nyuuzyou/wikis huggingface dataset:
Finite Union of Dilations of Balanced Set

== Theorem ==
Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a vector space over $\Bbb F$. 
Let $\lambda_1, \lambda_2, \ldots, \lambda_n$ be real numbers.
Let $W$ be a balanced set. 
Then: 
:$\ds \bigcup_{i \mathop = 1}^n \lambda_i W = \lambda W$
where $\lambda \in \set {\lambda_1, \lambda_2, \ldots, \lambda_n}$ is picked so that $\cmod \lambda$ is maximized.
== Proof ==
If $\lambda = 0$, then $\lambda_i = 0$ for each $i \in \set {1, 2, \ldots, n}$, so that:
:$\ds \bigcup_{i \mathop = 1}^n \lambda_i W = \set {\mathbf 0_X} = \lambda W$
Now suppose that $\lambda \ne 0$.
Then for each $i$ we have:
:$\ds \cmod {\frac {\lambda_i} \lambda} = \frac {\cmod {\lambda_i} } {\cmod \lambda} \le 1$
since we picked $\lambda$ among the $\lambda_i$ to maximise $\cmod \lambda$.
Since $W$ is balanced set, we then have:
:$\ds \frac {\lambda_i} \lambda W \subseteq W$
so that:
:$\lambda_i W \subseteq \lambda W$
for each $i$.
So, from Union of Subsets is Subset, we have:
:$\ds \bigcup_{i \mathop = 1}^n \lambda_i W \subseteq \lambda W$
From Set is Subset of Union, we also have:
:$\ds \lambda W \subseteq \bigcup_{i \mathop = 1}^n \lambda_i W$
From definition of set equality:
:$\ds \bigcup_{i \mathop = 1}^n \lambda_i W = \lambda W$
Category:Dilation of Subsets of Vector Spaces
Category:Balanced Sets
Finite Union of Finite Sets is Finite

== Theorem ==
Let $S$ be a finite set of finite sets.
Then the union of $S$ is finite.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 6$ Finite Sets: Exercise $6.3$
Category:Set Union
Category:Finite Sets
Category:Finite Union of Finite Sets is Finite
Finite Union of Finite Sets is Finite/Proof 1

== Theorem ==
== Proof ==
The proof proceeds by induction.
Let $S$ be a finite set with cardinality $n$.
If $n = 0$ then $S = \O$, so $\bigcup S = \O$, which is finite.
Suppose that an arbitrary finite set with cardinality $n$ of finite sets has a finite union.
Let $S$ have cardinality $n^+$.
Then there is a bijection $f: n^+ \to S$.
Then:
:$\ds \bigcup S = \bigcup_{k \mathop \in n^+} \map f k = \bigcup_{k \mathop \in n} \map f k \cup \map f n$
By Union of Finite Sets is Finite, $\bigcup S$ is finite.
== Sources ==
* : $1$: Set Theory and Logic: $\S 6$: Finite Sets: Corollary $6.8$
Category:Finite Union of Finite Sets is Finite
Finite Union of Finite Sets is Finite/Proof 2

== Theorem ==
== Proof ==
Let $S = \set {A_1, \ldots, A_n}$ such that $A_k$ is finite $\forall k = 1, \ldots, n$.
Set:
: $m = \max \set {\card {A_1}, \ldots, \card {A_n} }$
Then:
: $\ds \card {\bigcup_{k \mathop = 1}^n A_k} \le \sum_{k \mathop = 1}^n \card {A_k} \le \sum_{k \mathop = 1}^n m = n m$
Hence the result.
== Also see ==
* Inclusion-Exclusion Principle
Category:Finite Union of Finite Sets is Finite
Finite Union of Open Sets in Complex Plane is Open

