nyuuzyou/wikis huggingface dataset:
Supremum of Set of Real Numbers is at least Supremum of Subset/Proof 3

== Theorem ==
== Proof ==
$S$ is bounded above as $S$ has a supremum.
Therefore, $T$ is bounded above as $T$ is a subset of $S$.
Accordingly, $T$ admits a supremum by the Continuum Property as $T$ is non-empty.
We know that $\sup T$ and $\sup S$ exist.
Therefore by Suprema of two Real Sets:
:$\forall \epsilon \in \R_{>0}: \forall t \in T: \exists s \in S: t < s + \epsilon \iff \sup T \le \sup S$
We have:
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 4

== Theorem ==
== Proof ==
By definition $\sup S$ is an upper bound for $S$.
Thus:
:$\forall x \in S: x \le \sup S$
As $T \subseteq S$ we have by definition of subset that:
:$\forall x \in T: x \in S$
Hence:
:$\forall x \in T: x \le \sup S$
So by definition $\sup S$ is an upper bound for $T$.
So $\sup S$ is at least as big as the smallest upper bound for $T$
Thus by definition of supremum:
:$\sup T \le \sup S$
Category:Supremum of Set of Real Numbers is at least Supremum of Subset
Supremum of Simple Order Product

== Theorem ==
Let $\struct {S_1, \preceq_1}$ and $\struct {S_2, \preceq_2}$ be ordered sets.
Let $\struct {S_1 \times S_2, \precsim}$ be the simple order product of $\struct {S_1, \preceq_1}$ and $\struct {S_2, \preceq_2}$.
Let $X_1$ be a non-empty subset of $S_1$, $X_2$ be a non-empty subset of $S_2$ such that
:$X_1$ and $X_2$ admit suprema.
Then:
:$X_1 \times X_2$ admits a supremum
and:
:$\map \sup {X_1 \times X_2} = \tuple {\sup X_1, \sup X_2}$
== Proof ==
We will prove that:
:$\tuple {\sup X_1, \sup X_2}$ is upper bound for $X_1 \times X_2$
Let $\tuple {a, b} \in X_1 \times X_2$.
By definition of Cartesian product:
:$a \in X_1$ and $b \in X_2$
By definitions of supremum and upper bound:
:$a \preceq_1 \sup X_1$ and $b \preceq_2 \sup X_2$
Thus by definition of simple order product:
:$\tuple {a, b} \precsim \tuple {\sup X_1, \sup X_2}$
We will prove that:
:$\forall \tuple {a, b} \in S1 \times S_2: \tuple {a, b}$ is upper bound for $X_1 \times X_2 \implies \tuple {\sup X_1, \sup X_2} \precsim \tuple {a, b}$
Let $\tuple {a, b} \in S1 \times S_2$ such that:
:$\tuple {a, b}$ is upper bound for $X_1 \times X_2$
We will prove as sublemma that:
:$a$ is upper bound for $X_1$
Let $c \in X_1$.
By definition of non-empty set:
:$\exists d: d \in X_2$
By definition of Cartesian product:
:$\tuple {c, d} \in X_1 \times X_2$
By definition of upper bound:
:$\tuple {c, d} \precsim \tuple {a, b}$
Thus by definition of simple order product:
:$c \preceq_1 a$
This ends the proof of sublemma.
Analogically we have that:
:$b$ is upper bound for $X_2$
By definition of supremum:
:$\sup X_1 \preceq_1 a$ and $\sup X_2 \preceq_2 b$
Thus by definition of simple order product:
:$\tuple {\sup X_1, \sup X_2} \precsim \tuple {a, b}$
Thus by definition:
:$X_1 \times X_2$ admits a supremum
and:
:$\map \sup {X_1 \times X_2} = \tuple {\sup X_1, \sup X_2}$
== Sources ==
* 
* 
Category:Simple Order Product
Category:Suprema
Supremum of Singleton

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Then for all $a \in S$:
:$\sup \set a = a$
where $\sup$ denotes supremum.
== Proof ==
Since $a \preceq a$, $a$ is an upper bound of $\set a$.
Let $b$ be another upper bound of $\set a$.
Then necessarily $a \preceq b$.
It follows that indeed:
:$\sup \set a = a$
as desired.
== Also see ==
* Infimum of Singleton
Category:Suprema
Category:Singletons
Supremum of Subgroups in Lattice

