nyuuzyou/wikis huggingface dataset:
Existence of Inverse Elementary Row Operation/Add Scalar Product of Row to Another

== Theorem ==
Let $\map \MM {m, n}$ be a metric space of order $m \times n$ over a field $K$.
Let $\mathbf A \in \map \MM {m, n}$ be a matrix. 
Let $\map e {\mathbf A}$ be the elementary row operation which transforms $\mathbf A$ to a new matrix $\mathbf A' \in \map \MM {m, n}$.
Let $\map {e'} {\mathbf A'}$ be the inverse of $e$.
Then $e'$ is the elementary row operation:
:$e' := r'_k \to r'_k - \lambda r'_l$
== Proof ==
In the below, let:
:$r_k$ denote row $k$ of $\mathbf A$
:$r'_k$ denote row $k$ of $\mathbf A'$
:$r_k$ denote row $k$ of $\mathbf A$
for arbitrary $k$ such that $1 \le k \le m$.
By definition of elementary row operation:
:only the row or rows directly operated on by $e$ is or are different between $\mathbf A$ and $\mathbf A'$
and similarly:
:only the row or rows directly operated on by $e'$ is or are different between $\mathbf A'$ and $\mathbf A$.
Hence it is understood that in the following, only those rows directly affected will be under consideration when showing that $\mathbf A = \mathbf A$.
Let $\map e {\mathbf A}$ be the elementary row operation:
:$e := r_k \to r_k + \lambda r_l$
Then $r'_k$ is such that:
:$\forall a'_{k i} \in r'_k: a'_{k i} = a_{k i} + \lambda a_{l i}$
Now let $\map {e'} {\mathbf A'}$ be the elementary row operation which transforms $\mathbf A'$ to $\mathbf A$:
:$e' := r'_k \to r'_k - \lambda r'_l$
Applying $e'$ to $\mathbf A'$ we get:
It is noted that for $e'$ to be an elementary row operation, the only possibility is for it to be as defined.
Category:Existence of Inverse Elementary Row Operation
Existence of Inverse Elementary Row Operation/Exchange Rows

== Theorem ==
Let $\map \MM {m, n}$ be a metric space of order $m \times n$ over a field $K$.
Let $\mathbf A \in \map \MM {m, n}$ be a matrix. 
Let $\map e {\mathbf A}$ be the elementary row operation which transforms $\mathbf A$ to a new matrix $\mathbf A' \in \map \MM {m, n}$.
Let $\map {e'} {\mathbf A'}$ be the inverse of $e$.
Then $e'$ is the elementary row operation:
:$e' := r_k \leftrightarrow r_l$
That is:
:$e' = e$
== Proof ==
In the below, let:
:$r_k$ denote row $k$ of $\mathbf A$
:$r'_k$ denote row $k$ of $\mathbf A'$
:$r_k$ denote row $k$ of $\mathbf A$
for arbitrary $k$ such that $1 \le k \le m$.
By definition of elementary row operation:
:only the row or rows directly operated on by $e$ is or are different between $\mathbf A$ and $\mathbf A'$
and similarly:
:only the row or rows directly operated on by $e'$ is or are different between $\mathbf A'$ and $\mathbf A$.
Hence it is understood that in the following, only those rows directly affected will be under consideration when showing that $\mathbf A = \mathbf A$.
Let $\map e {\mathbf A}$ be the elementary row operation:
:$e := r_k \leftrightarrow r_l$
Thus we have:
Now let $\map {e'} {\mathbf A'}$ be the elementary row operation which transforms $\mathbf A'$ to $\mathbf A$:
:$e' := r'_k \leftrightarrow r'_l$
Applying $e'$ to $\mathbf A'$ we get:
It is noted that for $e'$ to be an elementary row operation, the only possibility is for it to be as defined.
Category:Existence of Inverse Elementary Row Operation
Existence of Inverse Elementary Row Operation/Scalar Product of Row

