hcju/proofwikiqa huggingface dataset:
=== Necessary Condition === Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a group monomorphism. By Homomorphism to Group Preserves Identity, $e_S \in \map \ker \phi$. If $\map \ker \phi$ contained another element $s \ne e_S$, then $\map \phi s = \map \phi {e_S} = e_T$ and $\phi$ would not be injective, thus not be a group monomorphism. So $\map \ker \phi$ can contain only one element, and that must be $e_S$, which is therefore the trivial subgroup of $S$. {{qed|lemma}} === Sufficient Condition === Now suppose $\map \ker \phi = \set {e_S}$. Then, for $x, y \in S$: {{begin-eqn}} {{eqn | l = \map \phi x       | r = \map \phi y       | c =  }} {{eqn | ll= \leadsto       | l = \map \phi x * \paren {\map \phi y}^{-1}       | r = \map \phi y * \paren {\map \phi y}^{-1}       | c = Group Axioms }} {{eqn | ll= \leadsto       | l = \map \phi {x \circ y^{-1} }       | r = e_T       | c = {{Defof|Morphism Property}} }} {{eqn | ll= \leadsto       | l = x \circ y^{-1}       | o = \in       | r = \map \ker \phi       | c = {{Defof|Kernel of Group Homomorphism}} }} {{eqn | ll= \leadsto       | l = x \circ y^{-1}       | r = e_S       | c = by hypothesis }} {{eqn | ll= \leadsto       | l = x       | r = e_S \circ y = y       | c = Group Axioms }} {{end-eqn}} Thus $\phi$ is injective, and therefore a group monomorphism.  {{qed}}
The proof for the ring monomorphism follows directly from: :Ring Homomorphism of Addition is Group Homomorphism and: :Kernel of Group Monomorphism is Trivial for the group monomorphism. {{qed}}
From Exponential of Sum we have: :$\forall x, y \in \R: \map \exp {x + y} = \exp x \cdot \exp y$ That is, $\exp$ is a group homomorphism. Then we have that Exponential is Strictly Increasing. From Strictly Monotone Real Function is Bijective, it follows that $\exp$ is a bijection. So $\exp$ is a bijective group homomorphism, and so a group isomorphism. {{qed}}
From Real Numbers under Addition form Abelian Group, $\struct {\R, +}$ is a group. From Strictly Positive Real Numbers under Multiplication form Uncountable Abelian Group, $\struct {\R_{> 0}, \times}$ is a group. We have that for all $y \in R_{> 0}$ there exists $x = \map \ln y \in R$ which satisfies $\map \exp x = y$. {{explain|Link to a result backing up the above claim.}} Thus $\exp$ is a surjection. Then from Exponential on Real Numbers is Injection: :$\exp$ is an injection. Therefore, $\exp$ is a bijection. Let $x, y \in R$. From Exponential of Sum: :$\map \exp {x + y} = \map \exp x \, \map \exp y$ So $\exp$ is a homomorphism and a bijection. It follows by definition that $\exp: \struct {\R, +} \to \struct {\R_{> 0}, \times}$ is an isomorphism. {{qed}}
Let $K = \map \ker \phi$. By Kernel is Normal Subgroup of Domain, $G_1 / K$ exists. We need to establish that the mapping $\theta: G_1 / K \to G_2$ defined as: :$\forall x \in G_1: \map \theta {x K} = \map \phi x$ is well-defined. That is, we need to ensure that: :$\forall x, y \in G: x K = y K \implies \map \theta {x K} = \map \theta {y K}$ Let $x, y \in G: x K = y K$. Then: {{begin-eqn}} {{eqn | l = x K       | r = y K       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x^{-1} y       | o = \in       | r = K = \map \ker \phi       | c = Left Cosets are Equal iff Product with Inverse in Subgroup }} {{eqn | ll= \leadstoandfrom       | l = \map \phi {x^{-1} y}       | r = e_{G_2}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \paren {\map \phi x}^{-1} \map \phi y       | r = e_{G_2}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \map \phi x       | r = \map \phi y       | c =  }} {{end-eqn}} Thus we see that $\theta$ is well-defined. Since we also have that: :$\map \phi x = \map \phi y \implies x K = y K$ it follows that: :$\map \theta {x K} = \map \theta {y K} \implies x K = y K$ So $\theta$ is injective. We also note that: :$\Img \theta = \set {\map \theta {x K}: x \in G}$ So: {{begin-eqn}} {{eqn | l = \Img \theta       | r = \set {\map \theta {x K}: x \in G}       | c =  }} {{eqn | r = \set {\map \phi x: x \in G}       | c =  }} {{eqn | r = \Img \phi       | c =  }} {{end-eqn}} We also note that $\theta$ is a homomorphism: {{begin-eqn}} {{eqn | ll= \forall x, y \in G:       | o =        | r = \map \theta {x K y K}       | c =  }} {{eqn | r = \map \theta {x y K}       | c =  }} {{eqn | r = \map \phi {x y}       | c =  }} {{eqn | r = \map \phi x \map \phi y       | c =  }} {{eqn | r = \map \theta {x K} \map \theta {y K}       | c =  }} {{end-eqn}} Thus $\theta$ is a monomorphism whose image equals $\Img \phi$. The result follows. {{Qed}}
From Ring Homomorphism whose Kernel contains Ideal, let $J = \map \ker \phi$. This gives the ring homomorphism $\mu: R / \map \ker \phi \to S$ as follows: :300px {{improve|Replace the above graphic with a creation in xyplot}} That is: :$\phi = \mu \circ \nu$ Then we have: :$\map \ker \mu = \map \ker \phi / \map \ker \phi$ This is the null subring of $R / \map \ker \phi$ by Quotient Ring Defined by Ring Itself is Null Ring. Then from Kernel is Trivial iff Monomorphism it follows that $\mu$ is a monomorphism. From $\phi = \mu \circ \nu$, we have: :$\Img \mu = \Img \phi$ It follows that $\mu$ is an isomorphism. {{qed}}
The fact that $N$ is normal, together with Intersection with Normal Subgroup is Normal, gives us that $N \cap H \lhd H$. Also, $N \lhd N H = \gen {H, N}$ follows from Subset Product with Normal Subgroup as Generator. Now we define a mapping $\phi: H \to H N / N$ by the rule: :$\map \phi h = h N$ Note that $N$ need not be a subset of $H$. Therefore, the coset $h N$ is an element of $H N / N$ rather than of $H / N$. Then $\phi$ is a homomorphism, as: :$\map \phi {x y} = x y N = \paren {x N} \paren {y N} = \map \phi x \map \phi y$ Then: {{begin-eqn}} {{eqn | l = \map \ker \phi       | r = \set {h \in H: \map \phi h = e_{H N / N} }       | c =  }} {{eqn | r = \set {h \in H: h N = N}       | c =  }} {{eqn | r = \set {h \in H: h \in N}       | c =  }} {{eqn | r = H \cap N       | c =  }} {{end-eqn}} Then we see that $\phi$ is a surjection because $h n N = h N \in H N / N$ is $\map \phi h$. The result follows from the First Isomorphism Theorem. {{qed}}
