nyuuzyou/wikis huggingface dataset:
Equivalence of Definitions of Component/Maximal Connected Set is Union of Connected Sets

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in T$.
Let $\tilde C$ be a maximal connected set of $T$ that contains $x$.
Then:
:$\tilde C = \bigcup \set {A \subseteq S : x \in A \land A \text{ is connected in } T}$.
== Proof ==
Let $\CC_x = \set {A \subseteq S : x \in A \land A \text{ is connected in } T}$
Let $C = \bigcup \CC_x$
=== Lemma ===
By definition:
:$\tilde C \in \CC_x$
From Set is Subset of Union:
:$\tilde C \subseteq C$
By maximality of $\tilde C$:
:$\tilde C = C$
== Also see ==
* Union of Connected Sets is Maximal Connected Set
Category:Connected Sets
Equivalence of Definitions of Component/Union of Connected Sets is Maximal Connected Set

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in T$.
Let $\CC_x = \set {A \subseteq S : x \in A \land A \text{ is connected in } T}$
Let $\ds C = \bigcup \CC_x$
Then $C$ is a maximal connected set of $T$.
== Proof ==
=== Lemma ===
Let $\tilde C$ be any connected set such that:
:$C \subseteq \tilde C$
Then $x \in \tilde C$.
Hence $\tilde C \in \CC_x$.
From Set is Subset of Union, 
:$\tilde C \subseteq C$.
Hence $\tilde C = C$.
It follows that $C$ is a maximal connected set of $T$ by definition.
== Also see ==
* Maximal Connected Set is Union of Connected Sets
Category:Connected Sets
Equivalence of Definitions of Composition Series

== Theorem ==
=== Definition $1$ ===
=== Definition $2$ ===
== Proof ==
=== Definition $(1)$ implies Definition $(2)$ ===
Let $\GG$ be a composition series by definition $1$:
Thus $\GG$ is a normal series for $G$:
:$\GG = \sequence {G_k}_{0 \mathop \le k \mathop \le n}$
such that:
:$\set e = G_0 \lhd G_1 \lhd \cdots \lhd G_n = G$
where $A \lhd B$ denotes that $A$ is a proper normal subgroup of $B$.
By definition, $\GG$ is a normal series for $G$ which has no proper refinement.
 that there exists some $i \in \set {0, 1, \ldots, n}$ such that $G_{i - 1}$ is not a proper maximal normal subgroup of $G_i$.
Then there exists $H \subseteq G$ such that:
:$G_{i - 1} \lhd H \lhd G_i$
Hence:
:$\set e = G_0 \lhd G_1 \lhd \cdots \lhd G_{i - 1} \lhd H \lhd G_i \lhd \cdots \lhd G_n = G$
But this is by definition a proper refinement of $\GG$.
This contradicts our assertion that $\GG$ has no proper refinement.
Hence it must be the case that no such $i$ exists.
It follows that $\GG$ is a sequence of normal subgroups of $G$ such that:
:for all $i \in \set {1, 2, \ldots, n}$, $G_{i - 1}$ is a proper maximal normal subgroup of $G_i$.
Thus $\GG$ is a composition series by definition $2$.
=== Definition $(2)$ implies Definition $(1)$ ===
Let $\GG$ be a composition series by definition $2$.
Then, by definition, $\GG$ is a sequence of normal subgroups of $G$ such that:
:for all $i \in \set {1, 2, \ldots, n}$, $G_{i - 1}$ is a proper maximal normal subgroup of $G_i$.
 $\GG$ has a proper refinement.
Then there exists some $i \in \set {0, 1, \ldots, n}$ such that $H \subseteq G: G_{i - 1} \subseteq H \subseteq G_i$ such that $G_{i - 1} \lhd H \lhd G_i$.
Hence $G_{i - 1}$ is not a maximal normal subgroup of $G_i$.
This contradicts our assertion that $G_{i - 1}$ is a proper maximal normal subgroup of $G_i$.
Hence it must be the case that no such $i$ exists.
It follows that $\GG$ has no proper refinement.
Thus $\GG$ is a composition series by definition $1$.
Category:Composition Series
Equivalence of Definitions of Composition of Mappings