== Theorem ==
Let $\map \MM {m, n}$ be a metric space of order $m \times n$ over a field $K$.
Let $\mathbf A \in \map \MM {m, n}$ be a matrix. 
Let $\map e {\mathbf A}$ be the elementary row operation which transforms $\mathbf A$ to a new matrix $\mathbf A' \in \map \MM {m, n}$.
Let $\map {e'} {\mathbf A'}$ be the inverse of $e$.
Then $e'$ is the elementary row operation:
:$e' := r_i \to \dfrac 1 \lambda r_i$
== Proof ==
In the below, let:
:$r_k$ denote row $k$ of $\mathbf A$
:$r'_k$ denote row $k$ of $\mathbf A'$
:$r_k$ denote row $k$ of $\mathbf A$
for arbitrary $k$ such that $1 \le k \le m$.
By definition of elementary row operation:
:only the row or rows directly operated on by $e$ is or are different between $\mathbf A$ and $\mathbf A'$
and similarly:
:only the row or rows directly operated on by $e'$ is or are different between $\mathbf A'$ and $\mathbf A$.
Hence it is understood that in the following, only those rows directly affected will be under consideration when showing that $\mathbf A = \mathbf A$.
Let $\map e {\mathbf A}$ be the elementary row operation:
:$e := r_k \to \lambda r_k$
where $\lambda \ne 0$.
Then $r'_k$ is such that:
:$\forall a'_{k i} \in r'_k: a'_{k i} = \lambda a_{k i}$
Now let $\map {e'} {\mathbf A'}$ be the elementary row operation which transforms $\mathbf A'$ to $\mathbf A$:
:$e' := r_k \to \dfrac 1 \lambda r_k$
Because it is stipulated in the definition of an elementary row operation that $\lambda \ne 0$, it follows by definition of a field that $\dfrac 1 \lambda$ exists.
Hence $e'$ is defined.
So applying $e'$ to $\mathbf A'$ we get:
It is noted that for $e'$ to be an elementary row operation, the only possibility is for it to be as defined.
Category:Existence of Inverse Elementary Row Operation
Existence of Inverse of Class Relation

== Theorem ==
Let $X$ be a class.
Then there is a class $Z$ such that, for all sets $u, v$:
:$\tuple {u, v} \in Z \iff \tuple {v, u} \in X$
where $\tuple {\cdot, \cdot}$ denotes the Kuratowski formalization of ordered pairs.
== Proof ==
Let $X$ be arbitrary.
By Axiom $\text B 5$, there is a class $Z_1$ such that:
:$\paren 1 \quad \forall a, w: \tuple {a, w} \in Z_1 \iff a \in X$
Now, by Axiom $\text B 7$, there is a class $Z_2$ such that:
:$\paren 2 \quad \forall u, v, w: \tuple {\tuple {v, w}, u} \in Z_2 \iff \tuple {\tuple {v, u}, w} \in Z_1$
Next, by Axiom $\text B 6$, there is a class $Z_3$ such that:
:$\paren 3 \quad \forall u, v, w: \tuple {\tuple {u, v}, w} \in Z_3 \iff \tuple {\tuple {v, w}, u} \in Z_2$
Finally, by Axiom $\text B 4$, there is a class $Z$ such that:
:$\paren 4 \quad \forall b: b \in Z \iff \exists w: \tuple {b, w} \in Z_3$
We will prove that the $Z$ as defined above satisfies the theorem.
First, choose $a := \tuple {v, u}$ in $\paren 1$, to get:
:$\paren {1'} \quad \forall u, v, w: \tuple {\tuple {v, u}, w} \in Z_1 \iff \tuple {v, u} \in X$
Also, choose $b := \tuple {u, v}$ in $\paren 4$, to get:
:$\paren {4'} \quad \forall u, v: \tuple {u, v} \in Z \iff \exists w: \tuple {\tuple {u, v}, w} \in Z_3$
By Biconditional is Transitive with $\paren 2$ and $\paren 3$, for any $u, v, w$:
:$\paren {2,3} \quad \tuple {\tuple {u, v}, w} \in Z_3 \iff \tuple {\tuple {v, u}, w} \in Z_1$
Now, we can proceed with the equivalence proof.
Suppose $\tuple {v, u} \in X$ for some sets $u, v$.
By the Axiom of Infinity, there exists at least one set $c$.
Then, by $\paren {1'}$:
:$\tuple {\tuple {v, u}, c} \in Z_1$
Thus, by $\paren {2,3}$:
:$\tuple {\tuple {u, v}, c} \in Z_3$
Hence, by $\paren {4'}$, choosing $w := c$:
:$\tuple {u, v} \in Z$
Lastly, suppose $\tuple {u, v} \in Z$ for some sets $u, v$.
By $\paren {4'}$, there is some set $w_0$ such that:
:$\tuple {\tuple {u, v}, w_0} \in Z_3$
Then, by $\paren {2,3}$:
:$\tuple {\tuple {v, u}, w_0} \in Z_1$
Thus, by $\paren {1'}$, choosing $w := w_0$:
:$\tuple {v, u} \in X$
Category:Class Theory
Category:Von Neumann-Bernays-Gödel Set Theory
Existence of Latin Squares

