nyuuzyou/wikis huggingface dataset:
Uniformly Continuous Function to Complete Metric Space has Unique Continuous Extension to Closure of Domain/Lemma 2

== Theorem ==
Let $\struct {X, d}$ be a metric space.
Let $\struct {Y, d'}$ be a complete metric space.
Let $A \subseteq X$. 
Let $f : A \to Y$ be a uniformly continuous function.
Let $\sequence {a_n}$ be a convergent sequence in $A$. 
Then the limit of $\sequence {\map f {a_n} }$ is dependent only on the limit of $\sequence {a_n}$. 
That is, there exists a function $L : A^- \to Y$ such that: 
:$\ds \lim_{n \mathop \to \infty} \map f {a_n} = \map L {\lim_{n \mathop \to \infty} a_n}$
for every convergent sequence $\sequence {a_n}$.
== Proof ==
Let $\sequence {a_n}$ and $\sequence {b_n}$ be sequences in $A$ such that $a_n \to a$ and $b_n \to a$, with: 
:$\map f {a_n} \to L_1$
and:
:$\map f {b_n} \to L_2$
We have, by the Triangle Inequality:
:$\map {d'} {\map f {a_n}, L_2} \le \map {d'} {\map f {a_n}, \map f {b_n} } + \map {d'} {\map f {b_n}, L_2}$
We want to show that we can find $K$ such that: 
:$\map {d'} {\map f {a_n}, L_2} < \epsilon$
for $n > K$. 
Note that since $f$ is uniformly continuous, we can find $\delta_1$ such that for: 
:$\map d {a_n, b_n} < \delta_1$
we have:
:$\map {d'} {\map f {a_n}, \map f {b_n} } < \dfrac \epsilon 2$
By the Triangle Inequality, we can write: 
:$\map d {a_n, b_n} \le \map d {a_n, a} + \map d {a, b_n}$
Since $a_n \to a$ and $b_n \to a$, we can find $K_1, K_2$ such that for $n > K_1$ we have: 
:$\map d {a_n, a} < \dfrac {\delta_1} 2$
and for $n > K_2$ we have: 
:$\map d {b_n, a} < \dfrac {\delta_1} 2$
So, for $n > \max \set {K_1, K_2}$, we have: 
:$\map d {a_n, b_n} < \delta$
and so:
:$\map {d'} {\map f {a_n}, \map f {b_n} } < \dfrac \epsilon 2$
for $n > \max \set {K_1, K_2}$.
Since $\map f {b_n} \to L_2$, we can pick $K_3$ such that: 
:$\map {d'} {\map f {b_n}, L_2} < \dfrac \epsilon 2$
for $n > K_3$. 
So, setting $K = \max \set {K_1, K_2, K_3}$, we have: 
:$\map {d'} {\map f {b_n}, L_2} < \dfrac \epsilon 2$
for $n > K$.
So:
:$\map {d'} {\map f {a_n}, L_2} < \epsilon$
for $n > K$. 
So $\map f {a_n} \to L_2$.
From Convergent Sequence in Metric Space has Unique Limit, we have:
:$L_1 = L_2$
So, if $a_n \to a$, the sequence $\sequence {\map f {a_n} }$ converges to some limit $\map L a$ dependent only on $a$.
Category:Uniformly Continuous Function to Complete Metric Space has Unique Continuous Extension to Closure of Domain
Uniformly Continuous Real Function/Examples/Square Function

== Example of Uniformly Continuous Real Function ==
Let $S$ be the open interval $S = \openint 0 1$.
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = x^2$
Then $f$ is uniformly continuous on $S$.
== Proof ==
Let $\epsilon \in \R_{>0}$.
Let $\delta = \dfrac \epsilon 2$.
Then when $\size {x_2 - x_1} > \delta$ we have:
Hence, by definition, $f$ is uniformly continuous on $S$.
== Sources ==
* 
* 
Category:Examples of Uniformly Continuous Real Functions
Uniformly Continuous Semigroup Bounded on Compact Intervals

