nyuuzyou/wikis huggingface dataset:
Closed Form for Number of Derangements on Finite Set/Proof 1

== Theorem ==
== Proof ==
Let $s_i$ be the $i$th element of set $S$.
Begin by defining set $A_m$, which is all of the permutations of $S$ which fixes $S_m$.
Then the number of permutations, $W$, with at least one element fixed, $m$, is:
:$\ds W = \size {\bigcup_{m \mathop = 1}^n A_m}$
Applying the Inclusion-Exclusion Principle:
Each value $A_{m_1} \cap \cdots \cap A_{m_p}$ represents the set of permutations which fix $p$ values $m_1, \ldots, m_p$.
Note that the number of permutations which fix $p$ values only depends on $p$, not on the particular values of $m$.
Thus from Cardinality of Set of Subsets there are $\dbinom n p$ terms in each summation. 
So:
$\size {A_1 \cap \cdots \cap A_p}$ is the number of permutations fixing $p$ elements in position.
This is equal to the number of permutations which rearrange the remaining $n - p$ elements, which is $\paren {n - p}!$.
Thus we finally get: 
:$W = \dbinom n 1 \paren {n - 1}! - \dbinom n 2 \paren {n - 2}! + \dbinom n 3 \paren {n - 3}! - \cdots + \paren {-1}^{p - 1} \dbinom n p \paren {n - p}! \cdots$
That is:
:$\ds W = \sum_{p \mathop = 1}^n \paren {-1}^{p - 1} \binom n p \paren {n - p}!$
Noting that $\dbinom n p = \dfrac {n!} {p! \paren {n - p}!}$, this reduces to:
:$\ds W = \sum_{p \mathop = 1}^n \paren {-1}^{p - 1} \dfrac {n!} {p!}$ 
== Historical Note ==
Category:Closed Form for Number of Derangements on Finite Set
Closed Form for Number of Derangements on Finite Set/Proof 2

== Theorem ==
== Proof ==
A derangement is a permutation that fixes no elements.
So, to compute the total number of derangements, we compute the number of permutations that fix elements, that is leave elements in the same place.
Let $\pi$ be a permutation of $S$.
Let $x, y \in S$.
Let $i$ be a position in a given permutation of set $S$, where $1 \le i \le n$.
Let $\map {P_i} \pi$ be the property that $\map \pi x = y$ at position $i$.
Let $\map N {P_i, \dots, P_j}$ be the number of permutations where $\map {P_i} \pi, \dots, \map {P_j} \pi$ hold.
If we consider all the permutations of a set where the property $P_i$ holds, we have fixed a single element of the set.
So there are $\paren {n - 1}!$ such possible permutations.
Furthermore, we could have chosen $i$ as any element in the set so there are $\dbinom n 1$ terms to choose from.
So $\map N {P_i} = \paren {n - 1}!$ and there are $\dbinom n 1$ terms.
Similarly, $\map N {P_i, P_j} = \paren {n - 2}!$ and there are $\dbinom n 2$ terms.
So now we can compute the total number of derangements:
== Historical Note ==
Category:Closed Form for Number of Derangements on Finite Set
Closed Form for Octagonal Numbers

== Theorem ==
The closed-form expression for the $n$th octagonal number is:
:$O_n = n \paren {3 n - 2}$
== Proof ==
Octagonal numbers are $k$-gonal numbers where $k = 8$.
From Closed Form for Polygonal Numbers we have that:
:$\map P {k, n} = \dfrac n 2 \paren {\paren {k - 2} n - k + 4}$
Hence:
Hence the result.
== Sources ==
* : $8$
Category:Octagonal Numbers
Category:Closed Forms
Closed Form for Pentagonal Numbers

== Theorem ==
The closed-form expression for the $n$th pentagonal number is:
:$P_n = \dfrac {n \paren {3 n - 1} } 2$
== Proof ==
Pentagonal numbers are $k$-gonal numbers where $k = 5$.
From Closed Form for Polygonal Numbers we have that:
:$\map P {k, n} = \dfrac n 2 \paren {\paren {k - 2} n - k + 4}$
Hence:
Hence the result.
== Sources ==
* : $22$
* : $22$
* 
* 
Category:Pentagonal Numbers
Category:Closed Forms
Closed Form for Pentatope Numbers

