nyuuzyou/wikis huggingface dataset:
Supremum of Absolute Value of Difference equals Supremum of Difference

== Theorem ==
Let $S$ be a non-empty real set.
Let $\ds \sup_{x, y \mathop \in S} \paren {x - y}$ exist.
Then $\ds \sup_{x, y \mathop \in S} \size {x - y}$ exists and:
:$\ds \sup_{x, y \mathop \in S} \size {x - y} = \sup_{x, y \mathop \in S} \paren {x - y}$
== Proof ==
Consider the set $\set {x - y: x, y \in S, x - y \le 0}$.
There is a number $x'$ in $S$ as $S$ is non-empty.
Therefore, $0 \in \set {x - y: x, y \in S, x - y \le 0}$ as $x = y = x'$ implies that $x - y = 0$, $x, y \in S$, and $x - y \le 0$.
Also, $0$ is an upper bound for $\set {x - y: x, y \in S, x - y \le 0}$ by definition.
Accordingly:
:$\ds \sup_{x, y \mathop \in S, x − y \mathop \le 0} \paren {x - y} = 0$
Consider the set $\left\{{x - y: x, y \in S, x - y \ge 0}\right\}$.
There is a number $x'$ in $S$ as $S$ is non-empty.
Therefore, $0 \in \left\{{x - y: x, y \in S, x - y \ge 0}\right\}$ as $x = y = x'$ implies that $x - y = 0$, $x, y \in S$, $x - y \ge 0$.
Accordingly:
:$\ds \sup_{x, y \mathop \in S, x − y \mathop \ge 0} \paren {x - y} \ge 0$
Category:Suprema
Category:Absolute Value Function
Supremum of Bounded Above Set of Reals is in Closure

== Theorem ==
Let $\R$ be the real number line under the Euclidean metric.
Let $H \subseteq \R$ be a bounded above subset of $\R$ such that $H \ne \O$.
Let $u = \sup H$ be the supremum of $H$.
Then:
:$u \in \map \cl H$
where $\map \cl H$ denotes the closure of $H$ in $\R$.
== Proof ==
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
Let $\map {B_\epsilon} u$ be the open $\epsilon$-ball of $u$ in $\R$.
From Distance from Subset of Real Numbers:
:$\map d {u, H} = 0$
Thus by definition of distance from subset:
:$\exists x \in H: \map d {u, x} < \epsilon$
Thus $x \in \map {B_\epsilon} u$.
As $x \in H$ and $x \in \map {B_\epsilon} u$, from the definition of intersection:
:$x \in H \cap \map {B_\epsilon} u$
The result follows from Condition for Point being in Closure.
== Also see ==
* Infimum of Bounded Below Set of Reals is in Closure
== Sources ==
* : $2.28$
* : $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Examples $3.7.13$
Category:Set Closures
Category:Real Analysis
Supremum of Continuous Bounded Real-Valued Function on Everywhere Dense Subset

== Theorem ==
Let $\struct {X, \tau}$ be a topological space.
Let $\DD$ be an everywhere dense subset of $\struct {X, \tau}$. 
Let $f : X \to \R$ be a continuous bounded real-valued function.
Then:
:$\ds \sup_{x \mathop \in \DD} \map f x = \sup_{x \mathop \in X} \map f x$
== Proof ==
Since $\DD \subseteq X$ we have:
:$\ds \sup_{x \mathop \in \DD} \map f x \le \sup_{x \mathop \in X} \map f x$
Conversely, from the definition of supremum, for each $\epsilon > 0$ there exists $y \in X$ such that:
:$\ds \sup_{x \mathop \in X} \map f x - \frac \epsilon 2 \le \map f y \le \sup_{x \mathop \in X} \map f x$
Since $f$ is continuous, there exists an open set $U$ such that:
:$\ds \size {\map f y - \map f w} < \frac \epsilon 2$
hence:
:$\ds \map f y - \frac \epsilon 2 < \map f w < \map f y + \frac \epsilon 2$ 
for $w \in U$.
Since $\DD$ is everywhere dense, we have $\DD \cap U \ne \O$. 
Picking $w \in \DD \cap U$, we have:
:$\ds \map f w > \map f y - \frac \epsilon 2 \ge \sup_{x \mathop \in X} \map f x - \epsilon$ 
while:
:$\ds \map f w \le \sup_{x \mathop \in \DD} \map f x$
since $w \in \DD$.
Hence for each $\epsilon > 0$ we have:
:$\ds \sup_{x \mathop \in X} \map f x \le \sup_{x \mathop \in \DD} \map f x + \epsilon$
Taking $\epsilon \to 0^+$, we then obtain:
:$\ds \sup_{x \mathop \in X} \map f x \le \sup_{x \mathop \in \DD} \map f x$
Hence we have:
:$\ds \sup_{x \mathop \in \DD} \map f x = \sup_{x \mathop \in X} \map f x$
Category:Continuous Real-Valued Functions
Category:Everywhere Dense
Supremum of Doubleton in Totally Ordered Set

