avewright/proofwiki-math huggingface dataset:
Theorem: Construction of Triangle from Given Lengths

Given three straight lines such that the sum of the lengths of any two of the lines is greater than the length of the third line, it is possible to construct a triangle having the lengths of these lines as its side lengths.
{{:Euclid:Proposition/I/22}}

Proof:
Since $F$ is the center of the circle with radius $FD$, it follows from {{EuclidDefLink|I|15|Circle}} that $DF = KF$, so $a = KF$ by Euclid's first common notion.
Since $G$ is the center of the circle with radius $GH$, it follows from {{EuclidDefLink|I|15|Circle}} that $GH = GK$, so $c = GK$ by Euclid's first common notion.
$FG = b$ by construction.
Therefore the lines $FK$, $FG$, and $GK$ are, respectively, equal to the lines $a$, $b$, and $c$, so $\triangle FGK$ is indeed the required triangle.
{{qed}}
{{Euclid Note|22|I}}
Note that the condition required of the lengths of the segments is the inequality shown in Proposition $20$: Sum of Two Sides of Triangle Greater than Third Side.  Thus, this is a necessary condition for the construction of a triangle.
When {{AuthorRef|Euclid}} first wrote the proof of this proposition in {{BookLink|The Elements|Euclid}}, he neglected to prove that the two circles described in the construction actually do intersect, just as he did in Proposition $1$: Construction of Equilateral Triangle.
Theorem: Connected Open Subset of Euclidean Space is Path-Connected

Let $\R^n$ be a Euclidean $n$-space.
Let $U$ be a connected open subset of $\R^n$.
Then $U$ is path-connected.

Proof:
Let $a \in U$.
Let $H \subseteq U$ be the subset of points in $U$ which can be joined to $a$ by a path in $U$.
Let $K = U \setminus H$.
Let $x \in H$.
Then:
:$\exists \epsilon > 0: \map {B_\epsilon} x \subseteq U$
where $\map {B_\epsilon} x$ is the open $\epsilon$-ball of $x$.
Given any $y \in \map {B_\epsilon} x$, there is a (straight line) path $g$ in $\map {B_\epsilon} x \subseteq U$ connecting $x$ and $y$.
But since $x \in H$, there is a path $f$ in $U$ joining $a$ to $x$.
From Joining Paths makes Another Path, traversing $f$, and then $g$ forms a path from $a$ to $y$.
It follows that $y \in H$, and therefore $\map {B_\epsilon} x \subseteq H$.
Thus $H$ is open.
By a similar argument, $K$ is also shown to be open:
If $x \in K$, then $\map {B_\epsilon} x \subseteq U$ for some $\epsilon > 0$.
If any point in $\map {B_\epsilon} x$ can be joined to $a$ by a path in $U$, then so could $x$.
It is clear that $H \cap K = \O$ and $H \cup K = U$ by definition of set difference.
As, trivially, $a \in H$, we have $H \ne \O$.
Knowing that $U$ is connected, it follows that $K = \O$, and $H = U$.
Hence the result.
{{qed}}
Theorem: Joining Paths makes Another Path

Let $T$ be a topological space.
Let $I \subseteq \R$ be the closed real interval $\closedint 0 1$.
Let $f, g: I \to T$ be paths in $T$ from $a$ to $b$ and from $b$ to $c$ respectively.
Let $h: I \to T$ be the mapping given by:
$\map h x = \begin{cases}
\map f {2x} & : x \in \closedint 0 {\dfrac 1 2} \\
\map g {2x - 1} & : x \in \closedint {\dfrac 1 2} 1
\end{cases}$
Then $h$ is a path in $T$.

