T$:
:$\CC = \set {\set {x, p}: x \in S, x \ne p}$
As $S$ is infinite, then so is $\CC$, as we can set up a bijection from $\phi: S \setminus \set p \leftrightarrow \CC$:
:$\forall x \in S \setminus \set p: \map \phi x = \set {x, p}$
Hence $\CC$ has no finite subcover.
The result follows by definition of compactness.
== Sources ==
* : Part $\text {II}$: Counterexamples: $9 \text { - } 10$. Infinite Particular Point Topology: $5$
Category:Compact Spaces
Category:Particular Point Topology
Infinite Particular Point Space is not Countably Metacompact

== Theorem ==
Let $T = \struct {S, \tau_p}$ be an infinite particular point space.
Then $T$ is not countably metacompact.
== Proof ==
Suppose $T$ is a countable particular point space.
Let $\CC$ be the open cover of $T$ defined as:
:$\CC = \set {\set {x, p}: x \in S}$
$\CC$ is countable and has no open refinement except $\CC$ itself.
But $\CC$ is not point finite because $\forall U \in \CC: p \in U$, and $\CC$ is (countably) infinite.
Now suppose $T$ is an uncountable particular point space.
Let $S' \subseteq S$ be a countable subset of $S$.
Let $\CC$ be the open cover of $T$ defined as:
:$\CC = \set {\set {x, p}: x \in S'} \cup \set {S \setminus S' \cup \set p}$
$\CC$ is also countable and has no open refinement except $\CC$ itself.
And similarly, $\CC$ is not point finite because $\forall U \in \CC: p \in U$, and $\CC$ is (countably) infinite.
Hence the result by definition of countably metacompact.
== Sources ==
* : Part $\text {II}$: Counterexamples: $9 \text { - } 10$. Infinite Particular Point Topology: $19$
Category:Countably Metacompact Spaces
Category:Particular Point Topology
Infinite Particular Point Space is not Countably Paracompact

== Theorem ==
Let $T = \struct {S, \tau_p}$ be an infinite particular point space.
Then $T$ is not countably paracompact.
== Proof ==
 $T$ is countably paracompact.
From Countably Paracompact Space is Countably Metacompact it follows that $T$ is countably metacompact.
But we have that Infinite Particular Point Space is not Countably Metacompact.
Hence the result by Proof by Contradiction.
Category:Countably Paracompact Spaces
Category:Particular Point Topology
Infinite Particular Point Space is not Metacompact

== Theorem ==
Let $T = \left({S, \tau_p}\right)$ be an infinite particular point space.
Then $T$ is not metacompact.
== Proof ==
 $T$ is metacompact.
From Metacompact Space is Countably Metacompact it follows that $T$ is countably metacompact.
But we have that Infinite Particular Point Space is not Countably Metacompact.
Hence the result by Proof by Contradiction.
Category:Metacompact Spaces
Category:Particular Point Topology
Infinite Particular Point Space is not Strongly Locally Compact

== Theorem ==
Let $T = \struct {S, \tau_p}$ be an infinite particular point space.
Then $T$ is not strongly locally compact.
== Proof ==
By definition, $T$ is strongly locally compact every point of $S$ is contained in an open set whose closure is compact.
Let $x \in S: x \ne p$.
Let $x \in U$ where $U$ is open in $T$.
From Closure of Open Set of Particular Point Space we have:
:$U^- = S$
where $U^-$ is the closure of $U$.
But from Infinite Particular Point Space is not Compact we have that $S$ is not compact.
Hence the result, by definition of strongly locally compact.
== Sources ==
* : Part $\text {II}$: Counterexamples: $9 \text { - } 10$. Infinite Particular Point Topology: $5$
* : Part $\text {II}$: Counterexamples: $9 \text { - } 10$. Infinite Particular Point Topology: $15$
Category:Strongly Locally Compact Spaces
Category:Particular Point Topology
Infinite Particular Point Space is not Weakly Countably Compact