== Theorem ==
Let $\GF \in \set {\R, \C}$.
Let $X$ be a Banach space over $\GF$. 
Let $\family {\map T t}_{t \ge 0}$ be a uniformly continuous semigroup. 
Let $\struct {\map B X, \norm {\, \cdot \,}_{\map B X} }$ be the space of bounded linear transformations equipped with the canonical norm.
Let $T > 0$ be a real number. 
Then there exists $M > 0$ such that: 
:$\norm {\map T t}_{\map B X} \le M$
for $t \in \closedint 0 T$. 
== Proof ==
Since $\family {\map T t}_{t \ge 0}$ is a uniformly continuous semigroup, we have: 
:$\ds \lim_{t \mathop \to 0^+} \norm {\map T t - I}_{\map B X} = 0$ 
From Sequence in Normed Vector Space Convergent to Limit iff Norm of Sequence minus Limit is Null Sequence and Modulus of Limit: Normed Vector Space, we have:
:$\norm {\map T t}_{\map B X} \to \norm I_{\map B X}$ as $t \to 0^+$
In particular: 
:$\norm {\map T t}_{\map B X} \to 1$
since $\norm I_{\map B X} = 1$ from Identity Mapping on Normed Vector Space is Bounded Linear Operator.
Then there exists $\delta > 0$ such that: 
:$\ds \norm {\map T t}_{\map B X} \le \frac 3 2$ for $t \in \closedint 0 \delta$.
For $t_0 \in \closedint 0 T$, write $t_0 = N \delta + r$ for $N \in \N$ and $r \in \hointr 0 \delta$.
Then we have: 
Now note that if $t_0 \in \closedint 0 T$ and $t_0 = N \delta + r$ for $N \in \Z_{\ge 0}$ and $r \in \hointr 0 \delta$, we have: 
:$\ds N \le \frac T \delta$ 
So we have: 
:$\ds \norm {\map T t}_{\map B X} \le \paren {\frac 3 2}^{\frac T \delta + 1}$
for $t \in \closedint 0 T$.
$\delta$ is independent of $t$, so we have the result. 
Category:Uniformly Continuous Semigroups
Uniformly Convergent Product Satisfies Uniform Cauchy Criterion

== Theorem ==
Let $X$ be a compact topological space.
Let $\struct {\mathbb K, \norm {\, \cdot \,} }$ be a valued field.
Let $\sequence {f_n}$ be a sequence of continuous mappings $f_n: X \to \mathbb K$.
Let the infinite product $\ds \prod_{n \mathop = 1}^\infty f_n$ converge uniformly on $X$.
Then $\ds \prod_{n \mathop = 1}^\infty f_n$ satisfies the uniform Cauchy condition for products.
== Proof ==
== Also see ==
* Tail of Uniformly Convergent Product Converges Uniformly to One
* Factors in Uniformly Convergent Product Converge Uniformly to One
Category:Uniform Convergence
Category:Infinite Products
Uniformly Convergent Sequence Evaluated on Convergent Sequence

== Theorem ==
Let $X = \struct {A, d_X}$ and $Y = \struct {B, d_Y}$ be metric spaces.
Let $K$ be a subspace of $X$.
Let $f: X \to Y$ be a mapping.
Let $\FF = \sequence {f_n}$ be a sequence of continuous mappings $f_n: X \to Y$ that converges to $f$ uniformly on $K$.
Let $\sequence {a_n}$ be a convergent sequence in $K$ with limit $a \in K$.
Then the sequence $\sequence {\map {f_n} {a_n} }$ is convergent such that:
:$\ds \lim_{n \mathop \to \infty} \map {f_n} {a_n} = \map f a$
== Proof ==
We want to show that:
:$\map {d_Y} {\map {f_n} {a_n} , \map f a} \to 0$ as $n \to \infty$
Let $\epsilon \in \R_{>0}$ be fixed.
From :
:$\map {d_Y} {\map {f_n} {a_n} , \map f a} \le \map {d_Y} {\map {f_n} {a_n} , \map f {a_n} } + \map {d_Y} {\map f {a_n} , \map f a}$
From the Uniform Limit Theorem, it follows that $f$ is continuous.
Since $a_n \to a$, Sequential Continuity is Equivalent to Continuity in Metric Space tells us that:
:$\exists M \in \N : \forall n \ge M : \map {d_Y} {\map f {a_n} , \map f a} < \dfrac \epsilon 2$
Since $\sequence {f_n}$ is uniformly convergent on $K$, it follows that:
:$\exists N \in \N : \forall n \ge N : \map {d_Y} {\map {f_n} {a_n} , \map f {a_n} } < \dfrac \epsilon 2$
So if $n \ge \max \set {M, N}$, it follows that:
which shows that $\map {d_Y} {\map {f_n} {a_n} , \map f a} \to 0$ as $n \to \infty$. 
Category:Metric Spaces
Category:Uniform Convergence
Uniformly Convergent Sequence Multiplied with Function