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $\mathbb G$ be the set of all subgroups of $G$.
Let $\struct {\mathbb G, \subseteq}$ be the complete lattice formed by $\mathbb G$ and $\subseteq$.
Let $H, K \in \mathbb G$.
Let either $H$ or $K$ be normal in $G$.
Then:
:$\sup \set {H, K} = H \circ K$
where $H \circ K$ denotes subset product.
== Proof ==
Recall that Set of Subgroups forms Complete Lattice.
Let $L = \sup \set {H, K}$.
Let either $H$ or $K$ be normal in $G$.
Since $L$ contains $H$ and $K$, then $L$ contains $H \circ K$.
The smallest subgroup of $G$ containing $H$ and $K$ is:
:$\gen {H, K}$
the subgroup generated by $H$ and $K$.
From Subset Product with Normal Subgroup as Generator:
:$\gen {H, K} = H \circ K$
when either $H$ or $K$ is normal.
The result follows.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Theorem $14.6$
Category:Complete Lattices
Category:Suprema
Category:Subgroups
Supremum of Subset

== Theorem ==
Let $\left({U, \preceq}\right)$ be an ordered set.
Let $S \subseteq U$.
Let $T \subseteq S$.
Let $S$ admit a supremum (in $U$).
If $T$ also admits a supremum (in $U$), then $\sup \left({T}\right) \preceq\sup \left({S}\right)$.
== Proof ==
Let $B = \sup \left({S}\right)$.
Then $B$ is an upper bound for $S$.
As $T \subseteq S$, it follows by the definition of a subset that $x \in T \implies x \in S$.
Because $x \in S \implies x \preceq B$ (as $B$ is an upper bound for $S$) it follows that $x \in T \implies x \preceq B$.
So $B$ is an upper bound for $T$.
Therefore $B$ succeeds the supremum of $T$ in $S$.
Hence the result.
Category:Order Theory
Supremum of Subset Product in Ordered Group

== Theorem ==
Let $\struct {G, \circ, \preceq}$ be an ordered group.
Suppose that subsets $A$ and $B$ of $G$ admit suprema in $G$.
Then:
:$\sup \paren {A \circ_\PP B} = \sup A \circ \sup B$
where $\circ_\PP$ denotes subset product.
== Proof ==
Let $a \in A$, $b \in B$.
Then:
Hence $\sup A \circ \sup B$ is an upper bound for $A \circ_\PP B$.
Suppose that $u$ is an upper bound for $A \circ_\PP B$.
Then:
Therefore:
:$\sup \paren {A \circ_\PP B} = \sup A \circ \sup B$
== Also see ==
* Infimum of Subset Product in Ordered Group
Category:Suprema
Category:Subset Products
Category:Ordered Groups
Supremum of Subset of Bounded Above Set of Real Numbers

== Theorem ==
Let $A$ and $B$ be sets of real numbers such that $A \subseteq B$.
Let $B$ be bounded above.
Then:
:$\sup A \le \sup B$
where $\sup$ denotes the supremum.
== Proof ==
Let $B$ be bounded above.
By the Continuum Property, $B$ admits a supremum.
By Subset of Bounded Above Set is Bounded Above, $A$ is also bounded above.
Hence also by the Continuum Property, $A$ also admits a supremum.
 $\sup A > \sup B$.
Then:
:$\exists y \in A: y > \sup B$
Thus by definition of supremum, $y \notin B$.
That is:
:$A \nsubseteq B$
which contradicts our initial assumption that $A \subseteq B$.
Hence the result by Proof by Contradiction.
== Sources ==
* : $1$: Review of some real analysis: Exercise $1.5: 1$
Category:Bounded Above Sets of Real Numbers
Category:Suprema
Supremum of Subset of Real Numbers/Examples

== Examples of Suprema of Subsets of Real Numbers ==
=== Example 1 ===
=== Example 2 ===
=== Example 3 ===
=== Example 4 ===
=== Example: $\openint \gets 0$ ===
=== Example: $\hointl 0 1$ ===
=== Example 5 ===
=== Example 6 ===
=== Example 7 ===
=== Example 8 ===
Category:Examples of Suprema
Supremum of Subset of Real Numbers/Examples/Example 1