The relations being defined can be illustrated by this commutative diagram: :600px $(1): \quad S + J$ is a subring of $R$ From Sum of All Ring Products is Additive Subgroup, $S + J$ is an additive subgroup of $R$. Suppose $s, s' \in S, j, j' \in J$. Then: :$\paren {s + j} \paren {s' + j'}$ {{begin-eqn}} {{eqn | l = \paren {s + j} \paren {s' + j'}       | r = s s' + b{s j' + s' j + j j'}       | c =  }} {{eqn | o = \in       | r = S + J       | c = as $J$ is an ideal of $R$ }} {{end-eqn}} so by the Subring Test $S + J$ is a subring of $R$. {{qed|lemma}} $(2): \quad J$ is an ideal of $S + J$ Let $s + j \in S + J$ and let $j \in J$. Then: {{begin-eqn}} {{eqn | l = \paren {s + j} j'       | r = s j + s j'       | c =  }} {{eqn | o = \in       | r = J       | c = as $s j, s j' \in J$ as $J$ is an ideal of $R$ }} {{end-eqn}} So $J$ is an ideal of $S + J$. {{qed|lemma}} $(3): \quad S \cap J$ is an ideal of $S$ Let $\nu: R \to R / J$ be the quotient epimorphism. Let $\nu'$ be the restriction of $\nu$ to $S$. Then $\nu': S \to R / J$ is a homomorphism. The image of $\nu'$ is the set of all cosets $s + J$ for $s \in S$: :$\image {\nu'} = \dfrac {S + J} J$ Now, the kernel of $\nu'$ is the set of all elements of $S$ which are sent to $0_{S/J}$ by $\nu$. That is, all the elements of $S$ which are also in $J$ itself, which is how the quotient ring behaves. That is: :$\ker \paren {\nu'} = S \cap J$ and so from Kernel of Ring Homomorphism is Ideal, $S \cap J$ is an ideal of $S$. $(4): \quad \dfrac S {S \cap J} \cong \dfrac {S + J} J$ This follows directly from the First Isomorphism Theorem. {{qed}}
We define a mapping: : $\phi: G / N \to G / H$ by $\map \phi {g N} = g H$ Since $\phi$ is defined on cosets, we need to check that $\phi$ is well-defined. Suppose $x N = y N \implies y^{-1} x \in N$. Then $N \le H \implies y^{-1} x \in H$ and so $x H = y H$. So $\map \phi {x N} = \map \phi {y N}$ and $\phi$ is indeed well-defined. Now $\phi$ is a homomorphism, from: {{begin-eqn}} {{eqn | l = \map \phi {x N} \map \phi {y N}       | r = \paren {x H} \paren {y H}       | c =  }} {{eqn | r = x y H       | c =  }} {{eqn | r = \map \phi {x y N}       | c =  }} {{eqn | r = \map \phi {x N y N}       | c =  }} {{end-eqn}} Also, since $N \subseteq H$, it follows that: :$\order N \le \order H$ So: :$\order {G / N} \ge \order {G / H}$, indicating $\phi$ is surjective. So: {{begin-eqn}} {{eqn | l = \map \ker \phi       | r = \set {g N \in G / N: \map \phi {g N} = e_{G / H} }       | c =  }} {{eqn | r = \set {g N \in G / N: g H = H}       | c =  }} {{eqn | r = \set {g N \in G / N: g \in H}       | c =  }} {{eqn | r = H / N       | c =  }} {{end-eqn}} The result follows from the First Isomorphism Theorem. {{qed}}
In Ring Homomorphism whose Kernel contains Ideal, take $\phi: R \to R / K$ to be the quotient epimorphism. Then (from the same source) its kernel is $K$. Thus we have that: :$\phi = \psi \circ \nu$ where $\psi : R / J \to R / K$ is a homomorphism. This can be illustrated by means of the following commutative diagram: ::$\begin{xy}\xymatrix@L+2mu@+1em{  R \ar[dr]_*{\phi}      \ar[r]^*{\nu} & R / J \ar@{-->}[d]^*{\psi} \\  &  R / K  }\end{xy}$ As $\phi$ is an epimorphism then from Surjection if Composite is Surjection we have that $\psi$ is a surjection. So $\Img \psi = \Img \phi = R / K$ and the First Isomorphism Theorem applies. {{qed}}