== Theorem ==
For each $n \in \Z_{>0}$ there exists at least one Latin square of order $n$.
== Proof ==
The Cayley table of a finite group of order $n$ is a Latin square, from Group has Latin Square Property.
For every $n \in \Z_{>0}$ there exists a cyclic group of order $n$.
It follows that for every $n \in \Z_{>0}$ the Cayley table of the cyclic group of order $n$ is a Latin square of order $n$.
Category:Latin Squares
Existence of Laurent Series

== Theorem ==
Let $z_0 \in \C$ be a complex number.
Let $R \in \R_{>0}$ be a real number.
Let $\map {B'} {z_0, R}$ be the open punctured disk at $z_0$ of radius $R$.
Let $f: \map {B'} {z_0, R} \to \C$ be holomorphic.
Then there exists a sequence $\sequence {a_n}_{n \mathop \in \Z}$ such that:
:$\ds \map f z = \sum_{n \mathop = -\infty}^\infty a_n \paren {z - z_0}^n$
for all $z \in B'(z_0, R)$.
== Proof ==
Choose circles $C_1$ and $C_3$ centered at $z_0$
Connect $C_2$ and $C_3$ by path $C_2$ such that $z$ is inside $C_1 + C_2 - C_3 - C_2$ as shown below:
:
This curve and its interior are contained in $B'$, so by Cauchy's Integral Formula:
Since $C_1$ contains $z$ we have for all $w$ on $C_1$:
:$\dfrac {\cmod {z - z_0} } {\cmod {w - z_0} } < 1$
Therefore:
Similarly, since $z$ lies outside $C_3$ we have for all $w$ on $C_3$:
:$\dfrac {\cmod {w - z_0} } {\cmod {z - z_0} } < 1$
Therefore:
Combining what has been shown above yields:
Category:Laurent Series
Category:Complex Analysis
Existence of Left Universal Product

