nyuuzyou/wikis huggingface dataset:
Existence of Right Universal Product

== Theorem ==
For each $m \in \N$, the following is a theorem of von Neumann-Bernays-Gödel set theory:
:$\forall X: \exists Z: \forall v_1, \dotsc, v_m: \forall x: \tuple {\tuple {\dots \tuple {x, v_1}, \dots}, v_m} \in Z \iff x \in X$
Informally, for any class $X$, there is some class $Z$ such that:
:$Z = \paren {\dotsm \paren {X \times V} \times \dotsm} \times V$
where there are $m$ copies of $V$, the universal class, combined with $X$ via the cartesian product.
Note, however, that the formal statement above allows for additional elements of $Z$ which are not of the form as written.
=== Corollary ===
== Proof ==
We will proceed by induction on $m$.
Importantly, this induction is not done in the context of NBG.
Rather, it is performed in the metamathematical system in which NBG is formalized.
=== Basis for the Induction ===
If $m = 0$, then the statement reduces to:
:$\forall X: \exists Z: \forall x: x \in Z \iff x \in X$
But that trivially holds by taking $Z := X$.
Thus, the basis for the induction holds.
=== Induction Hypothesis ===
As the induction hypothesis, suppose that:
:$\forall X: \exists Z: \forall v_1, \dotsc, v_m: \forall x: \tuple {\tuple {\dots \tuple {x, v_1}, \dots}, v_m} \in Z \iff x \in X$
for some arbitrary $m \in \N$.
We want to prove that:
:$\forall X: \exists Z: \forall v_1, \dotsc, v_m, v_{m + 1}: \forall x: \tuple {\tuple {\tuple {\dots \tuple {x, v_1}, \dots}, v_m}, v_{m + 1}} \in Z \iff x \in X$
=== Induction Step ===
Let $X$ be arbitrary.
By the induction hypothesis, there is some $Z$ such that:
:$\paren 1 \quad \forall v_1, \dotsc, v_m: \forall x: \tuple {\tuple {\dots \tuple {x, v_1}, \dots}, v_m} \in Z \iff x \in X$
By Axiom $\text B 5$, there is some $Z'$ such that:
:$\paren 2 \quad \forall u, v: \tuple {u, v} \in Z' \iff u \in Z$
Substituting $\tuple {\tuple {\dots \tuple {x, v_1}, \dots}, v_m}$ for $u$ in $\paren 2$:
:$\paren 3 \quad \forall v_1, \dotsc, v_m, v: \forall x: \tuple {\tuple {\tuple {\dots \tuple {x, v_1}, \dots}, v_m}, v} \in Z' \iff \tuple {\tuple {\dots \tuple {x, v_1}, \dots}, v_m} \in Z$
By $\paren 1, \paren 3$ and Biconditional is Transitive:
:$\forall v_1, \dotsc, v_m, v: \forall x: \tuple {\tuple {\tuple {\dots \tuple {x, v_1}, \dots}, v_m}, v} \in Z' \iff x \in X$
Clearly, $Z'$ satisfies the induction step.
Therefore, by the Principle of Mathematical Induction, the theorem holds.
Category:Class Theory
Category:Von Neumann-Bernays-Gödel Set Theory
Existence of Right Universal Product/Corollary

