nyuuzyou/wikis huggingface dataset:
Equivalence of Definitions of Boolean Algebra

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $\struct {S, \vee, \wedge, \neg}$ be an algebraic system which satisfies the criteria of Definition 1.
;:
From , we have that $S$ is closed under $\vee$, $\wedge$ and $\neg$.
That is:
;:
From , we have that $\vee$ and $\wedge$ are commutative on $S$.
That is:
;:
From Operations of Boolean Algebra are Associative:
;:
From Absorption Laws (Boolean Algebras):
The specific format in which these results are expressed in follow from : $\vee$ and $\wedge$ are commutative on $S$.
;:
From , we have that both $\vee$ and $\wedge$ distribute over the other.
Hence:
;:
We have:
All axioms are fulfilled.
=== Definition 2 implies Definition 1 ===
Let $\struct {S, \vee, \wedge, \neg}$ be an algebraic system which satisfies the criteria of Definition 2.
;:
From :
That is, $S$ is closed under $\vee$, $\wedge$ and $\neg$.
;:
From :
That is, $\vee$ and $\wedge$ are commutative on $S$.
;:
From :
From , we have that $\vee$ and $\wedge$ are commutative on $S$, and so:
That is, both $\vee$ and $\wedge$ distribute over the other.
;:
From Meet with Complement is Bottom:
:$\exists \bot \in S: \forall a \in S: a \wedge \neg a = \bot$
From Join with Complement is Top:
:$\exists \top \in S: \forall a \in S: a \vee \neg a = \top$
These elements $\bot$ and $\top$ are the only elements of $S$ which have these properties.
;:
We have demonstrated above.
Then:
From , we have that $\vee$ and $\wedge$ are commutative on $S$, and so:
That is:
:$\bot$ is an identity element for $\vee$
:$\top$ is an identity element for $\wedge$.
All axioms are fulfilled.
Hence the result.
Category:Boolean Algebras
Equivalence of Definitions of Boolean Lattice

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== Definition 1 $\implies$ Definition 2 ===
=== Definition 2 $\implies$ Definition 3 ===
=== Definition 3 $\implies$ Definition 1 ===
Category:Boolean Lattices
Equivalence of Definitions of Bottom of Lattice

== Theorem ==
Let $\struct {S, \vee, \wedge, \preceq}$ be a lattice.
Let $\bot$ be a bottom of $\struct {S, \vee, \wedge, \preceq}$.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
By definition, $\bot$ is the smallest element of $S$ for all $a \in S$:
:$\bot \preceq a$
By Ordering in terms of Join, this is equivalent to:
:$a \vee \bot = a$
If this equality holds for all $a \in S$, then by definition $\bot$ is an identity for $\vee$.
The result follows.
Category:Bottom of Lattice
Equivalence of Definitions of Boundary

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
=== Definition $1$: from Closure and Interior ===
=== Definition $2$: from Neighborhood ===
=== Definition $3$: from Intersection of Closure with Closure of Complement ===
=== Definition $4$: from Closure and Exterior ===
== Proof ==
=== Definition $1$ is equivalent to Definition $3$ ===
This is demonstrated in Boundary is Intersection of Closure with Closure of Complement.
=== Definition $2$ is equivalent to Definition $3$ ===
Let $x \in S$.
By definition of the closure:
:$x \in H^-$ every neighborhood $N$ of $x$ satisfies $H \cap N \ne \O$
:$x \in \paren{\overline H}^-$ every neighborhood $N$ of $x$ satisfies $\overline H \cap N \ne \O$
Therefore $x \in H^- \cap \paren{\overline H}^-$ every neighborhood $N$ of $x$ satisfies:
:$H \cap N \ne \O$
and
:$\overline H \cap N \ne \O$
Category:Set Boundaries
Equivalence of Definitions of Bounded Lattice

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== Definition 1 Definition 2 ===
From Supremum of Empty Set is Smallest Element, we have that:
:$\bot := \sup \O$
satisfies $\bot \preceq a$ for all $a \in S$.
From Join Semilattice has Smallest Element iff has Identity:
:$\bot$ is an identity element for $\vee$, and conversely.
That $\inf \O$ is an identity element for $\wedge$, and conversely, follows by the Duality Principle.
Hence the result. 
=== Definition 1 Definition 3 ===
Let $x \in S$.
From Supremum of Empty Set is Smallest Element:
:$x = \sup \O$ $x$ is the smallest element of $\struct{S, \preceq}$
From Ordered Set has Lower Bound iff has Smallest Element:
:$x$ is the smallest element of $\struct{S, \preceq}$ $x$ is a lower bound of $S$ in $\struct{S, \preceq}$
That $\inf \O$ is an upper bound for $S$, and conversely, follows by the Duality Principle.
Hence the result. 
== Also see ==
* Definition:Bounded Lattice
Category:Bounded Lattices
Equivalence of Definitions of Bounded Metric Space