== Theorem ==
Let $f_1: S_1 \to S_2$ and $f_2: S_2 \to S_3$ be mappings such that the domain of $f_2$ is the same set as the codomain of $f_1$.
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
Let $f_1: S_1 \to S_2$ and $f_2: S_2 \to S_3$ be mappings such that:
:$\Dom {f_2} = \Cdm {f_1}$
=== $(1)$ implies $(2)$ ===
Let $f_2 \circ f_1$ be a composite mapping by definition 1.
Then by definition:
:$\forall x \in S_1: \map {\paren {f_2 \circ f_1} } x := \map {f_2} {\map {f_1} x}$
We have that $f_1$ is a mapping, and so:
:$\forall x \in S_1: \exists \map {f_1} x \in S_2$
We have that $f_2$ is a mapping, and so:
:$\forall \map {f_1} x \in S_2: \exists \map {f_2} {\map {f_1} x} \in S_3$
Then:
and so:
:$f_2 \circ f_1 = \set {\tuple {x, z} \in S_1 \times S_3: \tuple {\map {f_1} x, z} \in f_2}$
Thus $f_2 \circ f_1$ is a composite mapping by definition 2.
=== $(2)$ implies $(1)$ ===
Let $f_2 \circ f_1$ be a composite mapping by definition 2.
Then by definition:
:$f_2 \circ f_1 := \set {\tuple {x, z} \in S_1 \times S_3: \tuple {\map {f_1} x, z} \in f_2}$
We have that $f_1$ is a mapping, and so:
:$\forall x \in S_1: \exists \map {f_1} x \in S_2$
and so by definition of a as a Relation:
Thus $f_2 \circ f_1$ is a composite mapping by definition 1.
=== $(2)$ implies $(3)$ ===
Let $f_2 \circ f_1$ be a composite mapping by definition 2.
Then by definition:
:$f_2 \circ f_1 := \set {\tuple {x, z} \in S_1 \times S_3: \tuple {\map {f_1} x, z} \in f_2}$
Because $f_1$ is a mapping, it follows that:
:$\forall x \in S_1: \exists y \in S_2: \map {f_1} x = y$
Similarly:
:$\forall y \in S_2: \exists z \in S_3: \map {f_2} y = z$
Hence:
:$\tuple {\map {f_1} x, z} \in f_2 \implies \exists y \in S_2: \map {f_1} x = y \land \map {f_2} y = z$
and so:
:$f_2 \circ f_1 := \set {\tuple {x, z} \in S_1 \times S_3: \exists y \in S_2: \map {f_1} x = y \land \map {f_2} y = z}$
Thus $f_2 \circ f_1$ is a composite mapping by definition 3.
=== $(3)$ implies $(2)$ ===
Let $f_2 \circ f_1$ be a composite mapping by definition 3.
Then by definition:
:$f_2 \circ f_1 := \set {\tuple {x, z} \in S_1 \times S_3: \exists y \in S_2: \map {f_1} x = y \land \map {f_2} y = z}$
We have that:
:$\forall x \in S_1: \exists y \in S_2: \map {f_1} x = y$
and that:
:$\forall y \in S_2: \exists z \in S_3: \map {f_2} y = z$
Hence:
:$\forall x \in S_1: \exists z \in S_3: \map {f_1} x = y \land \map {f_2} y = z$
Thus:
:$\forall x \in S_1: \exists z \in S_3: \tuple {\map {f_1} x, y} \in f_2$
and so:
:$f_2 \circ f_1 := \set {\tuple {x, z} \in S_1 \times S_3: \tuple {\map {f_1} x, z} \in f_2}$
Thus $f_2 \circ f_1$ is a composite mapping by definition 2.
== Sources ==
* : $\S 1$: Naive Set Theory: $\S 1.6$: Functions: Exercise $1.6.2$
Category:Composite Mappings
Equivalence of Definitions of Concave Real Function