== Corollary to Existence of Right Universal Product ==
For all $n \in \N_{> 0}$ and $m \in \N$, the following is a theorem of von Neumann-Bernays-Gödel set theory:
:$\forall X: \exists Z: \forall v_1, \dotsc, v_m: \forall x_1, \dotsc, x_n: \sequence {x_1, \dotsc, x_n, v_1, \dotsc, v_m} \in Z \iff \sequence{x_1, \dotsc, x_n} \in X$
where the notation $\sequence {\cdot, \dotsc, \cdot}$ is defined inductively as:
:$\sequence {y_1} := y_1$
:$\sequence {y_1, \dotsc, y_k, y_{k + 1}} := \tuple {\sequence {y_1, \dotsc, y_k}, y_{k + 1}}$
and $\tuple {\cdot, \cdot}$ denotes the Kuratowski formalization of ordered pairs.
== Proof ==
Let $X$ be arbitrary.
By Existence of Right Universal Product, let $Z$ be a class such that:
:$\paren 1 \quad \forall v_1, \dotsc, v_m: \forall x: \tuple {\tuple {\dots \tuple {x, v_1}, \dots}, v_m} \in Z \iff x \in X$
Let $v_1, \dotsc, v_m, x_1, \dotsc, x_n$ be arbitrary sets.
Substitute $\sequence {x_1, \dotsc, x_n}$ for $x$ in $\paren 1$ producing:
:$\paren 2 \quad \tuple {\tuple {\dots \tuple {\sequence {x_1, \dotsc, x_n}, v_1}, \dots}, v_m} \in Z \iff \sequence {x_1, \dotsc, x_n} \in X$
By definition of the notation $\sequence {\cdot, \dotsc, \cdot}$, this is equivalent to:
:$\paren 3 \quad \sequence {x_1, \dotsc, x_n, v_1, \dotsc, v_m} \in Z \iff \sequence {x_1, \dotsc, x_n} \in X$
As $v_1, \dotsc, v_m, x_1, \dotsc, x_n$ were arbitrary, the theorem holds.
Category:Class Theory
Category:Von Neumann-Bernays-Gödel Set Theory
Existence of Ring of Polynomial Forms in Transcendental over Integral Domain

== Theorem ==
Let $\struct {R, +, \circ}$ be a commutative ring with unity.
Let $\struct {D, +, \circ}$ be an integral subdomain of $R$ whose zero is $0_D$.
Let $X \in R$ be transcendental over $D$
Then the ring of polynomials $D \sqbrk X$ in $X$ over $D$ exists.
== Proof ==
Suppose that $D \sqbrk X$ exists.
Let $\ds \map P X = \sum_{k \mathop = 0}^n a_k X^k$, where $a_n \ne 0_D$, be an arbitrary element of $D \sqbrk X$.
Then $\map P X$ corresponds to, and is completely described by, the ordered tuple of coefficients $\tuple {a_0, a_1, \dotsc, a_n, 0_D, 0_D, 0_D, \dotsc}$.
Consider the set $S$ of infinite sequences of elements of $D$ which are eventually $0_D$.
That is, whose elements are of the form $\tuple {b_0, b_1, \dotsc, b_n, 0_D, 0_D, 0_D, \dotsc}$ where $b_0, \ldots, b_n \in D$.
Consider the polynomial ring over $S$ by defining the operations:
From Polynomial Ring of Sequences is Ring we have that $\struct {S, +, \circ}$ is a ring.
== Sources ==
* : $\S 64$. Polynomial rings over an integral domain: Footnote
Category:Polynomial Theory
Existence of Scalar for Vector Subspace Dimension One Less

== Theorem ==
Let $K$ be a field.
Let $M$ be a subspace of the $n$-dimensional vector space $K^n$.
Let $\map \dim M = n - 1$.
Let $\sequence {\beta_n}$ be a sequence of scalars such that:
:$M = \set {\tuple {\lambda_1, \ldots, \lambda_n} \in K^n: \beta_1 \lambda_1 + \cdots + \beta_n \lambda_n = 0}$
Then there is a non-zero scalar $\gamma$ such that:
:$\forall k \in \closedint 1 n: \beta_k = \gamma \alpha_k$
== Proof ==
Let $\sequence { {e_n}'}$ be the ordered basis of $\paren {K^n}^*$ dual to the standard ordered basis of $K^n$.
Let $M = \map \ker \psi$, where $\ds \psi = \sum_{k \mathop = 1}^n \beta_k {e_k}'$.
From Equivalent Statements for Vector Subspace Dimension One Less:
:$\psi = M^\circ$
As $M^\circ$ is one-dimensional and since $\psi \ne 0$, it follows that:
:$\exists \gamma \ne 0: \psi = \gamma \phi$
Therefore:
:$\forall k \in \closedint 1 n: \beta_k = \gamma \alpha_k$
== Sources ==
* : Chapter $\text V$: Vector Spaces: $\S 28$: Linear Transformations: Theorem $28.11$
Category:Linear Algebra
Existence of Semiregular Topological Space which is not Completely Hausdorff