== Theorem ==
The closed-form expression for the $n$th pentatope number is:
:$P_n = \dfrac {n \paren {n + 1} \paren {n + 2} \paren {n + 3} } {24}$
== Proof ==
== Sources ==
* : $56$
* : $56$
Category:Closed Forms
Category:Pentatope Numbers
Closed Form for Polygonal Numbers

== Theorem ==
Let $\map P {k, n}$ be the $n$th $k$-gonal number.
The closed-form expression for $\map P {k, n}$ is given by:
== Proof ==
By definition of the $n$th $k$-gonal number:
$\map P {k, n} = \begin{cases}
0 & : n = 0 \\
\map P {k, n - 1} + \paren {k - 2} \paren {n - 1} + 1 & : n > 0
\end{cases}$
Then:
:$\paren {\paren {k - 2} \paren {j - 1} + 1}$
is an arithmetic sequence.
Its initial term $a$ is $1$, and its common difference $d$ is $k - 2$.
Hence:
as required.
== Examples ==
== Sources ==
* : $45$
* 
* : $45$
Category:Polygonal Numbers
Category:Closed Forms
Closed Form for Polygonal Numbers/Examples

== Examples of Closed Form for Polygonal Numbers ==
Let $\map P {k, n}$ be the $n$th $k$-gonal number.
The closed-form expression for $\map P {k, n}$ for various $k$ can be expressed as:
and so on.
== Sources ==
* : $45$
* : $45$
Category:Polygonal Numbers
Category:Closed Forms
Closed Form for Sequence 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, ...

== Theorem ==
Let $a_1, a_2, a_3, \ldots$ be the integer sequence:
:$\sequence {a_n} = 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, \ldots$
Then:
:$a_n = \ceiling {\dfrac {\sqrt {1 + 8 n} - 1} 2}$
== Proof ==
From Closed Form for Triangular Numbers, for $n = 1, 3, 6, 10, \ldots$:
:$n = \dfrac {a_n \paren {a_n + 1} } 2$
Thus by the Quadratic Formula:
$a_n = \dfrac {-1 \pm \sqrt {1 + 8 n} } 2$
In this context it is the positive root that is required.
The result follows by definition of ceiling function.
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $41$
Category:Sequences
Category:Ceiling Function
Category:Closed Forms
Closed Form for Square Pyramidal Numbers

== Theorem ==
The closed-form expression for the $n$th square pyramidal number is:
:$S_n = \dfrac {n \paren {n + 1} \paren {2 n + 1} } 6$
== Proof ==
== Sources ==
* : $55$
* : $55$
Category:Closed Forms
Category:Pyramidal Numbers
Closed Form for Tetrahedral Numbers

== Theorem ==
The closed-form expression for the $n$th tetrahedral number is:
:$H_n = \dfrac {n \paren {n + 1} \paren {n + 2} } 6$
== Proof ==
== Sources ==
* : $56$
* : $56$
Category:Closed Forms
Category:Tetrahedral Numbers
Closed Form for Triangular Numbers

