nyuuzyou/wikis huggingface dataset:
Supremum Norm is Norm

== Theorem ==
Let $S$ be a set.
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space over $K \in \set {\R, \C}$.
Let $\BB$ be the set of bounded mappings $S \to X$.
Let $\norm {\, \cdot \,}_\infty$ be the supremum norm on $\BB$.
Then $\norm {\, \cdot \,}_\infty$ is a norm on $\BB$.
=== Continuous on Closed Interval Real-Valued Function ===
=== Space of Bounded Sequences ===
== Proof ==
First:
Now let $\lambda \in K, f \in \BB$
We have:
Finally let $f, g \in \BB$.
We have:
Thus $\norm {\, \cdot \,}_\infty$ has the defining properties of a norm.
Category:Examples of Norms
Category:Supremum Norm
Category:Supremum Norm is Norm
Supremum Norm is Norm/Continuous on Closed Interval Real-Valued Function

== Theorem ==
Let $I = \closedint a b$ be a closed interval.
Let $\struct {\map C I, +, \, \cdot \,}_\R$ be the vector space of real-valued functions, continuous on $I$.
Let $\map x t \in \map C I$ be a continuous real function.
Let $\size {\, \cdot \,}$ be the absolute value.
Let $\norm {\, \cdot \,}_\infty$ be the supremum norm on real-valued functions, continuous on $I$.
Then $\norm {\, \cdot \,}_\infty$ is a norm over $\struct {\map C I, +, \, \cdot \,}_\R$.
== Proof ==
=== Positive definiteness ===
Suppose $\norm x_\infty = 0$.
Then:
Therefore:
:$\forall t \in I : \map x t = 0$
=== Positive homogeneity === 
Let $\alpha \in \R$.
=== Triangle inequality ===
== Sources ==
* : $\S 1.2$: Normed and Banach spaces. Normed spaces
Category:Examples of Norms
Category:Supremum Norm is Norm
Supremum Norm is Norm/Space of Bounded Sequences

== Theorem ==
The supremum norm on the vector space of bounded sequences is a norm.
== Proof ==
=== ===
Let $x \in \ell^\infty$.
By definition of supremum norm:
:$\ds \norm {\mathbf x}_\infty = \sup_{n \mathop \in \N} \size {x_n}$
The complex modulus of $x_n$ is real and non-negative.
Hence, $\norm {\mathbf x}_\infty \ge 0$.
Suppose $\norm {\mathbf x}_\infty = 0$.
Then:
Thus is satisfied.
=== ===
Suppose $\alpha \in \C$.
Thus is satisfied.
=== ===
Let $\mathbf x = \sequence {x_n}_{n \mathop \in \N}, \mathbf y = \sequence {y_n}_{n \mathop \in \N} \in \ell^\infty$.
By Triangle Inequality for Complex Numbers:
:$\forall n \in \N : \size {x_n + y_n} \le \size {x_n} + \size {y_n}$
Then:
Thus is satisfied.
== Sources ==
* : $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces
Category:Supremum Norm is Norm
Supremum Norm on Vector Space of Real Matrices is Norm

== Theorem ==
Supremum Norm forms a norm on the vector space of real matrices.
== Proof ==
Let $M \in \R^{m \times n}: m, n \in \N_{>0}$ be a real matrix.
Denote the $\paren {i, j}$-th entry of $M$ by $a_{i j}$.
Note that the set of matrix elements of $M$ is a finite set of real numbers.
We have that:
:Real Numbers form Totally Ordered Field
:Finite Non-Empty Subset of Ordered Set has Maximal and Minimal Elements
Therefore, $M$ has the greatest element.
=== ===
Equality is obtained for $M$ being a zero matrix.
Suppose $\norm M_\infty = 0$.
Then: 
:$\ds \forall i, j: 1 \le i \le m, 1 \le j \le n: \size {a_{i j} } = 0$
In other words, $M$ is a zero matrix.
=== ===
=== ===
Let $P, Q \in \R^{m \times n}$.
Denote their $\paren {i, j}$-th matrix elements as $p_{i j}$ and $q_{i j}$ respectively.
Fix $i, j \in \N: 1 \le i \le m, 1 \le j \le n$.
We have that:
This holds for any $i, j$.
Hence:
All norm axioms are seen to be satisfied.
Hence the result.
== Sources ==
* : $\S 1.2$: Normed and Banach spaces. Normed spaces
Category:Examples of Norms
Supremum Operator Norm Need not be Attained