== Theorem ==
Let $S_1, S_2, \ldots, S_n$ be open sets of $\C$.
Then $\ds \bigcup_{k \mathop = 1}^n S_k$ is an open set of $\C$.
== Proof ==
Let $\ds z \in \bigcup_{k \mathop = 1}^n S_k$.
Then by definition of finite union:
:$\exists k \in \set {1, 2, \ldots, n}: z \in S_k$
By definition of open set:
:$\exists \epsilon \in \R_{>0}: \map {N_\epsilon} z \subseteq S_k$
where $\map {N_\epsilon} z$ is the $\epsilon$-neighborhood of $z$ for $\epsilon$.
By Set is Subset of Union:
:$S_k \subseteq S$
Thus by Subset Relation is Transitive:
:$\map {N_\epsilon} z \subseteq S$
As $z$ is any arbitrary element of $S$, it follows that $S$ is an open set of $\C$ by definition.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Point Sets: $125$
Category:Complex Plane
Category:Set Union
Finite Union of Regular Closed Sets is Regular Closed

== Theorem ==
Let $T$ be a topological space.
Let $n \in \N$.
Suppose that:
:$\forall i \in \set {1, 2, \dotsc, n}: H_i \subseteq T$
where all the $H_i$ are regular closed in $T$.
That is:
:$\forall i \in \set {1, 2, \dotsc, n}: H_i = H_i^{\circ -}$
where $H_i^{\circ -}$ denotes the closure of the interior of $H_i$
Then $\ds \bigcup_{i \mathop = 1}^n H_i$ is regular closed in $T$.
That is:
:$\ds \bigcup_{i \mathop = 1}^n H_i = \paren {\bigcup_{i \mathop = 1}^n H_i}^{\circ -}$
== Proof ==
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Closures and Interiors
Category:Regular Closed Sets
Finite Union of Sets in Additive Function

== Theorem ==
Let $\AA$ be an algebra of sets.
Let $f: \AA \to \overline {\R}$ be an additive function.
Let $A_1, A_2, \ldots, A_n$ be any finite collection of pairwise disjoint elements of $\AA$.
Then:
:$\ds \map f {\bigcup_{i \mathop = 1}^n A_i} = \sum_{i \mathop = 1}^n \map f {A_i}$
That is, for any collection of pairwise disjoint elements of $\AA$, $f$ of their union equals the sum of $f$ of the individual elements.
== Proof ==
Proof by induction:
In the below, we assume that $A_1, A_2, \ldots$ are all pairwise disjoint elements of $\AA$.
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\ds \map f {\bigcup_{i \mathop = 1}^n A_i} = \sum_{i \mathop = 1}^n \map f {A_i}$
$\map P 1$ is the case:
:$\ds \map f {\bigcup_{i \mathop = 1}^1 A_i} = \map f {A_1} = \sum_{i \mathop = 1}^1 \map f {A_i}$
Thus $\map P 1$ is seen to hold.
=== Basis for the Induction ===
$\map P 2$ is the case:
:$\map f {A_1 \cup A_2} = \map f {A_1} + \map f {A_2}$
which comes from the definition of an additive function.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\ds \map f {\bigcup_{i \mathop = 1}^k A_i} = \sum_{i \mathop = 1}^k \map f {A_i}$
Then we need to show:
:$\ds \map f {\bigcup_{i \mathop = 1}^{k + 1} A_i} = \sum_{i \mathop = 1}^{k + 1} \map f {A_i}$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \N_{>0}: \paren {\bigcup_{i \mathop = 1}^n A_i} = \sum_{i \mathop = 1}^n \map f {A_i}$
Category:Algebras of Sets
Category:Additive Functions
Category:Set Union
Category:Proofs by Induction
Finite Union of Sets in Subadditive Function