== Theorem ==
For each $m \in \N_{> 0}$, the following is a theorem of von Neumann-Bernays-Gödel set theory:
:$\forall X: \exists Z: \forall x: \forall v_1, \dotsc, v_m: \sequence {v_1, \dotsc, v_m, x} \in Z \iff x \in X$
where the notation $\sequence {\cdot, \dotsc, \cdot}$ is defined inductively as:
:$\sequence {y_1} := y_1$
:$\sequence {y_1, \dotsc, y_k, y_{k + 1}} := \tuple {\sequence {y_1, \dotsc, y_k}, y_{k + 1}}$
and $\tuple {\cdot, \cdot}$ denotes the Kuratowski formalization of ordered pairs.
== Proof ==
Let $X$ be arbitrary.
By Axiom $\text B 5$, there is a class $Z$ such that:
:$\forall x, v: \tuple {x, v} \in Z \iff x \in X$
By Existence of Inverse of Class Relation, there is a class $Z'$ such that:
:$\forall v, x: \tuple {v, x} \in Z' \iff \tuple {x, v} \in Z$
Thus, by Biconditional is Transitive:
:$\paren 1 \quad \forall v, x: \tuple {v, x} \in Z' \iff x \in X$
Now, let $v_1, \dotsc, v_m$ be arbitrary.
Substitute $\sequence {v_1, \dotsc, v_m}$ for $v$ in $\paren 1$ to get:
:$\forall x: \forall v_1, \dotsc, v_m: \tuple {\sequence {v_1, \dotsc, v_m}, x} \in Z \iff x \in X$
But, by definition of the notation $\sequence {\cdot, \dotsc, \cdot}$, that is equivalent to:
:$\forall x: \forall v_1, \dotsc, v_m: \sequence {v_1, \dotsc, v_m, x} \in Z \iff x \in X$
Category:Class Theory
Category:Von Neumann-Bernays-Gödel Set Theory
Existence of Lindelöf Space which is not Second-Countable

== Theorem ==
There exists at least one example of a Lindelöf space which is not also a second-countable space.
== Proof ==
Let $T$ be the Sorgenfrey line.
From Sorgenfrey Line is Lindelöf, $T$ is a Lindelöf space.
From Sorgenfrey Line is not Second-Countable, $T$ is not a second-countable space.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Countability Axioms and Separability
Category:Lindelöf Spaces
Category:Second-Countable Spaces
Existence of Lindelöf Space which is not Sigma-Compact

== Theorem ==
There exists at least one example of a $\sigma$-compact topological space which is not also a compact space.
== Proof ==
Let $T$ be the Sorgenfrey line.
From Sorgenfrey Line is Lindelöf, $T$ is a Lindelöf space.
From Sorgenfrey Line is not $\sigma$-Compact, $T$ is not a $\sigma$-compact space.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Global Compactness Properties
Category:Lindelöf Spaces
Category:Sigma-Compact Spaces
Existence of Lowest Common Multiple

== Theorem ==
Let $a, b \in \Z: a b \ne 0$.
The lowest common multiple of $a$ and $b$, denoted $\lcm \set {a, b}$, always exists.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : Chapter $1$: Properties of the Natural Numbers: $\S 23 \gamma$
* : Chapter $2$: Divisibility Theory in the Integers: $2.3$ The Euclidean Algorithm
Category:Lowest Common Multiple
Category:Existence of Lowest Common Multiple
Existence of Lowest Common Multiple/Proof 1

== Theorem ==
== Proof ==
We prove its existence thus:
$a b \ne 0 \implies \size {a b} \ne 0$
Also $\size {a b} = \pm a b = a \paren {\pm b} = \paren {\pm a} b$.
So the lowest common multiple definitely exists, and we can say that:
:$0 < \lcm \set {a, b} \le \size {a b}$
Now we prove it is the lowest.
That is:
:$a \divides n \land b \divides n \implies \lcm \set {a, b} \divides n$
Let $a, b \in \Z: a b \ne 0, m = \lcm \set {a, b}$.
Let $n \in \Z: a \divides n \land b \divides n$.
We have:
:$n = x_1 a = y_1 b$
:$m = x_2 a = y_2 b$
As $m > 0$, we have:
Since $r < m$, and $m$ is the smallest positive common multiple of $a$ and $b$, it follows that $r = 0$.
So:
:$\forall n \in \Z: a \divides n \land b \divides n: \lcm \set {a, b} \divides n$
That is, $\lcm \set {a, b}$ divides any common multiple of $a$ and $b$.
Category:Existence of Lowest Common Multiple
Existence of Lowest Common Multiple/Proof 2