== Theorem ==
Let $f$ be a real function which is defined on a real interval $I$.
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
Let $f$ be concave real function on $I$ according to definition 1.
That is:
:$\forall x, y \in I: \forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: \map f {\alpha x + \beta y} \ge \alpha \map f x + \beta \map f y$
Make the substitutions $x_1 = x, x_2 = \alpha x + \beta y, x_3 = y$.
As $\alpha + \beta = 1$, we have $x_2 = \alpha x_1 + \paren {1 - \alpha} x_3$.
Thus:
:$\alpha = \dfrac {x_3 - x_2} {x_3 - x_1}, \beta = \dfrac {x_2 - x_1} {x_3 - x_1}$
So:
Again from $(1)$:
Thus:
:$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} \ge \dfrac {\map f {x_3} - \map f {x_1} } {x_3 - x_1} \ge \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$
So:
:$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} \ge \dfrac {\map f {x_3} - \map f {x_1} } {x_3 - x_1}$
demonstrating that $f$ is concave on $I$ according to definition 3, and:
:$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} \ge \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$
demonstrating that $f$ is concave on $I$ according to definition 2.
As each step is an equivalence, the argument reverses throughout.
Category:Concave Real Functions
Equivalence of Definitions of Concentration of Complex Measure on Measurable Set

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
From Characterization of Null Sets of Variation of Complex Measure, we have that: 
:$\map {\size \mu} {E^c} = 0$ :
::for each $\Sigma$-measurable set $A \subseteq E^c$, we have $\map \mu A = 0$.
Hence the desired equivalence. 
Category:Concentration of Complex Measure on Measurable Set
Equivalence of Definitions of Concentration of Signed Measure on Measurable Set

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
From Characterization of Null Sets of Variation of Signed Measure, we have that: 
:$\map {\size \mu} {E^c} = 0$ :
::for each $\Sigma$-measurable set $A \subseteq E^c$, we have $\map \mu A = 0$.
Hence the desired equivalence. 
Category:Concentration of Signed Measure on Measurable Set
Equivalence of Definitions of Congruence

== Theorem ==
Let $z \in \R$.
=== Definition by Remainder after Division ===
=== Definition by Modulo Operation ===
=== Definition by Integer Multiple ===
== Proof ==
Let $x_1, x_2, z \in \R$.
Let $x_1 \equiv x_2 \pmod z$ as defined by an equivalence relation.
That is, let $\RR_z$ be the relation on the set of all $x, y \in \R$:
:$\RR_z = \set {\tuple {x, y} \in \R \times \R: \exists k \in \Z: x = y + k z}$
Let $\tuple {x_1, x_2} \in \RR_z$.
Then by definition, $\exists k \in \Z: x_1 = x_2 + k z$.
So, by definition of the modulo operation, we have:
So:
:$x_1 \equiv x_2 \pmod z$
in the sense of definition by modulo operation.
Now let $x_1 \equiv x_2 \pmod z$ in the sense of definition by modulo operation.
That is, :$x_1 \equiv x_2 \pmod z \iff x_1 \bmod z = x_2 \bmod z$.
Let $z = 0$.
Then by definition, $x_1 \bmod 0 = x_1$ and $x_2 \bmod 0 = x_2$.
So as $x_1 \bmod 0 = x_2 \bmod 0$ we have that $x_1 = x_2$.
So $x_1 - x_2 = 0 = 0.z$ and so $x_1 \equiv x_2 \pmod z$ in the sense of definition by integer multiple.
Now suppose $z \ne 0$.
Then from definition of the modulo operation:
:$x_1 \bmod z = x_1 - z \floor {\dfrac {x_1} z}$
:$x_2 \bmod z = x_2 - z \floor {\dfrac {x_2} z}$
Thus:
:$x_1 - z \floor {\dfrac {x_1} z} = x_2 - z \floor {\dfrac {x_2} z}$
and so:
:$x_1 - x_2 = z \paren {\floor {\dfrac {x_1} z} - \floor {\dfrac {x_2} z} }$
From the definition of the floor function, we see that both $\floor {\dfrac {x_1} z}$ and $\floor {\dfrac {x_2} z}$ are integers.
Therefore, so is $\floor {\dfrac {x_1} z} - \floor {\dfrac {x_2} z}$ an integer.
So $\exists k \in \Z: x_1 - x_2 = k z$.
Thus $x_1 - x_2 = k z$ and:
:$x_1 \equiv x_2 \pmod z$
in the sense of definition by integer multiple.
Now let $x_1 \equiv x_2 \pmod z$ in the sense of definition by integer multiple.
That is, $\exists k \in \Z: x_1 - x_2 = k z$.
Then $x_1 = x_2 + k z$ and so $\tuple {x_1, x_2} \in \RR_z$ where:
:$\RR_z = \set {\tuple {x, y} \in \R \times \R: \exists k \in \Z: x = y + k z}$
and so
:$x_1 \equiv x_2 \pmod z$
in the sense of definition by equivalence relation.
So all three definitions are equivalent: $(1) \implies (2) \implies (3) \implies (1)$.
== Also see ==
* Equivalence of Definitions of Integer Congruence for the integer case.
Category:Modulo Arithmetic
Equivalence of Definitions of Conic Section