== Theorem ==
Let $\Bbb K \in \set {\R, \C}$.
Let $\sequence {\lambda_n}_{n \mathop \in \N}$ be a bounded sequence in $\Bbb K$ such that:
:$\ds \forall n \in \N_{> 0} : \lambda_n = 1 - \frac 1 n$.
Let $\struct {\ell^2, \norm {\, \cdot \,}_2}$ be the normed $2$-sequence space.
Let $\mathbf x = \tuple {a_1, a_2, a_3, \ldots} \in \ell^2$.
Suppose $\Lambda : \ell^2 \to \ell^2$ is a diagonal operator such that:
:$\Lambda \tuple {a_1, a_2, a_3, \ldots} = \tuple {\lambda_1 a_1, \lambda_2 a_2, \lambda_3 a_3, \ldots}$
Then there is no $\mathbf x \in \ell^2$ such that:
:$\norm {\mathbf x}_2 \le 1$
and:
:$\norm {\Lambda \mathbf x}_2 = \norm \Lambda$
== Proof ==
Suppose $\ds \lambda_n = 1 - \frac 1 n$ with $n \in \N_{> 0}$.
By Supremum Operator Norm of Diagonal Operator over 2-Sequence Space:
Suppose:
:$\mathbf x = \tuple {a_n}_{n \mathop \in \N_{> 0} } \in \ell^2$ 
such that $\norm {\mathbf x}_2 \le 1$ and $\norm {\Lambda \mathbf x}_2 = 1$.
Suppose $0 = a_2 = a_3 = \ldots$.
Then 
Also $\norm {\Lambda \mathbf x}_2 = 0$.
This is in contradiction with the assumption $\norm {\Lambda \mathbf x}_2 = 1$.
Hence, at least one of the terms $a_2, a_3, \ldots$ must be nonzero.
However:
This is a contradiction.
Hence, there exists no nonzero $\mathbf x \in \ell^2$ such that $\norm {\mathbf x}_2 \le 1$ and $\norm {\Lambda \mathbf x}_2 = \norm \Lambda$.
== Sources ==
* : Chapter $\S 2.3$: The normed space $\map {CL} {X,Y}$. Operator norm and the normed space $\map {CL} {X, Y}$
Category:Supremum Norm
Category:Continuous Linear Transformations
Supremum Operator Norm as Universal Upper Bound

== Theorem ==
Let $\map {CL} {X, Y}$ be the continuous linear transformation space.
Let $T \in \map {CL} {X, Y}$.
Let $\norm {\, \cdot \,}$ be the supremum operator norm on $\map {CL} {X, Y}$.
Then:
:$\forall x \in X : \norm {Tx}_Y \le \norm T \norm x_X$
== Proof ==
Let $S = \set {\norm {Tx}_Y : x \in X : \norm x_X \le 1}$.
Suppose $x = \mathbf 0$.
Then:
Suppose $x \ne \mathbf 0$.
Let $\hat x \in X$ such that:
:$\ds \hat x = \frac x {\norm x_X}$
Then:
:$\ds \norm {\hat x}_X = \frac {\norm x_X}{\norm x_X} = 1$.
Thus:
:$\norm {T \hat x}_Y \in S$
and 
$\norm {T \hat x}_Y \le \sup S = \norm T$
Furthermore:
Rearranging, we get:
:$\norm {T x}_Y \le \norm T \norm x_X$
== Sources ==
* : Chapter $\S 2.3$: The normed space $\map {CL} {X,Y}$. Operator norm and the normed space $\map {CL} {X, Y}$
Category:Linear Transformations
Category:Continuous Linear Transformations
Category:Continuous Mappings
Category:Supremum Norm
Supremum Operator Norm is Norm