== Theorem ==
Let $X$ be a set.
Let $V$ be a normed vector space over $\mathbb K$.
Let $\sequence {f_n}$ be a sequence of mappings $f_n: X \to V$.
Let $\sequence {f_n}$ be uniformly convergent.
Let $g: X \to \mathbb K$ be bounded.
Then $\sequence {f_n g}$ is uniformly convergent.
=== Corollary ===
== Proof ==
Denote $\norm {\, \cdot \,}$ as the norm on $V$.
Let $\epsilon > 0$.
By boundedness of $g$:
:$\exists M \in \R: \forall x \in X: \norm {\map g x} < M$
By uniformly convergence of $\sequence {f_n}$:
:$\exists f: X \to \mathbb K: \exists N \in \R: \forall x \in X: \norm {\map {f_n} x - \map f x} < \dfrac \epsilon M$
Pick any $x \in X$. Then:
As the choice of $x$ is arbitrary, $\sequence {f_n g}$ uniformly converges to $\sequence {f g}$.
== Also see ==
* Dirichlet's Test for Uniform Convergence
Category:Uniform Convergence
Uniformly Convergent Sequence Multiplied with Function/Corollary

== Corollary to Uniformly Convergent Sequence Multiplied with Function ==
Let $X$ be a compact topological space.
Let $V$ be a normed vector space over $\mathbb K$.
Let $\left\langle{f_n}\right\rangle$ be a sequence of mappings $f_n: X \to V$.
Let $\left\langle{f_n}\right\rangle$ be uniformly convergent.
Let $g: X \to \mathbb K$ be continuous.
Then $\left\langle{f_n g}\right\rangle$ is uniformly convergent.
== Proof ==
Follows directly from:
: Continuous Function on Compact Subspace of Euclidean Space is Bounded
: Uniformly Convergent Sequence Multiplied with Function
Category:Uniform Convergence
Uniformly Convergent Sequence of Bounded Functions is Uniformly Bounded

== Theorem ==
Let $X = \left({A, d}\right)$ and $Y = \left({B, \rho}\right)$ be metric spaces.
Let $\left \langle{f_i}\right \rangle_{i \in I}$ be a uniformly convergent sequence of mappings $f_i: X \to Y$.
$\forall i \in I$, let $f_i$ be bounded.
Then $\left \langle{f_i}\right \rangle$ is uniformly bounded.
== Proof ==
Category:Metric Spaces
Uniformly Convergent Sequence of Continuous Functions Converges to Continuous Function

== Theorem ==
Let $S \subseteq \R$.
Let $x \in S$.
Let $\sequence {f_n}$ be a sequence of real functions $S \to \R$ converging uniformly to $f: S \to \R$.
Let $f_n$ be continuous at $x$ for all $n \in \N$.
Then $f$ is continuous at $x$.
=== Corollary ===
== Proof ==
Let $\epsilon \in \R_{> 0}$. 
Since $f_n \to f$ uniformly, there exists some $N \in \N$ such that: 
:$\size {\map {f_n} x - \map f x} < \dfrac \epsilon 3$
for all $x \in S$ and $n \ge N$.
Since $f_N$ is continuous at $x$, there exists some $\delta > 0$ such that: 
:for all $y$ with $\size {x - y} < \delta$, we have $\size {\map {f_N} x - \map {f_N} y} < \dfrac \epsilon 3$
Then for $y$ with $\size {x - y} < \delta$ we have: 
Since $\epsilon$ was arbitrary, $f$ is continuous at $x$.
== Sources ==
* : $\S 9.4$: Uniform Convergence and Continuity: Theorem $9.2$
Category:Uniform Convergence
Category:Continuous Functions
Uniformly Convergent Sequence of Continuous Functions Converges to Continuous Function/Corollary

