nyuuzyou/wikis huggingface dataset:
Serpent Climbing Out of a Well

== Classic Problem ==
A serpent lies at the bottom of a well $30$ length units deep.
It starts to climb out of the well.
Every day it climbs $\dfrac 2 3$ of a length unit, only to fall back $\dfrac 1 5$ of a length unit every night.
How long does it take to climb out of the well?
== Solution ==
:$63 \frac 9 {10}$ days.
== Proof ==
We have that:
:$\dfrac 2 3 - \dfrac 1 5 = \dfrac {10 - 3} {15} = \dfrac 7 {15}$
This is how far the serpent's efforts will take it after a day and a night.
The key point to note is that once the serpent has reached the top, it will not slip down again.
So the solution is more complicated than just calculating how long it will take to climb $30$ at $\dfrac 7 {15}$ per day.
We calculate how long it will take to climb to $30 - \dfrac 2 3$:
Now we examine the situation in detail.
At the end of $62$ days and nights, the serpent has climbed $62 \times \dfrac 7 {15} = 28 \frac {14} {15}$ from the top.
It has $1 \frac 1 {15}$ left to go.
After the end of the $63$rd day, it has climbed $28 \frac {14} {15} + \dfrac 2 3 = 29 \frac 3 5$.
At the end of the $63$rd night it has slipped back down to $29 \frac 3 5 - \dfrac 1 5 = 29 \frac 2 5$.
There remains $\dfrac 3 5$ of the way to climb.
This will take $\dfrac 3 5 \div \dfrac 2 3 = \dfrac 3 5 \times \dfrac 3 2 = \dfrac 9 {10}$ of the next day to reach the top.
Hence the result.
== Historical Note ==
== Sources ==
* : Serpent Climbing out of a Well: $92$
Category:Classic Problems
Set/Examples/Axiomatic Sets

== Example of Set ==
Sets constructed from axiomatic set theories vary depending on the formal system used. For example, some systems allow for urelements - elements that are not sets. Other systems, like ZFC only allow elements to be other sets or classes. Thus the natural numbers can be constructed from the empty set and axiom of infinity like so:
== Sources ==
* : $\S 4.6$ Axiomatic set theory: Other axiomatizations of set theory: Set theory with urelements
Category:Examples of Sets
Set/Examples/Positive Integers Less than 10

== Example of Set ==
The (strictly) positive integers less than $10$ form a set:
:$\set {1, 2, 3, 4, 5, 6, 7, 8, 9}$
== Sources ==
* : Chapter $1$: Basic Concepts: Lesson $2 \text A$: The Meaning of the Term Set
Category:Examples of Sets
Set Closure Preserves Set Inclusion

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $B \subseteq A \subseteq S$.
Then:
:$B^- \subseteq A^-$
where $A^-$ and $B^-$ are the set closures in $T$ of $A$ and $B$ respectively.
== Proof ==
By definition 1 of set closure:
:$B^- = B \cup B'$
where $B'$ is the derived set of $B$ in $T$. 
Similarly:
:$A^- = A \cup A'$
where $A'$ is the derived set of $A$ in $T$.
From Derived Set Preserves Set Inclusion:
:$B' \subseteq A'$
So, by Set Union Preserves Subsets:
:$B \cup B' \subseteq A \cup A'$
That is:
:$B^- \subseteq A^-$
Category:Set Closures
Set Closure as Intersection of Closed Sets