== Theorem ==
The closed-form expression for the $n$th triangular number is:
:$\ds T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$
Plainly stated: the sum of the first $n$ natural numbers is equal to $\dfrac {n \paren {n + 1} } 2$.
This formula pops up frequently in fields as differing as calculus and computer science, and it is elegant in its simplicity.
== Direct Proof ==
== Proof by Telescoping Sum ==
== Proof by Recursion ==
== Proof by Induction ==
== Proof by Arithmetic Sequence ==
== Proof by Polygonal Numbers ==
== Proof using Binomial Coefficients ==
== Proof using Cardinality of Set ==
== Proof using Odd Number Theorem ==
== Proof using Bernoulli Numbers ==
== Also presented as ==
== Also see ==
* Compare Integral of Power for $n = 1$:
::$\ds \int_0^b x \rd x = \frac {b^2} 2$
== Historical Note ==
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 18$: Induced $N$-ary Operations: Exercise $18.10 \ \text{(a)}$
* : $\S 19$: Arithmetic Series: $19.2$
* : $\text {1-1}$ Principle of Mathematical Induction: $\text{1-1-1}$
* : $\text {1-1}$ Principle of Mathematical Induction: Theorem $\text{1-1}$
* : $\S 1.17$: Finite Induction and Well-Ordering for Positive Integers: Exercise $2$
* : Appendix $\text{A}.6$: Mathematical Induction: Problem Set $\text{A}.6$: $35$
* : Chapter $1$: Some Preliminary Considerations: $1.3$ Early Number Theory: Problems $1.3$: $1 \ \text {(a)}$
* : Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers
* : Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers: Exercise $9$
* : $15$
* : Chapter $\text {A}.5$: Archimedes (ca. B.C.)
* : $15$
* : $\S 21$: Series of Constants: Arithmetic Series: $21.2.$
* 
Category:Triangular Numbers
Category:Sums of Sequences
Category:Closed Forms
Category:Closed Form for Triangular Numbers
Closed Form for Triangular Numbers/Also presented as

== Closed Form for Triangular Numbers: Also presented as ==
The closed-form expression for the $n$th triangular number can also be presented as:
:$\ds T_n = \sum_{i \mathop = 0}^n i = \frac {n \paren {n + 1} } 2$
This is seen to be equivalent to the given form by the fact that the first term $T_0$ evaluates as $\dfrac {0 \paren {0 + 1} } 2$ which is zero.
== Sources ==
* : Chapter $1$: Preliminaries: Exercises: $1.2 \ \text {a)}$
* : Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers
* : $\S 1.2.6$: Binomial Coefficients
Category:Closed Form for Triangular Numbers
Closed Form for Triangular Numbers/Combinatorial Proof

== Theorem ==
== Combinatorial Proof ==
Suppose we have $n + 1$ people who all shake hands with each other person once then we can count how many handshakes occur by counting the handshakes of each person, the first person shakes hands with all other $n$ people, then the second person shakes hands with the remaining $n - 1$ people, which continues until the $n$-th person can only shake hands with the $(n + 1)$-th person, so counting backwards this consists of $\sum_{i = 0}^n i$ handshakes. Similarly, we can count the number of choices of handshakes between any two of the $n + 1$ people, which will be ${n + 1 \choose 2} = \frac{n(n + 1)}{2}$ handshakes, which is our proposed solution to the sum.
Category:Closed Form for Triangular Numbers
Closed Form for Triangular Numbers/Direct Proof

== Theorem ==
== Proof ==
We have that:
:$\ds \sum_{i \mathop = 1}^n i = 1 + 2 + \cdots + n$
Consider $\ds 2 \sum_{i \mathop = 1}^n i$.
Then:
So:
== Historical Note ==
== Sources ==
* : $\S 1.4.2$: Mathematical induction
Category:Direct Proofs
Category:Closed Form for Triangular Numbers
Closed Form for Triangular Numbers/Direct Proof/Historical Note

== Historical Note on Closed Form for Triangular Numbers: Direct Proof ==
This direct proof is the method supposedly employed by who, when very young (according to the apocryphal story), calculated the sum of the numbers from $1$ to $100$ before the teacher had barely sat back down after setting the assignment.
Whether this story is actually true or not is the subject of speculation.
== Sources ==
* : $\S 1.4.2$: Mathematical induction
Category:Historical Notes
Closed Form for Triangular Numbers/Historical Note

== Historical Note on Closed Form for Triangular Numbers ==
Some sources report that the closed-form expression for the $n$th triangular number was proved by during the course of his proofs of the volumes of various solids of revolution in his .
Others suggest that it was known as far back as , circa $550$ B.C.E.
== Sources ==
* : Chapter $1$: Some Preliminary Considerations: $1.3$ Early Number Theory: Problems $1.3$: $1 \ \text {(a)}$
* : Chapter $\text {A}.5$: Archimedes (ca. B.C.)
Category:Historical Notes
Closed Form for Triangular Numbers/Proof by Induction

