nyuuzyou/wikis huggingface dataset:
Equivalence of Definitions of Join Semilattice

== Theoremm ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $\struct{S, \preceq}$ be an ordered set.
For all $a, b \in S$ let:
:$a \vee b \in S$
where $a \vee b$ is the join of $a$ and $b$ with respect to $\preceq$.
We have , $\vee$ is closed.
The other defining properties for a semilattice follow respectively from:
:Join is Commutative
:Join is Associative
:Join is Idempotent
Hence $\struct {S, \vee}$ is a semilattice.
From Successor is Supremum:
:$\forall a, b \in S : a \preceq b \iff b = a \vee b$
=== Definition 2 implies Definition 1 ===
Let $\struct{S, \vee}$ be a semilattice.
Let $\preceq$ be the relation on $S$ defined by:
:$a \preceq b \iff \paren {a \vee b} = b$
From Semilattice has Unique Ordering such that Operation is Supremum:
:$\preceq$ is the unique ordering on $S$ such that:
::$\forall a, b \in S : a \vee b$ is the join of $a$ and $b$ $\preceq$
Category:Join Semilattices
Equivalence of Definitions of Kernel of Morphism

== Theorem ==
Let $\mathbf C$ be a category with zero object $0$.
Let $f : A \to B$ be a morphism in $\mathbf C$.
Then the following definitions of kernel of $f$ are equivalent.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
By definition of Zero Object, $0$ is an initial object, so Definition 1 is possible.
Let $f : A \to B$ be a morphism in $\mathbf C$.
Let $k : K \to A$ be a pullback of $f$ along the zero morphism $0 : 0 \to B$.
We check the universal property of the equalizer of $f$ and the zero morphism $0 : A \to B$.
Suppose $T$ is any object and $h : T \to A$ is any morphism with $f \circ h$ = $0 \circ h$.
By Composition with Zero Morphism is Zero Morphism $f \circ h = 0 : T \to B$.
By Definition it follows, that $h : T \to A$ and $0 : T \to 0$ is a cone on the pullback diagram defined by $f$ and $0 : 0 \to B$.
There is a unique morphism $t : T \to K$ with $0 \circ t = 0 : T \to 0$ and $k \circ t = h$.
It follows, that $k$ is an equalizer $\map {\mathrm {Eq} } {f, 0}$.
=== Definition 2 implies Definition 1 ===
Let $f : A \to B$ be a morphism in $\mathbf C$.
Assume, that $k : K \to A$ is an equalizer of $f$ and $0 : A \to B$.
Since $0$ is a zero object, we have unique morphisms $\alpha : K \to 0$ and $\beta : 0 \to B$.
The composition $0 : K \to B$ is a zero morphism.
The following diagram is commutative:
::$\begin{xy}\xymatrix@L+2mu@+1em{
K \ar[r]^k \ar[d]^{\alpha} & A \ar[d]^f \\
0 \ar[r]^{\beta} & B
}\end{xy}$
Let $T$ be an object of $\mathbf C$.
Let $t_1 : T \to A$ and $t_2 : T \to 0$ be morphisms such that:
:$f \circ t_1 = \beta \circ t_2$
By definition of zero morphism $\beta \circ t_2$ is a zero morphism.
Thus $f \circ t_1$ is a zero morphism.
By Composition with Zero Morphism is Zero Morphism $0 \circ t_1$ is a zero morphism.
By Uniqueness of Zero Morphism $0 \circ t_1 = f \circ t_1$.
It follows, that $\struct {T, t_1}$ is a cone on the equalizer diagram given by $f$ and $0 : A \to B$.
By definition of equalizer, there exists a unique morphism $t : T \to K$ such that $k \circ t = t_1$.
By Uniqueness of Zero Morphism:
:$\alpha \circ t = t_2$
It follows that $\struct {K, k, \alpha}$ satisfies the definition of pullback.
Category:Category Theory
Equivalence of Definitions of Kuratowski Closure Operator