== Theorem ==
Let $S \subseteq \R$.
Let $\sequence {f_n}$ be a sequence of real functions $S \to \R$ converging uniformly to $f : S \to \R$.
Let $f_n$ be continuous for all $n \in \N$.
Then $f$ is continuous.
== Proof ==
Let $x \in S$. 
As $f_n$ is continuous at $x$ and $f_n \to f$ uniformly, we have by Uniformly Convergent Sequence of Continuous Functions Converges to Continuous Function:
:$f$ is continuous at $x$.
As $x \in S$ was arbitrary: 
:$f$ is continuous at all $x \in S$.
So $f$ is continuous.
Category:Uniform Convergence
Category:Continuous Functions
Uniformly Convergent Sequence on Dense Subset

== Theorem ==
Let $X$ be a metric space.
Let $Y \subset X$ be dense.
Let $V$ be a Banach space.
Let $\sequence {f_n}$ be a sequence of continuous mappings $f_n : X\to V$.
Let $\sequence {f_n}$ be uniformly convergent on $Y$.
Then $\sequence {f_n}$ is uniformly convergent on $X$.
== Proof ==
Let $\epsilon>0$.
Let $N \in\N$ be such that $\norm {f_n - f_m} < \epsilon$ for $n, m > N$ on $Y$.
Let $x \in X$.
Then there exists a sequence $\sequence {y_n} \in Y$ with $y_n \to x$.
By continuity:
:$\norm {\map {f_n} x - \map {f_m} x} \le \epsilon$
Because $V$ is complete, $\sequence {f_n}$ is uniformly convergent on $X$.
Category:Uniform Convergence
Uniformly Convergent Series of Continuous Functions Converges to Continuous Function

== Theorem ==
Let $S \subseteq \R$.
Let $x \in S$.
Let $\sequence {f_n}$ be a sequence of real functions.
Let $f_n$ be continuous at $x$ for all $n \in \N$.
Let the infinite series: 
:$\ds \sum_{n \mathop = 1}^\infty f_n$
be uniformly convergent to a real function $f : S \to \R$.
Then $f$ is continuous at $x$.
=== Corollary ===
== Proof ==
Let $\sequence {s_n}$ be sequence of real functions $S \to \R$ such that: 
:$\ds \map {s_n} x = \sum_{k \mathop = 1}^n \map {f_n} x$
for each $n \in \N$ and $x \in S$.
By Sum Rule for Continuous Real Functions:
:$s_n$ is continuous at $x$ for all $n \in \N$.
Since additionally $s_n \to f$ uniformly, we have by Uniformly Convergent Sequence of Continuous Functions Converges to Continuous Function: 
:$f$ is continuous at $x$.
== Sources ==
* : $\S 9.6$: Uniform Convergence of Infinite Series of Functions: Theorem $9.7$
Category:Uniform Convergence
Category:Continuous Real Functions
Uniformly Convergent Series of Continuous Functions Converges to Continuous Function/Corollary

== Theorem ==
Let $S \subseteq \R$.
Let $\sequence {f_n}$ be a sequence of real functions.
Let $f_n$ be continuous for all $n \in \N$.
Let the infinite series:
:$\ds \sum_{n \mathop = 1}^\infty f_n$
be uniformly convergent to a real function $f : S \to \R$.
Then $f$ is continuous.
== Proof ==
Let $x \in S$.
Then $f_n$ is continuous at $x$ for all $n \in \N$.
Since:
:$\ds \sum_{n \mathop = 1}^\infty f_n$
converges uniformly to $f$, we have by Uniformly Convergent Series of Continuous Functions Converges to Continuous Function:
:$f$ is continuous at $x$.
As $x \in S$ was arbitrary, we have that: 
:$f$ is continuous.
Category:Uniform Convergence
Category:Continuous Functions
Uniformly Convergent Series of Continuous Functions is Continuous

