nyuuzyou/wikis huggingface dataset:
Infimum of Set of Oscillations on Set

== Lemma ==
Let $f: D \to \R$ be a real function where $D \subseteq \R$.
Let $x$ be a point in $D$.
Let $S_x$ be a set of real sets that contain (as an element) $x$.
Let:
:$\map {\omega_f} x = \inf \set {\map {\omega_f} {I \cap D}: I \in S_x}$
where $\map {\omega_f} {I \cap D}$ denotes the oscillation of $f$ on the set $I \cap D$:
:$\map {\omega_f} {I \cap D} = \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$
Then:
:$\map {\omega_f} x \in \R$ $\set {\map {\omega_f} {I \cap D}: I \in S_x}$ contains a real number.
== Proof ==
Let:
:$S = \set {\map {\omega_f} {I \cap D}: I \in S_x}$
We observe that:
:$\inf S = \map {\omega_f} x$
=== Necessary Condition ===
Let $\inf S \in \R$.
We need to prove that $S$ contains a real number.
Note that $S$ is non-empty as the empty set does not admit an infimum (in $\R$).
Therefore, $S$ has at least one element.
Accordingly, there is an $I \in S_x$ such that $\map {\omega_f} {I \cap D} \in S$.
Let $I \in S_x$.
Therefore, $x \in I$.
From this follows by Oscillation on Set is an Extended Real Number that $\map {\omega_f} {I \cap D}$ is an extended real number.
Therefore $S$ is a set of extended real numbers as $S = \set {\map {\omega_f} {I \cap D}: I \in S_x}$.
Accordingly, $S$ contains a real number by Infimum of Subset of Extended Real Numbers is Arbitrarily Close as $\inf S \in \R$.
=== Sufficient Condition ===
Let $S$ contain a real number.
We need to prove that $\inf S \in \R$.
We have:
:$S \cap \R$ is non-empty as $S$ contains a real number.
Let $I \in S_x$.
Therefore, $x \in I$.
From this follows by Oscillation on Set is an Extended Real Number that $\map {\omega_f} {I \cap D} \in \overline \R_{\ge 0}$.
Therefore:
:$S$ is a subset of $\overline \R_{\ge 0}$ as $S = \set {\map {\omega_f} {I \cap D}: I \in S_x}$
Accordingly:
:$S$ is bounded below.
From this follows that:
:$S \cap \R$ is bounded below as $SR$ is a subset of $S$
We have:
:$S \cap \R$ is bounded below
:$S \cap \R$ is not empty
Therefore:
:$\inf S \cap \R \in \R$ Continuum Property
We have:
:$S$ is a set of extended real numbers as $S$ is a subset of $\overline \R_{\ge 0}$
:$S$ is bounded below
Therefore:
:$\inf S \in \R$ by Infimum of Real Subset as $\inf S \cap \R \in \R$
Category:Real Analysis
Category:Oscillation
Infimum of Set of Oscillations on Set is Arbitrarily Close

== Lemma ==
Let $f: D \to \R$ be a real function where $D \subseteq \R$.
Let $x$ be a point in $D$.
Let $S_x$ be a set of real sets that contain (as an element) $x$.
Let:
:$\map {\omega_f} x = \ds \inf \set {\map {\omega_f} {I \cap D}: I \in S_x}$
where $\map {\omega_f} {I \cap D}$ is the oscillation of $f$ on a real set $I \cap D$:
:$\map {\omega_f} {I \cap D} = \ds \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$
Let $\epsilon \in \R_{>0}$.
Let $\map {\omega_f} x \in \R$.
Then an $I \in S_x$ exists such that:
:$\map {\omega_f} {I \cap D} - \map {\omega_f} x < \epsilon$
== Proof ==
Let $\epsilon \in \R_{>0}$.
Let $\map {\omega_f} x \in \R$.
We need to prove that an $I \in S_x$ exists such that:
:$\map {\omega_f} {I \cap D} - \map {\omega_f} x < \epsilon$
By Oscillation on Set is an Extended Real Number:
:$\forall I \in S_x: \map {\omega_f} {I \cap D} \in \overline \R_{\ge 0}$
Therefore:
:$\set {\map {\omega_f} {I \cap D}: I \in S_x}$ is a subset of $\overline \R$
We have also:
:$\inf \set {\map {\omega_f} {I \cap D}: I \in S_x} \in \R$
as $\inf \set {\map {\omega_f} {I \cap D}: I \in S_x} = \map {\omega_f} x$
Therefore:
Category:Real Analysis
Category:Oscillation
Infimum of Set of Reciprocals of Positive Integers