== Theorem ==
Let $\struct {S, \preccurlyeq}$ be a totally ordered set.
Let $D = \set {a, b} \subseteq S$ be an arbitrary doubleton subset of $S$.
Then:
:$\map \sup D \in D$
where $\sup D$ denotes the supremum of $D$.
== Proof ==
Let $D = \set {a, b} \subseteq S$ as asserted.
As $S$ is a totally ordered set, either $a \preccurlyeq b$ or $b \preccurlyeq a$.
 let $a \preccurlyeq b$.
We have by definition of supremum that:
:$\forall x \in D: x \preccurlyeq \map \sup D$
:$\forall y \in S:$ if $y$ is an upper bound of $D$, then $\map \sup D \preccurlyeq y$.
Now we have:
:$\forall x \in D: x \preccurlyeq b$
Let $y \in S$ be an upper bound of $D$.
Then:
:$\forall x \in D: x \preccurlyeq y$
In particular:
:$b \preccurlyeq y$
Hence by definition:
:$\map \sup D = b$
Similarly, if $b \preccurlyeq a$, then :
:$\map \sup D = a$
In either case:
:$\map \sup D \in D$
Category:Suprema
Category:Doubletons
Supremum of Elements of Sublattice not necessarily Same as for Lattice

== Theorem ==
Let $\struct {S, \preceq}$ be a lattice.
Let $\struct {T, \preceq_T}$ be a sublattice of $S$.
Let $a, b \in T$.
Then it is not necessarily the case that:
:$\sup_S \set {a, b}$
is the same as:
:$\sup_T \set {a, b}$
== Proof ==
Proof by Counterexample:
Let $\struct {G, \circ}$ be a group.
Let $\mathbb G$ be the set of all subgroups of $G$.
Let $\powerset G$ denote the power set of $G$.
Let $\struct {\powerset G, \subseteq}$ be the complete lattice formed by $\powerset G$ and $\subseteq$.
From Power Set is Complete Lattice, $\struct {\powerset G, \subseteq}$ indeed forms a complete lattice.
Let $\struct {\mathbb G, \subseteq}$ be the complete lattice formed by $\mathbb G$ and $\subseteq$.
From Set of Subgroups forms Complete Lattice, $\struct {\mathbb G, \subseteq}$ indeed forms a complete lattice.
By definition, $\struct {\mathbb G, \subseteq}$ is a sublattice of $\struct {\powerset G, \subseteq}$.
Let $H, K \in \mathbb G$.
We have that:
:$H \cup K$ is the supremum of $H$ and $K$ in $\struct {\powerset G, \subseteq}$.
But from Union of Subgroups, $H \cup K$ is not a subgroup of $G$ element of $\mathbb G$.
However, from Supremum of Subgroups in Lattice, we do have that the subset product $H K$ in $G$ is a subgroup of $G$ such that:
:$\sup_{\mathbb G} \set {H, K} = H K$
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings
Category:Suprema
Category:Sublattices
Supremum of Empty Set is Smallest Element

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Then:
:the supremum of the empty set exists the smallest element of $S$ exists
in which case:
:$\map \sup \O$ is the smallest element of $S$ 
== Proof ==
Observe that, vacuously, any $s \in S$ is an upper bound for $\O$.
=== Necessary Condition ===
Let $\map \sup \O$ exist.
For any upper bound $s$ of $\O$, $\map \sup \O \preceq s$ by definition of supremum.
Hence:
:$\forall s \in S: \map \sup \O \preceq s$
=== Sufficient Condition ===
Let $t$ be the smallest element of $S$.
Then $t$ is an upper bound of $\O$.
For any upper bound $s$ of $\O$, $t \preceq s$ by definition of the smallest element.
By definition of the supremum:
:$t = \map \sup \O$
== Also see ==
* Infimum of Empty Set is Greatest Element
== Sources ==
* : Chapter $\text I$: Preliminaries, Definition $1.2$
Category:Suprema
Category:Empty Set
Supremum of Function is less than Supremum of Greater Function