== Theorem ==
Let $T$ be a topological space.
Let $H \subseteq T$.
Let the closure of $H$ (in $T$) be defined as:
:$H^- := H \cup H'$
where $H'$ is the derived set of $H$.
Let $\mathbb K$ be defined as:
:$\mathbb K := \leftset {K \supseteq H: K}$ is closed in $\rightset T$
That is, let $\mathbb K$ be the set of all closed sets of $T$ which contain $H$.
Then the closure of $H$ (in $T$) can be defined as:
:$\ds H^- := \bigcap \mathbb K$
that is, as the intersection of all the closed sets of $T$ which contain $H$.
== Proof ==
What needs to be proved here is:
:$\ds H^- = \bigcap \mathbb K$
where:
:$H^- = H \cup H'$
:$H'$ denotes the set of all limit points of $H$
:$\ds \bigcap \mathbb K$ is the intersection of all closed sets of $T$ which contain $H$.
Let $K \in \mathbb K$.
From Topological Closure of Subset is Subset of Topological Closure, we have:
:$H^- \subseteq K^-$
Thus from Closed Set Equals its Closure, we have:
:$K^- = K$
Thus:
:$\ds \forall K \in \bigcap \mathbb K: H^- \subseteq K$
since the choice of $K$ is arbitrary. 
So from Intersection is Largest Subset:
:$\ds H^- \subseteq \bigcap \mathbb K$
Conversely, from Topological Closure is Closed, $H^-$ is closed.
From Set is Subset of its Topological Closure, $H \subseteq H^-$.
So $H^-$ is, by definition, a closed set of $T$ which contains $H$.
But we have by its definition that $\ds \bigcap \mathbb K$ is the intersection of all closed sets in $T$ that contain $H$.
So from Intersection is Subset it follows that:
:$\ds \bigcap \mathbb K \subseteq H^-$
Finally, we have that:
:$\ds H^- \subseteq \bigcap \mathbb K$
:$\ds \bigcap \mathbb K \subseteq H^-$
So by definition of set equality:
:$\ds H^- = \bigcap \mathbb K$
which is what we needed to prove.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Proposition $3.7.16$
Category:Set Closures
Set Closure is Smallest Closed Set/Closure Operator

== Theorem ==
Let $S$ be a set.
Let $\cl: \powerset S \to \powerset S$ be a closure operator.
Let $T \subseteq S$.
Then $\map \cl T$ is the smallest closed set (with respect to $\cl$) containing $T$ as a subset.
== Proof ==
By definition, $\map \cl T$ is closed.
Let $C$ be closed.
Let $T \subseteq C$.
By the definition of closure operator, $\cl$ is $\subseteq$-increasing.
So:
:$\map \cl T \subseteq \map \cl C$
Since $C$ is closed, $\map \cl C = C$.
So:
:$\map \cl T \subseteq C$
Thus $\map \cl T$ is the smallest closed set containing $T$ as a subset.
Category:Closure Operators
Category:Set Closure is Smallest Closed Set
Set Closure is Smallest Closed Set/Normed Vector Space

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Let $S$ be a subset of $X$:
:$S \subseteq X$
Let $S^-$ be the closure of $S$.
Then $S^-$ is the smallest closed set which contains $S$.
== Proof ==
Let $F$ be a closed set in $X$.
Suppose $S \subseteq F$.
=== $S^-$ is contained in $F$===
Let $L$ be a limit point of $S$.
Then there exists a sequence $\sequence {x_n}_{n \mathop \in \N}$ in $S \setminus \set L$ which converges to $L$.
In other words:
:$\forall n \in \N : x_n \in S \setminus \set L$.
Furthermore:
:$S \setminus \set L \subseteq S \subseteq F$
Since $F$ is closed, $L \in F$.
So all limit points of $S$ belong to $F$.
Hence, $S^- \subseteq F$.
=== $S^-$ is closed ===
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $S^-$.
Let $\sequence {x_n}_{n \mathop \in \N}$ converge to $L$:
:$\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall n \in \N: n > N \implies \norm {x_n - L} < \epsilon$
We can have either $L \in S$ or $L \notin S$.
Suppose $L \in S$.
:By definition of closure, $L \in S^-$.
Suppose $L \notin S$.
Define a new sequence $x_n'$ using $x_n$ as follows:
:$(1): \quad$ if $x_n \in S$, then $x_n' := x_n$;
:$(2): \quad$ if $x_n \notin S$, then $x_n$ is a limit point of $S$.
::Then there is open ball $\ds \map {B_{\frac 1 n}} {x_n}$ which has an element of $S$.
::Define $x_n'$ such that $x_n' \in S$ and $x_n' \in \map {B_{\frac 1 n}} {x_n}$
Suppose $x_n \in S$.
Then:
:$\norm {x_n' - L} = \norm {x_n - L}$
Suppose $x_n \notin S$.
Then:
Thus, $\sequence {x_n'}_{n \mathop \in \N}$ is a sequence in $S \setminus \set L$ which converges to $L$.
So $L$ is a limit point of $S$:
:$L \in S^-$.
By definition, $S^-$ is closed.
=== $S^-$ is the smallest closed set containing $S$ ===
 there exists a closed set $Q$ smaller than $S^-$ which contains $S$.
$S^-$ differs from $S$ only by limit points of $S$.
If $Q$ is smaller than $S^-$, it has to contain fewer limit points of $S$ than $S^-$.
Hence, $Q$ would not contain all its limit points.
By definition, $Q$ would not be closed.
This is a contradiction.
== Sources ==
* : $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces
Category:Closed Sets
Category:Normed Vector Spaces
Category:Set Closure is Smallest Closed Set
Set Closure is Smallest Closed Set/Topology