== Theorem ==
Let $\AA$ be an algebra of sets.
Let $f: \AA \to \overline \R$ be a subadditive function.
Let $A_1, A_2, \ldots, A_n$ be any finite collection of elements of $\AA$.
Then:
:$\ds \map f {\bigcup_{i \mathop = 1}^n A_i} \le \sum_{i \mathop = 1}^n \map f {A_i}$
That is, for any finite collection of elements of $\AA$, $f$ of their union is less than or equal to the sum of $f$ of the individual elements.
== Proof ==
Proof by induction:
In the below, let $A_1, A_2, \ldots$ all be elements of $\AA$.
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\ds \map f {\bigcup_{i \mathop = 1}^n A_i} \le \sum_{i \mathop = 1}^n \map f {A_i}$
$\map P 1$ is trivially true, as this just says $\map f {A_1} \le \map f {A_1}$.
=== Basis for the Induction ===
$\map P 2$ is the case $\map f {A_1 \cup A_2} \le \map f {A_1} + \map f {A_2}$, which comes from the definition of a subadditive function.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\ds \map f {\bigcup_{i \mathop = 1}^k A_i} \le \sum_{i \mathop = 1}^k \map f {A_i}$
Then we need to show:
:$\ds \map f {\bigcup_{i \mathop = 1}^{k + 1} A_i} \le \sum_{i \mathop = 1}^{k + 1k + 1} \map f {A_i}$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \N_{>0}: \paren {\bigcup_{i \mathop = 1}^n A_i} \le \sum_{i \mathop = 1}^n \map f {A_i}$
Category:Algebras of Sets
Category:Subadditive Functions
Category:Set Union
Category:Proofs by Induction
Finite Weight Space has Basis equal to Image of Mapping of Intersections

== Theorem ==
Let $T = \struct {X, \tau}$ be a topological space with finite weight.
Then there exist a basis $\BB$ of $T$ and a mapping $f:X \to \tau$ such that:
:$\BB = \Img f$ and
:$\forall x \in X: \paren {x \in \map f x \land \forall U \in \tau: x \in U \implies \map f x \subseteq U}$
where $\Img f$ denotes the image of $f$.
== Proof ==
By definition of weight there exists a basis $\BB$ such that:
:$\card \BB = \map w T$
where:
:$\map w T$ denotes the weight of $T$
:$\card \BB$ denotes the cardinality of $\BB$.
By assumption that weight is finite:
:$\card \BB$ is finite
Then by Cardinality of Set is Finite iff Set is Finite:
:$\BB$ is finite
Define a mapping $f: X \to \powerset X$:
:$(1): \quad \forall x \in X: \map f x = \bigcap \set {U \in \BB: x \in U}$
By definition of subset:
:$\forall x \in X: \set {U \in \BB: x \in U} \subseteq \BB$
By Subset of Finite Set is Finite:
:$\forall x \in X: \set {U \in \BB: x \in U}$ is finite
Then by General Intersection Property of Topological Space:
:$\forall x \in X: \bigcap \set {U \in \BB: x \in U} \in \tau$
So:
:$f: X \to \tau$
We will prove that:
:$(2): \quad \forall x \in X: \paren {x \in \map f x \land \forall U \in \tau: x \in U \implies \map f x \subseteq U}$
Let $x \in X$.
By $(1)$:
:$\map f x = \bigcap \set {U \in \BB: x \in U}$
Thus by definition of intersection:
:$x \in \map f x$
Let $U$ be an open set of $T$.
Let $x \in U$.
By definition of basis:
:$\exists V \in \BB: x \in V \subseteq U$
Then:
:$V \in \set {U \in \BB: x \in U}$
Hence by Intersection is Subset:
:$\map f x \subseteq V$
Thus by Subset Relation is Transitive:
:$\map f x \subseteq U$
This ends the proof of $(2)$.
We will prove that $\Img f$ is a basis of $T$.
By $f: X \to \tau$ and definition of image:
:$\Img f \subseteq \tau$
Let $U$ be an open set of $T$.
Let $x$ be a point $x \in X$ such that:
:$x \in U$
By $(2)$:
:$\map f x \in \Img f \land x \in \map f x \subseteq U$
By definition of basis this ends the proof of basis.
Thus the result.
== Sources ==
* 
* 
Category:Topology
Finite iff Cardinality Less than Aleph Zero