== Theorem ==
Let $S$ be the subset of the set of real numbers defined as:
:$S = \set {\dfrac 1 n: n \in \Z_{>0} }$
Then:
:$\inf S = 0$
where $\inf S$ denotes the infimum of $S$.
== Proof ==
We have that $\Z_{>0}$ contains only (strictly) positive integers.
So it follows from Reciprocal of Strictly Positive Real Number is Strictly Positive that $S$ contains only (strictly) positive real numbers.
Hence $0$ is a lower bound for $S$.
 that $0$ is not the infimum of $S$.
Then $\exists h \in \R_{>0}$ such that $h$ is a lower bound for $S$.
Then:
:$\forall n \in \Z_{>0}: \dfrac 1 n \ge h$
Hence from Reciprocal Function is Strictly Decreasing:
:$\forall n \in \Z_{>0}: n \le \dfrac 1 h$
where $\dfrac 1 h \in \R$.
But from the Axiom of Archimedes:
:$\exists n \in \Z_{>0}: n > \dfrac 1 h$
This is a contradiction.
It follows by Proof by Contradiction that no such $h > 0$ exists.
Hence $0$ is the infimum of $S$.
== Sources ==
* : $\S 3$: Natural Numbers: $\S 3.4$: Example
Category:Reciprocals
Category:Infima
Infimum of Singleton

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Then for all $a \in S$:
:$\inf \set a = a$
where $\inf$ denotes infimum.
== Proof ==
Since $a \preceq a$, $a$ is a lower bound for $\set a$.
Let $b$ be another lower bound for $\set a$.
Then necessarily $b \preceq a$.
It follows that indeed:
:$\inf \set a = a$
as desired.
== Also see ==
* Supremum of Singleton
Category:Infima
Category:Singletons
Infimum 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 $H, K \in \mathbb G$.
Then:
:$\inf \set {H, K} = H \cap K$
== Proof ==
Let $H, K \in \mathbb G$.
From Set of Subgroups forms Complete Lattice:
:$\inf \set {H, K} = H \cap K$
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Theorem $14.6$
Category:Complete Lattices
Category:Infima
Category:Subgroups
Infimum of Subset

== Theorem ==
Let $\struct {U, \preceq}$ be an ordered set.
Let $S \subseteq U$.
Let $T \subseteq S$.
Let $\struct {S, \preceq}$ admit an infimum in $U$.
If $T$ also admits an infimum in $U$, then $\map \inf S \preceq \map \inf T$.
== Proof ==
Let $B = \map \inf S$.
Then $B$ is a lower bound for $S$.
As $T \subseteq S$, it follows by the definition of a subset that $x \in T \implies x \in S$.
Because $x \in S \implies B \preceq x$ (as $B$ is a lower bound for $S$) it follows that $x \in T \implies B \preceq x$.
So $B$ is a lower bound for $T$.
Therefore $B$ precedes the infimum of $T$ in $U$.
Hence the result.
== Also see ==
* Supremum of Subset
== Sources ==
* : $\S 2$: Continuum Property: Exercise $\S 2.13 \ (3)$
Category:Order Theory
Infimum of Subset Product in Ordered Group

== Theorem ==
Let $\struct {G, \circ, \preceq}$ be an ordered group.
Let subsets $A$ and $B$ of $G$ admit infima in $G$.
Then:
:$\map \inf {A \circ_\PP B} = \inf A \circ \inf B$
where $\circ_\PP$ denotes subset product.
== Proof ==
This follows from Supremum of Subset Product in Ordered Group and the Duality Principle.
== Also see ==
* Supremum of Subset Product in Ordered Group
Category:Infima
Category:Subset Products
Category:Ordered Groups
Infimum of Subset of Bounded Below Set of Real Numbers