== Theorem ==
Let $T = \struct {S, \tau_p}$ be an infinite particular point space.
Then $T$ is not weakly countably compact.
== Proof ==
By definition, $T$ is weakly countably compact every infinite subset of $S$ has a limit point in $S$.
Let $H \subseteq S$ be an infinite subset of $S$ where $p \notin H$.
$H$ is not open in $T$ by definition.
So from Subset of Particular Point Space is either Open or Closed, $H$ is closed in $T$.
Then we have that a Closed Set in Particular Point Space has no Limit Points.
The result follows from definition of weakly countably compact.
If $T = \struct {S, \tau_p}$ is a finite particular point space, then Finite Space Satisfies All Compactness Properties applies.
== Mistakes in Sources ==
See : Part $\text {II}$: Counterexamples: $8 \text { - } 10$. Infinite Particular Point Topology: $12$ where it is stated that:
Particular Point Space is not Weakly Countably Compact
== Sources ==
* : Part $\text {II}$: Counterexamples: $9 \text { - } 10$. Infinite Particular Point Topology: $12$
Category:Weakly Countably Compact Spaces
Category:Particular Point Topology
Infinite Product of Analytic Functions

== Theorem ==
Let $D \subset \C$ be an open connected set.
Let $\sequence {f_n}$ be a sequence of analytic functions $f_n: D \to \C$ that are not identically zero.
Let $\ds \sum_{n \mathop = 1}^\infty \paren {f_n - 1}$ converge locally uniformly absolutely on $D$.
Then:
:$(1): \quad f = \ds \prod_{n \mathop = 1}^\infty f_n$ converges locally uniformly absolutely on $D$
:$(2): \quad f$ is analytic
:$(3): \quad$ For each $z \in D$, $\map {f_n} z = 0$ for finitely many $n \in \N$
:$(4): \quad$ For each $z \in D$, $\map {\operatorname {mult}_z} f = \ds \sum_{n \mathop = 1}^\infty \map {\operatorname {mult}_z} {f_n}$
where $\operatorname {mult}$ denotes multiplicity.
== Proof ==
Because $\ds \sum_{n \mathop = 1}^\infty \paren {f_n - 1}$ converges locally uniformly absolutely on $D$, the series converges locally uniformly absolutely.
Thus $f = \ds \prod_{n \mathop = 1}^\infty f_n$ converges locally uniformly absolutely.
By Uniform Limit of Analytic Functions is Analytic, $f$ is analytic
The last claim follows from Zeroes of Infinite Product of Analytic Functions
== Also see ==
* Derivative of Infinite Product of Analytic Functions
Category:Infinite Products
Infinite Product of Analytic Functions is Analytic

== Theorem ==
Let $D \subset \C$ be an open set.
Let $\sequence {f_n}$ be a sequence of analytic functions $f_n: D \to \C$.
Let $\ds \prod_{n \mathop = 1}^\infty f_n$ converge locally uniformly to $f$.
Then $f$ is analytic.
== Proof ==
Follows directly from:
:Partial Products of Uniformly Convergent Product Converge Uniformly
:Uniform Limit of Analytic Functions is Analytic
== Also see ==
* Derivative of Infinite Product of Analytic Functions
== Sources ==
* $VII$: Compact and Convergence in the Space of Analytic Functions: $\S5$: Weierstrass Factorization Theorem: Theorem $5.9$
Category:Infinite Products
Infinite Product of Product of Sequence of n plus alpha over Sequence of n plus beta