== Theorem ==
Let $\sequence {f_n}$ be a sequence of real functions.
Let each of $\sequence {f_n}$ be continuous on the interval $\hointr a b$.
Let the series:
:$\ds \map f x := \sum_{n \mathop = 1}^\infty \map {f_n} x$
be uniformly convergent for all $x \in \closedint a b$.
Then $f$ is continuous on $\hointr a b$.
== Proof ==
== Sources ==
* : $\S 1.3.1$: Properties of uniformly convergent series: Theorem $1.9 \ \text{(a)}$
Category:Real Analysis
Category:Convergence
Category:Continuity
Category:Series
Uniformly Convergent iff Difference Under Supremum Metric Vanishes

== Theorem ==
Let $X$ be a set.
Let $\struct {Y, d}$ be a metric space.
Let $S$ be the set of all bounded mappings from $X$ to $Y$.
Let $\sequence {f_n}_{n \mathop \in \N}$ be a sequence in $S$.
Let $f \in S$.
Let $d_S: S \times S \to Y$ denote the supremum metric on $S$.
Then:
:$\sequence {f_n}$ converges uniformly to $f$ on $S$
:
:$\map {d_S} {f_n, f} \to 0$ as $n \to \infty$.
== Proof ==
Category:Metric Spaces
Category:Uniform Convergence
Uniformly Normal Neighborhood Lemma

== Theorem ==
Let $\struct {M, g}$ be a Riemannian manifold.
Let $U \subseteq M$ be a neighborhood of $p \in M$.
Then there is a uniformly normal subset which is a neighborhood of $p \in M$ and is contained in $U$.
== Proof ==
== Sources ==
* : $\S 6$: Geodesics and Distance. Uniformly Normal Neighborhoods
Category:Definitions/Riemannian Manifolds
Union Distributes over Intersection

== Theorem ==
Set union is distributive over set intersection:
:$R \cup \paren {S \cap T} = \paren {R \cup S} \cap \paren {R \cup T}$
=== Family of Sets ===
=== General Result ===
== Proof 1 ==
== Proof 2 ==
== Demonstration by Venn Diagram ==
== Examples ==
== Also known as ==
== Also see ==
* Intersection Distributes over Union
* Class Union Distributes over Class Intersection
== Sources ==
* : Chapter $0$: Subsets and Complements; Union and Intersection: Theorem $2 \ \text{(d)}$
* : $\S 4$: Unions and Intersections
* : A Note on Set-Theoretic Concepts: $(3)$
* : Exercise $1.1: \ 8 \ \text{(e)}$
* : Chapter $1$: Set Theory: $1.3$: Set operations
* : Chapter $\text I$: Algebraic Structures: $\S 3$: Unions and Intersections of Sets: Theorem $3.1$
* : Chapter $\text I$: Algebraic Structures: $\S 3$: Unions and Intersections of Sets: Exercise $3.2$
* : $\S 1.2$: Operations on sets: $(2)$
* : Chapter $1$: The Notation and Terminology of Set Theory: $\S 7 \ \text{(a)}$
* : $\S 1.6$: Set Identities and Other Set Relations: Exercise $2 \ \text{(g)}$
* : $\S 2$: Sets and functions: Operations
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: $\text{(d)}$
* : Chapter $1$: Theory of Sets: $\S 4$: Indexed Families of Sets: Exercise $2$
* : $\S 7.4 \ \text{(ii)}$: Unions and Intersections
* : Chapter $1$: Sets and mappings: $\S 1.2$: Sets: Exercise $1 \ \text{(viii)}$
* : $\S 1$: Naive Set Theory: $\S 1.2$: Operations on Sets: Exercise $1.2.1 \ \text{(v)}$
* : $1$: Set Theory and Logic: $\S 1$: Fundamental Concepts
* : $\S 2$
* : $\S 2$
* 
* : Appendix $\text{A}.2$: Theorem $\text{A}.10$
* : $\text {(vi)}$
Category:Union Distributes over Intersection
Category:Distributive Laws of Set Theory
Category:Set Union
Category:Set Intersection
Category:Examples of Distributive Operations
Union Distributes over Intersection/Also known as

== Union Distributes over Intersection: Also known as == 
The result Union Distributes over Intersection, along with Intersection Distributes over Union, are together known as:
:the distributive laws (of set theory)
:the distributive properties (of set theory).
== Sources ==
* : $1$: Set Theory and Logic: $\S 1$: Fundamental Concepts
Category:Union Distributes over Intersection
Union Distributes over Intersection/Examples/Arbitrary Integer Sets 1