== Theorem ==
== Proof ==
Proof by induction:
=== Basis for the Induction ===
When $n = 1$, we have:
:$\ds \sum_{i \mathop = 1}^1 i = 1$
Also:
:$\dfrac {n \paren {n + 1} } 2 = \dfrac {1 \cdot 2} 2 = 1$
This is our base case.
=== Induction Hypothesis ===
:$\forall k \in \N: k \ge 1: \ds \sum_{i \mathop = 1}^k i = \frac {k \paren {k + 1} } 2$
This is our induction hypothesis.
It is to be demonstrated that:
:$\ds \sum_{i \mathop = 1}^{k + 1} i = \frac {\paren {k + 1} \paren {k + 2} } 2$
=== Induction Step ===
This is our induction step:
Consider $n = k + 1$.
By the properties of summation:
:$\ds \sum_{i \mathop = 1}^{k + 1} i = \sum_{i \mathop = 1}^k i + k + 1$
Using the induction hypothesis this can be simplified to:
Hence the result by induction.
== Also see ==
This is usually the first proof by induction that a student mathematician encounters.
The second one is often Proof by Induction of Sum of Sequence of Squares.
== Sources ==
* : $\text {1-1}$ Principle of Mathematical Induction
* : Chapter $1$: Properties of the Natural Numbers: $\S 20 \beta$
* : $\S 3$: Natural Numbers: $\S 3.9$: Example
* : Chapter $1$: Preliminaries: Exercises: $1.2 \ \text {a)}$
* : Chapter $1$: Some Preliminary Considerations: $1.1$ Mathematical Induction: Problems $1.1$: $1 \ \text {(a)}$
* 
* : $\S 1.4.2$: Mathematical induction: Theorem $1.30$
* 
* 
Category:Proofs by Induction
Category:Closed Form for Triangular Numbers
Closed Form for Triangular Numbers/Proof by Polygonal Numbers

== Theorem ==
== Proof ==
Triangular numbers are $k$-gonal numbers where $k = 3$.
From Closed Form for Polygonal Numbers we have that:
:$P \left({k, n}\right) = \dfrac n 2 \left({\left({k - 2}\right) n - k + 4}\right)$
Hence:
Hence the result.
Category:Closed Form for Triangular Numbers
Closed Form for Triangular Numbers/Proof by Telescoping Sum

== Theorem ==
== Proof ==
Observe that:
Moreover, we have:
:$\paren {i + 1}^2 - i^2 = 2 i + 1$
And also:
:$\paren {n + 1}^2 - 1 = n^2 + 2 n$
Combining these results, we obtain:
This concludes the proof.
Category:Direct Proofs
Category:Closed Form for Triangular Numbers
Closed Form for Triangular Numbers/Proof using Bernoulli Numbers

== Theorem ==
== Proof ==
From Faulhaber's Formula:
where $B_k$ are the Bernoulli numbers.
Setting $p = 1$:
Hence the result.
== Sources ==
* : $\S 19$: Sums of Powers of Positive Integers: $19.9$
* : $\S 21$: Series of Constants: Sums of Powers of Positive Integers: $21.9.$
Category:Closed Form for Triangular Numbers
Closed Form for Triangular Numbers/Proof using Binomial Coefficients

== Theorem ==
== Proof ==
From Binomial Coefficient with One:
:$\forall k \in \Z, k > 0: \dbinom k 1 = k$
Thus:
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients
Category:Closed Form for Triangular Numbers
Closed Form for Triangular Numbers/Proof using Cardinality of Set

== Theorem ==
== Proof ==
Let $\N_n^* = \set {1, 2, 3, \cdots, n}$ be the initial segment of natural numbers.
Let $A = \set {\tuple {a, b}: a \le b, a, b \in \N_n^*}$
Let $B = \set {\tuple {a, b}: a \ge b, a, b, \in \N_n^*}$
Let $\phi: A \to B$ be the mapping:
:$\map \phi {x, y} = \tuple {y, x}$
By definition of dual ordering, $\phi$ is a bijection:
:$(1): \quad \size A = \size B$
We have:
Thus:
Combined with $\left({1}\right)$ this yields:
:$\size A = \dfrac {n^2 + n} 2 = \dfrac {n \paren {n + 1} } 2$
It remains to prove that:
:$T_n = \size A$
Category:Closed Form for Triangular Numbers
Closed Graph Theorem