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 2 implies Definition 1 ===
A closure operator, by definition, is inflationary and idempotent.
Thus it follows immediately that Definition 2 implies Definition 1.
=== Definition 1 implies Definition 2 ===
Let $X$ be a set.
Let $\cl$ be a Kuratowski closure operator on $X$ by Definition 1.
By definition of closure operator, it remains to be proved that $\cl$ is increasing.
Let $A \subseteq B \subseteq X$.
Then by Definition 1 and Union with Superset is Superset:
:$\map \cl B = \map \cl {A \cup B} = \map \cl A \cup \map \cl B$
By Set is Subset of Union:
:$\map \cl A \subseteq \map \cl A \cup \map \cl B = \map \cl B$
Category:Topology
Category:Closure Operators
Equivalence of Definitions of Lattice Filter

== Theorem ==
Let $\struct {S, \vee, \wedge, \preceq}$ be a lattice.
Let $F \subseteq S$ be a non-empty subset of $S$.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $F$ satisify the lattice filter axioms.
To show that $F$ is a meet semilattice filter it is sufficient to show:
Let $x \in F, y \in S : x \preceq y$.
By the lattice filter axioms, $F$ is a sublattice of $\struct {S, \vee, \wedge, \preceq}$, so:
:$x \vee y \in F$
From Preceding iff Join equals Larger Operand:
:$y = x \vee y$
Hence:
:$y \in F$
The result follows.
=== Definition 2 implies Definition 1 ===
Let $F$ be a meet semilattice filter of $\struct {S, \wedge, \preceq}$.
To show that $F$ is a lattice filter of $\struct {S, \vee, \wedge, \preceq}$ it is sufficient to show:
:$\forall x \in F, a \in S: x \vee a \in F$
Let $x \in F, a \in S$.
By definition of join:
:$x \preceq x \vee a$
By definition of meet semilattice filter, $F$ is an upper section, so:
:$x \vee a \in F$
The result follows.
Category:Lattice Filters
Equivalence of Definitions of Lattice Ideal

== Theorem ==
Let $\struct {S, \vee, \wedge, \preceq}$ be a lattice.
Let $I \subseteq S$ be a non-empty subset of $S$.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $I$ satisify the lattice ideal axioms.
To show that $I$ is a join semilattice ideal it is sufficient to show:
Let $x \in I, y \in S : y \preceq x$.
By the lattice ideal axioms, $F$ is a sublattice of $\struct {S, \vee, \wedge, \preceq}$, so:
:$x \wedge y \in I$
From Preceding iff Meet equals Less Operand:
:$y = x \wedge y$
Hence:
:$y \in I$
The result follows.
=== Definition 2 implies Definition 1 ===
Let $I$ be a join semilattice ideal of $\struct {S, \vee, \preceq}$.
To show that $I$ is a lattice ideal of $\struct {S, \vee, \wedge, \preceq}$ it is sufficient to show:
:$\forall x \in I, a \in S: x \wedge a \in I$
Let $x \in I, a \in S$.
By definition of meet:
:$x \wedge a \preceq x$
By definition of join semilattice ideal, $I$ is an lower section, so:
:$x \wedge a \in I$
The result follows.
Category:Lattice Ideals
Equivalence of Definitions of Left Quasi-Reflexive Relation

== Theorem ==
Let $\RR \subseteq S \times S$ be a relation in $S$.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== $(1)$ implies $(2)$ ===
Let $\RR$ be a left quasi-reflexive relation by definition $1$.
Then by definition:
:$\forall x, y \in S: \tuple {x, y} \in \RR \implies \tuple {x, x} \in \RR$
Let $x \in \Dom \RR$ be arbitrary.
Then by definition of domain:
:$\exists y \in S: \tuple {x, y} \in \RR$
Hence :
:$\tuple {x, x} \in \RR$
As $y$ is arbitrary:
:$\forall x \in \Dom \RR: \tuple {x, x} \in \RR$
Thus $\RR$ is a left quasi-reflexive relation by definition $2$.
=== $(2)$ implies $(1)$ ===
Let $\RR$ be a left quasi-reflexive relation by definition $2$.
Then by definition:
:$\forall x \in \Dom \RR: \tuple {x, x} \in \RR$
Let $x, y \in S$ be arbitrary such that $\tuple {x, y} \in \RR$.
Then by definition of image set:
:$x \in \Dom \RR$
Hence :
:$\tuple {x, x} \in \RR$
As $\tuple {x, y} \in \RR$ is arbitrary:
:$\forall x, y \in S: \tuple {x, y} \in \RR \implies \tuple {x, x} \in \RR$
Thus $\RR$ is a left quasi-reflexive relation by definition $1$.
Category:Left Quasi-Reflexive Relations
Equivalence of Definitions of Lemniscate of Bernoulli