== Theorem ==
=== Intersection of Plane with Cone ===
=== Focus-Directrix Property ===
== Proof ==
: 
Consider the cone in which $\alpha$ is half the opening angle.
Let the tilting angle of the slicing plane to the horizontal be $\beta$.
The conic section thus defined is given here as an ellipse, but the argument holds for the other cases also.
Inside the cone, let a sphere be inscribed which is tangent to the slicing plane at the point $F$.
Let the sphere be tangent to the cone along the circle $C$.
Let $d$ be the line in which the slicing plane intersects the plane of the circle $C$.
It is to be demonstrated that the conic section induced by the slicing plane has $F$ as its focus and $d$ as its directrix.
Let $P$ be a point on the conic section.
Let $PQ$ be constructed parallel to the axis of the cone such that $Q$ lies on the plane of the circle $C$.
Let $R$ be the point where the generatrix of the cone through $P$ intersects $C$.
Let a perpendicular be constructed from $P$ to $d$, intersecting $d$ at $D$.
Hence:
:$\beta = \angle PDQ$
We have that $PR$ and $PF$ are both tangents to the sphere from the same point $P$.
Thus $PF$ and $PR$ are the same length:
:$(1): \quad PR = PF$
From the right triangle $\triangle PQR$:
:$PQ = PR \cos \alpha$
From the right triangle $\triangle PQD$:
:$PQ = PD \sin \beta$
Thus:
Let us define:
:$e = \dfrac {\cos \gamma} {\cos \alpha}$
For a given slicing plane and cone this is constant.
It is noted that:
:$(1): \quad$ when $\gamma < \alpha$, that is, the tilting angle is less than half the opening angle of the cone, the conic section is an ellipse
:$(2): \quad$ when $\gamma = \alpha$, that is, the tilting angle equals half the opening angle of the cone, the conic section is a parabola
:$(3): \quad$ when $\gamma > \alpha$, that is, the tilting angle is greater than half the opening angle of the cone, the conic section is a hyperbola
:$(4): \quad$ when $\gamma = 0$, that is, the slicing plane is parallel to the plane of the circle $C$, the conic section is a circle and there is no directrix
The result follows.
== Sources ==
* : Chapter $\text {A}.8$: Pappus (fourth century A.D.): Appendix: The Focus-Directrix-Eccentricity Definitions of the Conic Sections
Category:Conic Sections
Equivalence of Definitions of Conjugate Point

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== Definition 2 implies Definition 3 ===
Let the extremal $y = \map y x$ satisfy:
:$\map y a = A$
Let the extremal $y^*_\alpha$ pass through $M = \tuple {a, A}$ and satisfy:
:$\map { {y^*}'_\alpha} a - \map {y'} a = \alpha$
Then the following is a valid expression for $y^*_\alpha$:
:$\map {y^*_\alpha} x = \map y x + \map h x \alpha + \epsilon$
where $\epsilon = k \alpha$ with:
:$\alpha \to 0 \implies k \to 0$
and:
:$\map h a = 0$
:$\map {h'} a = 1$ 
where $\map h x$ is an appropriate solution to Jacobi's equation.
Let:
:$\map h {\tilde a} = 0$
:$\beta = \sqrt {\dfrac \epsilon \alpha}$
Then:
:$\map h x \ne 0 \quad \forall x \in \openint a b \implies \map {h'} a \ne 0$
By Taylor's Theorem, the expression:
:$\map {y_\alpha} x - \map y x = \map h x \alpha + \epsilon$
takes values with different signs at the points $\tilde a - \beta$ and $\tilde a + \beta$.
Since:
:$\alpha \to 0 \implies \beta \to 0$
the limit of points of intersection of $y = \map {y_\alpha^*} x$ and $y = \map y x$, as $\alpha \to 0$ is $\map {\tilde M} {\tilde a, \map y {\tilde a} }$ 
=== Definition 3 implies Definition 2 ===
== Sources ==
* : $\S 5.27$: Jacobi's Necessary Condition. More on Conjugate Points
Category:Calculus of Variations
Equivalence of Definitions of Connected Manifold