== Theorem ==
== Proof ==
Either $a$ and $b$ are coprime or they are not.
Let:
:$a \perp b$
where $a \perp b$ denotes that $a$ and $b$ are coprime.
Let $a b = c$.
Then:
:$a \divides c, b \divides c$
where $a \divides c$ denotes that $a$ is a divisor of $c$.
Suppose both $a \divides d, b \divides d$ for some $d \in \N_{> 0}: d < c$.
Then:
:$\exists e \in \N_{> 0}: a e = d$
:$\exists f \in \N_{> 0}: b f = d$
Therefore:
:$a e = b f$
and from :
:$a : b = f : e$
But $a$ and $b$ are coprime.
From:
:
and:
:
it follows that $b \divides e$
Since:
:$a b = c$
and:
:$a e = d$
it follows from that:
:$b : e = c : d$
But $b \divides e$ and therefore:
:$c \divides d$
But $c > d$ which is impossible.
Therefore $a$ and $b$ are both the divisor of no number less than $c$.
Now suppose $a$ and $b$ are not coprime.
Let $f$ and $e$ be the least numbers of those which have the same ratio with $a$ and $b$.
Then from :
:$a e = b f$
Let $a e = c$.
Then $b f = c$.
Hence:
:$a \divides c$
:$b \divides c$
Suppose $a$ and $b$ are both the divisor of some number $d$ which is less than $c$.
Let:
:$a g = d$
and:
:$b h = d$
Therefore:
:$a g = b h$
and so by :
:$a : b = f : e$
Also:
:$f : e = h : g$
But $f, e$ are the least such.
From :
:$e \divides g$
Since $a e = c$ and $a g = d$, from :
:$e : g = c : d$
But:
:$e \divides g$
So $c \divides d$
But $c > d$ which is impossible.
Therefore $a$ and $b$ are both the divisor of no number less than $c$.
== Sources ==
* : Book $\text{VII}$. Propositions
Category:Existence of Lowest Common Multiple
Existence of Lowest Common Multiple/Proof 3

== Theorem ==
== Proof ==
Note that as Integer Divides Zero, both $a$ and $b$ are divisors of zero.
Thus by definition $0$ is a common multiple of $a$ and $b$.
Non-trivial common multiples of $a$ and $b$ exist.
Indeed, $a b$ and $-\paren {a b}$ are common multiples of $a$ and $b$.
Either $a b$ or $-\paren {a b}$ is strictly positive.
Let $S$ denote the set of strictly positive common multiples of $a$ and $b$.
By the Well-Ordering Principle, $S$ contains a smallest element.
This can then be referred to as the lowest common multiple of $a$ and $b$.
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.3$ The Euclidean Algorithm
Category:Existence of Lowest Common Multiple
Existence of Magma with no Proper Submagma

== Theorem ==
Let $n \in \Z_{>0}$ be a strictly positive integer.
Let $S$ be a set of cardinality $n$:
:$\card S = n$
Then there exists an operation $\circ$ on $S$ such that:
:$\struct {S, \circ}$ is a magma
:$\struct {S, \circ}$ has no submagma $\struct {T, \circ}$ such that $T$ is a non-empty proper subset of $S$.
== Proof ==
For $n = 1$ the result follows trivially: there are no non-empty proper subsets of a singleton.
Let $S = \set {s_1, s_2, \ldots, s_n}$.
Let $\circ$ be defined on $S$ such that:
:$\forall s_a \in S: s_a \circ s_a = \begin{cases} s_{a + 1} & : a < n \\ s_1 & : a = n \end{cases}$
For $a \ne b$ the operation $s_a \circ s_b$ can be arbitrary as long as $s_a \circ s_b \in S$.
Let $T$ be such that $\O \subsetneq T \subsetneq S$.
Then either:
:$\exists s_k \in T: s_{k + 1} \notin T$
or:
:$s_n \in T$ but $s_1 \notin T$
otherwise $T = S$.
Thus either:
:$s_k \circ s_k \notin T$
or:
:$s_n \circ s_n \notin T$
In either case, $\circ$ is not closed in $T$.
Thus by definition, $\struct {T, \circ}$ is not a submagma of $\struct {S, \circ}$.
As $T$ is arbitrary, the result follows.
== Sources ==
* : Chapter $5$: Subgroups: Exercise $2$
Category:Magmas
Existence of Matrix Logarithm