== Theorem ==
Let $f$ and $g$ be real functions.
Let $S$ be a subset of $\Dom f \cap \Dom g$.
Let $\map f x \le \map g x$ for every $x \in S$.
Let $\ds \sup_{x \mathop \in S} \map g x$ exist.
Then $\ds \sup_{x \mathop \in S} \map f x$ exists and:
:$\ds \sup_{x \mathop \in S} \map f x \le \sup_{x \mathop \in S} \map g x$.
== Proof ==
We have:
Supremum of Sum equals Sum of Suprema also gives that $\sup f$ and $\sup \paren {g - f}$ exist.
We have:
Category:Real Analysis
Category:Suprema
Supremum of Ideals is Increasing

== Theorem ==
Let $L = \left({S, \preceq}\right)$ be an up-complete ordered set.
Let $\mathit{Ids}\left({L}\right)$ be the set of all ideals in $L$.
Let $P = \left({\mathit{Ids}\left({L}\right), \precsim}\right)$ be an ordered set where $\mathord \precsim = \subseteq\restriction_{\mathit{Ids}\left({L}\right)\times \mathit{Ids}\left({L}\right)}$
Let $f: \mathit{Ids}\left({L}\right) \to S$ be a mapping such that
:$\forall I \in \mathit{Ids}\left({L}\right): f\left({I}\right) = \sup I$
Then $f$ is an increasing mapping.
== Proof ==
Let $I, J \in \mathit{Ids}\left({L}\right)$ such that
:$I \precsim J$
By definition of $\precsim$:
:$I \subseteq J$
By definition of up-complete:
:$I$ and $J$ admit suprema in $L$.
By Supremum of Subset:
:$\sup I \preceq \sup J$
Thus by definition of $f$:
:$f\left({I}\right) \preceq f\left({J}\right)$
Hence $f$ is an increasing mapping.
== Sources ==
* 
* 
Category:Increasing Mappings
Supremum of Ideals is Upper Adjoint

== Theorem ==
Let $L = \struct {S, \vee, \preceq}$ be a bounded below continuous join semilattice.
Let $\map {\mathit {Ids} } L$ be the set of all ideals in $L$.
Let $P = \struct {\map {\mathit {Ids} } L, \precsim}$ be an ordered set where $\mathord \precsim = \subseteq \restriction_{\map {\mathit {Ids} } L \times \map {\mathit {Ids} } L}$
Let $f: \map {\mathit {Ids} } L \to S$ be a mapping such that
:$\forall I \in \map {\mathit {Ids} } L: \map f I = \sup I$
Then $f$ is an upper adjoint of Galois connection.
== Proof ==
Define $d: S \to \map {\mathit {Ids} } L$
:$\forall t \in S: \map d t = \map \inf {f^{-1} \sqbrk {t^\succeq} }$
where
:$t^\succeq$ denotes the upper closure of $t$,
:$f^{-1} \sqbrk {t^\succeq} $ denotes the image of $t^\succeq$ over $f^{-1}$.
We will prove that
:$\forall t \in S: \map d t = \map \min {f^{-1} \sqbrk {t^\succeq} }$
Let $t \in S$.
By Continuous iff For Every Element There Exists Ideal Element Precedes Supremum:
:there exists an ideal $J$ in $L$ such that
::$t \preceq \sup J$ and for every ideal $K$ in $L$: $t \preceq \sup K \implies J \subseteq K$
We will prove that
:$\forall K \in \map {\mathit {Ids} } L: K$ is lower bound for $f^{-1} \sqbrk {t^\succeq} \implies K \precsim J$
Let $K \in \map {\mathit {Ids} } L$ such that
:$K$ is a lower bound for $f^{-1} \sqbrk {t^\succeq}$
By definition of $f$:
:$t \preceq \map f J$
By definition of upper closure of element:
:$\map f J \in t^\succeq$
By definition of image of set:
:$J \in f^{-1} \sqbrk {t^\succeq}$
Thus by definition of lower bound:
:$K \precsim J$
We will prove that:
:$J$ is a lower bound for $f^{-1} \sqbrk {t^\succeq}$
Let $K \in \map {\mathit {Ids} } L$ such that:
:$K \in f^{-1} \sqbrk {t^\succeq}$
By definition of image of set:
:$\map f K \in t^\succeq$
By definition of upper closure of element:
:$t \preceq \map f K$
By definition of $f$:
:$t \preceq \sup K$
Then
:$J \subseteq K$
Thus by definition of $\precsim$:
:$J \precsim K$
By definition of supremum:
:$t \preceq \map \sup {\map \inf {f^{-1} \sqbrk {t^\succeq} } }$
By definition of $f$:
:$t \preceq \map f {\map \inf {f^{-1} \sqbrk {t^\succeq} } }$
By definition of upper closure of element:
:$\map f {\map \inf {f^{-1} \sqbrk {t^\succeq} } } \in t^\succeq$
By definition of image of set:
:$\map \inf {f^{-1} \sqbrk {t^\succeq} } \in f^{-1} \sqbrk {t^\succeq}$
Thus by definition of smallest element:
:$\map d t = \map \min {f^{-1} \sqbrk {t^\succeq} }$
By Supremum of Ideals is Increasing:
:$f$ is an increasing mapping.
By Galois Connection is Expressed by Minimum:
:$\struct {f, d}$ is a Galois connection.
Hence $f$ is an upper adjoint of Galois connection.
== Sources ==
* 
* 
Category:Continuous Lattices
Category:Galois Connections
Supremum of Ideals is Upper Adjoint implies Lattice is Continuous