== Theorem ==
Let $M$ be a topological manifold.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $M$ be connected.
From Topological Manifold is Locally Path-Connected:
:$M$ is locally path-connected.
From Connected and Locally Path-Connected Implies Path Connected:
:$M$ is path-connected.
=== Definition 2 implies Definition 1 ===
Follows immediately from Path-Connected Space is Connected.
== Sources ==
* : Chapter $4.$ Connectedness and Compactness
Category:Connected Manifolds
Equivalence of Definitions of Connected Set (Complex Analysis)

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== $(1)$ implies $(2)$ ===
Let $D$ be a connected set by definition 1.
Then by definition:
:Every pair of points in $D$ can be joined by a staircase contour.
But a staircase contour is a polygonal path all of whose points are in $D$.
Thus $D$ is a connected set by definition 2.
=== $(2)$ implies $(1)$ ===
Let $D$ be a connected set by definition 2.
Then by definition:
:Every pair of points in $D$ can be joined by a polygonal path $P$ all points of which are in $D$.
Thus $D$ is a connected set by definition 1.
Category:Connected Sets (Complex Analysis)
Equivalence of Definitions of Connected Topological Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
=== Definition 4 ===
=== Definition 5 ===
=== Definition 6 ===
=== Definition 7 ===
== Proof ==
=== $(1) \iff (2)$: No Separation iff No Union of Closed Sets ===
=== $(1) \iff (4)$: No Separation iff No Clopen Sets ===
=== $(2) \implies (3)$: No Union of Closed Sets implies No Subsets with Empty Boundary ===
=== $(3) \implies (4)$: No Subsets with Empty Boundary implies No Clopen Sets ===
=== $(4) \implies (5)$: No Clopen Sets implies No Union of Separated Sets ===
=== $(5) \implies (6)$: No Union of Separated Sets implies No Continuous Surjection to Discrete Two-Point Space ===
=== $(6) \implies (1)$: No Continuous Surjection to Discrete Two-Point Space implies No Separation ===
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness
Category:Connected Spaces
Category:Equivalence of Definitions of Connected Topological Space
Equivalence of Definitions of Connected Topological Space/No Clopen Sets implies No Union of Separated Sets

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let the only clopen sets of $T$ be $S$ and $\O$.
Then there are no two non-empty separated sets of $T$ whose union is $S$.
== Proof ==
Suppose $A$ and $B$ are separated subsets of $T$ such that $A \cup B = S$.
By definition of separated sets:
:$A \cap B^- = \O$
Then:
Hence $A = S \setminus B^-$.
From Topological Closure is Closed, $B^-$ is closed in $T$.
Thus $A$ is open in $T$.
Also by definition of separated sets:
:$A^- \cap B = \O$
Hence, by the same reasoning, $B$ must also be open.
But:
:$A \cap B \subseteq A \cap B^- = \O$
and $A \cup B = S$, by assumption.
So:
:$A = S \setminus B$ and $B = S \setminus A$
and we conclude that both $A$ and $B$ are clopen.
Therefore, , one of them must be $S$ and the other must be $\O$.
That is, there are no two non-empty separated sets of $T$ whose union is $S$.
Category:Equivalence of Definitions of Connected Topological Space
Equivalence of Definitions of Connected Topological Space/No Continuous Surjection to Discrete Two-Point Space implies No Separation

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $T$ be such that there exists no continuous surjection from $T$ onto a discrete two-point space.
Then there exist no open sets $A, B \in \tau$ such that $A, B \ne \O$, $A \cup B = S$ and $A \cap B = \O$.
== Proof ==
Let $T = \struct {S, \tau}$ be a topological space such that there exists no continuous surjection from $T$ onto a discrete two-point space.
Let $D = \struct {\set {0, 1}, \tau}$ be the discrete two-point space on $\left\{{0, 1}\right\}$.
Let $A$ and $B$ be disjoint open sets of $T$ such that $A \cup B = S$.
The aim is to show that one of them is empty.
Let us define the mapping $f: S \to \set {0, 1}$ by:
:$\map f x = \begin{cases}
0 & : x \in A \\
1 & : x \in B
\end{cases}$
There are only four open sets in $\set {0, 1}$, namely:
: $\O$
: $\set 0$
: $\set 1$
: $\set {0, 1}$
We have that:
:$f^{-1} \sqbrk \O = \O$
:$f^{-1} \sqbrk {\set 0} = A$
:$f^{-1} \sqbrk {\set 1} = B$
:$f^{-1} \sqbrk {\set {0, 1} } = S$
where $f^{-1} \sqbrk X$ denotes the preimage of the set $X$.
But all of $\O, A, B, S$ are open sets of $T$.
So by definition $f$ is continuous.
Also , $f$ cannot be surjective.
It follows that $f$ must be constant.
So either $A$ or $B$ must be empty, and the other one must be $S$.
Hence the result.
== Sources ==
* : $6.2$: Connectedness: Proposition $6.2.3$
Category:Equivalence of Definitions of Connected Topological Space
Equivalence of Definitions of Connected Topological Space/No Separation iff No Clopen Sets