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
=== Definition 4 ===
== Proof ==
Let $M = \struct {X, d}$ be a metric space.
Let $M' = \struct {Y, d_Y}$ be a subspace of $M$.
=== Definition 1 implies Definition 2 ===
Let $M'$ be bounded according to Definition 1:
:$\exists a \in X, K \in \R: \forall x \in Y: \map d {x, a} \le K$
Let $x, y \in Y$.
Then:
Thus:
:$\map d {x, y} \le 2 K$
Thus, setting $r = 2 K$, $M'$ fulfils the conditions to be bounded according to Definition 2.
=== Definition 2 implies Definition 1 ===
Let $M'$ be bounded according to Definition 2:
:$\exists K \in \R: \forall x, y \in M': \map d {x, y} \le K$
Let $a = y$.
Then:
:$\map d {x, a} \le K$
and so:
:$\exists a \in Y, K \in \R: \forall x \in Y: \map d {x, a} \le K$
As $X \subseteq Y$ it follows by definition of subset that:
:$\exists a \in Y \implies \exists a \in X$
and so:
:$\exists a \in X, K \in \R: \forall x \in Y: \map d {x, a} \le K$
Thus $M'$ fulfils the conditions to be bounded according to Definition 1.
=== Definition 1 implies Definition 3 ===
Let $M'$ be bounded according to Definition 1:
:$\exists a \in X, K \in \R: \forall x \in Y: \map d {x, a} \le K$
Although not specified, $K \le 0$ for the definition to make sense, as $\map d {x, a} \ge 0$.
Let $\map {B_{K + 1} } a$ be the open $K + 1$-ball of $x$.
By definition of open ball:
:$\map {B_{K + 1} } a := \set {x \in M: \map d {x, a} < K + 1}$
Let $y \in S$.
Then by definition:
:$\map d {y, a} \le K < K + 1$
and so:
:$y \in \map {B_{K + 1} } a$
It follows by the definition of subset that:
:$Y \subseteq \map {B_{K + 1} } a$
and so $Y$ can be fitted inside an open ball.
Thus $M'$ fulfils the conditions to be bounded according to Definition 3.
=== Definition 3 implies Definition 1 ===
Let $M'$ be bounded according to Definition 3:
:$\exists x \in A, \epsilon \in \R_{>0}: B \subseteq \map {B_\epsilon} x$
where $\map {B_\epsilon} x$ is the open $\epsilon$-ball of $x$.
Then by definition:
:$\forall y \in Y: \map d {x, y} \le \epsilon$
and so the condition for boundedness in $M$ is fulfilled.
Thus $M'$ fulfils the conditions to be bounded according to Definition 1.
=== Definition 1 implies Definition 4 ===
Suppose:
:$\exists x \in M, \epsilon \in \R_{>0}: Y \subseteq \map {B_\epsilon} x$
where $\map {B_\epsilon} x$ is the open $\epsilon$-ball of $x$.
Then by definition:
:$\forall y \in Y: \map d {x, y} \le \epsilon$
and so the condition for boundedness in $M$ is fulfilled.
This follows from Element in Bounded Metric Space has Bound.
=== Definition 4 implies Definition 1 ===
This follows directly.
== Sources ==
* : Chapter $\text{III}$: Metric Spaces: Compactness
Category:Bounded Metric Spaces
Equivalence of Definitions of Bounded Real-Valued Function