== Theorem ==
Let $A$ and $B$ be sets of real numbers such that $A \subseteq B$.
Let $B$ be bounded below.
Then:
:$\inf A \ge \inf B$
where $\inf$ denotes the infimum.
== Proof ==
Let $B$ be bounded below.
By the Continuum Property, $B$ admits an infimum.
By Subset of Bounded Below Set is Bounded Below, $A$ is also bounded below.
Hence also by the Continuum Property, $A$ also admits a infimum.
 $\inf A < \inf B$.
Then:
:$\exists y \in A: y < \inf B$
Thus by definition of infimum, $y \notin B$.
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 $1.5: 3$
Category:Bounded Below Sets of Real Numbers
Infimum of Subset of Extended Real Numbers is Arbitrarily Close

== Theorem ==
Let $A \subseteq \overline \R$ be a subset of the extended real numbers.
Let $b$ be an infimum (in $\R$) of $A$.
Let $\epsilon \in \R_{>0}$.
Then:
:$\exists x \in A \cap \R: x - b < \epsilon$
== Proof ==
We have that:
:$A$ is a a set of extended real numbers
:$A$ is bounded below (in $\R$) as the real number $b$ is a lower bound for $A$.
From this follows by Infimum of Real Subset:
:$\map \inf {A \cap \R} \in \R$ as $\inf A \in \R$
Let $\epsilon \in \R_{>0}$.
Then:
Category:Extended Real Numbers‎
Infimum of Subset of Real Numbers/Examples/Example 1

== Example of Infimum of Subset of Real Numbers ==
The subset $S$ of the real numbers $\R$ defined as:
:$S = \set {1, 2, 3}$
admits an infimum:
:$\inf S = 1$
== Sources ==
* : $\S 2$: Continuum Property: $\S 2.5$: Examples: $\text{(i)}$
Category:Examples of Infima
Infimum of Subset of Real Numbers/Examples/Example 2

== Example of Infimum 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 an infimum:
:$\inf T = 1$
== Sources ==
* : $\S 2$: Continuum Property: $\S 2.5$: Examples: $\text{(ii)}$
Category:Examples of Infima
Infimum of Subset of Real Numbers/Examples/Example 3

== Example of Infimum of Subset of Real Numbers ==
The subset $V$ of the real numbers $\R$ defined as:
:$V := \set {x \in \R: x > 0}$
admits an infimum:
:$\inf V = 0$
== Sources ==
* : $\S 2$: Continuum Property: $\S 2.5$: Examples: $\text{(iii)}$
Category:Examples of Infima
Infimum 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 an infimum $m$.
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 $\inf S = 0$.
Thus $\inf S \notin S$.
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 $\inf T = 1$.
Thus $\inf T \in T$.
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Real Numbers: $1.35$. Example $\text{(a)}$
Category:Suprema
Infimum of Subset of Real Numbers is Arbitrarily Close

== Theorem ==
Let $A \subseteq \R$ be a subset of the real numbers.
Let $b$ be an infimum of $A$.
Let $\epsilon \in \R_{>0}$.
Then:
:$\exists x \in A: x - b < \epsilon$
== Proof ==
Note that $A$ is non-empty as the empty set does not admit an infimum (in $\R$).
Suppose $\epsilon \in \R_{>0}$ such that:
:$\forall x \in A: x - b \ge \epsilon$
Then:
:$\forall x \in A: b + \epsilon \le x$
and so $b + \epsilon$ would be a lower bound of $A$ which is greater than $b$.
But since $b$ is an infimum of $A$ there can be no such $b + \epsilon$.
From that contradiction it follows that:
:$\exists x \in A: x - b < \epsilon$
== Also see ==
* Supremum of Subset of Real Numbers is Arbitrarily Close
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 5$: Limits: Lemma $5.6$
Category:Real Analysis
Infimum of Union of Bounded Below Sets of Real Numbers