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded below up-complete lattice.
Let $\map {\mathit {Ids} } L$ be the set of all ideals in $L$.
Let $P = \struct {\map {\mathit {Ids} } L, \precsim}$ be an ordered set where $\mathord \precsim = \subseteq \restriction_{\map {\mathit {Ids} } L \times \map {\mathit {Ids} } L}$
Let $f: \map {\mathit {Ids} } L \to S$ be a mapping such that
:$\forall I \in \map {\mathit {Ids} } L: f \sqbrk I = \sup I$
Let $f$ be an upper adjoint of a Galois connection.
Then $L$ is continuous.
== Proof ==
We will prove that
:$\forall x \in S: \exists I \in \map {\mathit {Ids} } L: x \preceq \sup I \land \forall J \in \map {\mathit {Ids} } L: x \preceq \sup J \implies I \subseteq J$ 
Let $x \in S$.
Define $I := \map \inf {f^{-1} \sqbrk {x^\succeq} }$.
By definition of $P$:
:$I \in \map {\mathit {Ids} } L$
We will prove that
:$\forall J \in \map {\mathit {Ids} } L: x \preceq \sup J \implies I \subseteq J$
Let $J \in \map {\mathit {Ids} } L$ such that
:$x \preceq \sup J$
By definition of $f$:
:$x \preceq f \sqbrk J$
By definition of upper closure of element:
:$f \sqbrk J \in x^\succeq$
By definition of image of set:
:$J \in f^{-1} \sqbrk {x^\succeq}$
By definition of infimum:
:$I \precsim J$
Hence by definition of $\precsim$:
:$I \subseteq J$
By definition of upper adjoint of a Galois connection:
:there exists a mapping $d: S \to \map {\mathit {Ids} } L$: $\struct {f, d}$ is a Galois connection.
By Galois Connection is Expressed by Minimum
:$\map d x = \map \min {f^{-1} \sqbrk {x^\succeq} }$
By definition of smallest element:
:$I \in f^{-1} \sqbrk {x^\succeq}$
By definition of image of set:
:$f \sqbrk I \in x^\succeq$
By definition of upper closure of element:
:$x \preceq f \sqbrk I$
Thus by definition of $f$:
:$x \preceq \sup I$
Hence
:$\exists I \in \map {\mathit {Ids} } L: x \preceq \sup I \land \forall J \in \map {\mathit {Ids} } L: x \preceq \sup J \implies I \subseteq J$ 
Hence by Continuous iff For Every Element There Exists Ideal Element Precedes Supremum:
:$L$ is continuous.
== Sources ==
* 
* 
Category:Continuous Lattices
Category:Galois Connections
Supremum of Lower Closure of Element