Proof:
First we see that $h$ is well-defined, because on $\closedint 0 {\dfrac 1 2} \cap \closedint {\dfrac 1 2} 1 = \set {\dfrac 1 2}$ we have $\map f 1 = b = \map g 0$.
Now $\mathbin h {\restriction_{\closedint 0 {\frac 1 2} } } \mathop = f \circ k$ where $k: \closedint 0 {\dfrac 1 2} \to \closedint 0 1$ is given by $\map k x = 2 x$.
So by Composite of Continuous Mappings is Continuous, $\mathbin h {\restriction_{\closedint 0 {\frac 1 2} } }$ is continuous.
Similarly, $\mathbin h {\restriction_{\closedint {\frac 1 2} 1} }$ is continuous.
By Pasting Lemma for Finite Union of Closed Sets, it follows that $h$ is continuous.
Finally, $\map h 0 = \map f 0 = a$ and $\map h 1 = \map g 1 = c$.
{{qed}}
Theorem: Hinge Theorem

If two triangles have two pairs of sides which are the same length, the triangle with the larger included angle also has the larger third side.
{{:Euclid:Proposition/I/24}}

Proof:
:250px
Let $\triangle ABC$ and $DEF$ be two triangles in which $AB = DE$, $AC = DF$, and $\angle CAB > \angle FDE$.
Construct $\angle EDG$ on $DE$ at point $D$.
Place $G$ so that $DG = AC$.
Join $EG$ and $FG$.
Since $AB = DE$, $\angle BAC = \angle EDG$, and $AC = DG$, by Triangle Side-Angle-Side Congruence:
:$BC = GE$
By Euclid's first common notion:
:$DG = AC = DF$
Thus by Isosceles Triangle has Two Equal Angles:
:$\angle DGF = \angle DFG$
So by Euclid's fifth common notion:
:$\angle EFG \, > \, \angle DFG = \angle DGF \, > \, \angle EGF$
Since $\angle EFG > \angle EGF$, by Greater Angle of Triangle Subtended by Greater Side:
:$EG > EF$
Therefore, because $EG = BC$, $BC > EF$.
{{qed}}
Theorem: Equivalence of Definitions of Component

{{TFAE|def = Component (Topology)|view = component|context = Topology (Mathematical Branch)|contextview = topology}}
Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in T$.
=== Definition 1: Equivalence Class ===
{{:Definition:Component (Topology)/Definition 1}}
=== Definition 2: Union of Connected Sets ===
{{:Definition:Component (Topology)/Definition 2}}
=== Definition 3: Maximal Connected Set ===
{{Definition:Component (Topology)/Definition 3}}

Proof:
Let $\CC_x = \set {A \subseteq S : x \in A \land A \text{ is connected in } T}$
Let $C = \bigcup \CC_x$
Theorem: Converse Hinge Theorem

If two triangles have two pairs of sides which are the same length, the triangle in which the third side is longer also has the larger angle contained by the first two sides.
{{:Euclid:Proposition/I/25}}

Proof:
:300px
Let $\triangle ABC$ and $\triangle DEF$ be two triangles in which:
:$AB = DF$
:$AC = DE$
:$BC > EF$
{{AimForCont}} that $\angle BAC \not > \angle EDF$.
Then either:
:$\angle BAC = \angle EDF$
or:
:$\angle BAC < \angle EDF$
Let $\angle BAC = \angle EDF$.
Then by Triangle Side-Angle-Side Congruence:
:$BC = EF$
But we know this is not the case, so by Proof by Contradiction:
:$\angle BAC \ne \angle EDF$
Suppose $\angle BAC < \angle EDF$.
Then by Greater Angle of Triangle Subtended by Greater Side:
:$EF > BC$
But we know this is not the case, so by Proof by Contradiction:
:$\angle BAC \not < \angle EDF$
Thus:
:$\angle BAC > \angle EDF$
{{qed}}
Theorem: Sequentially Compact Metric Subspace is Sequentially Compact in Itself iff Closed

Let $M$ be a metric space.
Let $C \subseteq M$ be a subspace of $M$ which is sequentially compact in $M$.
Then $C$ is sequentially compact in itself {{iff}} $C$ is closed in $M$.