== Theorem ==
Let $T$ be a topological space.
Let $H \subseteq T$.
Let $H^-$ denote the closure of $H$ in $T$.
Then $H^-$ is the smallest superset of $H$ that is closed in $T$.
== Proof ==
Define:
:$\mathbb K := \leftset {K \supseteq H: K}$ is closed in $\rightset T$
That is, let $\mathbb K$ be the set of all supersets of $H$ that are closed in $T$.
The claim is that $H^-$ is the smallest set of $\mathbb K$.
From Set is Subset of its Topological Closure:
:$H \subseteq H^-$
From Topological Closure is Closed, $H^-$ is closed in $T$.
Thus $H^- \in \mathbb K$.
Let $K \in \mathbb K$.
From Set Closure as Intersection of Closed Sets:
:$\ds H^- = \bigcap \mathbb K$
Therefore, from Intersection is Subset: General Result:
:$H^- \subseteq K$
Thus by definition $H^-$ is the smallest set of $\mathbb K$.
== Sources ==
* : $2.27 c$
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Closures and Interiors
Category:Set Closures
Category:Set Closure is Smallest Closed Set
Set Coarser than Upper Section is Subset

== Theorem ==
Let $\struct {S, \preceq}$ be a preordered set.
Let $A, B$ be subsets of $S$ such that
:$A$ is coarser than $B$
and
:$B$ is an upper section.
Then:
:$A \subseteq B$
== Proof ==
Let $x \in A$.
By definition of coarser subset:
:$\exists y \in B: y \preceq x$
Thus by definition of upper section:
:$x \in B$
== Sources ==
* 
* 
Category:Preorder Theory
Category:Upper Sections
Set Complement/Examples/Positive Real Numbers in Complex Numbers

== Example of Set Complement ==
Let the universal set $\Bbb U$ be defined to be the set of real numbers $\C$.
Let the set of (strictly) positive real numbers be denoted by $\R_{>0}$.
Then:
:$\relcomp {} {\R_{>0} } = \set {x + i y: y \ne 0 \text { or } x \le 0}$
== Proof ==
== Sources ==
* : Introduction: Set-Theoretic Notation
Category:Examples of Set Complement
Set Complement/Examples/Positive Real Numbers in Real Numbers

== Example of Set Complement ==
Let the universal set $\Bbb U$ be defined to be the set of real numbers $\R$.
Let the set of (strictly) positive real numbers be denoted by $\R_{>0}$.
Then:
:$\relcomp {} {\R_{>0} } = \R_{\le 0}$
the set of non-negative real numbers.
== Proof ==
== Sources ==
* : Introduction: Set-Theoretic Notation
Category:Examples of Set Complement
Set Complement inverts Subsets