== Theorem ==
=== Geometric Definition ===
=== Cartesian Definition ===
=== Polar Definition ===
=== Parametric Definition ===
== Proof ==
=== Geometric Definition equivalent to Cartesian Definition ===
Let $M$ be a lemniscate of Bernoulli by the geometric definition.
Then by definition:
:
Let $P_1 = \tuple {a, 0}$ and $P_2 = \tuple {-a, 0}$.
Let $p = \tuple {x, y}$ be an arbitrary point of $M$.
We have:
Thus $M$ is a lemniscate of Bernoulli by the Cartesian definition.
=== Geometric Definition equivalent to Polar Definition ===
Let $M$ be a lemniscate of Bernoulli by the geometric definition.
Then by definition:
:
Let $M$ be embedded in a polar coordinate plane whose origin is at $O$ and such that $P_1 = \polar {a, 0}$ and $P_2 = \polar {a, \pi}$.
Consider an arbitrary point $p = \polar {r, \theta}$.
Let $d_1 = \size {P_1 p}$ and $d_2 = \size {P_2 p}$.
We have:
=== Parametric Definition equivalent to Cartesian Definition ===
Let $M$ be a lemniscate of Bernoulli by the parametric definition.
Then by definition:
:
We have:
Then:
Category:Lemniscate of Bernoulli
Equivalence of Definitions of Limit Inferior of Sequence of Sets

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
Let $\sequence {E_n : n \in \N}$ be a sequence of sets.
Let:
:$\ds B_n := \bigcap_{j \mathop = n}^\infty E_j$
Let:
:$\ds \liminf_{n \mathop \to \infty} E_n := \bigcup_{n \mathop = 0}^\infty B_n$
that is, $\ds \liminf_{n \mathop \to \infty} \ E_n$ according to definition 1.
Let:
:$E := \set {x: x \in E_i \text{ for all but finitely many } i}$
that is, $\ds \liminf_{n \mathop \to \infty} E_n$ according to definition 2.
By definition of set equality, it is enough to prove:
:$\ds \liminf_{n \mathop \to \infty} E_n \subseteq E$
and:
:$E \subseteq \ds \liminf_{n \mathop \to \infty} E_n$
From Set Complement inverts Subsets, we have:
:$\ds \map \complement E \subseteq \map \complement {\liminf_{n \mathop \to \infty} E_n} \iff \liminf_{n \mathop \to \infty} E_n \subseteq E$
The strategy will therefore be to prove:
:$\ds E \subseteq \liminf_{n \mathop \to \infty} E_n$
and:
:$\ds \map \complement E \subseteq \map \complement {\liminf_{n \mathop \to \infty} E_n}$
=== Definition $2$ is contained in Definition $1$ ===
Let $x \in E$.
By definition $x$ is in all but a finite number of $E_i$.
Let $m \in \Z_{\ge 0}$ be the largest integer such that $x \notin E_m$.
$m$ has to exist, because the set $\set {i \in \N: x \notin E_i}$ is a finite set.
Let $M \in \Z$ such that $m < M$.
Then as $x \in E_j$ for all $j > m$ it follows by definition of set intersection that:
:$x \in B_M = \ds \bigcap_{j \mathop = M}^\infty E_j$
From the definition of set union, it follows that:
:$x \in \ds \bigcup_{n \mathop = 0}^\infty B_n = \liminf_{n \mathop \to \infty} E_n$
Hence:
:$E \subseteq \ds \liminf_{n \mathop \to \infty} E_n$
=== Complement of Definition 2 is contained in Complement of Definition 1 ===
Let $x \in \map \complement E$
That is, $x \notin E$.
The definition of $E$ grants the existence of a strictly increasing sequence $\sequence {i_n}$ of natural numbers, such that:
:$\forall n \in \N: x \notin E_{i_n}$
It follows that for every $k \in \N$, there exists an $n \in \Z$ with $i_n > k$.
Subsequently, from the definition of set intersection:
:$B_k = \ds \bigcap_{j \mathop = k}^\infty E_j \subseteq E_{i_n}$
and hence $x \notin B_k$.
As $k$ was arbitrary, we have:
:$x \notin \ds \bigcup_{k \mathop = 0}^\infty B_k = \liminf_{n \mathop \to \infty} E_n$
It follows that:
:$\ds \map \complement E \subseteq \map \complement {\liminf_{n \mathop \to \infty} E_n}$
Therefore, we have established that
:$x \in \ds \liminf_{n \mathop \to \infty} E_n \iff x \in E$
From the definition of set equality, it follows that:
:$\ds \liminf_{n \mathop \to \infty} E_n = E$
== Also see ==
* Inner Limit of Sequence of Sets in Normed Space described via the point-to-set distance function induced by the norm of the space
* Inner Limit in Hausdorff Space by Open Neighborhoods
== Sources ==
* : Chapter $\text {I}$: Sets of Points: $1 \cdot 1$. The algebra of sets
* : $\S 1.1$: Rings and Algebras: Problem $1.1.2$
Category:Limits Inferior of Set Sequences
Equivalence of Definitions of Limit Ordinal