Proof:
Follows directly from Closure of Subset of Metric Space by Convergent Sequence.
{{qed}}
Category:Metric Subspaces
Category:Sequentially Compact Spaces
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Theorem: Invariance of Extremal Length under Conformal Mappings

Let $X, Y$ be Riemann surfaces (usually, subsets of the complex plane).
Let $\phi: X \to Y$ be a conformal isomorphism between $X$ and $Y$.
Let $\Gamma$ be a family of rectifiable curves (or, more generally, of unions of rectifiable curves) in $X$.
Let $\Gamma'$ be the family of their images under $\phi$.
Then $\Gamma$ and $\Gamma'$ have the same extremal length:
:$\map \lambda \Gamma = \map \lambda {\Gamma'}$

Proof:
Let $\rho'$ be a conformal metric on $Y$ in the sense of the definition of extremal length, given in local coordinates as:
:$\map {\rho'} z \size {\d z}$
Let $\rho$ be the metric on $X$ obtained as the pull-back of this metric under $\phi$.
That is, $\rho$ is given in local coordinates as:
:$\map {\rho'} {\map \phi w} \cdot \size {\map {\dfrac {\d \phi} {\d w} } w} \cdot \size {\d w}$
Then the area of $X$ with respect to $\rho$ and the area of $Y$ with respect to $\rho'$ are equal by definition:
:$\map A {\rho'} = \map A \rho$
Furthermore, if $\gamma \in \Gamma$ and $\gamma' := \map \phi \gamma$, then also:
:$\map L {\gamma, \rho} = \map L {\gamma', \rho'}$
and hence:
:$\map L {\Gamma, \rho} = \map L {\Gamma', \rho'}$
In summary, for any metric $\rho'$ on $Y$, there is a metric $\rho$ on $X$ such that:
:$\dfrac {\map L {\Gamma, \rho} } {\map A \rho} = \dfrac {\map L {\Gamma', \rho'} } {\map A {\rho'} }$
It thus follows from the definition of extremal length that:
:$\map \lambda \Gamma \ge \map \lambda {\Gamma'}$
The opposite inequality follows by exchanging the roles of $X$ and $Y$.
{{qed}}
Theorem: Real Number Line is Complete Metric Space

The real number line $\R$ with the usual (Euclidean) metric forms a complete metric space.

Proof:
From Real Number Line is Metric Space, the distance function defined as $\map d {x, y} = \size {x - y}$ is a metric on $\R$.
It remains to be shown that the metric space $\struct {\R, d}$ is complete.
By definition, this is done by demonstrating that every Cauchy sequence of real numbers has a limit.
This is demonstrated in Cauchy's Convergence Criterion.
Hence the result.
{{qed}}
Theorem: Series Law for Extremal Length

Let $X$ be a Riemann surface.
Let $\Gamma_1$, $\Gamma_2$ and $\Gamma$ be families of rectifiable curves (or, more generally, families of unions of rectifiable curves) on $X$.
Let every $\gamma \in \Gamma$ contain a $\gamma_1 \in \Gamma_1$ and a $\gamma_2 \in \Gamma_2$ such that $\gamma_1 \cap \gamma_2 = \O$.
Then the extremal lengths of $\Gamma_1$, $\Gamma_2$ and $\Gamma$ satisfy:
:$\map \lambda \Gamma \ge \map \lambda {\Gamma_1} + \map \lambda {\Gamma_2}$