== Theorem ==
:$\ds \prod_{n \mathop \ge 1} \dfrac {\paren {n + \alpha_1} \cdots \paren {n + \alpha_k} } {\paren {n + \beta_1} \cdots \paren {n + \beta_k} } = \dfrac {\map \Gamma {1 + \beta_1} \cdots \map \Gamma {1 + \beta_k} } {\map \Gamma {1 + \alpha_1} \cdots \map \Gamma {1 + \alpha_k} }$
where:
:$\alpha_1 + \cdots + \alpha_k = \beta_1 + \cdots + \beta_k$
:none of the $\beta$s is a negative integer.
== Proof ==
First we note that if any of the $\beta$s is a negative integer, the would have $0$ as its denominator, and so would be undefined.
We have from the Euler form of the Gamma function that:
:$\map \Gamma {1 + \beta_i} = \ds \lim_{m \mathop \to \infty} \dfrac {m^{1 + \beta_i} m!} {\paren {1 + \beta_i} \paren {2 + \beta_i} \cdots \paren {m + 1 + \beta_i} }$
and so the can be written as:
== Sources ==
* : $\S 1.2.5$: Permutations and Factorials: Exercise $17$
Category:Gamma Function
Infinite Product of Sigma-Compact Spaces is not always Sigma-Compact

== Theorem ==
Let $I$ be an indexing set with infinite cardinality.
Let $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be a family of topological spaces indexed by $I$.
Let $\ds \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$.
Let each of $\struct {S_\alpha, \tau_\alpha}$ be $\sigma$-compact.
Then it is not necessarily the case that $\struct {S, \tau}$ is also $\sigma$-compact.
== Proof ==
Let $T = \struct {\Z_{\ge 0}, \tau}$ be the topological space formed by the discrete topology on the set of positive integers.
Let $T' = \struct {\ds \prod_{\alpha \mathop \in \Z_{\ge 0} } \struct {\Z_{\ge 0}, \tau}_\alpha, \tau'}$ be the countable Cartesian product of $\struct {\Z_{\ge 0}, \tau}$ indexed by $\Z_{\ge 0}$ with the product topology $\tau'$.
From Countable Discrete Space is Sigma-Compact, $T$ is $\sigma$-compact.
From Countable Product of Countable Discrete Spaces is not Sigma-Compact, $T'$ is not $\sigma$-compact.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Invariance Properties
Category:Sigma-Compact Spaces
Category:Product Topology
Infinite Product of Weakly Locally Compact Spaces is not always Weakly Locally Compact

== Theorem ==
Let $I$ be an indexing set with infinite cardinality.
Let $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be a family of topological spaces indexed by $I$.
Let $\ds \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$.
Let each of $\struct {S_\alpha, \tau_\alpha}$ be weakly locally compact-compact.
Then it is not necessarily the case that $\struct {S, \tau}$ is also weakly locally compact.
== Proof ==
Let $T = \struct {\Z_{\ge 0}, \tau}$ be the topological space formed by the discrete topology on the set of positive integers.
Let $T' = \struct {\ds \prod_{\alpha \mathop \in \Z_{\ge 0} } \struct {\Z_{\ge 0}, \tau}_\alpha, \tau'}$ be the countable Cartesian product of $\struct {\Z_{\ge 0}, \tau}$ indexed by $\Z_{\ge 0}$ with the product topology $\tau'$.
From Discrete Space is Strongly Locally Compact, $T$ is strongly locally compact.
From Strongly Locally Compact Space is Weakly Locally Compact, $T$ is weakly locally compact.
From Countable Product of Countable Discrete Spaces is not Weakly Locally Compact, $T'$ is not weakly locally compact.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Invariance Properties
Category:Weakly Locally Compact Spaces
Category:Product Topology
Infinite Ramsey's Theorem implies Finite Ramsey's Theorem