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
Let $S$ be a set.
Let $f: S \to \R$ be a real-valued function.
=== Definition 1 implies Definition 2 ===
Let $f$ be bounded according to definition 1:
:bounded above in $S$ by $H \in \R$
:bounded below in $S$ by $L \in \R$
Thus by definition:
:$\forall x \in S: L \le \map f x$
:$\forall x \in S: \map f x \le H$
By Equivalence of Definitions of Bounded Subset of Real Numbers:
:$\exists K \in \R_{\le 0}: \forall x \in S: \size {\map f x} \le K$
So $f$ is bounded according to definition 2.
=== Definition 2 implies Definition 1 ===
Let $f$ be bounded according to definition 2:
:$\exists K \in \R_{\le 0}: \forall x \in S: \size {\map f x} \le K$
Then by Equivalence of Definitions of Bounded Subset of Real Numbers:
:$\exists K \in \R_{\le 0}: \forall x \in S: -K \le \map f x$
and so $f$ is bounded below in $S$ by $-K \in \R$
and
:$\exists K \in \R_{\le 0}: \forall x \in S: \map f x \le K$
and so $f$ is bounded above in $S$ by $K \in \R$.
So $f$ is bounded according to definition 2.
=== Definition 1 implies Definition 3 ===
Let $f$ be bounded according to definition 1:
:bounded above in $S$ by $H \in \R$
:bounded below in $S$ by $L \in \R$
Thus by definition:
:$\forall x \in S: L \le \map f x$
:$\forall x \in S: \map f x \le H$
That is:
:$\map f x \in \closedint L H$
Thus $f$ is bounded according to definition 3.
=== Definition 3 implies Definition 1 ===
Let $f$ be bounded according to definition 3:
:$\exists a, b \in \R_{\le 0}: \forall x \in S: \map f x \in \closedint a b$
Thus by definition:
:$\forall x \in S: a \le \map f x$
and so $f$ is bounded below in $S$ by $a \in \R$
:$\forall x \in S: \map f x \le b$
and so $f$ is bounded above in $S$ by $b \in \R$.
So $f$ is bounded according to definition 1.
Category:Bounded Real-Valued Functions
Equivalence of Definitions of Bounded Subset of Real Numbers

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $S$ be a bounded subset of the real numbers by definition 1.
Then by definition $S$ is bounded both above and below.
As $S$ is bounded below:
:$\exists L \in \R: \forall x \in S: L \le x$
As $S$ is bounded above:
:$\exists H \in \R: \forall x \in S: H \ge x$
Let:
:$K = \max \set {\size L, \size H}$
Then:
:$K \ge \size L$ and $K \ge \size H$
It follows from Negative of Absolute Value: Corollary 2 that $-K \le L \le K$ and $-K \le H \le K$.
In particular:
:$-K \le L$ and so $\forall x \in S: -K \le x$
:$H \le K$ and so $\forall x \in S: x \le K$
Thus:
:$\forall x \in S: -K \le x \le K$
and, by Negative of Absolute Value: Corollary 2:
:$\forall x \in S: \size x \le K$
Thus $S$ is a bounded subset of the real numbers by definition 2.
=== Definition 2 implies Definition 1 ===
Let $S$ be a bounded subset of the real numbers by definition 2.
Thus:
:$\exists K \in \R: \forall x \in S: \size x \le K$
Then by Negative of Absolute Value: Corollary 2:
:$\forall x \in S: -K \le x \le K$
Thus by definition, $S$ is both bounded above (by $K$) and bounded below (by $-K$).
Thus $S$ is a bounded subset of the real numbers by definition 1.
== Sources ==
* : $\S 2$: Continuum Property: $\S 2.3$
* : Appendix: $\S 18.2$: Bounded sets
Category:Bounded Sets of Real Numbers
Equivalence of Definitions of Bounded Variation for Real Function on Closed Bounded Interval

== Theorem ==
Let $a, b$ be real numbers with $a < b$.
Let $f : \closedint a b \to \R$ be a real function. 
=== Definition 1 ===
 
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Suppose that there exists a $M \in \R$ such that:
:$\map {V_f} {P ; \closedint a b} \le M$
for all finite subdivisions $P$.
Let $\SS$ be a finite non-empty subset of $\closedint a b$. 
Define: 
:$\SS^\ast = \SS \cup \set {a, b}$
and write: 
:$\SS^\ast = \set {x_0, x_1, \ldots, x_n}$
with: 
:$a = x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n = b$
Then $\SS^\ast$ is a finite subdivision of $\closedint a b$ and:
so: 
:$\map {V_f^\ast} {\SS; \closedint a b} \le M$ 
for all finite non-empty subset of $\closedint a b$.
=== Definition 2 implies Definition 1 ===
Suppose that there exists a $M \in \R$ such that:
:$\map {V_f^\ast} {\SS; \closedint a b} \le M$
for all finite non-empty subsets $\SS$ of $\closedint a b$.
Note that any finite subdivision of $\closedint a b$ is also a finite non-empty subset of $\closedint a b$. 
So, in particular, for any finite subdivision $P$ of $\closedint a b$, we have: 
:$\map {V_f} {P ; \closedint a b} = \map {V_f^\ast} {P ; \closedint a b} \le M$
as required.
Category:Bounded Variation
Equivalence of Definitions of Cantor Set