== Example of Supremum of Subset of Real Numbers ==
The subset $S$ of the real numbers $\R$ defined as:
:$S = \set {1, 2, 3}$
admits a supremum:
:$\sup S = 3$
== Sources ==
* : $2$: Continuum Property: $\S 2.5$: Examples: $\text{(i)}$
Category:Examples of Suprema
Supremum of Subset of Real Numbers/Examples/Example 2

== Example of Supremum of Subset of Real Numbers ==
The subset $T$ of the real numbers $\R$ defined as:
:$T = \set {x \in \R: 1 \le x \le 2}$
admits a supremum:
:$\sup T = 2$
== Sources ==
* : $\S 2$: Continuum Property: $\S 2.5$: Examples: $\text{(ii)}$
Category:Examples of Suprema
Supremum of Subset of Real Numbers/Examples/Example 3

== Example of Supremum of Subset of Real Numbers ==
The subset $V$ of the real numbers $\R$ defined as:
:$V := \set {x \in \R: x > 0}$
does not admit a supremum.
== Proof ==
 $x \in \R$ is a supremum for $V$.
Then we have that:
:$x + 1 \in V$
and it is seen that $x$ is not a supremum after all.
== Sources ==
* : $\S 2$: Continuum Property: $\S 2.5$: Examples: $\text{(iii)}$
Category:Examples of Suprema
Supremum of Subset of Real Numbers/Examples/Example 4

== Example of Supremum of Subset of Real Numbers ==
Consider the set $A$ defined as:
:$A = \set {3, 4}$
Then the supremum of $A$ is $4$.
However, $A$ contains no element $x$ such that:
:$3 < x < 4$.
== Sources ==
* : $\S 2$: Continuum Property: Exercise $\S 2.10 \ (5)$
Category:Examples of Suprema
Supremum of Subset of Real Numbers/Examples/Example 5

== Example of Supremum of Subset of Real Numbers ==
The subset $S$ of the real numbers $\R$ defined as:
:$S = \set {x \in \R: x^2 \le 2 x - 1}$
admits a supremum:
:$\sup S = 1$
such that $\sup S \in S$.
== Proof ==
But as $\forall x \in \R: \paren {x - 1}^2 \ge 0$ it follows that:
:$\paren {x - 1}^2 \le 0 \implies x - 1 = 0$
and so:
:$\set {x \in \R: x^2 \le 2 x - 1} = \set 1$
and so trivially:
:$\sup \set {x \in \R: x^2 \le 2 x - 1} = 1$
and as $1 \in \set 1$ it follows that $\sup S \in S$.
== Sources ==
* : $1$: Review of some real analysis: Exercise $1.5: 4$
Category:Examples of Suprema
Supremum of Subset of Real Numbers/Examples/Example 6

== Example of Supremum of Subset of Real Numbers ==
The subset $S$ of the real numbers $\R$ defined as:
:$S = \set {x \in \R: x^2 + 2 x \le 1}$
admits a supremum:
:$\sup S = -1 + \sqrt 2$
such that $\sup S \in S$.
== Proof ==
When $x < -1 - \sqrt 2$ we have that both $x + 1 - \sqrt 2 < 0$ and $x + 1 + \sqrt 2 < 0$, and so $x^2 + 2 x - 1 > 0$.
When $x > -1 + \sqrt 2$ we have that both $x + 1 - \sqrt 2 > 0$ and $x + 1 + \sqrt 2 > 0$, and so $x^2 + 2 x - 1 > 0$.
When $-1 - \sqrt x \le x \le -1 + \sqrt 2$, we have that $x + 1 - \sqrt 2 \ge 0$ and $x + 1 + \sqrt 2 \le 0$, and so $x^2 + 2 x - 1 < 0$
Hence:
:$S = \set {x \in \R: -1 - \sqrt x \le x \le -1 + \sqrt 2}$
and it follows that:
:$\sup \set {x \in \R: x^2 + 2 x \le 1} = -1 + \sqrt 2$
and it follows that $\sup S \in S$.
== Sources ==
* : $1$: Review of some real analysis: Exercise $1.5: 4$
Category:Examples of Suprema
Supremum of Subset of Real Numbers/Examples/Example 7