== Theorem ==
Let $\map {CL} {X, Y}$ be the continuous linear transformation space.
Let $\norm {\, \cdot \,} : \map {CL} {X, Y} \to \R$ be the supremum operator norm such that:
:$\forall T \in \map {CL} {X, Y} : \norm T := \sup \set {\norm {Tx}_Y : x \in X : \norm x_X \le 1}$
Then $\norm {\, \cdot \,}$ is a norm.
== Proof ==
=== Positive definiteness ===
Let $T \in \map {CL} {X, Y}$.
By definition of norm:
:$\norm {Tx}_Y \ge 0$
Then:
:$\ds \norm T = \sup_{\begin{split} x \mathop \in X \\ \norm x \mathop \le 1 \end{split} } \norm {T x}_Y \ge 0$
Suppose $\norm T = 0$.
Then:
Hence:
:$\norm {T x}_Y = 0$
By Norm Axiom $N1$: 
:$\forall x \in X : Tx = \mathbf 0_Y$
where $\mathbf 0_Y$ is the zero vector in the normed vector space $Y$.
Therefore:
:$T = \mathbf 0$.
=== Positive homogeneity ===
Let $\alpha \in \set{\R, \C}$.
Let $T \in \map {CL} {X, Y}$.
Then:
=== Triangle inequality ===
Take the supremum of both sides with the condition that $\forall x \in X : \norm x_X \le 1$.
Then:
:$\norm {T + S} \le \norm T + \norm S$
== Sources ==
* : Chapter $\S 2.3$: The normed space $\map {CL} {X,Y}$. Operator norm and the normed space $\map {CL} {X, Y}$
Category:Linear Transformations
Category:Continuous Linear Transformations
Category:Continuous Mappings
Category:Supremum Norm
Category:Examples of Norms
Supremum Operator Norm is Well-Defined

== Theorem ==
Let $K$ be a field.
Let $X, Y$ be normed vector spaces over $K$.
Let $\map {CL} {X, Y}$ be the continuous linear transformation space.
Let $\norm {\, \cdot \,}$ be an operator norm on $\map {CL} {X, Y}$ defined as:
:$\norm T := \map \sup {\norm {Tx} : x \in X : \norm x \le 1}$
Then $\norm {\, \cdot \,}$ is well-defined.
== Proof ==
Let $S = \set {\norm {T x} : x \in X : \norm x \le 1}$
By definition of the norm:
:$S \subseteq \R$
=== $S$ is non-empty ===
By definition, $X$ is a normed vector space, and thus a vector space.
Hence, by Zero Vector is Unique, the zero vector exists in $X$, and $X$ is non-empty. 
Let $x = \mathbf 0_X \in X$ be the zero vector.
Then:
and:
Thus, by definition of $S$:
:$0 \in S$
and:
:$S \ne \O$
=== $S$ is bounded above ===
Let $\map {CL} {X, Y}$.
Because:
:$\exists 0_X \in X: \norm 0_X \le 1$
let $x \in X : \norm x \le 1$.
Then:
By the least upper bound property of $\R$, a supremum of $S$ exists.
Hence:
:$\forall T \in \map {CL} {X, Y} : \norm T := \sup \set {\norm {Tx} : x \in X : \norm x \le 1} < \infty$
== Sources ==
* : Chapter $\S 2.3$: The normed space $\map {CL} {X,Y}$. Operator norm and the normed space $\map {CL} {X, Y}$
Category:Supremum Norm
Category:Linear Transformations
Category:Continuous Linear Transformations
Category:Continuous Mappings
Category:Examples of Well-Defined Mappings‎
Supremum Operator Norm of Cesàro Summation Operator

== Theorem ==
Let $A : \ell^\infty \to \ell^\infty$ be the Cesàro summation operator.
Let $\norm {\, \cdot \,}$ be the supremum operator norm.
Then $\norm A = 1$.
== Proof ==
By Cesàro Summation Operator is Continuous Linear Transformation:
:$\norm {A \mathbf x}_\infty \le \norm {\mathbf x}_\infty$
where $\norm {\, \cdot \,}_\infty$ denotes the supremum norm.
By Universal Upper Bound greater than Supremum Operator Norm:
:$\norm A \le 1$
Let $\mathbf 1 := \tuple {1, 1, \ldots} \in \ell^\infty$
Then:
Altogether, $\norm A = 1$.
== Sources ==
* : Chapter $\S 2.3$: The normed space $\map {CL} {X,Y}$. Operator norm and the normed space $\map {CL} {X, Y}$
Category:Cesàro Summation Operator
Category:Operator Theory
Category:Linear Transformations
Category:Continuous Transformations
Supremum Operator Norm of Diagonal Operator over 2-Sequence Space