== Theorem ==
There exists at least one example of a semiregular topological space which is not a completely Hausdorff space.
== Proof ==
Let $T$ be a simplified Arens square.
From Simplified Arens Square is Semiregular, $T$ is a semiregular space.
From Simplified Arens Square is not Completely Hausdorff, $T$ is not a $T_3$ space.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Additional Separation Properties
Category:Completely Hausdorff Spaces
Category:Semiregular Spaces
Existence of Semiregular Topological Space which is not T3

== Theorem ==
There exists at least one example of a semiregular topological space which is not a $T_3$ space.
== Proof ==
Let $T$ be a simplified Arens square.
From Simplified Arens Square is Semiregular, $T$ is a semiregular space.
From Simplified Arens Square is not $T_3$, $T$ is not a $T_3$ space.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Additional Separation Properties
Category:T3 Spaces
Category:Semiregular Spaces
Existence of Semiregular Topological Space which is not Urysohn Space

== Theorem ==
There exists at least one example of a semiregular topological space which is not a Urysohn space.
== Proof ==
Let $T$ be an Arens square.
From Arens Square is Semiregular, $T$ is a semiregular space.
From Arens Square is not Urysohn Space, $T$ is not a Urysohn space.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Additional Separation Properties
Category:Semiregular Spaces
Category:Urysohn Spaces
Existence of Separable Space which is not Second-Countable

== Theorem ==
There exists at least one example of a separable topological space which is not also a second-countable space.
== Proof ==
Let $T$ be the Sorgenfrey line.
From Sorgenfrey Line is Separable, $T$ is a separable space.
From Sorgenfrey Line is not Second-Countable, $T$ is not a second-countable space.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Countability Axioms and Separability
Category:Separable Spaces
Category:Second-Countable Spaces
Existence of Sequence in Set of Real Numbers whose Limit is Infimum

== Theorem ==
Let $A \subseteq \R$ be a non-empty subset of the real numbers.
Let $b$ be an infimum of $A$.
Then there exists a sequence $\sequence {a_n}$ in $\R$ such that:
:$(1): \quad \forall n \in \N: a_n \in A$
:$(2): \quad \ds \lim_{n \mathop \to \infty} a_n = b$
== Proof ==
From Infimum of Subset of Real Numbers is Arbitrarily Close:
For $\epsilon = \dfrac 1 n$ there exists an $a_n \in A$ such that:
:$a_n - b < \dfrac 1 n$
Since $b$ is an infimum of $A$:
:$0 \le a_n - b$
Therefore:
:$\ds \lim_{n \mathop \to \infty} a_n = b$
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 5$: Limits: Corollary $5.7$
Category:Real Analysis
Category:Limits of Sequences
Existence of Sequence in Subset of Metric Space whose Limit is Infimum

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $a \in A$.
Let $S \subseteq A$ be a non-empty subset of $A$.
Then there exists a sequence $\sequence {a_n}$ of points of $S$ such that:
:$\ds \lim_{n \mathop \to \infty} \map d {a, a_n} = \map d {a, S}$
== Proof ==
From Existence of Sequence in Set of Real Numbers whose Limit is Infimum:
:$\ds \lim_{n \mathop \to \infty} \map d {a, a_n} = b$
where $b$ is an infimum of $\map d {a, a_n}$.
Hence the result by definition of distance to a subset of a metric space.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 5$: Limits: Corollary $5.9$
Category:Metric Spaces
Category:Limits of Sequences
Existence of Set is Equivalent to Existence of Empty Set