== Theorem ==
Let $S$ and $T$ be sets.
Then:
:$S \subseteq T \iff \map \complement T \subseteq \map \complement S$
where:
:$S \subseteq T$ denotes that $S$ is a subset of $T$
:$\complement$ denotes set complement.
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Sources ==
* : Chapter $0$: Subsets and Complements; Union and Intersection: Theorem $1$
* : $\S 5$: Complements and Powers
* : Chapter $\text{I}$: Sets and Functions: Exercise $\text{B x}$
* : $\S 1.6$: Set Identities and Other Set Relations: Exercise $1 \ \text{(b)}$
* : Chapter $1$: Theory of Sets: $\S 3$: Set Operations: Union, Intersection and Complement: Exercise $1 \ \text{(e)}$
Category:Subsets
Category:Set Complement
Category:Set Complement inverts Subsets
Set Complement inverts Subsets/Corollary

== Corollary to Set Complement inverts Subsets ==
Let $S$ and $T$ be sets.
Then:
:$S \subseteq \map \complement T \iff T \subseteq \map \complement S$
where:
:$S \subseteq \map \complement T$ denotes that $S$ is a subset of the set complement of $T$.
== Proof ==
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 3$: Set Operations: Union, Intersection and Complement: Exercise $1 \ \text{(f)}$
Category:Subsets
Category:Set Complement inverts Subsets
Set Complement inverts Subsets/Proof 3

== Theorem ==
== Proof ==
By definition of set complement:
:$\map \complement T := \relcomp {\mathbb U} T$
where:
:$\mathbb U$ is the universal set
:$\relcomp {\mathbb U} T$ denotes the complement of $T$ relative to $\mathbb U$.
Thus the statement can be expressed as:
:$S \subseteq T \iff \relcomp {\mathbb U} T \subseteq \relcomp {\mathbb U} S$
The result then follows from Relative Complement inverts Subsets.
Category:Set Complement inverts Subsets
Set Consisting of Empty Set is not Empty

== Theorem ==
Let $S$ be the set defined as:
:$S = \set \O$
Then $S$ is not the empty set.
That is:
:$\O \ne \set \O$
== Proof ==
We have:
:$\O \in \set \O$
and so:
:$\neg \paren {\forall x: x \notin \O}$
The result follows by definition of the empty set.
== Sources ==
* : $\S 2$. Sets of sets
* : Part $1$: Not Entirely Naive Set Theory: Chapter $1$: Pairs, Relations, and Functions
Category:Empty Set
Set Contained in Smallest Transitive Set

== Theorem ==
Let $S$ be a set.
Then there exists a transitive set $G$ such that:
:$S \subseteq G$
and:
:if $Q$ is any transitive set such that $S \subseteq Q$, then $G \subseteq Q$.
== Proof ==
=== Construction of $G$ ===
Let $U$ be the class of all sets.
Define the mapping $F: \N \to U$ recursively:
:$\map F 0 = S$
:$\map F {n + 1} = \bigcup \map F n$
Applying the axiom of union inductively, $\map F n$ is a set for each $n \in \N$.
Let $\ds G = \bigcup_{i \mathop = 0}^\infty \map F i$.
By the , $G$ is a set.
=== Transitivity ===
It is to be proved that $G$ is transitive.
That is:
:$a \in b, b \in G \implies a \in G$
Let $a \in b$ and $b \in G$.
By the definition of $G$, there exists $n \in \N$, $b \in \map F n$.
By the definition of $F$:
:$\map F {n + 1} = \bigcup \map F n$
Then by the definition of union:
:$a \in \map F {n + 1}$
Thus by the definition of $G$:
:$a \in G$
=== Minimality ===
It is to be proved that if $Q$ is a transitive set and $S \subseteq Q$ then $G \subseteq Q$.
Let $Q$ be transitive and $S \subseteq Q$.
Define $F$ as above.
Prove by induction that $\map F n \subseteq Q$ for each $n$.
Set Containing Only Zero is Primitive Recursive