== Theorem ==
Let $\Bbb K = \set {\R, \C}$.
Let $\sequence {\lambda_n}_{n \mathop \in \N_{> 0} }$ be a bounded sequence in $\Bbb K$.
Let $\struct {\ell^2, \norm {\, \cdot \,}_2}$ be the normed $2$-sequence space.
Let $\mathbf x = \tuple {a_1, a_2, a_3, \ldots} \in \ell^2$.
Suppose $\Lambda : \ell^2 \to \ell^2$ be a diagonal operator such that:
:$\Lambda \tuple {a_1, a_2, a_3, \ldots} = \tuple {\lambda_1 a_1, \lambda_2 a_2, \lambda_3 a_3, \ldots}$
Let $\norm {\, \cdot \,}$ be the supremum operator norm.
Then $\ds \Lambda \in \map {CL} {\ell^2}$ and $\ds \norm \Lambda = \sup_{n \mathop \in \N_{> 0} } \size {\lambda_n}$
== Proof ==
From Diagonal Operator over 2-Sequence Space is Continuous Linear Transformation:
:$\ds \norm {\Lambda \mathbf x }_2^2 \le \paren {\sup_{n \mathop \in \N_{> 0} } \size{\lambda_n} }^2 \norm {\mathbf x}_2^2$
Therefore:
:$\ds \norm {\Lambda \mathbf x}_2 \le \sup_{n \mathop \in \N_{> 0} } \size{\lambda_n} \norm {\mathbf x}_2$
Suppose $\norm {\mathbf x}_2 \le 1$.
Then:
:$\ds \norm {\Lambda \mathbf x}_2 \le \sup \norm {\Lambda \mathbf x}_2 \le \sup_{n \mathop \in \N_{> 0} } \size{\lambda_n}$
By definition of supremum operator norm:
:$\norm \Lambda = \sup \set {\norm {\Lambda \mathbf x}_2 : \forall \mathbf x \in \ell^2 : \norm {\mathbf x}_2 \le 1 }$
Hence:
:$\ds \norm {\Lambda} \le \sup_{n \mathop \in \N_{> 0} } \size {\lambda_n}$
Let $\mathbf e_n \in \ell^2$ such that:
:$\mathbf e_n = \tuple {\underbrace{0, \ldots, 0}_{n - 1}, 1, 0, \ldots}$
Then:
Take the supremum of both sides over $n \in N$.
On the left-hand dize nothing happens because it is an $n$-independent real number.
On the left-hand dize we have $\ds \sup_{n \mathop \in \N_{> 0} } \size {\lambda_n}$
Hence:
:$\ds \norm \Lambda \ge \sup_{n \mathop \in \N_{> 0} } \size {\lambda_n}$
Combining the results yields:
:$\ds \norm \Lambda = \sup_{n \mathop \in \N_{> 0} } \size {\lambda_n}$
== Sources ==
* : Chapter $\S 2.3$: The normed space $\map {CL} {X,Y}$. Operator norm and the normed space $\map {CL} {X, Y}$
Category:Supremum Norm
Category:Examples of Norms
Supremum Operator Norm of Linear Transformation is Bounded Above by Hilbert-Schmidt Norm

== Theorem ==
Let $T_A : \R^n \to \R^m$ be the linear transformation such that:
:$\forall \mathbf x \in \R^n : T_A \mathbf x := A \mathbf x$
Let $\norm {\, \cdot \,}$ be the supremum operator norm.
Let $\norm {\, \cdot \,}_{HS}$ be the Hilbert Schmidt norm.
Then:
:$\norm {T_A} \le \norm A_{HS}$
== Proof ==
We have that Norms on Finite-Dimensional Real Vector Space are Equivalent.
Choose the Euclidean norm.
Let $X = \struct {\R^n, \norm {\, \cdot \,}_2}$ and $Y = \struct {\R^m, \norm {\, \cdot \,}_2}$ be normed vector spaces.
Let the matrix $A \in \R^{m \times n}$ be given by:
:$A = \begin {bmatrix}
a_{1 1} & \cdots & a_{1 n} \\
 \vdots & \ddots & \vdots \\
a_{m 1} & \cdots & a_{m n} \\
\end{bmatrix}$
We have that Set of Linear Transformations is Isomorphic to Matrix Space.
Let $T_A : \R^n \to \R^m$ be the linear transformation such that:
:$\forall \mathbf x \in \R^n : T_A \mathbf x := A \mathbf x$
From Linear Transformations between Finite-Dimensional Normed Vector Spaces are Continuous:
:$\ds \norm{T_A \mathbf x}_2 \le \norm{\mathbf x}_2 \sqrt {\sum_{i \mathop = 1}^m \sum_{j \mathop = 1}^n a_{ij}^2 }$
Take the supremum of both sides with the condition $\forall \mathbf x \in \R^n : \norm {\mathbf x}_2 \le 1$.
On the left we are left with $\norm {T_A}$.
On the right we have:
Hence:
:$\norm {T_A} \le \norm {A}_{HS}$
== Sources ==
* : Chapter $\S 2.3$: The normed space $\map {CL} {X,Y}$. Operator norm and the normed space $\map {CL} {X, Y}$
Category:Inequalities
Supremum Operator Norm on Continuous Linear Transformation Space is Submultiplicative