== Theorem ==
Let $V$ be a basic universe.
Let $P$ be the axiom:
:$V$ has at least one element.
Then $P$ is equivalent to the :
:The empty class $\O$ is a set.
== Proof ==
=== Necessary Condition ===
Let the hold.
That is:
:$\O$ is a set.
By definition of a basic universe, $V$ is a universal class.
Hence, by definition, every set is an element of $V$.
We have that $\O$ is a set .
Thus:
:$\O \in V$
and by definition of empty class it is seen that $V$ is not empty.
=== Sufficient Condition ===
Let $V$ have at least one element.
Let $x \in V$ be an element of $V$.
Then by the $x$ is also a class.
By Empty Class is Subclass of All Classes:
:$\O \subseteq x$
By the , every subclass of $x$ is an element of $V$.
That includes $\O$.
Thus:
:$\O \in V$
By definition of a basic universe, $V$ is a universal class.
Hence all of its elements are by definition sets.
Hence it follows that $\O$ is a set.
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 3$ Axiom of the empty set
Category:Empty Class
Category:Basic Universe
Existence of Set of Ordinals leads to Contradiction

== Theorem ==
The existence of the set of all ordinals leads to a contradiction.
== Proof ==
Suppose that the collection of all ordinals is a set.
Let this set be denoted as $\On$.
From Ordinals are Well-Ordered: Corollary, it is seen that $\Epsilon {\restriction_{\On} }$ is a strict well-ordering on $\On$.
By Element of Ordinal is Ordinal, it is seen that $\On$ is transitive.
And so $\On$ is itself an ordinal.
Since $\On$ is an ordinal, it follows that $\On \in \On$.
Since $\On$ is an ordinal, it follows from Ordinal is not Element of Itself that $\On \notin \On$.
This is a contradiction.
== Also known as ==
This theorem is sometimes presented as the Burali-Forti Paradox, but strictly speaking it is in fact a resolution to it.
== Also see ==
* The class of all ordinals, $\On$
* Russell's Paradox, another paradox in naive set theory.
== Sources ==
* : $\S 7.13$
Category:Ordinals
Category:Naive Set Theory
Existence of Set with Singleton Intersections with Disjoint Collection

== Theorem ==
Let $\CC$ be a set of sets all of which are pairwise disjoint.
Then:
:there exists a set $A$ such that $\forall S \in \CC: A \cap S$ is a singleton
:the axiom of choice holds.
== Proof ==
== Sources ==
* : $\S 15$: The Axiom of Choice
Category:Set Intersection
Category:Singletons
Category:Disjoint Sets
Existence of Sigma-Compact Space which is not Compact

== Theorem ==
There exists at least one example of a $\sigma$-compact topological space which is not also a compact space.
== Proof ==
Let $T = \struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
From Real Number Line is Sigma-Compact, $T$ is a $\sigma$-compact space.
From Real Number Line is not Countably Compact, $T$ is not a countably compact space.
The result follows from Compact Space is Countably Compact.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Global Compactness Properties
Category:Sigma-Compact Spaces
Category:Compact Spaces
Existence of Singleton Set

== Theorem ==
Let $a$ be a set.
Then the singleton set $\set a$ may be constructed such that:
:$a \in \set a$
== Proof ==
Let $a$ be a set.
From the Axiom of Pairing the set $\set {a, a}$ may be formed.
From the Axiom of Extension it follows that:
:$\set {a, a} = \set a$
== Sources ==
* : $\S 3$: Unordered Pairs
* : Introduction to Set Theory: $1$. Elementary Operations on Sets
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems
* : Chapter $1$: Pairing
Category:Axiomatic Set Theory
Category:Singletons
Existence of Smooth Orthonormal Frames for Pseudo-Riemannian Manifolds

== Theorem ==
Let $\struct {M, g}$ be a pseudo-Riemannian manifold.
Then for each $p \in M$ there exists a smooth orthonormal frame on a neighborhood of $p$.
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Pseudo-Riemannian Metrics
Category:Pseudo-Riemannian Geometry
Existence of Smooth Transition Function in One Dimension