== Theorem ==
Let $\struct {X, \norm {\,\cdot\,}_X}$ and $\struct {Y, \norm {\,\cdot\,}_Y}$ be Banach spaces.
Let $T : X \to Y$ be a linear transformation.
Let $G_T \subseteq X \times Y$ be the graph of $T$. 
Suppose that: 
:$G_T$ is closed in the direct product $X \times Y$ equipped with the direct product norm $\norm \cdot_{X \times Y}$. 
Then: 
:$T$ is bounded.
== Proof ==
Let $\norm {\,\cdot\,}_{G_T}$ be the restriction of $\norm \cdot_{X \times Y}$ to $G_T$. 
By Closed Subspace of Banach Space forms Banach Space, $\struct {G_T, \norm \cdot_{G_T} }$ is a Banach space.
Define $\pi_X : G_T \to X$ by: 
:$\map {\pi_X} {x, y} = x$ for each $\tuple {x, y} \in X \times Y$. 
Note that by the definition of $G_T$, we have $\tuple {x, y} \in G_T$ $y = T x$.
So, we have that $\pi_X$ is a bijection with inverse $\pi_X^{-1} : X \to G_T$ with: 
:$\map {\pi_X^{-1} } x = \tuple {x, T x}$ for each $x \in X$. 
From Projections on Direct Product of Normed Vector Spaces define Bounded Linear Transformations, $\pi_X$ is a bounded linear transformation.
By the Banach Isomorphism Theorem: 
:since $\pi_X$ is a bijective bounded linear transformation between Banach spaces, its inverse $\pi_X^{-1}$ is a bounded linear transformation.
So there exists $C > 0$ such that: 
:$\norm {\pi_X^{-1} x}_{G_T} \le C \norm x_X$ 
for each $x \in X$.
That is: 
:$\norm {\tuple {x, T x} }_{X \times Y} \le C \norm x_X$ for each $x \in X$. 
From the definition of the direct product norm, we have: 
:$\max \set {\norm x_X, \norm {T x}_Y} \le C \norm x_X$ for each $x \in X$.
From the definition of the max operation, we have: 
:$\norm {T x}_Y \le \max \set {\norm x_X, \norm {T x}_Y}$
So:
:$\norm {T x}_Y \le C \norm x_X$ for all $x \in X$.
So:
:$T$ is bounded.
Category:Linear Transformations on Banach Spaces
Closed Ideal of C*-Algebra is Self-Adjoint

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra.
Let $I$ be a closed ideal of $A$.
Then for each $x \in I$, we have $x^\ast \in I$. 
That is, $I$ is self-adjoint.
== Proof 1 ==
== Proof 2 ==
Category:C*-Algebras
Category:Closed Ideal of C*-Algebra is Self-Adjoint
Closed Ideal of C*-Algebra is Self-Adjoint/Proof 1