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $s$ be an element of $S$.
Then:
:$\map \sup {s^\preceq} = s$
where $s^\preceq$ denotes the lower closure of $s$.
== Proof ==
== Sources ==
* 
* 
Category:Lower Closures
Supremum of Lower Closure of Set

== Theorem ==
Let $\left({S, \preceq}\right)$ be an ordered set.
Let $T \subseteq S$.
Let $L = T^\preceq$ be the lower closure of $T$ in $S$.
Let $s \in S$
Then $s$ is the supremum of $T$ it is the supremum of $L$.
== Proof ==
By Supremum and Infimum are Unique we need only show that $s$ is a supremum of $L$ it is a supremum of $T$. 
=== Forward Implication ===
Let $s$ be a supremum of $T$.
$s$ is an upper bound of $L$:
Let $l \in L$.
Then by the definition of lower closure, there is a $t \in T$ such that $l \preceq t$.
By the definition of supremum, $s$ is an upper bound of $T$.
Thus $t \preceq s$.
Since $\preceq$ is transitive, $l \preceq s$.
Since this holds for all $l \in L$, $s$ is an upper bound of $L$.
Suppose that $u$ is an upper bound of $L$.
Then since $T \subseteq L$, $u$ is an upper bound of $T$.
Thus by the definition of supremum, $s \preceq u$.
So:
: $s$ is an upper bound of $L$
: $s$ precedes all upper bounds of $L$
Thus it follows that $s$ is the supremum of $L$.
=== Reverse Implication ===
Let $s$ be a supremum of $L$.
Then $s$ is an upper bound of $L$.
Since $T \subseteq L$, $s$ is an upper bound of $T$.
Let $u$ be any upper bound of $T$ and let $l \in L$.
Then by the definition of lower closure, there is a $t \in T$ such that $l \preceq t$.
Then since $u$ is an upper bound of $T$, $t \preceq u$.
Thus since $\preceq$ is transitive, $l \preceq u$.
Thus $u$ is an upper bound of $L$.
By the definition of supremum, $s \preceq u$.
Thus $s$ is an upper bound of $T$ which precedes every upper bound of $T$.
Therefore $s$ is the supremum of $T$.
Category:Lower Closures
Supremum of Lower Sums Never Greater than Upper Sum

== Theorem ==
Let $\closedint a b$ be a closed real interval .
Let $f$ be a bounded real function defined on $\closedint a b$.
Let $S$ be a finite subdivision of $\closedint a b$.
Let $\map U S$ be the upper Darboux sum of $f$ on $\closedint a b$ with respect to $S$.
Let $\map L P$ be the lower Darboux sum of $f$ on $\closedint a b$ with respect to a finite subdivision $P$.
Then:
:$\sup_P \map L P \le \map U S$
== Proof ==
From Upper Sum Never Smaller than Lower Sum for any Pair of Subdivisions, $\map U S$ is an upper bound for the real set:
:$T = \leftset {\map L P: P}$ is a finite subdivision of $\rightset {\closedint a b}$
Since $\sup_P \map L P$ is the supremum of $T$:
:$\sup_P \map L P \le \map U S$
Hence the result.
Category:Real Analysis
Supremum of Meet Image of Directed Set