== Example of Use of Union Distributes over Intersection ==
Let:
Then:
:$A \cup \paren {B \cap C} = \set {1, 2, 3, 4, 6, 8, \dotsc} = \paren {A \cup B} \cap \paren {A \cup C}$
== Proof ==
== Sources ==
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Exercise $1$
Category:Examples of Use of Union Distributes over Intersection
Union Distributes over Intersection/Examples/Arbitrary Integer Sets 2

== Example of Use of Union Distributes over Intersection ==
Let:
Then:
:$B \cup \paren {A \cap C} = \set {1, 2, 3, 4, 5, 7, \dotsc} = \paren {B \cup A} \cap \paren {B \cup C}$
== Proof ==
== Sources ==
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Exercise $1$
Category:Examples of Use of Union Distributes over Intersection
Union Distributes over Intersection/Family of Sets

== Theorem ==
Let $I$ be an indexing set.
Let $\family {A_\alpha}_{\alpha \mathop \in I}$ be an indexed family of subsets of a set $S$.
Let $B \subseteq S$.
Then:
:$\ds \map {\bigcap_{\alpha \mathop \in I} } {A_\alpha \cup B} = \paren {\bigcap_{\alpha \mathop \in I} A_\alpha} \cup B$
where $\ds \bigcap_{\alpha \mathop \in I} A_\alpha$ denotes the intersection of $\family {A_\alpha}_{\alpha \mathop \in I}$.
=== Corollary ===
== Proof ==
By definition of subset:
:$\ds \map {\bigcap_{\alpha \mathop \in I} } {A_\alpha \cup B} \subseteq \paren {\bigcap_{\alpha \mathop \in I} A_\alpha} \cup B$
By definition of subset:
:$\ds \paren {\bigcap_{\alpha \mathop \in I} A_\alpha} \cup B \subseteq \map {\bigcap_{\alpha \mathop \in I} } {A_\alpha \cup B}$
By definition of set equality:
:$\ds \map {\bigcap_{\alpha \mathop \in I} } {A_\alpha \cup B} = \paren {\bigcap_{\alpha \mathop \in I} A_\alpha} \cup B$
== Also see ==
* Intersection Distributes over Union of Family
== Sources ==
* : $\S 9$: Families
* : Chapter $1$: Theory of Sets: $\S 4$: Indexed Families of Sets: Exercise $1 \ \text{(f)}$
* : $\S 1$: Naive Set Theory: $\S 1.4$: Sets of Sets: Exercise $1.4.4 \ \text{(vi)}$
Category:Union Distributes over Intersection
Category:Indexed Families
Union Distributes over Intersection/Family of Sets/Corollary

== Theorem ==
Let $I$ and $J$ be indexing sets.
Let $\family {A_\alpha}_{\alpha \mathop \in I}$ and $\family {B_\beta}_{\beta \mathop \in J}$ be indexed families of subsets of a set $S$.
Then:
:$\ds \bigcap_{\tuple{\alpha, \beta} \mathop \in I \times J} \paren {A_\alpha \cup B_\beta} = \paren {\bigcap_{\alpha \mathop \in I} A_\alpha} \cup \paren {\bigcap_{\beta \mathop \in J} B_\beta}$
where $\ds \bigcap_{\alpha \mathop \in I} A_\alpha$ denotes the intersection of $\family {A_\alpha}_{\alpha \mathop \in I}$.
== Proof ==
== Sources ==
* : $\S 9$: Families
Category:Union Distributes over Intersection
Union Distributes over Intersection/General Result

== Theorem ==
Set union is distributive over set intersection:
Let $S$ and $T$ be sets.
Let $\powerset T$ be the power set of $T$.
Let $\mathbb T$ be a subset of $\powerset T$.
Then:
:$\ds S \cup \bigcap \mathbb T = \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$
== Proof ==
== Also see ==
* Intersection Distributes over Union/General Result
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Unions and Intersections: Theorem $2 \ \text{(ii)}$
Category:Union Distributes over Intersection
Union Distributes over Intersection/General Result/Proof