Proof:
Let $\rho_1 = \map {\rho_1} z \size {\d z}$ and $\rho_2 = \map {\rho_2} z \size {\d z}$ be conformal metrics as in the definition of extremal length.
It can be assumed that these are normalized:
:$\map A {\rho_j} = \map L {\Gamma_j, \rho_j}$ for $j \in \set {1, 2}$. 
We define another metric $\rho = \map \rho z \size {\d z}$ by:
:$\map \rho z := \map \max {\map {\rho_1} z, \map {\rho_2} z}$
Note that this is a well-defined metric.
By definition, the area form $\map {\rho^2} z \size {\d z}$ satisfies:
:$\map {\rho^2} z \size {\d z}^2 = \map \max {\map {\rho_1} z^2, \map {\rho_2} z^2} \size {\d z}^2 \le \paren {\map {\rho_1} z^2 + \map {\rho_2} z^2} \size {\d z}^2$
Hence:
:$\map A \rho \le \map A {\rho_1} + \map A {\rho_2} = \map L {\Gamma_1, \rho_1} + \map L {\Gamma_2, \rho_2}$
On the other hand, let $\gamma \in \Gamma$.
Let $\gamma_1$, $\gamma_2$ be as in the assumption.
Then:
{{begin-eqn}}
{{eqn | l = \map L {\gamma, \rho}
      | o = \ge
      | r = \map L {\gamma_1, \rho} + \map L {\gamma_2, \rho}
      | c = as $\gamma_1$ and $\gamma_2$ are disjoint
}}
{{eqn | o = \ge
      | r = \map L {\gamma_1, \rho_1} + \map L {\gamma_2, \rho_2}
      | c = Definition of $\rho$
}}
{{eqn | o = \ge
      | r = \map L {\Gamma_1, \rho_1} + \map L {\Gamma_2, \rho_2}
      | c = Definition of $\map L {\Gamma_j, \rho_j}$
}}
{{end-eqn}}
Thus 
:$\map L {\Gamma, \rho} \ge \map L {\Gamma_1, \rho_1} + \map L {\Gamma_2, \rho_2}$
Combining this with the inequality for the area:
:$\dfrac {\map L {\Gamma, \rho}^2} {\map A \rho} \ge \dfrac {\paren {\map L {\Gamma_1, \rho_1} + \map L {\Gamma_2, \rho_2} }^2} {\map L {\Gamma_1, \rho_1} + \map L {\Gamma_2, \rho_2} } = \map L {\Gamma_1, \rho_1} + \map L {\Gamma_2, \rho_2}$
Taking the supremum over all metrics $\rho_1$ and $\rho_2$ as above:
:$\map L \Gamma \ge \map L {\Gamma_1} + \map L {\Gamma_2}$
as claimed.
{{qed}}
Theorem: Parallel Law for Extremal Length

Let $X$ be a Riemann surface.
Let $\Gamma_1, \Gamma_2$ be families of rectifiable curves (or, more generally, families of disjoint unions of rectifiable curves) on $X$.
Let $\Gamma_1$ and $\Gamma_2$ be disjoint, in the sense that:
:there exist disjoint Borel subsets $A_1, A_2 \subseteq X$ such that:
::for any $\gamma_1 \in \Gamma_1$ and $\gamma_2 \in \Gamma_2$, we have $\gamma_1 \subseteq A_1$ and $\gamma_2 \subseteq A_2$.
Let $\Gamma$ be a third curve family, with the property that every element $\Gamma_1$ and every element of $\Gamma_2$ contains some element of $\Gamma$.
Then the extremal length of $\Gamma$ satisfies:
:$\dfrac 1 {\lambda \left({\Gamma}\right)} \ge \dfrac 1 {\lambda \left({\Gamma_1}\right)} + \dfrac 1 {\lambda \left({\Gamma_2}\right)}$

Proof:
The assumption means that every element of $\Gamma_1 \cup \Gamma_2$ contains some element of $\Gamma$.
Hence:
{{begin-eqn }}
{{eqn | l =\frac 1 {\lambda \left({\Gamma}\right)}
      | o = \ge
      | r = \frac 1 {\lambda \left({\Gamma_1 \cup \Gamma_2}\right)}
      | c = Comparison Principle for Extremal Length
}}
{{eqn | r = \frac 1 {\lambda \left({\Gamma_1}\right)} + \frac 1 {\lambda \left({\Gamma_2}\right)}
      | c = Extremal Length of Union
}}
{{end-eqn}}
{{qed}}
{{MissingLinks|Some results and definitions need to be linked.}}
Theorem: Comparison Principle for Extremal Length

Let $X$ be a Riemann surface.
Let $\Gamma_1$ and $\Gamma_2$ be families of rectifiable curves (or, more generally, families of unions of rectifiable curves) on $X$.
Let every element of $\Gamma_1$ contain some element of $\Gamma_2$.
Then the extremal lengths of $\Gamma_1$ and $\Gamma_2$ are related by:
:$\lambda \left({\Gamma_1}\right) \ge \lambda \left({\Gamma_2}\right)$
More precisely, for every conformal metric $\rho$ as in the definition of extremal length, we have:
:$L \left({\Gamma_1, \rho}\right) \ge L \left({\Gamma_2, \rho}\right)$