== Theorem ==
Let $a, b, c, d \in \R$ be real numbers with $a < b < c < d$. 
Then there exists a smooth function $h : \R \to \closedint 0 1$ such that:
:$\map h x = 1$ for $x \in \closedint b c$
and:
:$\map h x = 0$ for $x \in \R \setminus \openint a d$
== Proof ==
First define $f : \R \to \R$ by:
:$\ds \map f x = \begin{cases}e^{-1/x} & x \ge 0 \\ 0 & x < 0\end{cases}$
for each $x \in \R$.
Note that $\map f x \ge 0$ for all $x \in \R$. 
We first show that $f$ is smooth.
=== Lemma ===
Note that for all $x \in \R$ we have $\map f x \ge 0$. 
Now define $g : \R \to \R$ by:
:$\ds \map g x = \frac {\map f x} {\map f x + \map f {1 - x} }$
for each $x \in \R$.
Note that we cannot simultaneously have $x \le 0$ and $1 - x \le 0$, and we have $\map f x > 0$ for $x > 0$, so this is well-defined.
By repeated application of the Quotient Rule, $g$ is smooth.
For $x \ge 1$, we have $\map f {1 - x} = 0$ and so $\map g x = 1$.
For $x \le 0$, we have $\map f x = 0$ and so $\map g x = 0$.
Note that for $0 < x < 1$, we have $\map f x > 0$ and hence:
:$\ds 0 \le \frac {\map f x} {\map f x + \map f {1 - x} } \le \frac {\map f x} {\map f x} = 1$
Now, define:
:$\ds \map h x = \map g {\frac {x - a} {b - a} } \map g {\frac {d - x} {d - c} }$
By the Product Rule and Chain Rule, $h$ is smooth.
We show that $h$ is our desired function.
Note that:
:$\ds \frac {x - a} {b - a} \ge 1$
is equivalent to $x \ge b$, and:
:$\ds \frac {x - a} {b - a} \le 0$
is equivalent to $x \le a$.
Note also that:
:$\ds \frac {d - x} {d - c} \ge 1$
is equivalent to $x \le c$, and:
:$\ds \frac {d - x} {d - c} \le 0$
is equivalent to $x \ge d$.
Hence if $x \le a$ we have:
:$\ds \frac {x - a} {b - a} \le 0$
and so:
:$\ds \map g {\frac {x - a} {b - a} } = 0$
If $x \ge d$, then:
:$\ds \frac {d - x} {d - c} \le 0$
and so:
:$\ds \map g {\frac {d - x} {d - c} } = 0$
Hence $\map h x = 0$ for $x \in \R \setminus \openint a d$.
If $b \le x \le c$, then we have:
:$\ds \frac {x - a} {b - a} \ge 1$
and:
:$\ds \frac {d - x} {d - c} \ge 1$
So:
:$\ds \map g {\frac {d - x} {d - c} } = \map g {\frac {x - a} {b - a} } = 1$
Hence $\map h x = 1$ for $x \in \closedint b c$.
We have already shown that $0 \le \map g x \le 1$ for each $x \in \R$.
Hence $0 \le \map h x \le 1$ for each $x \in \R$.
So $h$ is our desired function.
Category:Existence of Smooth Transition Function in One Dimension
Existence of Solution of 2nd Order Linear ODE

== Theorem ==
Let $\map P x$, $\map Q x$ and $\map R x$ be continuous real functions on a closed interval $I = \closedint a b$.
Let $x_0 \in I$, and let $y_0 \in \R$ and $y_0' \in \R$ be arbitrary.
Then the initial value problem:
:$\ds \frac {\d^2 y}{\d x^2} + \map P x \frac {\d y}{\d x} + \map Q x y = \map R x, \map y {x_0} = y_0, \map {y'} {x_0} = y_0'$
has one and only one solution $y = \map y x$ on the interval $I$.
== Proof ==
Let us introduce the variable $z = \dfrac {\d y}{\d x}$.
Then the initial ODE can be written:
:$\begin{cases}
\dfrac {\d y}{\d x} = z & : \map y {x_0} = y_0 \\
& \\
\dfrac {\d z}{\d x} = - \map P x z - \map Q x y + \map R x & : \map z {x_0} = y_0'
\end{cases}$
The converse is also true.
The result follows from Existence of Solution to System of First Order ODEs.
Category:Differential Equations
Existence of Solution to System of First Order ODEs