== Theorem ==
The subset $\left\{{0}\right\} \subset \N$ is primitive recursive.
== Proof ==
We note that:
:$1 \mathop {\dot -} n = \begin{cases}
1 & : n = 0 \\
0 & : n > 0
\end{cases}$
and so the characteristic function $\chi_{\left\{{0}\right\}}$ is given by $\chi_{\left\{{0}\right\}} \left({n}\right) = 1 \mathop {\dot -} n$.
So $\chi_{\left\{{0}\right\}}$ is obtained by substitution from the primitive recursive function $1 \mathop {\dot -} n$ using constants, which are primitive recursive.
Hence the result.
Category:Primitive Recursive Functions
Set Definition by Predicate/Examples

== Examples of Set Definition by Predicate ==
=== Men ===
=== University Professors ===
=== Musical Mathematicians ===
=== Set of Integers $x$ such that $2 \le x$ ===
=== Set of Integers $x$ such that $x \le 5$ ===
=== Set Indexed by Natural Numbers between $1$ and $100$ ===
=== Set Indexed by Natural Numbers between $1$ and $10$ ===
=== Even Integers ===
=== People who Love Romeo ===
=== Cube Numbers ===
=== Sums of Two Squares ===
=== United States Senate ===
Category:Examples of Sets
Set Definition by Predicate/Examples/Cube Numbers

== Examples of Set Definition by Predicate ==
Let $C$ be the set defined as:
:$C := \set {n: n = k^3 \text { and } k = 1, 2, \dotsc}$
Then $C$ is the set of all cube numbers.
== Sources ==
* : Introduction to Set Theory: $1$. Elementary Operations on Sets
Category:Examples of Sets
Set Definition by Predicate/Examples/Even Integers

== Examples of Set Definition by Predicate ==
Let $S$ be the set defined as:
:$S := \set {i: \text {$i$ is an integer and there exists an integer $j$ such that $i = 2 j$} }$
Then $S$ is the set of all even integers.
== Sources ==
* : Chapter $1$: Preliminaries: $1.4$ Set Notation
Category:Examples of Sets
Set Definition by Predicate/Examples/Integers Greater Than or Equal to 2

== Example of Set Definition by Predicate ==
Let $S$ be the set defined as:
:$S := \set {x \in \Z: 2 \le x}$
Then $S$ is the set of all integers greater than or equal to $2$:
:$S = \set {2, 3, 4, \ldots}$
== Sources ==
* : Chapter $1$. Sets: Exercise $1$
Category:Examples of Sets
Set Definition by Predicate/Examples/Integers Less Than or Equal to 5

== Examples of Set Definition by Predicate ==
Let $S$ be the set defined as:
:$S := \set {x \in \Z: x \le 5}$
Then $S$ is the set of all integers less than or equal to $5$:
:$S = \set {\ldots, 2, 3, 4, 5}$
== Sources ==
* : Chapter $1$. Sets: Exercise $1$
Category:Examples of Sets
Set Definition by Predicate/Examples/Musical Mathematicians

== Example of Set Definition by Predicate ==
Let $M$ denote the set of all the mathematicians in the world.
Let $I$ denote the set of all people who can play a musical instrument.
Let $S$ denote the set of all mathematicians who can play a musical instrument.
Then we can define $S$ as:
:$S := \set {x: x \in M \text { and } x \in I}$
or as:
:$S := \set {x \in M: x \in I}$
== Sources ==
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems
Category:Examples of Sets
Set Definition by Predicate/Examples/Set Indexed by Natural Numbers between 1 and 10

== Examples of Set Definition by Predicate ==
Let $U$ be the set defined as:
:$U := \set {u_i: 1 < i < 10, i \in \N}$
Then $U$ has exactly $8$ elements:
:$U = \set {u_2, u_3, u_4, u_5, u_6, u_7, u_8, u_9}$
== Sources ==
* : Appendix $\text{A}.1$: Sets and Subsets: Problem Set $\text{A}.1$: $1$
Category:Examples of Sets
Set Definition by Predicate/Examples/Set Indexed by Natural Numbers between 1 and 100