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
Let $x \in \On$ be an element of the class of all ordinals.
From Categories of Elements under Well-Ordering, $x$ falls into one of the following $3$ categories:
:$(1): \quad x$ is the smallest element of $\On$ the subset relation
:$(2): \quad x$ is the immediate successor of another element $y \in \On$ the subset relation
:$(3): \quad x$ is a limit element of $\On$ under the subset relation.
We have that Zero is Smallest Ordinal.
The result follows.
Category:Limit Ordinals
Equivalence of Definitions of Limit Point in Metric Space

== Theorem ==
Let $M = \struct {S, d}$ be a metric space.
Let $\tau$ be the topology induced by the metric $d$.
Let $A \subseteq S$ be a subset of $S$.
Let $\alpha \in S$. 
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== $(1) \implies (2)$ ===
Let:
:$\forall \epsilon \in \R_{>0}: \paren {\map {B_\epsilon} \alpha \setminus \set \alpha} \cap A \ne \O$
=== $(2) \implies (3)$ ===
Let there exist a sequence $\sequence{\alpha_n}$ in $A \setminus \set \alpha$ such that $\sequence{\alpha_n}$ converges to $\alpha$ in $S$.
=== $(3) \implies (1)$ ===
Let $\alpha$ be a limit point in the topological space $\struct{S, \tau}$.
Category:Limit Points in Metric Spaces
Category:Equivalence of Definitions of Limit Point in Metric Space
Equivalence of Definitions of Limit Point in Metric Space/Definition 1 implies Definition 2