== Theorem ==
== Proof ==
=== Union Subset of Intersection ===
Let $\ds x \in S \cup \bigcap \mathbb T$.
We need to show that:
:$\ds x \in \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$
and then by definition of subset we will have shown that:
:$\ds S \cup \bigcap \mathbb T \subseteq \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$.
So, we have that $\ds x \in S \cup \bigcap \mathbb T$.
By definition of set union, $x \in S$ or $\ds x \in \bigcap \mathbb T$.
So there are two cases to consider:
$(1): \quad$ Suppose $x \in S$.
Then by definition of set union, $\forall X \in \mathbb T: x \in S \cup X$.
So:
:$\ds x \in \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$
$(2): \quad$ Suppose $\ds x \in \bigcap \mathbb T$.
Then by definition of set intersection:
:$\forall X \in \mathbb T: x \in X$
So by definition of set union:
:$\forall X \in \mathbb T: x \in S \cup X$
So:
:$\ds x \in \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$
In both cases we see that:
:$\ds x \in \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$
so by Proof by Cases, we have that:
:$\ds S \cup \bigcap \mathbb T \subseteq \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$
=== Intersection Subset of Union ===
Let $\ds x \in \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$.
We need to show that:
:$\ds x \in S \cup \bigcap \mathbb T$
and then by definition of subset we will have shown that:
:$\ds \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X} \subseteq S \cup \bigcap \mathbb T$.
So, we have that:
:$\ds x \in \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$
By definition of set intersection:
:$(A): \quad \forall X \in \mathbb T: x \in S \cup X$
There are two cases to consider:
:$(1): \quad \forall X \in \mathbb T: x \in X$
Then by definition of set intersection:
:$\ds x \in \bigcap_{X \mathop \in \mathbb T} X$
and so by definition of set union:
:$\ds x \in S \cup \bigcap \mathbb T$
:$(2): \quad \exists X \in \mathbb T: x \notin X$
From $(A)$ we have that $x \in S \cup X$.
But as $x \notin X$ it follows that $x \in S$.
Then by definition of set union:
:$\ds x \in S \cup \bigcap \mathbb T$
In both cases we see that:
:$\ds x \in S \cup \bigcap \mathbb T$
so by Proof by Cases, we have that:
:$\ds \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X} \subseteq S \cup \bigcap \mathbb T$
So we have that:
:$S \cup \ds \bigcap \mathbb T \subseteq \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$
and
:$\ds \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X} \subseteq S \cup \bigcap \mathbb T$
and so by definition of set equality:
:$S \cup \ds \bigcap \mathbb T = \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 3$: Unions and Intersections of Sets: Exercise $3.6 \ \text{(b)}$
* : Chapter $\text{I}$: Sets and Functions: Exercise $\text{C}$
Category:Union Distributes over Intersection
Union Distributes over Intersection/Proof 2

== Theorem ==
== Proof ==
From Intersection Distributes over Union:
:$R \cap \paren {S \cup T} = \paren {R \cap S} \cup \paren {R \cap T}$
From the Duality Principle for Sets, exchanging $\cup$ for $\cap$ throughout, and vice versa, reveals the result:
:$R \cup \paren {S \cap T} = \paren {R \cup S} \cap \paren {R \cup T}$
== Sources ==
* : $\S 1.5$. The algebra of sets: Example $18$
Category:Union Distributes over Intersection
Union Distributes over Intersection/Venn Diagram

== Theorem ==
== Proof ==
Demonstration by Venn diagram:
: 
In the left hand diagram, $R$ is depicted in blue and $S \cap T$ is depicted in yellow.
Their intersection, where they overlap, is depicted in green.
Their union $R \cup \paren {S \cap T}$ is the total shaded area: yellow, blue and green.
In the right hand diagram, $\paren {R \cup S}$ is depicted in yellow and $\paren {R \cup T}$ is depicted in blue.
Their intersection, where they overlap, is depicted in green.
As can be seen by inspection, the areas are the same.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 3$: Unions and Intersections of Sets: Exercise $3.2$
* : Chapter $1$: Elementary, my dear Watson: $\S 1.2$: Elements, my dear Watson: Ponderable $1.2.1 \ \text{(b)}$
Category:Union Distributes over Intersection
Category:Examples of Venn Diagrams