== Theorem ==
Let $X$ be a set.
Then $X$ is finite $\card X < \aleph_0$
where:
:$\card X$ denotes the cardinality of $X$
:$\aleph_0 = \card \N$ by Aleph Zero equals Cardinality of Naturals.
== Proof ==
=== Sufficient Condition ===
Let $X$ be finite.
By definition of finite set:
:$\exists n \in \N: X \sim \N_n$
where:
:$\sim$ denotes the set equivalence
:$\N_n$ denotes the initial segment of natural numbers less than $n$.
By the von Neumann construction of natural numbers:
:$\N_n = n$
By definition of cardinality:
:$\card X = n$
By the von Neumann construction of natural numbers:
:$\forall i \in \N: i \subseteq \N$
Then:
:$n + 1 \subseteq \N$
By Subset implies Cardinal Inequality:
:$n + 1 = \card {n + 1} \le \card \N = \aleph_0$
Also:
:$n < n + 1$
Thus:
:$\card X < \aleph_0$
=== Necessary Condition ===
Let $\card X < \aleph_0$.
By definition of aleph mapping:
:$\aleph_0 = \omega$
By the von Neumann construction of natural numbers:
:$\N = \omega$
By definition of ordinal:
:$\card X \in \N$
By definition of cardinal:
:$\exists n \in \N: X \sim n$
By the von Neumann construction of natural numbers:
:$\exists n \in \N: X \sim \N_n$
Thus by definition:
:$X$ is finite.
== Sources ==
* 
Category:Cardinals
Category:Aleph Mapping
Finitely Generated Algebraic Extension is Finite

== Theorem ==
Let $L/K$ be a field extension and $\alpha_1,\ldots,\alpha_n \in L$ algebraic over $K$.
Then $K \left({\alpha_1, \ldots, \alpha_n}\right) / K$ is a finite field extension.
== Proof ==
Let $S = \left\{{\alpha_1, \ldots, \alpha_n}\right\}$.
We show by induction on $n$ that $K \left({S}\right) / K$ is finite.
Clearly $K$ is finite over itself, so the result holds when $n = 0$.
Now suppose that for all sets $T \subseteq L$ with $\left|{T}\right| \leq n - 1$ and each element of $T$ algebraic over $K$, $K \left({T}\right) / K$ is finite. We have:
:$K \left({S}\right) = K \left({\alpha_1, \ldots, \alpha_{n-1}}\right) \left({\alpha_n}\right)$
By the induction hypothesis $K \left({\alpha_1, \ldots, \alpha_{n-1}}\right) / K$ is finite.
By the Structure of Simple Algebraic Field Extension $K \left({S}\right) / K \left({\alpha_1, \ldots, \alpha_{n-1}}\right)$ is finite.
Therefore, by the Tower Law, $K \left({S}\right) / K$ is finite.
Category:Field Extensions
Finitely Generated Vector Space has Finite Basis

== Theorem ==
Let $K$ be a division ring.
Let $V$ be a finitely generated vector space over $K$.
Then $V$ has a finite basis.
== Proof ==
This follows from Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set.
It suffices to find:
:A linearly independent subset $L\subset V$
:A finite generator $S\subset V$
with $L\subset S$.
By Empty Set is Linearly Independent, we make take $L = \O$ and $S$ any finite generator, which exists because $V$ is finitely generated.
== Also see ==
* Vector Space has Basis, which does not assume that $V$ is finitely generated
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Theorem $27.8$
Category:Bases of Vector Spaces
Finitely Many Reduced Associated Quadratic Irrationals