== Examples of Set Definition by Predicate ==
Let $V$ be the set defined as:
:$V := \set {v_i: 1 \le i \le 100, i \in \N}$
Then $V$ is the set of the $100$ elements:
:$V = \set {v_1, v_2, \ldots, v_{100} }$
and can also be written:
:$V := \set {v_i: i = 1, 2, \ldots, 100}$
or even:
:$V := \set {v_i: 1 \le i \le 100}$
as it is understood that the domain of $i$ is the set of natural numbers.
== Sources ==
* : Appendix $\text{A}.1$: Sets and Subsets
Category:Examples of Sets
Set Definition by Predicate/Examples/Sums of Two Squares

== Examples of Set Definition by Predicate ==
Let $S$ be the set defined as:
:$S := \set {n: n = x^2 + y^2 \text { and } x, y \in \Z}$
The $C$ is the set of all integers which can be written as the sum of two squares.
== Sources ==
* : Introduction to Set Theory: $1$. Elementary Operations on Sets
Category:Examples of Sets
Set Difference/Examples/1, 2, 3 less 2, 4, 5, 6

== Example of Set Difference ==
Let $S$ and $T$ be sets such that:
:$S = \set {1, 2, 3}$
:$T = \set {2, 4, 5, 6}$
Let $\setminus$ denote set difference.
Then:
:$S \setminus T = \set {1, 3}$
while:
:$T \setminus S = \set {4, 5, 6}$
It can immediately be seen that $\setminus$ is not commutative.
== Also see ==
* Set Difference is Anticommutative
== Sources ==
* : $\S 6$: Subsets
Category:Examples of Set Difference
Set Difference/Examples/Arbitrary Example 2

== Example of Set Difference ==
Let $S$ and $T$ be sets such that:
:$A = \set {1, 2, 3}$
:$B = \set {1, 2, 4, 5}$
Let $\setminus$ denote set difference.
Then:
:$A \setminus B = \set 3$
== Sources ==
* 
* 
Category:Examples of Set Difference
Set Difference Equals First Set iff Empty Intersection

== Theorem ==
:$S \setminus T = S \iff S \cap T = \O$
== Proof ==
Assume $S, T \subseteq \Bbb U$ where $\Bbb U$ is a universal set.
== Sources ==
* : Chapter $1$: The Notation and Terminology of Set Theory: $\S 8 \ \text{(b)}$
* : $\S 1.5$: Complementation
Category:Set Difference
Category:Set Intersection
Category:Empty Set
Set Difference Intersection with First Set is Set Difference

== Theorem ==
The intersection of the set difference with the first set is the set difference.
Let $S, T$ be sets.
Then:
:$\paren {S \setminus T} \cap S = S \setminus T$
== Proof 1 ==
== Proof 2 ==
Category:Set Difference
Category:Set Intersection
Category:Set Difference Intersection with First Set is Set Difference
Set Difference Intersection with Second Set is Empty Set

== Theorem ==
Let $S$ and $T$ be sets.
The intersection of the set difference of $S$ and $T$ with $T$ is the empty set:
:$\paren {S \setminus T} \cap T = \O$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 1.1$: Theorem $1.5$
* : $\S 1$: Naive Set Theory: $\S 1.2$: Operations on Sets: Exercise $1.2.5 \ \text{(iii)}$
Category:Set Difference
Category:Set Intersection
Category:Set Difference Intersection with Second Set is Empty Set
Set Difference Then Union Equals Union Then Set Difference

== Theorem ==
Let $S, A, B$ be sets.
Let $A \cap B = \O$.
Then:
:$\paren {S \setminus A} \cup B = \paren {S \cup B} \setminus A$
=== Corollary ===
== Proof ==
We have:
Category:Set Difference
Category:Set Union
Set Difference Then Union Equals Union Then Set Difference/Corollary

== Corollary of Set Difference Then Union Equals Union Then Set Difference ==
Let $S, T$ be sets.
Let $A \subseteq S \setminus T$.
Let $B \subseteq T \setminus S$.
Then:
:$\paren{S \setminus A} \cup B = \paren{S \cup B} \setminus A$
== Proof ==
From Set Difference is Disjoint with Reverse:
:$\paren{S \setminus T} \cap \paren{T \setminus S} = \O$
From Subsets of Disjoint Sets are Disjoint:
:$A \cap B = \O$
From Set Difference Then Union Equals Union Then Set Difference:
:$\paren{S \setminus A} \cup B = \paren{S \cup B} \setminus A$
Category:Set Difference
Category:Set Union
Set Difference Union First Set is First Set