Let $V \subset S$ be an open set. By assumption, we have that, for all $i \in I$, $U_i = \left({f \restriction_{C_i} }\right)^{-1} \left({V}\right)$ is also open. From the definition of a restriction, we have that $U_i = C_i \cap f^{-1} \left({V}\right)$. Therefore, we can compute: {{begin-eqn}} {{eqn|l = \left({f \restriction_C}\right)^{-1} \left({V}\right)      |r = C \cap f^{-1} \left({V}\right)      |c = Definition of restriction }} {{eqn|r = \left({ \bigcup_{i \mathop \in I} C_i }\right) \cap f^{-1} \left({V}\right)      |c = Definition of $C$ }} {{eqn|r = \bigcup_{i \mathop \in I} \left({ C_i \cap f^{-1} \left({V}\right) }\right)      |c = Intersection Distributes over Union }} {{eqn|r = \bigcup_{i \mathop \in I} U_i      |c = Definition of $U_i$ }} {{end-eqn}} That is, $U = \left({f \restriction_C}\right)^{-1} \left({V}\right)$ is a union of open sets. Therefore, $U$ is itself open by definition of a topology. It follows that $f \restriction_C$ is also continuous by the definition of continuity. {{qed}}
So as to alleviate possible confusion over notation, let the composite of $\phi$ and $\psi$ be denoted $\psi \bullet \phi$ instead of the more usual $\psi \circ \phi$. Then what we are trying to prove is denoted: :$\left({\psi \bullet \phi}\right): \left({S_1, \circ_1, \circ_2, \ldots, \circ_n}\right) \to \left({S_3, \oplus_1, \oplus_2, \ldots, \oplus_n}\right)$ is a homomorphism. To prove the above is the case, we need to demonstrate that the morphism property is held by each of the operations $\circ_1, \circ_2, \ldots, \circ_n$ under $\psi \bullet \phi$. Let $\circ_k$ be one of these operations. We take two elements $x, y \in S_1$, and put them through the following wringer: {{begin-eqn}} {{eqn | l=\left({\psi \bullet \phi}\right) \left({x \circ_k y}\right)       | r=\psi \left({\phi \left({x \circ_k y}\right)}\right)       | c=Definition of composition of mappings }} {{eqn | r=\psi \left({\phi \left({x}\right) *_k \phi \left({y}\right)}\right)       | c=Morphism Property applied to $\circ_k$ under $\phi$ }} {{eqn | r=\psi \left({\phi \left({x}\right)}\right) \oplus_k \psi \left({\phi \left({y}\right)}\right)       | c=Morphism Property applied to $*_k$ under $\psi$ }} {{eqn | r=\left({\psi \bullet \phi}\right) \left({x}\right) \oplus_k \left({\psi \bullet \phi}\right) \left({y}\right)       | c=Definition of composition of mappings }} {{end-eqn}} Disentangling the confusing and tortuous expressions above, we (eventually) see that this shows that the morphism property is indeed held by $\circ_k$ under $\psi \bullet \phi$. As this holds for any arbitrary operation $\circ_k$ in $\left({S_1, \circ_1, \circ_2, \ldots, \circ_n}\right)$, it follows that it holds for all of them. Thus $\left({\psi \bullet \phi}\right): \left({S_1, \circ_1, \circ_2, \ldots, \circ_n}\right) \to \left({S_3, \oplus_1, \oplus_2, \ldots, \oplus_n}\right)$ is indeed a homomorphism. {{qed}}
=== Subset of Normalizer === First we show that $H$ is a subset of $\map {N_G} H$. This follows directly from Left Coset Equals Subgroup iff Element in Subgroup: :$x \in H \implies x H = H$ As $x \in H \implies x^{-1} \in H$ it also follows that $x \in H \implies H x^{-1} = H$. Thus: :$x \in H \implies x H x^{-1} = H^x = H$ and so: :$x \in \map {N_G} H$ So: :$H \subseteq \map {N_G} H$ as we wanted to show. {{qed|lemma}} === Subgroup of Normalizer === By hypothesis, $H$ is a subgroup of $G$. Thus $H$ is itself a group. So by definition of subgroup: : $(1): \quad H \subseteq \map {N_G} H$ : $(2): \quad $ is a group it follows that $H$ is a subgroup of $\map {N_G} H$. {{qed}}