== Lemma ==
Let $n \in \Z: n > 0$.
There are finitely many reduced quadratic irrationals which are associated to $n$.
== Proof ==
By definition, we can write an arbitrary reduced irrational as $\alpha = \dfrac{\sqrt n + P} Q$.
By definition, $\alpha > 1$ and its conjugate $\tilde{\alpha} > -1$.
So we know $\alpha + \tilde{\alpha} = \dfrac {2P} Q > 0$.
Hence with the assumption $Q > 0$ we have $P > 0$.
Since $\tilde{\alpha} < 0$ we also have $P < \sqrt n$.
Also, since $\alpha > 1$ by assumption, we have $Q < P + \sqrt n < 2 \sqrt n$.
Thus there are finitely many choices for both $P$ and $Q$, forcing finitely many reduced quadratic irrationals associated to a fixed $n$.
This amount is strictly bounded above by $2n$.
Category:Quadratic Irrationals
Finitely Satisfiable Theory has Maximal Finitely Satisfiable Extension

== Theorem ==
Let $T$ be a finitely satisfiable $\LL$-theory.
Then there exists a finitely satisfiable $\LL$-theory $T'$ which contains $T$ as a subset such that:
:for all $\LL$-sentences $\phi$, either $\phi \in T'$ or $\neg \phi \in T'$.
=== Lemma ===
== Proof 1 ==
== Proof 2 ==
Category:Model Theory for Predicate Logic
Finitely Satisfiable Theory has Maximal Finitely Satisfiable Extension/Lemma

== Theorem ==
Let $\LL$ be a language of predicate logic.
Let $T$ be a finitely satisfiable $\LL$-theory.
Let $\phi$ be an $\LL$-sentence.
Then either:
:$T \cup \set \phi$
or:
:$T \cup \set {\neg \phi}$
is finitely satisfiable.
== Proof ==
, suppose that $T \not\models \phi$ and $T \not\models \neg \phi$.
 that $T \cup \set \phi$ is not finitely satisfiable.
Then by definition there must be a finite subset $K$ of $T \cup \set \phi$ which is not satisfiable.
Since $T$ is finitely satisfiable, $\phi \in K$.
Therefore $\Delta = K \setminus \set \phi$ is a finite subset of $T$.
Thus, for every $\LL$-structure $\MM$ that models $\Delta$, $\phi$ is not valid.
Hence, for every $\LL$-structure $\MM$ that models $\Delta$, $\neg \phi$ is valid.
By definition, then, $\Delta \models \neg \phi$
:where $\models$ denotes semantic consequence.
Let $\Sigma$ be a finite subset of $T$.
Since $\Delta \cup \Sigma$ is a finite subset of $T$, it is satisfiable.
But $\Delta \models \neg \phi$.
Hence $\set {\neg \phi} \cup \Sigma$ is satisfiable.
But any finite subset of $T \cup \set {\neg \phi}$ is either in $T$ or of the form $\Sigma \cup \set {\neg \phi}$.
This demonstrates that $T \cup \set {\neg \phi}$ is finitely satisfiable, contradicting our assumption.
Category:Model Theory for Predicate Logic
Finitely Satisfiable Theory has Maximal Finitely Satisfiable Extension/Proof 1

== Theorem ==
=== Lemma ===
== Proof ==
The set of all finitely satisfiable $\LL$-theories containing $T$ forms an ordered set using subset inclusion as the ordering.
Let $C$ be a non-empty chain in this ordered set.
Let $\ds T_C = \bigcup_{\Sigma \mathop \in C} \Sigma$.
Let $\Delta$ be a finite subset of $T_C$.
Then there exists a single $\Sigma$ in $C$ which contains $\Delta$.
Since this $\Sigma$ is finitely satisfiable by definition, this means that $\Delta$ is satisfiable.
Hence $T_C$ is finitely satisfiable.
Since each $\Sigma \in C$ is contained in $T_C$, this means that $T_C$ is an upper bound for $C$ in the ordered set.
Thus, by Zorn's Lemma, there is a finitely satisfiable $\LL$-theory $T'$ containing $T$ such that $T'$ contains all other such theories.
Let $\phi$ be an $\LL$-sentence.
Let $\phi \nsubseteq T'$.
 $T' \cup \set \phi$ were finitely satisfiable.
Then by definition of $T'$, $T'$ would contain $T' \cup \set \phi$ as a subset.
This would mean that $T'$ contains $\phi$, which contradicts the assumption.
Thus $T' \cup \set \phi$ is not finitely satisfiable.
By the lemma, $T' \cup \set {\neg \phi}$ is finitely satisfiable.
Thus, by definition of $T'$, $T'$ contains $T' \cup \set {\neg \phi}$ as a subset.
Hence $T'$ contains $\neg \phi$.
Category:Model Theory for Predicate Logic
Finitely Satisfiable Theory has Maximal Finitely Satisfiable Extension/Proof 2