== Theorem ==
Consider the system of initial value problems:
:$\begin{cases}
\dfrac {\d y} {\d x} = \map f {x, y, z} & : \map y {x_0} = y_0 \\
& \\
\dfrac {\d z} {\d x} = \map g {x, y, z} & : \map z {x_0} = z_0 \\
\end{cases}$
where $\map f {x, y, z}$ and $\map g {x, y, z}$ are continuous real functions in some region of space $x y z$ that contains the point $\tuple {x_0, y_0, z_0}$.
Then this system of equations has a unique solution which exists on some interval $\size {x - x_0} \le h$.
== Proof ==
Category:Differential Equations
Existence of Square Roots of Positive Real Number

== Theorem ==
Let $r \in \R_{\ge 0}$ be a positive real number.
Then:
:$\exists y_1 \in \R_{\ge 0}: {y_1}^2 = r$
:$\exists y_2 \in \R_{\le 0}: {y_2}^2 = r$
== Proof ==
Let $S = \set {x \in \R: x^2 < r}$.
As $0 \in S$, it follows that $S$ is non-empty.
To show that $S$ is bounded above, we note that $r + 1$ is an upper bound:
:$y > r + 1 \implies y^2 > r^2 + 2 r + 1 > r$
and so $y \notin S$.
Thus $x \in S \implies x < r + 1$.
By the Completeness Axiom, $S$ has a supremum, say:
:$u = \sup S$
We already have that $u \ge 0$, as $0 \in S$ as seen.
It remains to demonstrate that $u^2 = r$.
 $u^2 \ne r$.
Then either $u^2 > r$ or $u^2 < r$.
Suppose that $u^2 > r$.
Then:
:$\dfrac {u^2 - r} {2 u} > 0$
So there exists $n \in \N$ such that:
:$0 < \dfrac 1 n < \dfrac {u^2 - r} {2 u}$
We note that:
Then:
Hence:
which contradicts the leastness of $u$.
Suppose instead that $u^2 < r$.
Then $\exists n \in \N$ such that:
:$0 < \dfrac 1 n \le \dfrac {r - u^2} {4 u}$
and:
:$\dfrac 1 n < 2 u$
Then:
Hence:
:$u + \dfrac 1 n \in S$
which contradicts the fact that $u$ is an upper bound of $S$.
Hence by Proof by Contradiction it follows that $u^2 = r$.
Hence let:
:$y_1=u$
and:
:$y_2=-u$
== Examples ==
== Sources ==
* : Chapter $\text{IX}$: Analysis Incarnate
* 
Category:Roots of Numbers
Category:Square Roots
Category:Existence of Square Roots of Positive Real Number
Existence of Square Roots of Positive Real Number/Examples/Square Root of 2

== Examples of Use of Existence of Square Roots of Positive Real Number ==
There exists $u \in \R$ such that $u^2 = 2$.
== Proof ==
A special case of Existence of Square Roots of Positive Real Number.
== Sources ==
* : $1$: Review of some real analysis: $\S 1.1$: Real Numbers: Example $1.1.8$
Category:Existence of Square Roots of Positive Real Number
Existence of Strongly Locally Compact Space which is not Weakly Sigma-Locally Compact