== Theorem ==
The union of a set difference with the first set is the set itself:
Let $S, T$ be sets.
Then:
:$\paren {S \setminus T} \cup S = S$
== Proof ==
Consider $S, T \subseteq \mathbb U$, where $\mathbb U$ is considered as the universal set.
Category:Set Difference
Category:Set Union
Set Difference Union Intersection

== Theorem ==
:$S = \paren {S \setminus T} \cup \paren {S \cap T}$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Exercise $\text{B ix}$
* : $1$: Set Theory and Logic: $\S 1$: Fundamental Concepts: Exercise $1.2 \ \text{(i)}$
* : Appendix $\text{A}.2$: Theorem $\text{A}.11$
Category:Set Difference
Category:Set Union
Category:Set Intersection
Category:Set Difference Union Intersection
Set Difference Union Intersection/Proof 3

== Theorem ==
== Proof ==
By Set Difference is Subset:
:$S \setminus T \subseteq S$
By Intersection is Subset:
:$S \cap T \subseteq S$
Hence from Union is Smallest Superset:
:$\paren {S \setminus T} \cup \paren {S \cap T} \subseteq S$
Let $s \in S$.
Either:
:$s \in T$, in which case $s \in S \cap T$ by definition of set intersection
or
:$s \notin T$, in which case $s \in S \setminus T$ by definition of set difference.
That is, by definition of set union:
:$s \in \paren {S \setminus T} \cup \paren {S \cap T}$
and so by definition of subset:
:$S \subseteq \paren {S \setminus T} \cup \paren {S \cap T}$
Hence the result by definition of set equality.
== Sources ==
* : $\S 1.1$: Theorem $1.2$
Category:Set Difference Union Intersection
Set Difference Union Second Set is Union

== Theorem ==
The union of a set difference with the second set is the union of the two sets.
That is, let $S, T$ be sets.
Then:
:$\paren {S \setminus T} \cup T = S \cup T$
== Proof ==
Consider $S, T \subseteq \mathbb U$, where $\mathbb U$ is considered as the universal set.
== Sources ==
* : $\S 1.1$: Theorem $1.3$
* : $\S 1$: Sets and Functions: Problem $2$
:: This problem is worded: Show that in general $A - B \cup B \ne A$.
Category:Set Difference
Category:Set Union
Set Difference and Intersection are Disjoint

== Theorem ==
Let $S$ and $T$ be sets.
Then: 
:$S \setminus T$ and $S \cap T$ are disjoint 
where $S \setminus T$ denotes set difference and $S \cap T$ denotes set intersection.
== Proof ==
From Set Difference Intersection with Second Set is Empty Set:
:$\paren {S \setminus T} \cap T = \O$
and hence immediately from Intersection with Empty Set:
:$\paren {S \setminus T} \cap \paren {S \cap T} = \O$
So $S \setminus T$ and $S \cap T$ are disjoint.
Category:Set Difference
Category:Set Intersection
Set Difference and Intersection form Partition

== Theorem ==
Let $S$ and $T$ be sets such that:
:$S \setminus T \ne \O$
:$S \cap T \ne \O$
where $S \setminus T$ denotes set difference and $S \cap T$ denotes set intersection.
Then $S \setminus T$ and $S \cap T$ form a partition of $S$.
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
From Set Difference and Intersection are Disjoint:
:$S \setminus T$ and $S \cap T$ are disjoint.
Next from Set Difference Union Intersection:
:$S = \paren {S \setminus T} \cup \paren {S \cap T}$
Thus by definition $S \setminus T$ and $S \cap T$ form a partition of $S$.
Category:Set Difference
Category:Set Intersection
Category:Set Partitions
Category:Set Difference and Intersection form Partition