Proof:
We have:
{{begin-eqn}}
{{eqn | l = L \left({\Gamma_1, \rho}\right)
      | r = \inf_{\gamma \mathop \in \Gamma_1} L \left({\gamma, \rho}\right)
      | c = by definition
}}
{{eqn | o = \ge
      | r = \inf_{\gamma \mathop \in \Gamma_2} L \left({\gamma, \rho}\right)
      | c = since every curve of $\Gamma_1$ contains a curve of $\Gamma_2$
}}
{{eqn | r = L \left({\Gamma_2, \rho}\right)
      | c = by definition
}}
{{end-eqn}}
This proves the second claim.
The second claim implies the first by definition.
{{qed}}
Theorem: Reverse Triangle Inequality

Let $M = \struct {X, d}$ be a metric space.
Then:
:$\forall x, y, z \in X: \size {\map d {x, z} - \map d {y, z} } \le \map d {x, y}$
=== Normed Division Ring ===
{{:Reverse Triangle Inequality/Normed Division Ring}}
=== Normed Vector Space ===
{{:Reverse Triangle Inequality/Normed Vector Space}}
=== Real and Complex Numbers ===
{{:Reverse Triangle Inequality/Real and Complex Fields}}

Proof:
Let $M = \struct {X, d}$ be a metric space.
By {{Metric-space-axiom|2}}, we have:
:$\forall x, y, z \in X: \map d {x, y} + \map d {y, z} \ge \map d {x, z}$
By subtracting $\map d {y, z}$ from both sides:
:$\map d {x, y} \ge \map d {x, z} - \map d {y, z}$
Now we consider 2 cases. 
;Case $1$: Suppose $\map d {x, z} - \map d {y, z} \ge 0$.
Then:
:$\map d {x, z} - \map d {y, z} = \size {\map d {x, z} - \map d {y, z} }$
and so:
:$\map d {x, y} \ge \size {\map d {x, z} - \map d {y, z} }$
;Case 2: Suppose $\map d {x, z} - \map d {y, z} < 0$.
Applying {{Metric-space-axiom|2}} again, we have:
:$\forall x, y, z \in X: \map d {y, x} + \map d {x, z} \ge \map d {y, z}$
Hence:
:$\map d {x, y} \ge \map d {y, z} - \map d {x, z}$
Since we assumed $\map d {x, z} - \map d {y, z} < 0$, we have that:
:$\map d {y, z} - \map d {x, z} > 0$
and so:
:$\map d {y, z} - \map d {x, z} = \size {\map d {y, z} - \map d {x, z} }$
Thus we obtain:
:$\map d {x, y} \ge \size {\map d {x, z} - \map d {y, z} }$
Since these cases are exhaustive, we have shown that:
:$\forall x, y, z \in X: \map d {x, y} \ge \size {\map d {x, z} - \map d {y, z} }$
{{qed}}
Theorem: Reverse Triangle Inequality

Let $M = \struct {X, d}$ be a metric space.
Then:
:$\forall x, y, z \in X: \size {\map d {x, z} - \map d {y, z} } \le \map d {x, y}$
=== Normed Division Ring ===
{{:Reverse Triangle Inequality/Normed Division Ring}}
=== Normed Vector Space ===
{{:Reverse Triangle Inequality/Normed Vector Space}}
=== Real and Complex Numbers ===
{{:Reverse Triangle Inequality/Real and Complex Fields}}

Proof:
From the Reverse Triangle Inequality:
:$\cmod {x - y} \ge \cmod {\cmod x - \cmod y}$
By the definition of both absolute value and complex modulus:
:$\cmod {\cmod x - \cmod y} \ge 0$
As:
:$\cmod x - \cmod y = \pm \cmod {\cmod x - \cmod y}$
it follows that:
:$\cmod {\cmod x - \cmod y} \ge \cmod x - \cmod y$
Hence the result.
{{qed}}
Theorem: Reverse Triangle Inequality

Let $M = \struct {X, d}$ be a metric space.
Then:
:$\forall x, y, z \in X: \size {\map d {x, z} - \map d {y, z} } \le \map d {x, y}$
=== Normed Division Ring ===
{{:Reverse Triangle Inequality/Normed Division Ring}}
=== Normed Vector Space ===
{{:Reverse Triangle Inequality/Normed Vector Space}}
=== Real and Complex Numbers ===
{{:Reverse Triangle Inequality/Real and Complex Fields}}