== Theorem ==
Let $T$ be a finitely satisfiable $\LL$-theory.
There is a finitely satisfiable $\LL$-theory $T'$ which contains $T$ as a subset such that for all $\LL$-sentences $\phi$, either $\phi \in T'$ or $\neg\phi \in T'$.
=== Lemma ===
== Proof ==
Let $\AA$ be the set of finitely satisfiable extensions of $T$.
By the lemma, for each element $S$ of $\AA$ and each $\LL$-sentence $\phi$, either $S \cup \set \phi \in \AA$ or $S \cup \set {\neg \phi} \in \AA$.
$\AA$ has finite character, by the following argument:
Let $S \in \AA$.
Let $F$ be a finite subset of $S$.
Then $S$ is satisfiable and hence finitely satisfiable.
Thus in $\AA$.
Let $S$ be a theory on $\LL$.
Let every finite subset of $S$ be finitely satisfiable.
Then every finite subset of $S$ is satisfiable.
Therefore $S$ is finitely satisfiable.
Thus $\AA$ has finite character.
By the Restricted Tukey's Theorem, $\AA$ has an element $T'$ such that:
:for each $\LL$-sentence $\phi$, either $\phi \in T'$ or $\neg \phi \in T'$.
Category:Model Theory for Predicate Logic
First-Countability is Hereditary

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which is first-countable.
Let $T_H = \struct {H, \tau_H}$, where $\O \subset H \subseteq S$, be a subspace of $T$.
Then $T_H$ is first-countable.
== Proof ==
From the definition of first-countable, every point in $S$ has a countable local basis in $T$.
The intersection of $H$ with the countable local basis of $S$ provides a countable local basis for $H$.
As every point in $H$ is also a point in $S$, the result follows from Basis for Topological Subspace‎.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Countability Properties
* : $\S 3$: New Spaces From Old: Subspaces
Category:First-Countable Spaces
Category:Topological Subspaces
First-Countability is Preserved under Open Continuous Surjection

== Theorem ==
Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a surjective open mapping which is also continuous.
If $T_A$ is first-countable, then $T_B$ is also first-countable.
== Proof ==
Let $\phi$ be surjective, continuous and open.
Let $T_A$ be first countable.
Let $b \in S_B$.
Since $\phi$ is surjective there is a point $a \in S_A$ such that:
:$\map \phi a = b$
From the first-countability of $T_A$, there is a local base $\BB$, say, of $a$ which is countable.
Let $\BB = \set {V_n: n \in \N}$.
We need to show that $\set {\phi \sqbrk {V_n}: n \in \N}$ is a local base for $b$.
Let $U$ be an open set of $T_B$ that contains $b$.
As $b = \map \phi a$ we have that:
:$a \in \phi^{-1} \sqbrk U$
From the continuity of $\phi$, we have that $\phi^{-1} \sqbrk U$ is open.
As $\BB$ is a local base, there is an open set $V_n \subseteq \phi^{-1} \sqbrk U$ such that $a \in V_n$.
$\phi$ is surjective, so from Surjection iff Right Inverse we have that:
:$\phi \sqbrk {\phi^{-1} \sqbrk U} = U$
So, applying $\phi$ to $V_n$, from Image of Subset under Mapping is Subset of Image we obtain:
:$\phi \sqbrk {V_n} \subseteq U$
such that $b \in \phi \sqbrk {V_n}$.
This means that $\set {\phi \sqbrk {V_n}: n \in \N}$ is a local base for $b$.
Thus, $T_B$ is first countable.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Invariance Properties
Category:First-Countable Spaces
Category:Open Mappings
Category:Continuous Mappings
Category:Surjections
First-Countability is not Continuous Invariant