=== Definition by Open Sets implies Definition by Closed Neighborhoods === Let $T$ satisfy the definition by open sets of a $T_4$ space. Let $A$ be a closed set in $T$, and let $U_A$ be an open neighborhood of $A$. By definition of open neighborhood: :$A \subseteq U_A$ Let $B := \relcomp S {U_A}$. By Intersection with Complement is Empty iff Subset: :$A \cap B = \O$ As $B$ is the complement of $U_A$ in $S$, it is by definition closed in $T$. Thus $A$ and $B$ are disjoint closed sets in $T$. Also, from Set Complement inverts Subsets: :$B \subseteq \relcomp S A$ By assumption, there exist disjoint open sets $U$ and $V$ such that $A \subseteq U$ and $B \subseteq V$. From Set Intersection Preserves Subsets: :$B \subseteq \map \complement A \cap V$ Note that the latter set, being an intersection of open sets, is itself open. Then, from Complement of Complement, De Morgan's Laws: Complement of Union and Set Complement inverts Subsets: :$A \cup \map \complement V \subseteq U_A$ From Subset of Union we also have: :$A \subseteq A \cup \map \complement V$ Since $\relcomp S A \cap V$ is open, $A \cup \relcomp S V$ is closed. Hence we have found a closed neighborhood for $A$ in $U_A$, as desired. Hence it is concluded that $T$ satisfies the definition by closed neighborhoods of a $T_4$ space. {{qed|lemma}} === Definition by Closed Neighborhoods implies Definition by Open Sets === Let $T$ satisfy the definition by closed neighborhoods of a $T_4$ space. Let $A$ and $B$ be disjoint closed sets in $T$. Then from Empty Intersection iff Subset of Complement, we have: :$A \subseteq \map \complement B$ and the latter is open in $T$. Applying the assumption, we find a closed neighborhood $C_A$ of $A$ contained in $\map \complement B$. From Empty Intersection iff Subset of Complement and Set Complement inverts Subsets we establish: :$A \cap \map \complement {C_A} = \O$ :$B \subseteq \map \complement {C_A}$ Similarly, we find a closed neighborhood $C_B$ of $B$ contained in $\map \complement {C_A}$. Then from Intersection with Complement is Empty iff Subset: :$B \cap \map \complement {C_B} = \O$ But from Set Complement inverts Subsets, we have: :$\map \complement {C_A} \subseteq \map \complement A$ and so from Subset Relation is Transitive: :$C_B \subseteq \map \complement A$ Finally, another application of Set Complement inverts Subsets shows: :$A \subseteq \map \complement {C_B}$ Since $C_B \subseteq \map \complement {C_A}$, Empty Intersection iff Subset of Complement shows that $C_A$ and $C_B$ are disjoint sets. They are also open sets, being the complement of closed sets. Above, we established also that: :$A \subseteq \map \complement {C_B}$ :$B \subseteq \map \complement {C_A}$ Hence it is concluded that $T$ satisfies the definition by open sets as well. {{qed}}
=== Necessary Condition === This is shown in Factor Spaces of Hausdorff Product Space are Hausdorff. {{qed|lemma}} === Sufficient Condition === This is shown in Product of Hausdorff Factor Spaces is Hausdorff:General Result. {{qed}}
=== Necessary Condition === {{:Product Space is Completely Hausdorff iff Factor Spaces are Completely Hausdorff/Necessary Condition}}{{qed|lemma}} === Sufficient Condition === {{:Product Space is Completely Hausdorff iff Factor Spaces are Completely Hausdorff/Sufficient Condition}}{{qed}}