Proof:
By the Triangle Inequality:
:$\cmod {x + y} - \cmod y \le \cmod x$
Let $z = x + y$.
Then $x = z - y$ and so:
:$\cmod z - \cmod y \le \cmod {z - y}$
Renaming variables as appropriate gives:
:$\cmod {x - y} \ge \cmod x - \cmod y$
{{qed}}
Theorem: Reverse Triangle Inequality

Let $M = \struct {X, d}$ be a metric space.
Then:
:$\forall x, y, z \in X: \size {\map d {x, z} - \map d {y, z} } \le \map d {x, y}$
=== Normed Division Ring ===
{{:Reverse Triangle Inequality/Normed Division Ring}}
=== Normed Vector Space ===
{{:Reverse Triangle Inequality/Normed Vector Space}}
=== Real and Complex Numbers ===
{{:Reverse Triangle Inequality/Real and Complex Fields}}

Proof:
Let $z_1$ and $z_2$ be represented by the points $A$ and $B$ respectively in the complex plane.
From Geometrical Interpretation of Complex Subtraction, we can construct the parallelogram $OACB$ where:
:$OA$ and $OB$ represent $z_1$ and $z_2$ respectively
:$BA$ represents $z_1 - z_2$.
:400px
But $OA$, $OB$ and $BA$ form the sides of a triangle.
The result then follows directly from Sum of Two Sides of Triangle Greater than Third Side.
{{qed}}
Theorem: Reverse Triangle Inequality

Let $M = \struct {X, d}$ be a metric space.
Then:
:$\forall x, y, z \in X: \size {\map d {x, z} - \map d {y, z} } \le \map d {x, y}$
=== Normed Division Ring ===
{{:Reverse Triangle Inequality/Normed Division Ring}}
=== Normed Vector Space ===
{{:Reverse Triangle Inequality/Normed Vector Space}}
=== Real and Complex Numbers ===
{{:Reverse Triangle Inequality/Real and Complex Fields}}

Proof:
Let $X$ denote either $\R$ or $\C$ as appropriate.
From Real Number Line is Metric Space and Complex Plane is Metric Space the distance function $d: X \times X \to \R$ can be defined as:
:$\map d {x, y} = \size {x - y}$
From the Reverse Triangle Inequality as applied to metric spaces:
:$(1): \quad \forall x, y, z \in X: \size {\map d {x, z} - \map d {y, z} } \le \map d {x, y}$
Let $z = 0$.
Then $(1)$ translates to:
:$\forall x, y, z \in X: \size {\size {x - 0} - \size {y - 0} } \le \size {x - y}$
Hence the result.
{{qed}}
Theorem: Reverse Triangle Inequality

Let $M = \struct {X, d}$ be a metric space.
Then:
:$\forall x, y, z \in X: \size {\map d {x, z} - \map d {y, z} } \le \map d {x, y}$
=== Normed Division Ring ===
{{:Reverse Triangle Inequality/Normed Division Ring}}
=== Normed Vector Space ===
{{:Reverse Triangle Inequality/Normed Vector Space}}
=== Real and Complex Numbers ===
{{:Reverse Triangle Inequality/Real and Complex Fields}}

Proof:
From proof $2$ of corollary $1$ to this result, which is derived independently:
:$\size {x - y} \ge \size x - \size y$
There are two cases:
$(1): \quad \size x \ge \size y$
We have :
:$\size {\size x - \size y} = \size x - \size y$
and the proof is finished.
{{qed|lemma}}
$(2): \quad \size y \ge \size x$
We have:
:$\size {y - x} \ge \size y - \size x = \size {\size y - \size x}$
But:
:$\size {y - x} = \size {x - y}$
and:
:$\size {\size y - \size x} = \size {\size x - \size y}$
From this we have:
:$-\size {\size x - \size y} \ge -\size {x - y}$
Since, by Negative of Absolute Value, we have that:
:$\size x - \size y \ge -\size {\size x - \size y}$
it follows that:
:$-\size {x - y} \le \size x - \size y \le \size {x - y}$
The result follows.
{{qed}}
Theorem: Inverse Completion of Integral Domain Exists

Let $\struct {D, +, \circ}$ be an integral domain whose zero is $0_D$ and whose unity is $1_D$.
Then an inverse completion of $\struct {D, \circ}$ can be constructed.