== Theorem ==
Let $T$ be a square matrix of order $n$.
Then there exists a real matrix $S$ such that $e^S = T$ :
:$(1): \quad T$ is not a singular matrix
and:
:$(2): \quad $for every negative eigenvalue $\lambda$ of $T$ and for every positive integer $k$, the Jordan form of $T$ has an even number of $k \times k$ blocks associated with $\lambda$.
Existence of Maximal Commutative Subalgebra

== Theorem ==
Let $K$ be a field. 
Let $A$ be an algebra over $K$.
Let $S \subseteq A$ be a non-empty set such that:
:for all $x, y \in S$ we have $x y = y x$. 
Then there exists a commutative subalgebra of $A$ containing $S$ that is maximal (among all subalgebras) with respect to set inclusion.
=== Unital ===
== Proof ==
Let $\SS$ be the set commutative subalgebras of $A$ containing $S$. 
We look to apply Zorn's Lemma to $\struct {\SS, \subseteq}$. 
From Subalgebra Generated by Commuting Elements is Commutative, we have $\SS \ne \O$. 
Let $\CC$ be a chain in $\struct {\SS, \subseteq}$.
From Union of Chain of Subalgebras is Subalgebra, $\bigcup \CC$ is a commutative subalgebras of $A$ containing $S$.
We can see that $\bigcup \CC$ is an upper bound for $\CC$ since $A \subseteq \bigcup \CC$ for each $A \in \CC$. 
Hence we have that every chain in $\struct {\SS, \subseteq}$ has an upper bound.
So by Zorn's Lemma, $\struct {\SS, \subseteq}$ has a maximal element $B$.
Now suppose that $C$ is a commutative subalgebras of $A$ with $B \subseteq C$.
Then $C$ in particular contains $S$, so $C \in \CC$. 
Then $C = B$. 
So $B$ is indeed maximal among all subalgebras.
Category:Algebras
Category:Existence of Maximal Commutative Subalgebra
Existence of Maximal Commutative Subalgebra/Unital

== Theorem ==
Let $K$ be a field. 
Let $A$ be an unital algebra over $K$.
Let $S \subseteq A$ be a non-empty set such that:
:for all $x, y \in S$ we have $x y = y x$. 
Then there exists a commutative unital subalgebra of $A$ containing $S$ that is maximal (among all subalgebras) with respect to set inclusion.
== Proof ==
Let ${\mathbf 1}_A$ be the identity element of $A$. 
Note that ${\mathbf 1}_A$ commutes with every element of $S$.
Hence $S' = S \cup \set { {\mathbf 1}_A}$ is such that:
:for all $x, y \in S'$ we have $x y = y x$. 
Hence applying Existence of Maximal Commutative Subalgebra to $S'$, there exists a commutative subalgebra of $A$ containing $S'$ that is maximal (among all subalgebras) with respect to set inclusion.
In particular:
:there exists a commutative unital subalgebra of $A$ containing $S$ that is maximal (among all subalgebras) with respect to set inclusion.
Category:Existence of Maximal Commutative Subalgebra
Existence of Maximal Compact Topological Space which is not Hausdorff

== Theorem ==
Let $S$ be a set.
Let $\tau$ be the minimal subset of the power set $\powerset S$ such that $\struct {S, \tau}$ is a compact topological space.
Then it is not necessarily the case that $\struct {S, \tau}$ is a Hausdorff space.
== Proof ==
Let $T = \struct {S, \tau}$ be the canonical maximal compact non-Hausdorff space.
This space has been so named on in order to allow reference to it without needing to describe it whenever it is mentioned.
By Canonical Maximal Compact Non-Hausdorff Space is Maximal Compact, $\tau$ is the maximal subset of the power set $\powerset S$ such that $T$ is compact.
By Canonical Maximal Compact Non-Hausdorff Space is not Hausdorff, $T$ is not a Hausdorff space.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Compactness Properties and the $T_i$ Axioms
Category:Hausdorff Spaces
Category:Compact Spaces
Existence of Maximum and Minimum of Bounded Sequence