== Theorem ==
== Proof ==
From Intersection of Ideal of *-Algebra with its Star is Self-Adjoint Ideal, $I \cap I^\ast$ is a self-adjoint ideal of $A$, hence $\ast$-subalgebra of $A$.
From Star of Closed Subset of Banach *-Algebra is Closed, $I^\ast$ is closed. 
Hence as the intersection of closed sets, $I \cap I^\ast$ is closed.
From Closed *-Subalgebra of C*-Algebra is C*-Algebra, $I \cap I^\ast$ is a $\text C^\ast$-algebra.
From Existence of Approximate Identity of C*-Algebra arising from Dense Ideal, there exists an approximate identity for $I \cap I^\ast$.
That is, there exists a directed set $\struct {\Lambda, \preceq}$ and net $\family {e_\lambda}_{\lambda \mathop \in \Lambda}$ in $I \cap I^\ast$ such that:
:$\family {x e_\lambda}_{\lambda \mathop \in \Lambda}$ converges to $x$
for all $x \in I \cap I^\ast$.
Now take $x \in I$.
We have, for each $\lambda \in \Lambda$:
Since $I$ is an ideal, we have $x x^\ast \in I$.
Further since $x x^\ast$ is Product of Element in *-Star Algebra with its Star is Hermitian, we have $x x^\ast \in I^\ast$. 
So $x x^\ast \in I \cap I^\ast$.
Let $\epsilon > 0$. 
Since $\family {e_\lambda}_{\lambda \mathop \in I}$ is an approximate identity for $I \cap I^\ast$, there exists $b \in \Lambda$ such that $\lambda \succeq b$ implies that:
:$\norm {x x^\ast - x x^\ast e_\lambda} < \dfrac {\epsilon^2} 2$
hence:
:$\norm {x^\ast - x^\ast e_\lambda} < \epsilon$ 
Since $\epsilon > 0$ was arbitrary, the net $\family {x^\ast e_\lambda}_{\lambda \mathop \in \Lambda}$ converges to $x^\ast$. 
Further, since $e_\lambda \in I$ for each $\lambda \in \Lambda$, we have $x^\ast e_\lambda \in I$ for each $\lambda \in \Lambda$. 
Since $I$ is closed, we have $x^\ast \in I$ from Characterization of Closedness in terms of Nets.
Hence for $x \in I$ we have $x^\ast \in I$, and $I$ is therefore self-adjoint.
Category:Closed Ideal of C*-Algebra is Self-Adjoint
Closed Ideal of C*-Algebra is Self-Adjoint/Proof 2

== Theorem ==
== Proof ==
Let $x \in I$.
Since $I$ is an ideal, we have $x^\ast x \in I$. 
From Generalized Polar Decomposition in C*-Algebra, there exists $u \in A$ such that $x = u \paren {\paren {x^\ast x}^{1/2} }^{1/2}$.
From Power of Power of Positive Element of Unital C*-Algebra, we have $x = u \paren {x^\ast x}^{1/4}$. 
From Continuous Function applied to Normal Element of Unital C*-Algebra is Positive iff Function is Non-Negative, $\paren {x^\ast x}^{1/4}$ is positive.
Hence we have by $(\text C^\ast 3)$ in the definition of a $\text C^\ast$-algebra:
:$x^\ast = \paren {x^\ast x}^{1/4} u^\ast$
From Real-Valued Continuous Function Vanishing at Zero applied to Hermitian Element of Closed Ideal of Unital C*-Algebra is contained in Ideal, we have $\paren {x^\ast x}^{1/4} \in I$.
Hence since $I$ is an ideal, we have $x^\ast = \paren {x^\ast x}^{1/4} u^\ast \in I$.
Hence if $x \in I$ then $x^\ast \in I$.
So $I$ is self-adjoint.
== Sources ==
* : Theorem $5.4$
Category:Closed Ideal of C*-Algebra is Self-Adjoint
Closed Image of Closure of Set under Continuous Mapping equals Closure of Image

== Theorem ==
Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $H \subseteq S_1$ be a subset of $S_1$.
Let $\map \cl H$ denote the closure of $H$.
Let $f: T_1 \to T_2$ be a continuous mapping.
Let $f \sqbrk {\map \cl H}$ be closed in $T_2$.
Then:
:$f \sqbrk {\map \cl H} = \map \cl {f \sqbrk H}$
== Proof ==
By Continuity Defined by Closure:
:$f \sqbrk {\map \cl H} \subseteq \map \cl {f \sqbrk H}$
The proof follows by definition of set equality.
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 26$
Category:Set Closures
Category:Continuous Mappings
Closed Interval Defined by Absolute Value