== Example of Supremum of Subset of Real Numbers ==
The subset $S$ of the real numbers $\R$ defined as:
:$S = \set {x \in \R: x^3 < 8}$
admits a supremum:
:$\sup S = 2$
such that $\sup S \notin S$.
== Proof ==
Hence:
:$S = \set {x \in \R: x < 2}$
and it follows that:
:$\sup \set {x \in \R: x^3 < 8} = 2$
and it follows that $\sup S \notin S$.
== Sources ==
* : $1$: Review of some real analysis: Exercise $1.5: 4$
Category:Examples of Suprema
Supremum of Subset of Real Numbers/Examples/Example 8

== Example of Supremum of Subset of Real Numbers ==
The subset $S$ of the real numbers $\R$ defined as:
:$S = \set {x \in \R: x \sin x < 1}$
does not admit a supremum.
== Proof ==
Let $x = n \pi$.
From Sine of Multiple of Pi, we have that:
:$\sin x = 0$
and so:
:$x \sin x = 0$
That is:
:$\forall n \in \Z: n \pi \sin n \pi < 1$
and $n \pi$ is not bounded above.
== Sources ==
* : $1$: Review of some real analysis: Exercise $1.5: 4$
Category:Examples of Suprema
Supremum of Subset of Real Numbers/Examples/Left Half Open Interval from 0 to 1

== Example of Supremum of Subset of Real Numbers ==
Let $\hointl 0 1$ denote the left half-open real interval:
:$\hointl 0 1 := \set {x \in \R: 0 < x \le 1}$
Then the supremum of $\R_{<0}$ is $1$.
== Proof ==
We have that $1$ is an upper bound of $\hointl 0 1$.
Let $0 < x \le 1$.
Then $x \in I$.
Then:
:$x < \dfrac {1 + x} 2 < 1$
and so $x$ is not an upper bound of $\R_{<0}$.
Hence the result.
Here we see that the supremum of $\hointl 0 1$ is in this case an element of $\hointl 0 1$.
== Sources ==
* : $1$: Review of some real analysis: $\S 1.1$: Real Numbers: Example $1.1.3$
Category:Examples of Suprema
Supremum of Subset of Real Numbers/Examples/Strictly Negative Real Numbers

== Example of Supremum of Subset of Real Numbers ==
Let $\R_{<0}$ be the (strictly) negative real numbers:
:$\R_{<0} := \openint \gets 0$
Then the supremum of $\R_{<0}$ is $0$.
== Proof ==
We have that $0$ is an upper bound of $\R_{<0}$.
Let $x < 0$.
Then $x \in I$.
Then
:$\dfrac x 2 > x$
while:
:$\dfrac x 2 \in \R_{<0}$
and so $x$ is not an upper bound of $\R_{<0}$.
Hence the result.
Note that the supremum of $\R_{<0}$ is in this case not actually an element of $\R_{<0}$.
== Sources ==
* : $1$: Review of some real analysis: $\S 1.1$: Real Numbers: Example $1.1.3$
Category:Examples of Suprema
Supremum of Subset of Real Numbers May or May Not be in Subset

== Theorem ==
Let $S \subset \R$ be a proper subset of the set $\R$ of real numbers.
Let $S$ admit a supremum $M$.
Then $M$ may or may not be an element of $S$.
== Proof ==
Consider the subset $S$ of the real numbers $\R$ defined as:
:$S = \set {\dfrac 1 n: n \in \Z_{>0} }$
It is seen that:
:$S = \set {1, \dfrac 1 2, \dfrac 1 3, \ldots}$
and hence $\sup S = 1$.
Thus $\sup S \in S$.
Consider the subset $T$ of the real numbers $\R$ defined as:
:$T = \set {-\dfrac 1 n: n \in \Z_{>0} }$
It is seen that:
:$T = \set {-1, -\dfrac 1 2, -\dfrac 1 3, \ldots}$
and hence $\sup T = 0$.
Thus $\sup T \notin T$.
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Real Numbers: $1.35$. Example $\text{(a)}$
Category:Suprema
Supremum of Subset of Real Numbers is Arbitrarily Close