== Theorem ==
There exists at least one example of a strongly locally compact topological space which is not also a weakly $\sigma$-locally compact space.
== Proof ==
Let $T$ be an uncountable discrete space.
From Discrete Space is Strongly Locally Compact, $T$ is a strongly locally compact space.
From Uncountable Discrete Space is not $\sigma$-Compact, $T$ is not a weakly $\sigma$-compact space.
By definition, it follows that $T$ is not a weakly $\sigma$-locally compact space.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Localized Compactness Properties
Category:Strongly Locally Compact Spaces
Category:Weakly Sigma-Locally Compact Spaces
Existence of Subdivision with Small Norm

== Theorem ==
Let $\closedint a b$ be a closed real interval.
Let $\epsilon > 0$ be a positive real number.
Then, there exists a finite subdivision $P$ of $\closedint a b$ such that:
:$\norm P < \epsilon$
where $\norm P$ denotes the norm of $P$.
== Proof ==
By the Axiom of Archimedes, choose $N \in \N$ such that:
:$N > \dfrac {b - a} \epsilon > 0$
For each $k \in \set {0, 1, \dotsc, N - 1, N}$, define:
:$x_k = a + \dfrac k N \paren {b - a}$
We have:
Let $P = \set {x_0, x_1, \dotsc, x_{N - 1}, x_N}$.
From:
:$x_0 = a$
:$x_N = b$
:$\forall k \in \set {1, \dotsc, N}: x_{k - 1} < x_k$
it follows that $P$ is a finite subdivision of $\closedint a b$.
Now:
Category:Subdivisions (Real Analysis)
Existence of Subfamily of Cardinality not greater than Weight of Space and Unions Equal

== Theorem ==
Let $T$ be a topological space.
Let $\FF$ be a set of open sets of $T$.
There exists a subset $\GG \subseteq \FF$ such that:
:$\ds \bigcup \GG = \bigcup \FF$
and:
:$\card \GG \le \map w T$
where:
:$\map w T$ denotes the weight of $T$
:$\card \GG$ denotes the cardinality of $\GG$.
== Proof ==
By definition of weight of $T$ there exists a basis $\BB$ of $T$ such that:
:$(1): \quad \card \BB = \map w T$
Let:
:$\BB_1 = \set {W \in \BB: \exists U \in \FF: W \subseteq U}$
By definition of subset:
:$\BB_1 \subseteq \BB$
Then by Subset implies Cardinal Inequality:
:$(2): \quad \card {\BB_1} \le \card \BB$
By definition of set $\BB_1$:
:$\forall W \in \BB_1: \exists U \in \FF: W \subseteq U$
Then by the Axiom of Choice there exists a mapping $f$ from $\BB_1$ into $\FF$ such that
:$(3): \quad \forall U \in \BB_1: U \subseteq f \sqbrk U$
Let:
:$\GG = \Img f$
where $\Img f$ denotes the image of $f$.
By hypothesis:
:$\GG \subseteq \FF$
Then by Union of Subset of Family is Subset of Union of Family:
:$\ds \bigcup \GG \subseteq \bigcup \FF$
By definition of set equality, to prove $\ds \bigcup \GG = \bigcup \FF$ it is sufficient to show:
:$\ds \bigcup \FF \subseteq \bigcup \GG$
Let $\ds x \in \bigcup \FF$.
By definition of union there exists a set $A$ such that:
:$A \in \FF$ and $x \in A$
Because $A$ is open, then by definition of basis there exists $U \in \BB$ such that:
:$x \in U \subseteq A$
By definition of the set $\BB_1$:
:$U \in \BB_1$
Because $\GG = \Img f$:
:$f \sqbrk U \in \GG$
By $(3)$ and Set is Subset of Union:
:$\ds U \subseteq f \sqbrk U \subseteq \bigcup \GG$
Thus by definition of subset:
:$\ds x \in \bigcup \GG$
This ends the proof of inclusion.
By Cardinality of Image of Mapping not greater than Cardinality of Domain:
:$\card {\Img f} \le \card {\BB_1}$
Thus by $(1)$ and $(2)$:
:$\card \GG \le \map w T$
== Sources ==
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Category:Topology