== Theorem ==
Let $A$ and $B$ be sets of real numbers.
Let $A$ and $B$ both be bounded below.
Then:
:$\map \inf {A \cup B} = \min \set {\inf A, \inf B}$
where $\inf$ denotes the infimum.
== Proof ==
Let $A$ and $B$ both be bounded below.
By the Continuum Property, $A$ and $B$ both admit an infimum.
Let $x \in A \cup B$.
Then either $x \ge \inf A$ or $x \ge \inf B$ by definition of infimum.
Hence:
:$x \ge \min \set {\inf A, \inf B}$
and so $\min \set {\inf A, \inf B}$ is certainly a lower bound of $A \cup B$.
It remains to be shown that $\min \set {\inf A, \inf B}$ is the greatest lower bound.
 there exists $m \in \R$ such that:
:$m > \min \set {\inf A, \inf B}$
and:
:$\forall x \in A \cup B: x \ge m$
, let $\inf A \le \inf B$.
Then:
:$\min \set {\inf A, \inf B} = \sup A$
and so:
:$m > \sup A$
But then by definition of infimum:
:$\exists a \in A: a < m$
and so $m$ is not a lower bound of $A \cup B$.
This contradicts our assumption that $m$ is an infimum of $A \cup B$.
It follows by Proof by Contradiction that $\min \set {\inf A, \inf B}$ is the infimum of $A \cup B$.
The same argument shows, mutatis mutandis, that if $\inf A \le \inf B$, $\min \set {\inf A, \inf B}$ is the infimum of $A \cup B$.
== Sources ==
* : $1$: Review of some real analysis: Exercise $1.5: 3$
Category:Bounded Below Sets of Real Numbers
Category:Infima
Category:Set Union
Infimum of Upper Closure of Element

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $s$ be an element of $S$.
Then:
:$\map \inf {s^\succeq} = s$
where $s^\succeq$ denotes the upper closure of $s$.
== Proof ==
== Sources ==
* 
* 
Category:Upper Closures
Infimum of Upper Closure of Set

== Theorem ==
Let $\left({S, \preceq}\right)$ be an ordered set.
Let $T \subseteq S$.
Let $U = T^\succeq$ be the upper closure of $T$ in $S$.
Let $s \in S$.
Then $s$ is the infimum of $T$ it is the infimum of $U$.
== Proof ==
This follows by mutatis mutandis of the proof of Supremum of Lower Closure of Set.
== Sources ==
* 
* 
Category:Upper Closures
Infimum of Upper Sums Never Smaller than Lower 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 $\map L S$ be the lower Darboux sum of $f$ on $\closedint a b$ with respect to $S$.
Let $\map U P$ be the upper Darboux sum of $f$ on $\closedint a b$ with respect to a finite subdivision $P$.
Then:
:$\inf_P \map U P \ge \map L S$
== Proof ==
From Upper Sum Never Smaller than Lower Sum for any Pair of Subdivisions, $\map L S$ is a lower bound for the real set:
:$T = \leftset {\map U P: P}$ is a finite subdivision of $\rightset {\closedint a b}$
Since $\inf_P \map U P$ is the infumum of $T$:
:$\inf_P \map U P \ge \map L S$
Hence the result.
Category:Real Analysis
Infinite-Dimensional Banach Space has Uncountable Dimension

== Theorem ==
Let $\Bbb F \in \set {\R, \C}$. 
Let $X$ be an infinite-dimensional Banach space over $\Bbb F$. 
Let $B$ be a Hamel basis for $X$.
Then $B$ is uncountable.
That is, $\dim X > \aleph_0$.
== Proof ==
Suppose that $B$ is countable. 
Let: 
:$B = \set {e_n : n \in \N}$
For each $n \in \N$, let: 
:$F_n = \span \set {e_k : k \le n}$
Since $\dim F_n = n$, $F_n$ clearly has finite dimension. 
We show that: 
:$\ds X = \bigcup_{n \mathop = 1}^\infty F_n$ 
Let $x \in X$. 
Then there exists $n_1, \ldots, n_k \in \N$ and $\alpha_1, \ldots, \alpha_n \in \Bbb F$ such that: 
:$\ds x = \sum_{j \mathop = 1}^k \alpha_j e_{n_j}$
Setting:
:$\ds N = \max_{1 \le j \le k} n_j$ 
we have that $x \in F_N$, hence:
:$\ds X = \bigcup_{n \mathop = 1}^\infty F_n$ 
From Finite Dimensional Subspace of Normed Vector Space is Closed, $F_n$ is closed for each $n \in \N$. 
From Interior of Proper Subspace of Normed Vector Space is Empty, $F_n$ has empty interior for each $n \in \N$.
So each $F_n$ is nowhere dense.
Since $X$ is a complete metric space, it is a Baire space.
However, we have written $X$ as a countable union of nowhere dense sets.
This contradicts the Baire Category Theorem.
So $B$ is uncountable.
Category:Banach Spaces
Infinite Cyclic Group is Isomorphic to Integers