== Theorem ==
Let $A \subseteq \R$ be a subset of the real numbers.
Let $b$ be a supremum of $A$.
Let $\epsilon \in \R_{>0}$.
Then:
:$\exists x \in A: b − x < \epsilon$
== Proof ==
Note that $A$ is non-empty as the empty set does not admit a supremum (in $\R$).
Suppose $\epsilon \in \R_{>0}$ such that:
:$\forall x \in A: b − x \ge \epsilon$
Then:
:$\forall x \in A: b − \epsilon \ge x$
and so $b − \epsilon$ would be an upper bound of $A$ which is less than $b$.
But since $b$ is a supremum of $A$ there can be no such $b − \epsilon$.
From that contradiction it follows that:
:$\exists x \in A: b − x < \epsilon$
== Also see ==
* Infimum of Subset of Real Numbers is Arbitrarily Close
Category:Real Analysis
Supremum of Subset of Union Equals Supremum of Union

== Theorem ==
Let $S$ be a non-empty real set.
Let $S$ have a supremum.
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 $\bigcup S_i = S$.
Then there exists a $j$ in $\set {1, 2, \ldots, n}$ such that:
:$\sup S_j = \sup S$
== Proof ==
If $S$ equals $S_j$ for a $j$ in $\set {1, 2, \ldots, n}$, it is trivially true that $\sup S = \sup S_j$.
Now assume that $S$ is unequal to $S_i$ for every $i$ in $\left\{{1, 2, \ldots, n}\right\}$.
By Supremum of Set of Real Numbers is at least Supremum of Subset, $\sup S \ge \sup S_i$ for every $i$ in $\set{1, 2, \ldots, n}$.
There are two alternatives; either:
:$\sup S > \sup S_i$ for every $i$ in $\set {1, 2, \ldots, n}$
or:
:$\sup S = \sup S_j$ for at least one $j$ in $\set {1, 2, \ldots, n}$.
Suppose that:
:$\sup S > \sup S_i$ for every $i$ in $\set {1, 2, \ldots, n}$
Let $\epsilon = \sup S - \map \max {\sup S_1, \sup S_2, \ldots, \sup S_n}$.
We note that $\epsilon > 0$.
By Supremum of Subset of Real Numbers is Arbitrarily Close, $S$ has an element $x$ that satisfies:
:$x > \sup S - \epsilon$
We have:
Therefore:
:$x > \map \max {\sup S_1, \sup S_2, \ldots, \sup S_n}$
This means that $x > \sup S_i$ for every $i$ in $\set {1, 2, \ldots, n}$.
However, $x$ must be an element of $S_j$ for some $j$ in $\set {1, 2, \ldots, n}$ as $x \in S$ and $S = \bigcup S_i$.
Accordingly, it is not true that $\sup S > \sup S_i$ for every $i$ in $\set {1, 2, \ldots, n}$.
We just concluded that the alternative:
:$\sup S > \sup S_i$ for every $i$ in $\set {1, 2, \ldots, n}$
is not true.
Therefore, the other alternative:
:$\sup S = \sup S_j$ for a $j$ in $\set {1, 2, \ldots, n}$
is true.
Category:Real Analysis
Supremum of Sum equals Sum of Suprema