== Theorem ==
Let $M = \struct {S, d}$ be a metric space.
Let $\tau$ be the topology induced by the metric $d$.
Let $A \subseteq S$ be a subset of $S$.
Let $\alpha \in S$. 
Let:
:$\forall \epsilon \in \R_{>0}: \paren {\map {B_\epsilon} \alpha \setminus \set \alpha} \cap A \ne \O$
Then:
:there is a sequence $\sequence{\alpha_n}$ in $A$ such that $\alpha$ is a limit of the sequence $\sequence{\alpha_n}$, considered as sequence in $S$.
== Proof ==
A sequence $\sequence {\alpha_n}$ is constructed using finite recursion.
Let $\alpha_0$ be some arbitrary point in $A \cap \paren{\map {B_1} \alpha \setminus \set \alpha}$.
$\alpha_{n + 1}$ is some arbitrary point in $A \cap \paren{\map {B_{\frac {\map d {\alpha, \alpha_n} } 2} } \alpha \setminus \set \alpha}$ where $\map {B_{\frac {\map d {\alpha, \alpha_n} } 2} } \alpha$ denotes the open $\dfrac {\map d {\alpha, \alpha_n} } 2$-ball of $\alpha$.
We can be assured that such a point exists .
Then:
:$\map d {\alpha_n, \alpha} < 2^{-n} \map d {\alpha_0, \alpha}$
Since $\map d {\alpha_0, \alpha}$ is fixed:
:$\map d {\alpha_n, \alpha} < 2^{-n} \map d {\alpha_0, \alpha} < r$
for some $n$ because the natural numbers are Archimedean.
Therefore $\alpha$ is the limit of the sequence $\sequence {\alpha_n}$ by definition.
== Sources ==
* : $3.2d$
* : Chapter $2$: Metric Spaces: $\S 6$: Open Sets and Closed Sets: Definition $6.6$
Category:Equivalence of Definitions of Limit Point in Metric Space
Equivalence of Definitions of Limit Point in Metric Space/Definition 2 implies Definition 3

== Theorem ==
Let $M = \struct {S, d}$ be a metric space.
Let $\tau$ be the topology induced by the metric $d$.
Let $A \subseteq S$ be a subset of $S$.
Let $\alpha \in S$. 
Let there exist a sequence $\sequence{\alpha_n}$ in $A \setminus \set \alpha$ such that $\sequence{\alpha_n}$ converges to $\alpha$ in $S$.
Then:
:$\alpha$ is a limit point in the topological space $\struct{S, \tau}$.
== Proof ==
Let $U \in \tau$ be an arbitrary open set:
:$\alpha \in U$
By :
:$\exists \epsilon \in \R_{>0} : \map {B_\epsilon} \alpha \subseteq U$
By :
:$\exists N \in \N : \forall n \ge N : d(\alpha, \alpha_n) < \epsilon$
By :
:$\alpha_N \in \map {B_\epsilon} \alpha$
By hypothesis:
:$\alpha_N \in A \setminus \set \alpha$
We have:
Hence:
:$A \cap \paren{U \setminus \set \alpha} \ne \O$
Since $U$ was arbitrary, it follows that $\alpha$ is a limit point in the topological space $\struct{S, \tau}$ by definition.
Category:Equivalence of Definitions of Limit Point in Metric Space
Equivalence of Definitions of Limit Point in Metric Space/Definition 3 implies Definition 1

== Theorem ==
Let $M = \struct {S, d}$ be a metric space.
Let $\tau$ be the topology induced by the metric $d$.
Let $A \subseteq S$ be a subset of $S$.
Let $\alpha \in S$. 
Let $\alpha$ be a limit point in the topological space $\struct{S, \tau}$.
Then:
:$\forall \epsilon \in \R_{>0}: \paren {\map {B_\epsilon} \alpha \setminus \set \alpha} \cap A \ne \O$
== Proof ==
From Open Ball is Open Set:
:$\forall \epsilon \in \R_{>0}: \map {B_\epsilon} \alpha$ is open set in $M$
By :
:$\forall \epsilon \in \R_{>0}: \map {B_\epsilon} \alpha$ is open set in $\struct{S,\tau}$
By :
:$\forall \epsilon \in \R_{>0}: \paren {\map {B_\epsilon} \alpha \setminus \set \alpha} \cap A \ne \O$
Category:Equivalence of Definitions of Limit Point in Metric Space
Equivalence of Definitions of Limit Point of Filter Basis