== Theorem ==
Let $G$ be an infinite cyclic group.
Then $G$ is isomorphic to the additive group of integers: $G \cong \struct {\Z, +}$.
=== Corollary ===
== Proof ==
From the definition of an infinite cyclic group, we have:
:$G = \gen a = \set {a^k: k \in \Z}$
Let us define the mapping:
:$\phi: \Z \to G: \map \phi k = a^k$.
We now show that $\phi$ is an isomorphism.
From Mapping from Additive Group of Integers to Powers of Group Element is Homomorphism, $\phi$ is a homomorphism.
Now we show that $\phi$ is a surjection.
As $G$ is cyclic, every element of $G$ is apower of $a$ for some $a \in G$ such that $G = \gen a$.
Thus:
:$\forall x \in G: \exists k \in \Z: x = a^k$
By the definition of $\phi$:
:$\map \phi k = a^k = x$
Thus $\phi$ is surjective.
Now we show that $\phi$ is an injection.
This follows directly from Powers of Infinite Order Element, where:
:$\forall m, n \in \Z: m \ne n \implies a^m \ne a^n$
Thus $\phi$ is an injective, surjective homomorphism, thus:
: $G \cong \struct {\Z, +}$
as required.
== Comment ==
Now that as we have, in a sense, defined an infinite cyclic group with reference to the additive group of integers that we painstakingly constructed in the definition of integers, it naturally follows that we should use $\struct {\Z, +}$ as an "archetypal" infinite cyclic group.
== Sources ==
* : $\S 3.4$: Cyclic groups
Category:Infinite Cyclic Group
Category:Additive Group of Integers
Category:Examples of Group Isomorphisms
Infinite Cyclic Group is Unique up to Isomorphism

== Theorem ==
All infinite cyclic groups are isomorphic.
That is, up to isomorphism, there is only one infinite cyclic group.
== Proof ==
Let $G_1$ and $G_2$ be infinite cyclic groups.
From Infinite Cyclic Group is Isomorphic to Integers we have:
:$G_1 \cong \struct {\Z, +} \cong G_2$
where $\struct {\Z, +}$ is the additive group of integers.
From Isomorphism is Equivalence Relation it follows that:
:$G_1 \cong G_2$
== Comment ==
Now that as we have, in a sense, defined an infinite cyclic group with reference to the additive group of integers that we painstakingly constructed in the definition of integers, it naturally follows that we should use $\struct {\Z, +}$ as an "archetypal" infinite cyclic group.
Category:Infinite Cyclic Group
Category:Group Isomorphisms
Infinite Discontinuity/Examples/Example 1

== Examples of Infinite Discontinuities ==
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = \dfrac 1 x$
Then $f$ has an infinite discontinuity at $x = 0$.
== Sources ==
* 
* 
Category:Examples of Infinite Discontinuities
Infinite Group has Infinite Number of Subgroups

== Theorem ==
Let $\struct {G, \circ}$ be an infinite group.
Then $\struct {G, \circ}$ has an infinite number of distinct subgroups.
== Proof ==
There are two cases to consider: either $\struct {G, \circ}$ has an infinite cyclic subgroup, or it does not.
=== Case 1 ===
Suppose that $\struct {G, \circ}$ has an infinite cyclic subgroup denoted as $H$.
Let $a \in G$ be the element of $G$ such that $\gen a = H$.
Then by Distinct Subgroups of Infinite Cyclic Group, $\sequence {\gen {a^k} }_{k \mathop \in \N^*}$ is a sequence of distinct subgroups of $H$.
So, by Subgroup of Subgroup is Subgroup, it follows that $\struct {G, \circ}$ has an infinite number of subgroups.
=== Case 2 ===
Suppose that $\struct {G, \circ}$ does not have an infinite cyclic subgroup.
Then for every $x \in G$, we have that $\gen x$ is a finite group.
Let $a, b \in G$.
Let $\sim$ be the equivalence relation on $G$ defined as:
:$a \sim b \iff \gen a = \gen b$
Let $\eqclass a \sim$ denote the equivalence class of $a$ under $\sim$.
Let $x \sim a$.
Then:
:$\gen x = \gen a$
Since $x \in \gen x$, it follows that:
:$x \in \gen a$
It has been shown that:
:$\eqclass a \sim \subseteq \gen a$
By Subset of Finite Set is Finite, every equivalence class under $\sim$ is finite.
By Equivalence Class holds Equivalent Elements:
:$\eqclass a \sim \ne \eqclass b \sim \implies a \not \sim b \implies \gen a \ne \gen b$
So every equivalence class generates a unique subgroup of $\struct {G, \circ}$.
By Union of Equivalence Classes is Whole Set, the union of all equivalence classes under $\sim$ must equal $G$.
From Finite Union of Finite Sets is Finite, it follows that this must be a (countably) infinite union.
Thus there must exist an infinite number of equivalence classes, and hence an infinite number of subgroups.
The result then follows from Proof by Cases.
Category:Infinite Groups
Infinite Intersection of Open Sets of Metric Space may not be Open