== Theorem ==
Let $A$ and $B$ be non-empty sets of real numbers.
Let $A + B$ be $\set {x + y: x \in A, y \in B}$.
Let either $A$ and $B$ have suprema or $A + B$ have a supremum.
Then all $\sup A$, $\sup B$, and $\sup \paren {A + B}$ exist and:
:$\sup \paren {A + B} = \sup A + \sup B$
== Proof ==
Assume first that $A$ and $B$ have suprema.
We have:
:$x \le \sup A$ for an arbitrary $x$ in $A$
:$y \le \sup B$ for an arbitrary $y$ in $B$
Adding these inequalities, we get:
:$x + y \le \sup A + \sup B$
The number $x + y$ is an arbitrary element of $A + B$ as $x$ and $y$ are arbitrary elements of $A$ and $B$ respectively.
Therefore, $\sup A + \sup B$ is an upper bound for $A + B$.
$A + B$ is non-empty as $A$ and $B$ are non-empty.
Accordingly, $A + B$ has a supremum by the Continuum Property.
Next, assume that $\sup \paren {A + B}$ has a supremum.
We need to prove that $A$ and $B$ have suprema.
Let $y$ be a point in $B$.
We have:
Therefore, $A$ has an upper bound as $x$ is an arbitrary point in $A$.
Also, we have that $A$ is non-empty.
Accordingly, $A$ has a supremum by the Continuum Property.
A similar argument gives that $B$ has a supremum.
So, we have shown that all $\sup A$, $\sup B$, and $\sup \paren {A + B}$ exist.
We proceed to show that $\sup \paren {A + B} = \sup A + \sup B$.
We have $\sup \paren {A + B} \le \sup A + \sup B$ as $\sup A + \sup B$ is an upper bound for $A + B$.
Accordingly, either:
:$\sup \paren {A + B} < \sup A + \sup B$
or:
:$\sup \paren {A + B} = \sup A + \sup B$.
:
:$\sup \paren {A + B} < \sup A + \sup B$.
Let $\epsilon = \sup A + \sup B - \sup \paren {A + B}$.
We note that $\epsilon > 0$.
Since $\sup A$ is the least upper bound of $A$, there is an element $x$ in $A$ such that:
:$x > \sup A - \dfrac \epsilon 2$ by Supremum of Subset of Real Numbers is Arbitrarily Close
Since $\sup B$ is the least upper bound of $B$, there is an element $y$ in $B$ such that:
:$y > \sup B - \dfrac \epsilon 2$ by Supremum of Subset of Real Numbers is Arbitrarily Close
Adding these inequalities, we get:
which is impossible since the number $x + y$ is an element of $A + B$ as $x \in A$ and $y \in B$.
We have found that:
:$\sup \paren {A + B} < \sup A + \sup B$ is not true.
Therefore:
:$\sup \paren {A + B} = \sup A + \sup B$ as $\sup \paren {A + B} \le \sup A + \sup B$.
Category:Real Analysis
Supremum of Suprema

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $\mathbb T \subseteq \powerset S$, where $\powerset S$ is the power set of $S$.
Suppose all $T \in \mathbb T$ admit a supremum $\sup T$ in $S$.
Then:
:$\sup \bigcup \mathbb T = \sup {\set {\sup T: T \in \mathbb T} }$
if one of these two quantities exists (in $S$).
== Proof ==
Suppose that $s = \sup \bigcup \mathbb T \in S$.
By Set is Subset of Union, $T \subseteq \bigcup \mathbb T$ for all $T \in \mathbb T$.
Hence by Supremum of Subset:
:$\forall T \in \mathbb T: \sup T \preceq s$
Suppose now that $a \in S$ satisfies:
:$\forall T \in \mathbb T: \sup T \preceq a$
Then by transitivity of $\preceq$:
:$\forall t \in T: t \preceq a$
Since this holds for any $T \in \mathbb T$, also:
:$\forall t \in \bigcup \mathbb T: t \preceq a$
Hence $s \preceq a$, by definition of supremum.
That is, $s = \sup {\set {\sup T: T \in \mathbb T} }$.
Suppose now that $r = \sup {\set {\sup T: T \in \mathbb T} } \in S$.
By definition of supremum, for all $T \in \mathbb T$ and $t \in T$:
:$t \preceq \sup T$
By transitivity of $\preceq$:
:$\forall T \in \mathbb T: \forall t \in T: t \preceq r$
Hence for all $t \in \bigcup \mathbb T$:
:$t \preceq r$
Suppose that $a \in S$ satisfies:
:$\forall t \in \bigcup \mathbb T: t \preceq a$
In particular, for any $T \in \mathbb T$, since $T \subseteq \bigcup \mathbb T$:
:$\sup T \preceq a$
and therefore by definition of supremum, also:
:$r \preceq a$
That is, $r = \sup \bigcup \mathbb T$.
Category:Order Theory
Supremum of Suprema over Overlapping Domains

== Theorem ==
Let $\family {a_i}_{i \mathop \in I}$ be a family of elements of the non-negative real numbers $\R_{\ge 0}$ indexed by $I$.
Let $\map R i$ and $\map S i$ be propositional functions of $i \in I$.
Let $\ds \sup_{\map R i} a_i$ and $\ds \sup_{\map S i} a_i$ be the indexed suprema on $\family {a_i}$ over $\map R i$ and $\map S i$ respectively.
Then:
:$\ds \map \sup {\sup_{\map R i} a_i, \sup_{\map S i} a_i} = \sup_{\map R i \mathop \lor \map S i} a_i$
== Proof ==
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $35$
Category:Suprema
Supremum of Union of Bounded Above Sets of Real Numbers