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $\FF$ be a filter on the underlying set $S$ of $T$.
Let $\BB$ be a filter basis of $\FF$.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
Let $\FF$ be a filter on $S$.
Let $\BB$ be a filter basis of $\FF$.
=== $(1)$ implies $(2)$ ===
Let $x \in S$ be a limit Point of $\BB$ by definition $1$.
Then by definition $\FF$ converges on $x$.
By definition of convergent filter:
:$\forall N_x \subseteq S: N_x \in \FF$
where $N_x$ is a neighborhood of $x$.
That is, every neighborhood of $x$ is an element of $\FF$.
By definition of filter basis:
:$\forall U \in \FF: \exists V \in \BB: V \subseteq U$
But we have shown that $N_x \in \FF$.
Therefore:
:$\exists V \in \BB: V \subseteq N_x$
That is, every neighborhood of $x$ contains a set of $\BB$.
Thus $x \in S$ is a limit Point of $\BB$ by definition $2$.
=== $(2)$ implies $(1)$ ===
Let $x \in S$ be a limit Point of $\BB$ by definition $2$.
Then by definition every neighborhood of $x$ contains a set of $\BB$.
By definition of filter basis, $\FF := \set {V \subseteq X: \exists U \in \BB: U \subseteq V}$ is a filter on $S$ :
:$\forall V_1, V_2 \in \BB: \exists U \in \BB: U \subseteq V_1 \cap V_2$
But then letting $V_1 = V_2 = V$, we have:
:$\forall V \in \BB: \exists U \in \BB: U \subseteq V$
and so $V \in \FF$.
Thus every neighborhood of $x$ is an element of $\FF$.
Thus, by definition, $\FF$ converges on $x$.
Thus $x \in S$ is a limit Point of $\BB$ by definition $1$.
Category:Limit Points of Filter Bases
Equivalence of Definitions of Limit Point of Set

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $A \subseteq S$.
=== Definition from Open Neighborhood===
=== Definition from Closure===
=== Definition from Adherent Point===
=== Definition from Relative Complement===
== Proof ==
=== Definition from Open Neighborhood $\iff$ Definition from Closure ===
The closure of $A$ is defined as the union of the set of all isolated points of $A$ and the set of all limit points of $A$.
The rest then follows directly from Equivalence of Definitions of Isolated Point.
=== Definition from Closure $\iff$ Definition from Adherent Point ===
Follows directly from Equivalence of Definitions of Adherent Point.
=== Definition from Open Neighborhood $\iff$ Definition from Relative Complement ===
== Also see ==
* Definition:Limit Point of Set
Category:Equivalence of Definitions of Limit Point of Set
Category:Limit Points
Equivalence of Definitions of Limit Point of Set/Definition from Open Neighborhood iff Definition from Relative Complement

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $A \subseteq S$.
=== Definition from Open Neighborhood===
=== Definition from Relative Complement===
== Proof 1 ==
== Proof 2 ==
Category:Equivalence of Definitions of Limit Point of Set
Equivalence of Definitions of Limit Point of Set/Definition from Open Neighborhood iff Definition from Relative Complement/Proof 2

== Theorem ==
== Proof ==
The following equivalence holds:
There exists an open neighborhood $U$ of $x$ such that $A \cap \paren {U \setminus \set x} = \O$
By , $\map \complement A \cup \set x$ is a neighborhood of $x$
The result follows from the Rule of Transposition.
Category:Equivalence of Definitions of Limit Point of Set
Equivalence of Definitions of Limit Superior of Sequence of Sets

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
Begin by defining:
:$\ds B_n := \bigcup_{j \mathop = n}^\infty E_j$
Then by definition 1:
:$\ds \limsup_{n \mathop \to \infty} E_n = \bigcap_{n \mathop = 0}^\infty B_n$
=== First Direction ===
Let $x$ belong to $E_i$ for infinitely many $i \in \N$.
Let $\map \phi n$ be the sequence consisting of these numbers in increasing order.
Then for any number $k$, there exists a number $a$ such that $\map \phi a \ge k$.
Hence:
:$\ds x \in E_{\map \phi a} \subseteq \bigcup_{j \mathop = k}^\infty E_j = B_k$
Since $k$ was arbitrary, it follows that $x \in B_n$ for each $n$.
So:
:$\ds x \in \limsup_{n \mathop \to \infty} E_n$
=== Second Direction ===
Let:
:$\ds x \in \bigcap_{n \mathop = 0}^\infty B_n$
 $x$ occurs in only finitely many $E_i$.
Then there is a largest value of $i$ (call it $i_0$) for which the membership holds.
Hence:
:$x \notin \paren {E_{i_0 + 1} \cup E_{i_0 + 2} \cup \ldots} = B_{i_0 + 1}$
Therefore:
:$\ds x \notin \bigcap_{n \mathop = 0}^\infty B_n$
This contradicts our assumption about $x$.
Hence $x$ belongs to infinitely many terms of the sequence.
== Sources ==
* : Chapter $\text {I}$: Sets of Points: $1 \cdot 1$. The algebra of sets
* : $\S 1.1$: Rings and Algebras: Problem $1.1.2$
Category:Limits Superior of Set Sequences
Equivalence of Definitions of Limit of Mapping between Metric Spaces