== Theorem ==
Let $\sequence {x_n}$ be a bounded sequence in $\R$ (which may or may not be convergent).
Let $L$ be the set of all real numbers which are the limit of some subsequence of $\sequence {x_n}$.
Then $L$ has both a maximum and a minimum.
== Proof ==
From the Bolzano-Weierstrass Theorem:
:$L \ne \O$
From Lower and Upper Bounds for Sequences, $L$ is a bounded subset of $\R$.
Thus $L$ does have a supremum and infimum in $\R$.
The object of this proof is to confirm that:
:$\overline l := \map \sup L \in L$
and:
:$\underline l := \map \inf L \in L$
that is, that these points do actually belong to $L$.
First we show that $\overline l \in L$.
To do this, we show that:
:$\exists \sequence {x_{n_r} }: x_{n_r} \to \overline l$ as $n \to \infty$
where $\sequence {x_{n_r} }$ is a subsequence of $\sequence {x_n}$.
Let $\epsilon > 0$.
Then $\dfrac \epsilon 2 > 0$.
Since $\overline l = \map \sup L$, and therefore by definition the smallest upper bound of $L$, $\overline l - \dfrac \epsilon 2$ is not an upper bound of $L$.
Hence:
:$\exists l \in L: \overline l \ge l > \overline l - \dfrac \epsilon 2$
Therefore:
:$\sequence {l - \overline l} < \dfrac \epsilon 2$
Now because $l \in L$, we can find $\sequence {x_{m_r} }$, a subsequence of $\sequence {x_n}$, such that $x_{m_r} \to l$ as $n \to \infty$.
So:
:$\exists R: \forall r > R: \size {x_{m_r} - \overline l} < \dfrac \epsilon 2$
So, for any $r > R$:
Thus we have shown that:
:$\forall r > R: \size {x_{m_r} - \overline l} < \epsilon$
So far, what has been shown is that, given any $\epsilon > 0$, there exists an infinite set of terms of $\sequence {x_n}$ which satisfy $\size {x_n - \overline l} < \epsilon$.
Next it is shown how to construct a subsequence:
:$\sequence {x_{n_r} }$ such that $x_{n_r} \to \overline l$
as $n \to \infty$.
Take $\epsilon = 1$ in the above.
Then:
:$\exists n_1: \size {x_{n_1} - \overline l} < 1$
Now take $\epsilon = \dfrac 1 2$ in the above.
Then:
:$\exists n_2 > n_1: \size {x_{n_2} - \overline l} < \dfrac 1 2$
In this way a subsequence is contructed:
:$\sequence {x_{n_r} }$ satisfying $\size {x_{n_r} - \overline l} < \dfrac 1 r$
But $\dfrac 1 r \to 0$ as $r \to \infty$ from the Sequence of Reciprocals is Null Sequence.
From the Squeeze Theorem for Real Sequences, it follows that:
:$\size {x_{n_r} - \overline l} \to 0$ as $r \to \infty$
Thus $\overline l \in L$ as required.
A similar argument shows that the infimum $\underline l$ of $L$ is also in $L$.
== Note ==
From Limit of Subsequence equals Limit of Real Sequence we know that if $\sequence {x_n}$ is convergent then $L$ has exactly one element.
== Sources ==
* : $\S 5$: Subsequences: Lim sup and lim inf: $\S 5.13$
* : Appendix: $\S 18.4$: Subsequences
Category:Limits of Sequences
Existence of Metacompact Space which is not Paracompact

== Theorem ==
There exists at least one example of a metacompact topological space which is not also a paracompact space.
== Proof ==
Let $T$ be the Dieudonné plank.
From Dieudonné Plank is Metacompact, $T$ is a metacompact space.
From Dieudonné Plank is not Paracompact, $T$ is not a paracompact space.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Paracompactness
Category:Paracompact Spaces
Category:Metacompact Spaces