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,}_X}$ and $\struct {Y, \norm {\, \cdot \,}_Y}$, $\struct {Z, \norm {\, \cdot \,}_Z}$ be normed vector spaces.
Let $A : Y \to Z$ and $B : X \to Y$ be continuous linear transformations.
Let $\norm {\, \cdot \,}$ be the supremum operator norm.
Let $\circ$ denote the composition.
Then $\norm {\, \cdot \,}$ is submultiplicative:
:$\norm {A \circ B} \le \norm A \cdot \norm B$
== Proof ==
== Sources ==
Category:Continuous Linear Transformations
Category:Supremum Operator Norm
Category:Submultiplicative Functions
Supremum Plus Constant

== Theorem ==
Let $S$ be a subset of the set of real numbers $\R$.
Let $S$ be bounded above.
Let $\xi \in \R$.
Then:
:$\ds \map {\sup_{x \mathop \in S} } {x + \xi} = \xi + \map {\sup_{x \mathop \in S} } x$
where $\sup$ denotes supremum.
== Proof ==
Let $B = \sup S$.
Let $T = \set {x + \xi: x \in S}$.
Since $\forall x \in S: x \le B$ it follows that:
:$\forall x \in S: x + \xi \le B + \xi$
Hence $\xi + B$ is an upper bound for $T$.
If $C$ is the supremum for $T$ then $C \le \xi + B$.
On the other hand:
:$\forall y \in T: y \le C$
Therefore:
:$\forall y \in T: y - \xi \le C - \xi$
Since $S = \set {y - \xi: y \in T}$ it follows that $C - \xi$ is an upper bound for $S$ and so $B \le C - \xi$.
So we have shown that $C \le \xi + B$ and $C \ge \xi + B$, hence the result.
== Sources ==
* : $\S 2$: Continuum Property: Exercise $\S 2.13 \ (2)$
Category:Real Analysis
Supremum and Infimum are Unique

== Theorem ==
=== Supremum is Unique ===
=== Infimum is Unique ===
== Sources ==
* : Chapter $0$: Orderings
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Theorem $14.3$
* : $\S 7$
Category:Suprema
Category:Infima
Supremum by Suprema of Directed Set in Simple Order Product