== Theorem ==
:
=== $(1)$: As a Limit of Intersections ===
=== $(2)$: From Ternary Representation ===
=== $(3)$: As a Limit of a Decreasing Sequence ===
== Proof ==
Let $\CC_n$ be defined as in $(1)$.
Let $x \in \closedint 0 1$.
We need to show that:
:$x$ can be written in base $3$ without using the digit $1$ :
::$\forall n \in \Z, n \ge 1: x \in C_n$
First we note that from Sum of Infinite Geometric Sequence:
:$\ds 1 = \sum_{n \mathop = 0}^\infty \frac 2 3 \paren {\frac 1 3}^n$
that is:
:$1 = 0.2222 \ldots_3$
Thus any real number which, expressed in base $3$, ends in $\ldots 10000 \ldots$ can be expressed as one ending in $\ldots 02222 \ldots$ by dividing the above by an appropriate power of $3$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $29$. The Cantor Set: $1$
Category:Cantor Set
Equivalence of Definitions of Cardinality of Finite Class

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
Let $A_1$ be the class which has a bijection $\phi_1$ from $A_1$ to $n$.
Let $A_2$ be the class which has a bijection $\phi_2$ from $A_2$ to $n^+ \setminus \set 0$.
Consider the mapping $\phi: A_1 \to A_2$ defined as:
:$\forall k \in n: \map {\phi_1} k = k^+$
$\phi$ is trivially a bijection.
The result follows.
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 6$ Finite Sets
Category:Cardinality
Equivalence of Definitions of Cartesian Product of Indexed Family

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== $(1)$ implies $(2)$ ===
Let $\ds \prod_{i \mathop \in I} S_i$ be a Cartesian product of $\family {S_i}_{i \mathop \in I}$ by definition $1$.
Then by definition:
:$\ds \prod_{i \mathop \in I} S_i$ is the set of all families $\family {s_i}_{i \mathop \in I}$ with $s_i \in S_i$ for each $i \in I$
By definition, a indexed family of sets is a mapping from an indexing set $I$ to a set of sets $\SS$ such that:
:$\forall i \in I: S_i \subseteq \SS$
Let $f: I \to \SS$ be an indexed family of sets.
For each $i \in I$, $\map f i$ is an element of $S_i$.
That is:
:$\map f i \in S_i$
Hence:
:$\Img f = f \sqbrk I \subseteq \ds \bigcup_{i \mathop \in I} S_i$
Thus:
:$\ds \prod_{i \mathop \in I} S_i = \set {f \in \paren {\bigcup_{i \mathop \in I} S_i}^I : \forall i \in I: \paren {\map f i \in S_i} }$
Hence $\ds \prod_{i \mathop \in I} S_i$ is a Cartesian product of $\family {S_i}_{i \mathop \in I}$ by definition $2$.
=== $(2)$ implies $(1)$ ===
Let $\ds \prod_{i \mathop \in I} S_i$ be a Cartesian product of $\family {S_i}_{i \mathop \in I}$ by definition $2$.
Then by definition:
:$\ds \prod_{i \mathop \in I} S_i := \set {f \in \paren {\bigcup_{i \mathop \in I} S_i}^I : \forall i \in I: \paren {\map f i \in S_i} }$
where $\ds \paren {\bigcup_{i \mathop \in I} S_i}^I$ denotes the set of all mappings from $I$ to $\ds \bigcup_{i \mathop \in I} S_i$.
Let $f \in \ds \prod_{i \mathop \in I} S_i$ be arbitrary.
Then:
:$\forall i \in I: \map f i = s_i$ for some $s_i \in S_i$
That is, $\ds \prod_{i \mathop \in I} S_i$ is the set of all mapping from an indexing set $I$ to a set of sets $\ds \bigcup_{i \mathop \in I} S_i$.
Thus by definition of indexed family of sets:
:$\ds \prod_{i \mathop \in I} S_i$ is the set of all families $\family {s_i}_{i \mathop \in I}$ with $s_i \in S_i$ for each $i \in I$
Thus $\ds \prod_{i \mathop \in I} S_i$ is a Cartesian product of $\family {S_i}_{i \mathop \in I}$ by definition $1$.
Category:Cartesian Product
Category:Indexed Families
Equivalence of Definitions of Ceiling Function