== Theorem ==
Let $A$ and $B$ be sets of real numbers.
Let $A$ and $B$ both be bounded above.
Then:
:$\map \sup {A \cup B} = \max \set {\sup A, \sup B}$
where $\sup$ denotes the supremum.
== Proof ==
Let $A$ and $B$ both be bounded above.
By the Continuum Property, $A$ and $B$ both admit a supremum.
Let $x \in A \cup B$.
Then either $x \le \sup A$ or $x \le \sup B$ by definition of supremum.
Hence:
:$x \le \max \set {\sup A, \sup B}$
and so $\max \set {\sup A, \sup B}$ is certainly an upper bound of $A \cup B$.
It remains to be shown that $\max \set {\sup A, \sup B}$ is the smallest upper bound.
 there exists $m \in \R$ such that:
:$m < \max \set {\sup A, \sup B}$
and:
:$\forall x \in A \cup B: x \le m$
, let $\sup A \ge \sup B$.
Then:
:$\max \set {\sup A, \sup B} = \sup A$
and so:
:$m < \sup A$
But then by definition of supremum:
:$\exists a \in A: a > m$
and so $m$ is not an upper bound of $A \cup B$.
This contradicts our assumption that $m$ is a supremum of $A \cup B$.
It follows by Proof by Contradiction that $\max \set {\sup A, \sup B}$ is the supremum of $A \cup B$.
The same argument shows, mutatis mutandis, that if $\sup A \le \sup B$, $\max \set {\sup A, \sup B}$ is the supremum of $A \cup B$.
== Sources ==
* : $1$: Review of some real analysis: Exercise $1.5: 2$
Category:Bounded Above Sets of Real Numbers
Category:Suprema
Category:Set Union
Surface/Examples/Cylindrical Surface

== Example of Surface ==
The locus of the parametric equations in Cartesian $3$-space:
:$\begin {cases} x & = & r \cos \theta \\ y & = & r \sin \theta \\ z & = & \lambda \end {cases}$
is a cylindrical surface whose cross-section is a circle with radius $r$.
== Sources ==
* 
* 
Category:Examples of Surfaces
Surface Area of Cuboid

== Theorem ==
Let $\CC$ be a cuboid whose edges are of length $a$, $b$ and $c$.
The surface area $S$ of $\CC$ is given as:
:$S = 2 \paren {a b + b c + a c}$
== Proof ==
Each of the faces of $\CC$ are rectangles:
:$2$ of these faces are adjacent to edges of length $a$ and $b$
:$2$ of these faces are adjacent to edges of length $b$ and $c$
:$2$ of these faces are adjacent to edges of length $a$ and $c$.
Hence from Area of Rectangle:
:there are $2$ faces with area $a b$
:there are $2$ faces with area $a c$
:there are $2$ faces with area $b c$.
Hence the result.
== Sources ==
* : $\S 4$: Geometric Formulas: Rectangular Parallelepiped of Length $a$, Height $l$, Width $c$: $4.27$
* 
* 
* : $\S 7$: Geometric Formulas: Rectangular Parallelepiped of Length $a$, Height $b$, Width $c$: $7.27.$
Category:Cuboids
Category:Area Formulas
Surface Area of Sphere

== Theorem ==
The surface area $A$ of a sphere whose radius $a$ is given by:
:$A = 4 \pi a^2$
== Proof ==
== Historical Note ==
== Sources ==
* : $\S 4$: Geometric Formulas: Sphere of Radius $r$: $4.30$
* 
* 
* : $\S 7$: Geometric Formulas: Sphere of Radius $r$: $7.30.$
* 
Category:Spheres
Category:Area Formulas
Surface Area of Sphere/Historical Note

== Historical Note to Surface Area of Sphere ==
The Surface Area of Sphere was demonstrated by 
His proof appears as Proposition $34$ in his .
He also discusses this result in his :
:From this theorem, to the effect that a sphere is four times as great as the cone with a great circle of the sphere as base and height equal to the radius of the sphere, I conceived the notion that the surface of any sphere is four times as great as a great circle in it; for, judging from the fact that any circle is equal to a triangle with base equal to the circumference and height equal to the radius of the circle, I apprehended that, in like manner, a sphere is equal to a cone with base equal to the surface of the sphere and height equal to the radius.
== Sources ==
* : Chapter $\text {A}.5$: Archimedes (ca. B.C.)
Category:Historical Notes