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $\sequence {U_n}_{n \mathop \in \N}$ be an infinite sequence of open sets of $M$.
Then it is not necessarily the case that $\ds \bigcap_{n \mathop \in \N} U_n$ is itself an open set of $M$.
== Proof ==
Consider the open real interval $\openint {-\dfrac 1 n} {\dfrac 1 n} \subseteq \R$.
From Open Real Interval is Open Set, $\openint {-\dfrac 1 n} {\dfrac 1 n}$ is open in $\R$ for all $n \in \N_{> 0}$.
But:
:$\ds \bigcap_{n \mathop \in \N_{> 0}} \openint {-\dfrac 1 n} {\dfrac 1 n} = \set 0 = \closedint 0 0$
which is a closed interval of $\R$.
The result follows from Closed Real Interval is not Open Set.
== Also see ==
* Finite Intersection of Open Sets of Metric Space is Open
== Sources ==
* : $2.25$
* : $2$: Continuity generalized: metric spaces: $2.3$: Open sets in metric spaces: Example $2.3.18$
Category:Open Sets (Metric Spaces)
Category:Set Intersection
Infinite Lexicographic Order on Well-Ordered Sets is not Well-Ordering

== Theorem ==
Let $\struct {S, \preccurlyeq}$ be a well-ordered set.
Let $\preccurlyeq_l$ be the lexicographic order on the set $T$ of all ordered tuples of $S$:
:$\tuple {x_1, x_2, \ldots, x_m} \prec \tuple {y_1, y_2, \ldots, y_n}$ :
::$\exists k: 1 \le k \le \map \min {m, n}$ such that $\forall 1 \le j < k: x_j = y_j$ but $x_k \prec y_k$ in $S$
:or:
::$m < n$ and $\forall 1 \le j < m: x_j = y_j$.
Then $\preccurlyeq_l$ is not a well-ordering on $T$.
== Proof ==
It is straightforward to show that $\preccurlyeq_l$ is a total ordering on $T$.
It remains to investigate the nature of $\preccurlyeq_l$ as to whether or not it is a well-ordering.
Consider a set $S = \set {a, b}$ such that $a \prec b$.
Then the set $\set {\tuple b, \tuple {a, b}, \tuple {a, a, b}, \tuple {a, a, a, b}, \ldots}$ has no minimal element by $\preccurlyeq_l$.
Thus $T$ is not well-ordered by $\preccurlyeq_l$.
== Sources ==
* : $\S 1.2.1$: Mathematical Induction: Exercise $15 \ \text{(e)}$
Category:Lexicographic Order
Category:Well-Orderings
Infinite Limit Operator is Linear Mapping

== Theorem ==
Let $c$ be the space of convergent sequences.
Let $\R$ be the set of real numbers.
Let $L : c \to \R$ be the infinite limit operator.
Then $L$ is a linear mapping.
== Proof ==
Let $\mathbf x = \sequence {x_n}_{n \mathop \in \N}, \mathbf y = \sequence {y_n}_{n \mathop \in \N} \in c$.
Suppose $\mathbf x$ and $\mathbf y$ converge to $x$ and $y$ respectively.
Let $\alpha \in \R$.
=== Distributivity ===
=== Positive homogenity ===
== Sources ==
* : Chapter $\S 2.1$: Continuous and linear maps. Linear transformations
Category:Linear Transformations
Category:Operator Theory