== Theorem ==
Let $\struct {S, \preceq}$ be an up-complete meet semilattice.
Let $\struct {S \times S, \precsim}$ be the simple order product of $\struct {S, \preceq}$ and $\struct {S, \preceq}$.
Let $D$ be a directed subset of $S \times S$.
Then:
:$\sup D = \tuple {\map \sup {\map {\pr_1^\to} D}, \map \sup {\map {\pr_2^\to} D} }$
where
:$\pr_1$ denotes the first projection on $S \times S$
:$\pr_2$ denotes the second projection on $S \times S$
:$\map {\pr_1^\to} D$ denotes the image of $D$ under $\pr_1$
== Proof ==
By Up-Complete Product:
:$\struct {S \times S, \precsim}$ is up-complete.
By definition of up-complete:
:$D$ admits a supremum.
By definition of Cartesian product:
:$\exists d_1, d_2 \in S: \sup D = \tuple {d_1, d_2}$
By Up-Complete Product/Lemma 2:
:$D_1 := \map {\pr_1^\to} D$ is directed
and
:$D_2 := \map {\pr_2^\to} D$ is directed.
By definition of up-complete:
:$D_1$ admits a supremum
and
:$D_2$ admits a supremum
We will prove that
:$d_2$ is upper bound for $D_2$
Let $x \in D_2$.
By definition of image of set:
:$\exists \tuple {a, b} \in D: \map {\pr_2} {a, b} = x$
By definition of second projection:
$b = x$
By definition of supremum:
:$\tuple {d_1, d_2}$ is upper bound for $D$.
By definition of upper bound:
:$\tuple {a, x} \precsim \tuple {d_1, d_2}$
Thus by definition of simple order product:
:$x \preceq d_2$
Analogically we have that
:$d_1$ is upper bound for $D_1$
By definition of supremum:
:$\sup D_1 \preceq d_1$ and $\sup D_2 \preceq d_2$
By definition of simple order product:
:$\tuple {\sup D_1, \sup D_2} \precsim \sup D$
By Up-Complete Product/Lemma 1:
:$D_1 \times D_2$ is directed.
By definition of up-complete:
:$D_1 \times D_2$ admits a supremum.
By definition of subset:
:$D \subseteq D_1 \times D_2$
By Supremum of Subset:
:$\sup D \precsim \map \sup {D_1 \times D_2}$
By Supremum of Simple Order Product:
:$\sup D \precsim \tuple {\sup D_1, \sup D_2}$
Thus by definition of antisymmetry:
:$\sup D = \tuple {\sup D_1, \sup D_2}$
== Sources ==
* 
* 
Category:Up-Complete Semilattices
Category:Simple Order Product
Supremum does not Precede Infimum

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $T \subseteq S$ admit both a supremum $M$ and an infimum $m$.
Then $m \preceq M$.
== Proof ==
By definition of supremum:
:$\forall a \in T: a \preceq M$
By definition of infimum:
:$\forall a \in T: m \preceq a$
The result follows from transitivity of ordering.
== Sources ==
* : Chapter $\text I$: Functions and Limits: $\S 3$: Bounds of a Function
Category:Suprema
Category:Infima
Supremum in Ordered Subset

== Theorem ==
Let $L = \struct {S, \preceq}$ be an ordered set.
Let $R = \struct {T, \preceq'}$ be an ordered subset of $L$.
Let $X \subseteq T$ such that
:$X$ admits an supremum in $L$.
Then $\sup_L X \in T$ 
:$X$ admits an supremum in $R$ and $\sup_R X = \sup_L X$
== Proof ==
This follows by of the proof of Infimum in Ordered Subset.
== Sources ==
* 
Category:Ordered Sets
Category:Suprema
Supremum is Coproduct in Order Category

== Theorem ==
Let $\mathbf P$ be an order category with ordering $\preceq$.
Let $p, q \in P_0$, and suppose they have some supremum $r = \sup \left\{{p, q}\right\}$.
Then $r$ is the coproduct of $p$ and $q$ in $\mathbf P$.
== Proof ==
Let $\mathbf P^{\text{op}}$ be the dual category of $\mathbf P$.
From Dual of Order Category, it is the order category corresponding to the dual ordering $\succeq$.
From Dual Pairs (Order Theory), it follows that in $\mathbf P^{\text{op}}$:
:$r = \inf \left\{{p, q}\right\}$
where $\inf$ denotes infimum.
By Infimum is Product in Order Category, $r$ is the product of $p$ and $q$ in $\mathbf P^{\text{op}}$.
By Dual Pairs (Category Theory), $r$ is the coproduct of $p$ and $q$ in $\mathbf P$.
== Also see ==
* Infimum is Product in Order Category, the dual result
== Sources ==
* : $\S 3.2$: Example $3.7$
Category:Order Categories
Supremum is Dual to Infimum

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $a \in S$ and $T \subseteq S$.
The following are dual statements:
:$a$ is a supremum for $T$
:$a$ is an infimum for $T$
== Proof ==
By definition, $a$ is a supremum for $T$ :
:$a$ is an upper bound for $T$
:$a \preceq b$ for all upper bounds $b$ of $T$
The dual of this statement is:
:$a$ is a lower bound for $T$
:$b \preceq a$ for all lower bounds $b$ of $T$
by Dual Pairs (Order Theory).
By definition, this means $a$ is an infimum for $T$.
The converse follows from Dual of Dual Statement (Order Theory).
== Also see ==
* Duality Principle (Order Theory)
Category:Suprema
Category:Infima
Category:Dual Pairs (Order Theory)
Supremum is Increasing relative to Product Ordering