Proof:
From the definition of an integral domain:
:All elements of $D^* = D \setminus \set {0_D}$ are cancellable
:$\struct {D^*, \circ}$ is a commutative semigroup.
So by the Inverse Completion Theorem, there exists an inverse completion of $\struct {D, \circ}$.
From Construction of Inverse Completion, this is done as follows:
Let $\ominus$ be the congruence relation defined on $D \times D^*$ by:
:$\tuple {x_1, y_1} \ominus \tuple {x_2, y_2} \iff x_1 \circ y_2 = x_2 \circ y_1$
The fact that this is a congruence relation is proved in Equivalence Relation on Semigroup Product with Cancellable Elements.
Let $\struct {D \times D^*, \otimes}$ be the external direct product of $\struct {D, \circ}$ with $\struct {D^*, \circ}$, where $\otimes$ is the operation on $D \times D^*$ induced by $\circ$.
Let the quotient structure defined by $\ominus$ be $\struct {\dfrac {D \times D^*} \ominus, \otimes_\ominus}$.
where $\otimes_\ominus$ is the operation induced on $\dfrac {D \times D^*} \ominus$ by $\otimes$.
Let us use $D'$ to denote the quotient set $\dfrac {D \times D^*} {\ominus}$.
Let us use $\circ'$ to denote the operation $\otimes_\ominus$.
Thus $\struct {D', \circ'}$ is the inverse completion of $\struct {D, \circ}$.
An element of $D'$ is therefore an equivalence class of the congruence relation $\ominus$.
As the Inverse Completion is Unique up to isomorphism, it follows that we can ''define'' the structure $\struct {K, \circ}$ which is isomorphic to $\struct {D', \circ'}$.
An element of $D'$ is therefore an equivalence class of the congruence relation $\ominus$.
So an element of $K$ is the isomorphic image of an element $\eqclass {\paren {x, y} } \ominus$ of $\dfrac {D \times D^*} \ominus$.
Hence every element of $\struct {K, \circ}$ is of the form $x \circ y^{-1}$, where $x \in D$ and $y \in D^*$.
Alternatively, from the definition of division product, of the form $\dfrac x y$.
Hence we can therefore interpret any element of $\struct {K, \circ}$ as equivalence classes of elements of the form $\dfrac x y$.
{{qed}}
Category:Integral Domains
Category:Inverse Completions
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Theorem: Inverse Completion Less Zero of Integral Domain is Closed

Let $\struct {D, +, \circ}$ be an integral domain whose zero is $0_D$ and whose unity is $1_D$.
Let $\struct {K, \circ}$ be the inverse completion of $\struct {D, \circ}$.
Then $\struct {K^*, \circ}$ is closed, where $K^* = K \setminus \set {0_K}$.

Proof:
Let $\struct {K, \circ}$ be the inverse completion of $\struct {D, \circ}$.
We define $\struct {K, \circ}$ of $\struct {D, \circ}$ by means of the technique used in Inverse Completion of Integral Domain Exists.
The structure of $\struct {K, \circ}$ is such that each element of $\struct {K, \circ}$ is of the form $x \circ y^{-1}$, where $x \in D$ and $y \in D^*$.
From Zero of Inverse Completion of Integral Domain, $0_K$ is all elements of $D \times D^*$ of the form $\dfrac {0_D} x$.
Therefore the elements of $K^*$ are those of the form $x \circ y^{-1}$ where $x, y \in D^*$.
By Product of Division Products:
:$\dfrac a b \circ \dfrac c d = \dfrac {a \circ c} {b \circ d}$
As $\struct {D, +, \circ}$ is an integral domain, none of its non-zero elements are zero divisors.
Therefore:
:$\forall x, y \in D^*: x \circ y \ne 0_D$
So:
:$\forall \dfrac a b, \dfrac c d \in K^*: \dfrac {a \circ c} {b \circ d} \in K^*$
Hence the result.
{{qed}}
Category:Integral Domains
Category:Inverse Completions
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