== Theorem ==
Let $\struct {S, \preceq}$ be an up-complete meet semilattice.
Let $f: S \times S \to S$ be a mapping such that:
:$\forall \tuple {x, y} \in S \times S: \map f {x, y} = x \wedge y$
Let $D$ be directed subset of $S \times S$ in the simple order product $\struct {S \times S, \precsim}$ of $\struct {S, \preceq}$ and $\struct {S, \preceq}$.
Then:
:$\map \sup {\map {f^\to} D} = \sup \set {x \wedge y: x \in \map {\pr_1^\to} D, y \in \map {\pr_2^\to} D}$
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 definition of image of set:
:$\map {f^\to} D = \set {x \wedge y: \tuple {x, y} \in D}$
By definition of subset:
:$\map {f^\to} D \subseteq \set {x \wedge y: x \in \map {\pr_1^\to} D, y \in \map {\pr_2^\to} D}$
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 Meet of Directed Subsets is Directed:
:$\set {x \wedge y: x \in D_1, y \in D_2}$ is directed.
By definition of up-complete:
:$\set {x \wedge y: x \in D_1, y \in D_2}$ admits a supremum.
By Meet is Increasing:
:$f$ is increasing mapping.
By Image of Directed Subset under Increasing Mapping is Directed:
:$\map {f^\to} D$ is directed.
By definition of up-complete:
:$\map {f^\to} D$ admits a supremum.
By Supremum of Subset:
:$\map \sup {\map {f^\to} D} \preceq \sup \set {x \wedge y: x \in D_1, y \in D_2}$
We will prove that
:$\set {x \wedge y: x \in D_1, y \in D_2} \subseteq \paren {\map {f^\to} D}^\preceq$
where
:$\paren {\map {f^\to} D}^\preceq$ denotes the lower closure of $\map {f^\to} D$.
Let $z \in \set {x \wedge y: x \in D_1, y \in D_2}$.
Then
:$\exists x \in D_1, y \in D_2: z = x \wedge y$
By definition of image of set:
:$\exists \tuple {a, b} \in D: \map {\pr_1} {a, b} = x$
and
:$\exists \tuple {c, d} \in D: \map {\pr_2} {c, d} = y$:
By definition of first projection and second projection:
:$a = x$ and $d = y$
By definition of directed subset:
:$\exists \tuple {g, h} \in D: \tuple {x, b} \precsim \tuple {g, h} \land \tuple {c, y} \precsim \tuple {g, h}$
By definition of simple order product:
:$x \preceq g$ and $y \preceq h$
By Meet Semilattice is Ordered Structure:
:$x \wedge y \preceq g \wedge h \in \set {x \wedge y: \tuple {x, y} \in D}$
Thus by definition of lower closure:
:$z \in \paren {\map {f^\to} D}^\preceq$
By Supremum of Lower Closure of Set:
:$\paren {\map {f^\to} D}^\preceq$ admits a supremum
and
:$\map \sup {\map {f^\to} D}^\preceq = \map \sup {\map {f^\to} D}$
By Supremum of Subset:
:$\sup \set {x \wedge y: x \in D_1, y \in D_2} \preceq \map \sup {\map {f^\to} D}$
Thus by definition of antisymmetry:
:$\map \sup {\map {f^\to} D} = \sup \set {x \wedge y: x \in D_1, y \in D_2}$
== Sources ==
* 
* 
Category:Up-Complete Semilattices
Supremum of Power Set

== Theorem ==
Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Let $\struct {\powerset S, \subseteq}$ be the relational structure defined on $\powerset S$ by the relation $\subseteq$.
(From Subset Relation on Power Set is Partial Ordering, this is an ordered set.)
Then the supremum of $\struct {\powerset S, \subseteq}$ is the set $S$.
== Proof ==
By the definition of the power set:
:$\forall X \in \powerset S: X \subseteq S$
The result then follows from the definition of supremum.
Category:Power Set
Category:Order Theory
Supremum of Set Equals Maximum of Suprema of Subsets