== Theorem ==
Let $T_A = \struct {A, \tau_A}$ and $T_B = \struct {B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a continuous mapping.
If $T_A$ is a first-countable space, then it does not necessarily follow that $T_B$ is also first-countable.
== Proof ==
Let $T_S = \struct {S, \tau_S}$ be the Arens-Fort space.
Let $T_D = \struct {S, \tau_D}$ be the discrete space, also on $S$.
Let $I_S: S \to S$ be the identity mapping on $S$.
From Mapping from Discrete Space is Continuous, we have that $I_S$ is a continuous mapping.
Then we have:
:Discrete Space is First-Countable
:Arens-Fort Space is not First-Countable
Thus we have demonstrated a continuous mapping from a first-countable space to a space which is not first-countable space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $26$. Arens-Fort Space: $6$
Category:First-Countable Spaces
Category:Continuous Invariants
First-Countable Space is Hausdorff iff All Convergent Sequences have Unique Limit

== Theorem ==
Let $T = \struct {S, \tau}$ be a first-countable topological space.
Then $T$ is Hausdorff all convergent sequences on $T$ have a unique limit.
== Proof ==
=== Sufficient Condition ===
This is shown in Convergent Sequence in Hausdorff Space has Unique Limit.
Note that it does not require first-countability.
=== Necessary Condition ===
We prove the contrapositive.
Suppose $T$ is not Hausdorff.
Then by definition:
:$\exists x, y \in S, x \ne y: \forall U, V \in \tau: x \in U, y \in V: U \cap V \ne \O$
By first-countability, $x$ and $y$ has countable local basis.
Let $\set {X_n: n \in \N}$ and $\set {Y_n: n \in \N}$ be countable local bases for $x$ and $y$ respectively.
Let $U_n = \ds \bigcap_{k \mathop = 1}^n X_k$ and $V_n = \ds \bigcap_{k \mathop = 1}^n Y_k$.
We have for all $n \in \N$:
:$x \in U_n, y \in V_n$
:$U_m \subseteq U_n, V_m \subseteq V_n$ for all $m \ge n$
We have that:
:$\forall n \in \N: U_n \cap V_n \ne \O$
hence we can pick any $z_n \in U_n \cap V_n$.
We show that $\sequence {z_n}$ converges to both $x$ and $y$.
Let $A$ be an open set such that $x \in A$.
By definition of a local basis:
:$\exists N \in \N: X_N \subseteq A$
We also have:
:$\forall n \ge N: U_n \subseteq U_N \subseteq X_N \subseteq A$
Therefore for any $n \ge N$:
:$z_n \in U_n \cap V_n \subseteq U_n \subseteq A$
By definition of convergence, $\sequence {z_n}$ converges to $x$.
Similarly $\sequence {z_n}$ converges to $y$.
Hence $\sequence {z_n}$ does not have unique limit.
Category:First-Countable Spaces
Category:Hausdorff Spaces
First-Countable Space is Sequentially Compact iff Countably Compact

== Theorem ==
A first-countable topological space is sequentially compact it is countably compact.
== Proof ==
Follows directly from:
:Sequentially Compact Space is Countably Compact
:Countably Compact First-Countable Space is Sequentially Compact
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Countability Axioms and Separability
Category:Sequentially Compact Spaces
Category:Countably Compact Spaces
Category:First-Countable Spaces
First-Order Reaction

== Theorem ==
Let a substance decompose spontaneously in a first-order reaction.
Let $x_0$ be a measure of the quantity of that substance at time $t = 0$.
Let the quantity of the substance that remains after time $t$ be $x$.
Then:
:$x = x_0 e^{-k t}$
where $k$ is the rate constant.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Half-Life of Radioactive Substance
Category:First-Order Reactions
Category:Decay Equation