== Theorem ==
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $c$ be a limit point of $M_1$.
Let $f: A_1 \to A_2$ be a mapping from $A_1$ to $A_2$ defined everywhere on $A_1$ except possibly at $c$.
Let $L \in M_2$.
=== $\epsilon$-$\delta$ Condition ===
=== $\epsilon$-Ball Condition ===
</onlyinclude>
== Proof ==
=== $\epsilon$-$\delta$ Condition implies $\epsilon$-Ball Condition ===
Suppose that $f$ satisfies the $\epsilon$-$\delta$ condition:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: 0 < \map {d_1} {x, c} < \delta \implies \map {d_2} {\map f x, L} < \epsilon$
Let $y \in f \sqbrk {\map {B_\delta} {c; d_1} \setminus \set c}$.
By definition of open $\epsilon$-ball, this means:
:$\exists x \in A_1: 0 < \map {d_1} {x, c} < \delta: y = \map f x$
By hypothesis, it follows that $\map {d_2} {y, L} < \epsilon$
That is:
:$y \in \map {B_\epsilon} {L; d_2}$
By definition of subset:
:$f \sqbrk {\map {B_\delta} {c; d_1} \setminus \set c} \subseteq \map{B_\epsilon} {L; d_2}$
Thus it follows that $f$ satisfies the $\epsilon$-ball condition.
=== $\epsilon$-Ball Condition implies $\epsilon$-$\delta$ Condition ===
Suppose that $f$ satisfies the $\epsilon$-ball condition:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: f \sqbrk {\map {B_\delta} {c; d_1} \setminus \set c} \subseteq \map {B_\epsilon} {L; d_2}$
Let $0 < \map {d_1} {x, c} < \delta$.
By definition of open $\epsilon$-ball, this means:
:$x \in \map {B_\delta} {c; d_1} \setminus \set c$
By hypothesis, it follows that:
:$\map f x \in \map {B_\epsilon} {L; d_2}$
Thus by definition of open $\epsilon$-ball:
:$\map {d_2} {\map f x, L} < \epsilon$
That is, $f$ satisfies the $\epsilon$-$\delta$ condition.
== Notes ==
:$(1): \quad c$ does not need to be a point in $A_1$. Therefore $\map f c$ need not be defined. And even if $c \in A_1$, in may be that $\map f c \ne L$.
:$(2): \quad$ It is essential that $c$ be a limit point of $A_1$.
::Otherwise there would exist $\delta > 0$ such that $\set {z: 0 < \map {d_1} {z, c} < \delta}$ contains no points of $A_1$.
::In this case the $\epsilon$-$\delta$ condition would be vacuously true for any $L \in A_2$, which would not do.
Category:Limits of Mappings between Metric Spaces
Equivalence of Definitions of Limit of Real Function

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
By definition of deleted $\delta$-neighborhood of $c$:
:$x \in \map {N_\delta} c \setminus \set c$
:
:$0 < \size {x - c} < \delta$
By definition of $\epsilon$-neighborhood of $L$:
:$\map f x \in \map {N_\epsilon} L$
:
:$\size {\map f x - L} < \epsilon$
The result follows by comparison of the definitions.
Category:Limits of Real Functions