== Theorem ==
Let $x$ be a real number.
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== Definition 1 equals Definition 2 ===
Follows from Infimum of Set of Integers equals Smallest Element.
=== Definition 1 equals Definition 3 ===
Let $S$ be the set:
:$S = \left\{ {m \in \Z: m \ge x}\right\}$
Let $n = \inf S$.
By Infimum of Set of Integers is Integer, $n \in \Z$.
By Infimum of Set of Integers equals Smallest Element, $n\in S$.
Because $n\in S$, we have $n \geq x$.
Because $n-1 < n$, we have by definition of supremum:
:$n-1 \notin S$
Thus $n-1 < x$.
Thus $n$ is an integer such that:
:$n-1 < x \leq n$
So $n$ is the ceiling function by definition 3.
=== Definition 3 equals Definition 2 ===
Let $n$ be an integer such that:
:$n-1 < x \leq n$
We show that $n$ is the smallest element of the set:
:$S = \left\{ {m \in \Z: m \ge x}\right\}$
Let $m \in \Z$ such that $m \ge x$.
We show that $n\leq m$.
 $m < n$.
By Weak Inequality of Integers iff Strict Inequality with Integer minus One:
:$m \le n - 1$
and so from the definition of $g$ it follows that $m < x$.
By Proof by Contradiction it follows that $m \ge n$.
Because $m \in S$ was arbitrary, $n$ is the smallest element of $S$.
Thus $n$ is the ceiling function by definition 2.
== Also see ==
* Equivalence of Definitions of Floor Function
Category:Ceiling Function
Equivalence of Definitions of Change of Basis Matrix

== Theorem ==
Let $R$ be a ring with unity.
Let $G$ be a finite-dimensional unitary $R$-module.
Let $A = \sequence {a_n}$ and $B = \sequence {b_n}$ be ordered bases of $G$.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
It will be shown that the two matrices defined are equal column-wise.
Let $\ds b_i = \sum_{j \mathop = 1}^n c_{i j} a_j$ for $i$ ranging from $1$ to $n$, where $c_{i j}$'s are scalars.
The uniqueness of the above expression is justified by Expression of Vector as Linear Combination from Basis is Unique.
Then by definition of coordinate vector, the $i$th column of the matrix defined in definition 1 is:
:$\begin {bmatrix} c_{i 1} & c_{i 2} & \ldots & c_{in} \end {bmatrix}^\intercal$
We also have:
:$\ds \map {I_G} {b_i} = b_i = \sum_{j \mathop = 1}^n c_{i j} a_j$
So by definition of relative matrix, the $i$th column of the matrix defined in definition 2 is:
:$\begin {bmatrix} c_{i 1} & c_{i 2} & \ldots & c_{i n} \end {bmatrix}^\intercal$
The two matrices are equal, so the two definitions are equivalent.
Category:Linear Algebra
Category:Matrix Theory
Equivalence of Definitions of Characteristic of Ring

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== Definition 1 is equivalent to Definition 3 ===
By definition of order, the order of $1_R$ is the smallest $p \in \Z_{> 0}$ such that $p \cdot 1_R = 0_R$, the identity of the additive group $\struct {R, +}$ of $\struct {R, +, \circ}$.
Hence if the order of $1_R$ is finite, the definitions coincide.
If the order of $1_R$ is infinite, there is no such $p$.
Then both definitions give $\Char R = 0$, so they coincide as well.
=== Definition 1 is equivalent to Definition 2 ===
Let $g: \Z \to R$ be the initial homomorphism, with $\map g n = n \cdot 1_R$.
By Kernel of Ring Homomorphism is Ideal, $\ker g$ is an ideal of $R$.
By Ring of Integers is Principal Ideal Domain, there exists a unique $p \in \Z_{\ge 0}$ such that $\ker g$ is the principal ideal $\ideal p$.
Suppose there is a smallest $m$ such that $m \cdot 1_R = 0_R$.
Then for any $k \in \Z$:
:$\map g {k m} = k m \cdot 1_R = k \cdot 0_R = 0_R$.
This gives $\ideal m \subseteq \ker g = \ideal p$.
Let $a \in \ker g$.
Then by the Division Theorem:
:$\exists q, r \in \Z, 0 \le r < m: a = m q + r$
As $a, m \in \ker g = \ideal p$ so does $r = a - m q$.
Since $m$ is the smallest (strictly) positive integer such that $m \cdot 1_R = 0_R$:
:$\forall n \in \Z: 0 < n < m \implies \map g n = n \cdot 1_R \ne 0_R$
This forces $r = 0$.
Thus $a = m q \in \ideal m$.
Hence we have $\ideal m = \ideal p$, giving $p = m$.
Now consider the case where $m$ does not exist.
Then for any $k \ne 0$:
:$\map g k = k \cdot 1_R \ne 0_R$
And we have:
:$\map g 0 = 0 \cdot 1_R = 0_R$
giving $\ker g = \set 0 = \ideal 0$.
Both definitions give $\Char R = 0$.
Therefore the definitions are equivalent in each case.
Category:Characteristics of Rings
Equivalence of Definitions of Closed Element