=== Necessary Condition === {{:Product Space is T3 iff Factor Spaces are T3/Product Space is T3 implies Factor Spaces are T3}}{{qed|lemma}} === Sufficient Condition === {{:Product Space is T3 iff Factor Spaces are T3/Factor Spaces are T3 implies Product Space is T3}}{{qed}} {{AoC}}
=== Necessary Condition === {{:Product Space is T3 1/2 iff Factor Spaces are T3 1/2/Product Space is T3 1/2 implies Factor Spaces are T3 1/2}}{{qed|lemma}} === Sufficient Condition === {{:Product Space is T3 1/2 iff Factor Spaces are T3 1/2/Factor Spaces are T3 1/2 implies Product Space is T3 1/2}}{{qed}}
Let $T$ be a $T_5$ space. Let $\struct {S_\alpha, \tau_\alpha}$ be arbitrary. By Subspace of Product Space Homeomorphic to Factor Space: :$\struct {S_\alpha, \tau_\alpha}$ is homeomorphic to a subspace of $T$. By $T_5$ Property is Hereditary, this subspace is also $T_5$. Finally, by $T_5$ Space is Preserved under Homeomorphism: :$\struct {S_\alpha, \tau_\alpha}$ is a $T_5$ space. {{qed}}
Let $T$ be the Tychonoff plank. Let $T'$ be the deleted Tychonoff plank. By definition, $T'$ is a subspace of $T$. From Tychonoff Plank is Normal, $T$ is a normal space. From Deleted Tychonoff Plank is Not Normal, $T'$ is not a normal space. Thus it is seen that the property of being a normal space is not inherited by a subspace. Hence the result. {{qed}}
{{:Graph of Arcsecant Function}} Let $y = \arcsec x$ where $\size x > 1$. Then: {{begin-eqn}} {{eqn | l = y       | r = \arcsec x       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = \sec y       | c = where $y \in \closedint 0 \pi \land y \ne \dfrac \pi 2$ }} {{eqn | ll= \leadsto       | l = \frac {\d x} {\d y}       | r = \sec y \ \tan y       | c = Derivative of Secant Function }} {{eqn | ll= \leadsto       | l = \frac {\d y} {\d x}       | r = \dfrac 1 {\sec y \tan y}       | c = Derivative of Inverse Function }} {{eqn | ll= \leadsto       | l = \paren {\frac {\d y} {\d x} }^2       | r = \frac 1 {\sec^2 y \ \tan^2 y}       | c = squaring both sides }} {{eqn | r = \frac 1 {\sec^2 y \paren {\sec^2 y - 1} }       | c = Difference of Squares of Secant and Tangent }} {{eqn | r = \frac 1 {x^2 \paren {x^2 - 1} }       | c = Definition of $x$ }} {{eqn | ll= \leadsto       | l = \size {\dfrac {\d y} {\d x} }       | r = \dfrac 1 {\size x \sqrt {x^2 - 1} }       | c = squaring both sides }} {{end-eqn}} Since $\dfrac {\d y} {\d x} = \dfrac 1 {\sec y \tan y}$, the sign of $\dfrac {\d y} {\d x}$ is the same as the sign of $\sec y \tan y$. Writing $\sec y \tan y$ as $\dfrac {\sin y} {\cos^2 y}$, it is evident that the sign of $\dfrac {\d y} {\d x}$ is the same as the sign of $\sin y$. From Sine and Cosine are Periodic on Reals, $\sin y$ is never negative on its domain ($y \in \closedint 0 \pi \land y \ne \pi/2$). However, by definition of the arcsecant of $x$: :$0 < \arcsec x < \dfrac \pi 2 \implies x > 1$ :$\dfrac \pi 2 < \arcsec x < \pi \implies x < -1$ Thus: :$\dfrac {\map \d {\arcsec x} } {\d x} = \dfrac 1 {\size x \sqrt {x^2 - 1} } = \begin{cases} \dfrac {+1} {x \sqrt {x^2 - 1} } & : 0 < \arcsec x < \dfrac \pi 2 \ (\text {that is: $x > 1$})  \\ \dfrac {-1} {x \sqrt {x^2 - 1} } & : \dfrac \pi 2 < \arcsec x < \pi \ (\text {that is: $x < -1$}) \\ \end{cases}$ Hence the result. {{qed}}
By definition of connected space, $T$ admits no separation. Therefore, vacuously, every partition has one open containing $t_1, t_2 \in T$, for all $t_1, t_2 \in T$. That is, for all $t_1, t_2 \in T$, $T$ is connected between $t_1$ and $t_2$. {{qed}}