== Theorem ==
Let $S$ be a non-empty real set.
Let $\set {S_i: i \in \set {1, 2, \ldots, n} }$, $n \in \N_{>0}$, be a set of non-empty subsets of $S$.
Let $S = \bigcup S_i$.
Then:
:$S_i$ has a supremum for every $i$ in $\set {1, 2, \ldots, n}$
:
:$S$ has a supremum
and, in either case:
:$\sup S = \max \set {\sup S_1, \sup S_2, \ldots, \sup S_n}$
== Proof ==
=== Necessary Condition ===
Let:
:$S$ have a supremum.
We need to show that:
:$(1): \quad S_i$ has a supremum for every $i$ in $\set {1, 2, \ldots, n}$
:$(2): \quad \sup S = \max \set {\sup S_1, \sup S_2, \ldots, \sup S_n}$
By Supremum of Set of Real Numbers is at least Supremum of Subset, $\sup S_i$ exists for every $i$ in $\set {1, 2, \ldots, n}$.
This proves $(1)$.
Also by Supremum of Set of Real Numbers is at least Supremum of Subset, $\sup S \ge \sup S_i$ for every $i$ in $\set {1, 2, \ldots, n}$.
Therefore:
:$\sup S \ge \max \set {\sup S_1, \sup S_2, \ldots, \sup S_n}$
In addition, by Supremum of Subset of Union Equals Supremum of Union:
:$\sup S = \sup S_j$ for a $j$ in $\set {1, 2, \ldots, n}$
Therefore:
:$\sup S \le \max \set {\sup S_1, \sup S_2, \ldots, \sup S_n}$
Accordingly:
:$\sup S = \max \set {\sup S_1, \sup S_2, \ldots, \sup S_n}$
This proves $(2)$.
=== Sufficient Condition ===
Let:
:$S_i$ have a supremum for every $i$ in $\set {1, 2, \ldots, n}$.
We need to show that:
:$(3): \quad S$ has a supremum
:$(4): \quad \sup S = \max \set {\sup S_1, \sup S_2, \ldots, \sup S_n}$
Every $S_i$ has a supremum.
Therefore, $\max \set {\sup S_1, \sup S_2, \ldots, \sup S_n}$ exists as a real number.
The number $\max \set {\sup S_1, \sup S_2, \ldots, \sup S_n}$ is an upper bound for every $S_i$.
Therefore, $\max \set {\sup S_1, \sup S_2, \ldots, \sup S_n}$ is an upper bound for $S$ as $S = \bigcup S_i$.
By the Continuum Property $S$ has a supremum as $S$ is a non-empty real set with an upper bound.
This proves $(3)$.
We have proved that $S$ has a supremum.
By Supremum of Set of Real Numbers is at least Supremum of Subset:
:$\sup S \ge \sup S_i$ for every $i$ in $\set {1, 2, \ldots, n}$
Therefore:
:$\sup S \ge \max \set {\sup S_1, \sup S_2, \ldots, \sup S_n}$
In addition, by Supremum of Subset of Union Equals Supremum of Union:
:$\sup S = \sup S_j$ for a $j$ in $\set {1, 2, \ldots, n}$
Therefore:
:$\sup S \le \max \set {\sup S_1, \sup S_2, \ldots, \sup S_n}$
Accordingly:
:$\sup S = \max \set {\sup S_1, \sup S_2, \ldots, \sup S_n}$
This proves $(4)$.
== Also see ==
* Supremum of Suprema (a more general result)
Category:Real Analysis
Supremum of Set of Integers equals Greatest Element

== Theorem ==
Let $S \subset \Z$ be a non-empty subset of the set of integers.
Let $S$ be bounded above in the set of real numbers $\R$.
Then $S$ has a greatest element, and it is equal to the supremum $\sup S$.
== Proof ==
By Set of Integers Bounded Above by Real Number has Greatest Element, $S$ has a greatest element, say $n \in S$.
By Greatest Element is Supremum, $n$ is the supremum of $S$.
== Also see ==
* Infimum of Set of Integers equals Smallest Element
* Supremum of Set of Integers is Integer
* Supremum is Unique
Category:Integers
Supremum of Set of Integers is Integer

== Theorem ==
Let $S \subset \Z$ be a non-empty subset of the set of integers.
Let $S$ be bounded above in the set of real numbers.
Then its supremum $\sup S$ is an integer.
== Proof ==
By Supremum of Set of Integers equals Greatest Element, $S$ has a greatest element $n \in \Z$, that is equals to the supremum of $S$.
== Also see ==
* Infimum of Set of Integers is Integer
* Supremum is Unique
Category:Integers
Supremum of Set of Real Numbers is at least Supremum of Subset

== Theorem ==
Let $S$ be a set of real numbers.
Let $S$ have a supremum.
Let $T$ be a non-empty subset of $S$.
Then $\sup T$ exists and:
:$\sup T \le \sup S$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Sources ==
* : $\S 2$: Continuum Property: Exercise $\S 2.13 \ (1)$
Category:Real Analysis
Category:Supremum of Set of Real Numbers is at least Supremum of Subset
Supremum of Set of Real Numbers is at least Supremum of Subset/Proof 1

== Theorem ==
== Proof ==
The number $\sup S$ is an upper bound for $S$.
Therefore, $\sup S$ is an upper bound for $T$ as $T$ is a non-empty subset of $S$.
Accordingly, $T$ has a supremum by the Continuum Property.
The number $\sup S$ is an upper bound for $T$.
Therefore, $\sup S$ is greater than or equal to $\sup T$ as $\sup T$ is the least upper bound of $T$.
Category:Supremum of Set of Real Numbers is at least Supremum of Subset