== Theorem ==
Let $\xi, \delta \in \R$ be real numbers.
Let $\delta > 0$.
Then:
:$\set {x \in \R: \size {\xi - x} \le \delta} = \closedint {\xi - \delta} {\xi + \delta}$
where $\closedint {\xi - \delta} {\xi + \delta}$ is the closed real interval between $\xi - \delta$ and $\xi + \delta$.
== Proof ==
By definition of closed real interval:
:$\closedint {\xi - \delta} {\xi + \delta} = \set {x \in \R: \xi - \delta \le x \le \xi + \delta}$
Hence the result.
== Also presented as ==
:$\set {x \in \R: \size {x - \xi} \le \delta} = \closedint {\xi - \delta} {\xi + \delta}$
which is immediate from:
:$\size {x - \xi} = \size {\xi - x}$
== Also see ==
* Open Interval Defined by Absolute Value
* Complement of Closed Interval Defined by Absolute Value
* Complement of Open Interval Defined by Absolute Value
Category:Real Intervals
Category:Absolute Value Function
Closed Interval in Complete Lattice is Complete Lattice

== Theorem ==
Let $\struct {L, \preceq}$ be a complete lattice.
Let $a, b \in L$ with $a \preceq b$.
Let $I = \closedint a b$ be the closed interval between $a$ and $b$.
Then $\struct {I, \preceq}$ is also a complete lattice.
== Proof ==
Let $S \subseteq I$.
If $S = \O$, then it has a supremum in $I$ of $a$ and an infimum in $I$ of $b$.
Let $S \ne \O$.
Since $S \subseteq I$, $a$ is a lower bound of $S$ and $b$ is an upper bound of $S$.
Since $L$ is a complete lattice, $S$ has an infimum, $p$, and a supremum, $q$, in $L$.
Thus by the definitions of infimum and supremum:
:$a \preceq p$ and $q \preceq b$
Let $x \in S$.
Since an infimum is a lower bound:
:$p \preceq x$
Since a supremum is an upper bound:
:$x \preceq q$
Thus $a \preceq p \preceq x \preceq q \preceq b$.
Since $\preceq$ is an ordering, it is transitive, so by Transitive Chaining:
:$a \preceq p \preceq b$ and $a \preceq q \preceq b$.
That is, $p, q \in I$.
Thus $p$ and $q$ are the infimum and supremum of $S$ in $I$.
As every subset of $I$ has a supremum and infimum in $I$, $I$ is a complete lattice.
== Remark ==
Although $\struct {I, \preceq}$ is a complete lattice, it is only a complete sublattice of $\struct {L, \preceq}$ if $a = \inf L$ and $b = \sup L$. That is, if it equals $\struct {L, \preceq}$.
== Sources ==
* : $\S 1$
Category:Complete Lattices
Closed Interval in Reals is Uncountable

== Theorem ==
Let $a, b$ be extended real numbers such that $a < b$.
Then the closed interval $\set {x \in \R : a \le x \le b} \subseteq \R$ is uncountable.
== Proof ==
First suppose that $a, b \in \R$.
We have that the unit interval is uncountable.
Let $f: \closedint 0 1 \to \closedint a b$ such that $\map f x = a + \paren {b - a} x$.
Then if $\map f {x_1} = \map f {x_2}$, we have:
:$a + \paren {b - a} x_1 = a + \paren {b - a} x_2$
Because $b > a$:
:$b - a > 0$
so this implies:
:$x_1 = x_2$
Therefore $f$ is injective.
Now if $\closedint a b$ were countable, there would be an injective function $g : \closedint a b \to \N$.
By Composite of Injections is Injection, this implies that $g \circ f$ is an injection, so $\closedint 0 1$ is countable, a contradiction.
If one or both of $a, b$ is (are) not real, then we can pick a closed interval $\closedint c d \subseteq \closedint a b$ with $c, d \in \R, c < d$.
We know by the above that $\closedint c d$ is uncountable.
Let $S = \set {x \in \R : a \le x \le b}$. 
The identity function $I_\R : \closedint c d \to S$ is trivially injective.
If $S$ were uncountable, we would have an injective function $g : S \to \N$.
Then by Composite of Injections is Injection we would have an injection $\closedint c d \to \N$ given by $g \circ I_\R$.
This was shown above to be false, so $S$ is uncountable.
== Sources ==
* : Introduction: Review of Algebra, Geometry, and Trigonometry: $\text{0-1}$: The Real Numbers
Category:Uncountable Sets