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $I$ be a set.
Let $f, g: I \to S$.
Let $f \sqbrk I$ denote the image of $I$ under $f$.
Let:
:$\forall i \in I: \map f i \preceq \map g i$
That is, let $f \preceq g$ in the product ordering.
Let $f \sqbrk I$ and $g \sqbrk I$ admit suprema.
Then:
:$\sup f \sqbrk I \preceq \sup g \sqbrk I$
== Proof ==
Let $x \in f \sqbrk I$.
Then:
:$\exists j \in I: \map f j = x$
Then:
:$\map f j \prec \map g j$
By the definition of supremum:
:$\sup g \sqbrk I$ is an upper bound of $g \sqbrk I$
Thus:
:$\map g j \preceq \sup g \sqbrk I$
Since $\preceq$ is transitive:
:$x = \map f j \preceq \sup g \sqbrk I$
Since this holds for all $x \in f \sqbrk I$, $\sup g \sqbrk I$ is an upper bound of $f \sqbrk I$.
Thus by the definition of supremum:
:$\sup f \sqbrk I \preceq \sup g \sqbrk I$
Category:Order Theory
Category:Increasing Mappings
Supremum is Unique

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $T$ be a non-empty subset of $S$.
Then $T$ has at most one supremum in $S$.
== Proof ==
Let $c$ and $c'$ both be suprema of $T$ in $S$.
From the definition of supremum, $c$ and $c'$ are upper bounds of $T$ in $S$.
By that definition:
:$c$ is an upper bound of $T$ in $S$ and $c'$ is a supremum of $T$ in $S$ implies that $c' \preceq c$
:$c'$ is an upper bound of $T$ in $S$ and $c$ is a supremum of $T$ in $S$ implies that $c \preceq c'$.
So:
:$c' \preceq c \land c \preceq c'$
and thus by the antisymmetry of the ordering $\preceq$:
:$c = c'$
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Theorem $14.3$
* : Part $\text {II}$: Foundations: Chapter $8$: Least Upper Bounds
* : $1$: Review of some real analysis: $\S 1.1$: Real Numbers
* : Chapter $\text I$: Preliminaries, Definition $1.2$
Category:Order Theory
Supremum is not necessarily Greatest Element

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $T$ admit a supremum in $S$.
Then the supremum of $T$ in $S$ is not necessarily the greatest element of $T$.
== Proof ==
== Also see ==
* Greatest Element is Supremum
== Sources ==
* : $\S 2$: Continuum Property: $\S 2.7$: Maximum and Minimum
Category:Suprema
Category:Order Theory
Category:Supremum is not necessarily Greatest Element
Supremum is not necessarily Greatest Element/Proof

== Theorem ==
== Proof ==
Proof by Counterexample:
Consider the subset $T$ of the set of real numbers $\R$:
:$T := \set {x \in \R: 1 \le x < 2}$
The number $2$ cannot be the greatest element of $T$ as $2 \notin T$.
However, $2$ is the supremum of $T$ in $S$.
Indeed, by definition:
:$\forall x \in T: x < 2$
So, let $x < 2$.
Then consider $y = \dfrac x 1 + \dfrac 2 1 = \dfrac {x + 2} 2$.
We have that $x < y < 2$ by Mediant is Between.
Thus $y \in T$ but $y > x$ and so $x$ cannot be the greatest element of $T$.
By the same construction, neither can $y$ be the supremum of $T$ in $S$.
The conclusion is that there is no greatest element of $T$.
Hence the result.
== Sources ==
* : $\S 2$: Continuum Property: $\S 2.8$: Example $\text{(ii)}$
Category:Supremum is not necessarily Greatest Element
Supremum of Absolute Value of Difference equals Difference between Supremum and Infimum

== Theorem ==
Let $f$ be a real function.
Let $S$ be a subset of the domain of $f$.
Let $\ds \sup_{x \mathop \in S} \set {\map f x}$ and $\ds \inf_{x \mathop \in S} \set {\map f x}$ exist.
Then $\ds \sup_{x, y \mathop \in S} \set {\size {\map f x - \map f y} }$ exists and:
:$\ds \sup_{x, y \mathop \in S} \set {\size {\map f x - \map f y} } = \sup_{x \mathop \in S} \set {\map f x} - \inf_{x \mathop \in S} \set {\map f x}$
== Proof ==
Category:Real Analysis