== Theorem ==
:$\forall l, n, r \in \N: \exists m \in \N: m \to \left({l}\right)_r^n$
where $\alpha \to \left({\beta}\right)^n_r$ means that:
:for any assignment of $r$-colors to the $n$-subsets of $\alpha$
::there is a particular color $\gamma$ and a subset $X$ of $\alpha$ of size $\beta$ such that all $n$-subsets of $X$ are $\gamma$.
== Proof ==
 there is a $l$ such that:
:$\forall m \in \N: m \nrightarrow \left({l}\right)_r^n$
Let $\hat{K_i}$ denote a hypergraph on $i$ vertices where all possible $n$-subsets of the vertices are the hyperedges.
Let $G$ be a hypergraph with vertices $V = \left\{ {v_i: i \in \N}\right\}$.
Let the hyperedges of $G <$ be enumerated by:
:$E = \left\{ {E_i: E_i \subset \N, \left\vert{E_i}\right\vert = n}\right\}$
We construct a (rooted) tree $T$ as follows:
$(1): \quad$ First introduce a root node $r t$.
$(2): \quad$ Each node is allowed to have at most $r<$ children which correspond to the $r$-colors, subject to it satisfying the criteria:
:::A child is always labeled by one among the $r$-colors.
:::Let the colors be denoted $c_1, c_2, \ldots, c_r$.
$(3): \quad$ A child $c_i$ is permitted its introduction creates a path of some finite length $k$ starting from the root.
:::So, if the hyperedges $E_1, E_2, \ldots, E_k$ are colored by the colors used in the path in the same order, then the corresponding subgraph of $G$ does not contain a monochromatic $\hat{K_l}$.
:::For example: if the introduction of a child $c_i$ creates the $k$ length path $r t, c_a, c_b, \ldots, c_i$ and the hyperedges $E_1, E_2, \ldots, E_k$ when colored $c_a, c_b, \ldots, c_i$ do not contain a monochromatic $\hat{K_l}$, the child $c_i$ is permitted to be added to $T$.
Note that for all $m$, there always exists a coloring of $\hat{K_m}$ such that no monochromatic $\hat{K_l}$ exists within.
Thus the situation that a child cannot be added to any vertex at a given level $k$ cannot arise.
For we can always take a coloring of $\hat{K_{k+n}}$ containing no monochromatic $\hat{K_l}$.
Since any $k$ hyperedges in it would yield a sequence of colors already existing in $T$, we know which vertex to add the child to.
We give the child the color corresponding to any other edge.
Hence we can forever keep adding children and so $T$ is infinite.
It is also obvious that each level $k$ of $T$ has at most $r^k<$ vertices.
So each level is finite.
By König's Tree Lemma there will be an infinite path $P$ in $T$.
$P$ provides a $r$-coloring of $G$ that contains no monochromatic $\hat{K_i}$.
Hence $P$ contains no monochromatic infinite hypergraph.
This contradicts the Infinite Ramsey's Theorem.
The result follows by Proof by Contradiction.
Category:Ramsey Theory
Infinite Sequence Property of Strictly Well-Founded Relation

== Theorem ==
Let $\struct {S, \RR}$ be a relational structure.
Then $\RR$ is a strictly well-founded relation there is no infinite sequence $\sequence {a_n}$ of elements of $S$ such that:
:$\forall n \in \N: a_{n + 1} \mathrel \RR a_n$
== Proof ==
=== Reverse Implication ===
=== Forward Implication ===
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Theorem $2$
Category:Well-Founded Relations
Category:Infinite Sequence Property of Strictly Well-Founded Relation
Infinite Sequence Property of Strictly Well-Founded Relation/Forward Implication

== Theorem ==
Let $\struct {S, \RR}$ be a relational structure.
Let $\RR$ be a strictly well-founded relation.
Then there exists no infinite sequence $\sequence {a_n}$ of elements of $S$ such that:
:$\forall n \in \N: a_{n + 1} \mathrel \RR a_n$
== Proof ==
Let $\RR$ be a strictly well-founded relation.
 there exists an infinite sequence $\sequence {a_n}$ in $S$ such that:
:$\forall n \in \N: a_{n + 1} \mathrel \RR a_n$
Let $T = \set {a_0, a_1, a_2, \ldots}$.
Let $a_k \in T$ be a strictly minimal element of $T$.
That is:
:$\forall y \in T: y \not \mathrel \RR a_k$
But we have that:
:$a_{k + 1} \mathrel \RR a_k$
So $a_k$ is not a strictly minimal element.
It follows by Proof by Contradiction that such an infinite sequence cannot exist.
Category:Infinite Sequence Property of Strictly Well-Founded Relation