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $\cl$ be a closure operator on $S$.
Let $x \in S$.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $\cl: S \to S$ be a closure operator on $S$.
Let $x \in S$.
By the definition of closure operator, $\cl$ is idempotent.
Thus by Fixed Point of Idempotent Mapping:
:An element of $S$ is a fixed point of $\cl$ it is in the image of $\cl$.
Thus the above definitions are equivalent.
Category:Closed Elements
Equivalence of Definitions of Closed Set

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
=== Definition 1 implies Definition 2 ===
Let $H$ be a closed set of $T$ by definition 1.
Let $H^\complement$ denote the relative complement of $H$ in $S$.
By definition of closed set in $T$:
:$H^\complement$ is open in $T$
From Set is Open iff Neighborhood of all its Points:
:$\forall x \in S: x \notin H \implies H^\complement$ is a neighborhood of $x$.
By definition of limit point:
:$\forall x \in S: x \notin H \implies x$ is not a limit point of $H$
Thus $H$ is a closed set of $T$ by definition 2.
=== Definition 2 implies Definition 1 ===
Let $H$ be a closed set of $T$ by definition 2.
Then by definition:
$\forall x \in S: x \notin H \implies x$ is not a limit point of $H$
By definition of limit point:
$\forall x \in S: x \notin H \implies H^\complement$ is a neighborhood of $x$
By Set is Open iff Neighborhood of all its Points:
$H^\complement$ is open in $T$
Thus $H$ is a closed set of $T$ by definition 1.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Limit Points
* : $2.23$
Category:Closed Sets
Category:Limit Points
Equivalence of Definitions of Closed Set in Metric Space

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $H \subseteq A$.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
Let $H'$ denote the set of limit points of $H$.
=== Definition 1 implies Definition 2 ===
Let $H$ be closed in $M$ by definition 1.
Then by definition $A \setminus H$ is open in $M$.
Let $b \in M: b \in A \setminus H$.
Then by definition of open set:
:$\exists \delta \in \R_{>0}: \map {B_\delta} {b; d} \subseteq A \setminus H$
where $\map {B_\delta} {b; d}$ denotes the open $\delta$-ball of $b$.
From Intersection with Complement is Empty iff Subset:
:$\map {B_\delta} {b; d} \cap H = \O$
So by definition of limit point:
:$b \notin H'$
Hence:
:$A \setminus H \cap H' = \O$
By Intersection with Complement is Empty iff Subset:
:$H' \subseteq H$
Thus $H$ contains all its limit points.
That is, $H$ be closed in $M$ by definition 2.
=== Definition 2 implies Definition 1 ===
Let $H$ be closed in $M$ by definition 2.
That is:
:$H' \subseteq H$
Then by Set Complement inverts Subsets:
:$A \setminus H \subseteq A \setminus H'$
Let $b \in A \setminus H$.
Then by definition of relative complement:
:$b \notin H'$
Then by definition of limit point:
:$\exists \delta \in \R_{>0}: \map {B_\delta} {b; d} \cap H = \O$
and so:
:$\map {B_\delta} {b; d} \subseteq A \setminus H$
As $b$ is arbitrary, it follows that $A \setminus H$ is open in $M$.
Hence by definition, $H$ is closed in $M$ by definition 1.
== Also see ==
* Complement of Closed Set in Complex Plane is Open
* Complement of Open Set in Complex Plane is Closed
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 6$: Open Sets and Closed Sets: Theorem $6.7$
Category:Closed Sets