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
=== Definition by Separation ===
=== Definition by No Clopen Sets ===
== Proof ==
=== Definition by No Clopen Sets implies Definition by Separation ===
Let $T$ be connected by having no clopen sets.
 $T$ admits a separation, $A \mid B$ say.
Then both $A$ and $B$ are clopen sets of $T$, neither of which is either $S$ or $\O$.
From this contradiction it follows that $T$ can admit no separation
=== Definition by Separation implies Definition by No Clopen Sets ===
Let $T$ be connected by admitting no separation.
Suppose $\exists H \subseteq S$ which is clopen.
Then $\relcomp S H$ is also clopen.
Hence $H \mid \relcomp S H$ is a separation of $T$.
From this contradiction it follows that $T$ can have no non-empty proper subsets which are clopen.
== Sources ==
* : Chapter $\text {I}$: Topological Spaces: $1$. Open Sets and Closed Sets
* : $6.2$: Connectedness: Corollary $6.2.4$
Category:Equivalence of Definitions of Connected Topological Space
Equivalence of Definitions of Connected Topological Space/No Separation iff No Union of Closed Sets

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
=== Definition by Separation ===
=== Definition by Union of Disjoint Closed Sets ===
== Proof ==
From Biconditional Equivalent to Biconditional of Negations it follows that the statement can be expressed as:
:$T$ admits a separation
:
:there exist two closed sets of $T$ which form a (set) partition of $S$.
By definition, a separation of $T$ is a (set) partition of $S$ by $A, B$ which are open in $T$.
From Complements of Components of Two-Component Partition form Partition:
:$A \mid B$ is a (set) partition of $S$ $\relcomp S A \mid \relcomp S B$ is a (set) partition of $S$.
Hence the result by definition of closed set.
Category:Equivalence of Definitions of Connected Topological Space
Equivalence of Definitions of Connected Topological Space/No Subsets with Empty Boundary implies No Clopen Sets

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $T$ be such that the only subsets of $S$ whose boundary is empty are $S$ and $\O$.
Then the only clopen sets of $T$ are $S$ and $\O$.
== Proof ==
Let $H \subseteq S$ be a clopen set of $T$.
From Set is Clopen iff Boundary is Empty, $H$ has an empty boundary.
We have that $H = S$ or $H = \O$.
That is, the only clopen sets of $T$ are $S$ and $\O$.
Category:Equivalence of Definitions of Connected Topological Space
Equivalence of Definitions of Connected Topological Space/No Union of Closed Sets implies No Subsets with Empty Boundary

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $T$ have no two disjoint non-empty closed sets whose union is $S$.
Then the only subsets of $S$ whose boundary is empty are $S$ and $\O$.
== Proof ==
Let $H \subseteq S$ be a non-empty subset whose boundary $\partial H$ is empty.
Thus:
From Topological Closure is Closed, both $H^-$ and $\paren {S \setminus H}^-$ are closed sets of $T$.
From Union of Closure with Closure of Complement is Whole Space:
:$H^- \cup \paren {S \setminus H}^- = S$
Thus $H^-$ and $\paren {S \setminus H}^-$ are two disjoint closed sets of $T$ whose union is $S$.
Hence, , one of them must be empty.
Suppose $H$ is not empty.
It must therefore follow that:
: $S \setminus H = \O$
Therefore $H = S$.
Thus the only subsets of $S$ whose boundary is empty are $S$ and $\O$.
Category:Equivalence of Definitions of Connected Topological Space