This proof depends on Leibniz's law: :$x = y \dashv \vdash \map P x \iff \map P y$ We are trying to prove $a = a$. Our assertion, then, is: :$a = a \dashv \vdash \map P a \iff \map P a$ From Law of Identity, $\map P a \iff \map P a$ is a tautology. Thus $a = a$ is also tautologous, and the theorem holds. {{qed}}
Definition 1 implies definition 2 by the Nested Sphere Theorem. Let $M = \struct {A, d}$ be a complete metric space by definition 2. Let $\sequence {x_n}$ be a Cauchy sequence in $M$. By the definition of a Cauchy sequence, for each $k \in \N_{>0}$, there is an $N_k \in \N_{>0}$ such that for all $n, m \in \N$: :$n, m \ge N_k \implies \map d {x_n, x_m} < \dfrac 1 {2^k}$ For each $k \in \N_{>0}$, let $C_k = \map { {B_{1/2^k} }^-} {x_{N_k} }$. {{finish}}
=== $(1)$ implies $(2)$ === Assume $\left({D, +, \circ}\right)$ is an integral domain in sense $1$. As $\left({D, +, \circ}\right)$ is already a commutative ring, it remains to show that $\left({D^*, \circ}\right)$ is a monoid. Because $\circ$ is a ring product, and $\left({D, +, \circ}\right)$ has no zero divisors, we conclude Closure and Associativity. Furthermore, $\left({D, +, \circ}\right)$ is non-null, hence $0_D \ne 1_D$, and we conclude $1_D \in D^*$. Therefore, we also have an Identity for $\left({D^*, \circ}\right)$, and hence it is a monoid. It remains to show that all elements of $\left({D^*, \circ}\right)$ are cancellable. As $\left({D, +, \circ}\right)$ has no zero divisors, this follows from Ring Element is Zero Divisor iff not Cancellable. {{qed|lemma}} === $(2)$ implies $(1)$ === Assume $\left({D, +, \circ}\right)$ is an integral domain in sense $2$. $\left({D, +, \circ}\right)$ is already a commutative ring. Furthermore, as $\left({D^*, \circ}\right)$ is a monoid, it is nonempty. Also, we conclude that $\left({D, +, \circ}\right)$ is a non-null ring with unity. It remains to show that $\left({D, +, \circ}\right)$ has no zero divisors. We know all elements of $\left({D^*, \circ}\right)$ are cancellable. From Ring Element is Zero Divisor iff not Cancellable, we conclude that $\left({D, +, \circ}\right)$ cannot have zero divisors. {{qed}}
Let $n \in \N$. Let $\displaystyle F_{\frac 1 n} = \bigcup \limits_{x \mathop \in F} B \left({x, \dfrac 1 n}\right)$, where $B \left({x, \dfrac 1 n}\right)$ is the open ball around $x$ with radius $\dfrac 1 n$. $F_{\frac 1 n}$ is an open set by definition of open ball. Also, $F_{\frac 1 n}$ contains $F$ as it is the union of open balls around every element of $F$. $\displaystyle F \subseteq \bigcap \limits_{n \mathop = 1}^{\infty} F_{\frac 1 n}$ because $F \subseteq F_{\frac 1 n}$ for each $n \in \N$. Note that: :$\displaystyle \lim_{n \to \infty} B \left({x, \dfrac 1 n}\right) = \left\{{x}\right\}$ so the above statement holds even in the limit. Let $\displaystyle y \in \bigcap \limits_{n \mathop = 1}^\infty F_{\frac 1 n}$. Let $\epsilon > 0$ be given. There exists $n \in \N$ such that $\dfrac 1 n < \epsilon$. As $y \in \bigcap \limits_{n \mathop = 1}^\infty F_{\frac 1 n}$, there exists $x \in F$ such that $y \in B \left({x, \dfrac 1 n}\right)$. Therefore: : $d \left({x, y}\right) < \dfrac 1 n < \epsilon$ which shows that $y$ is a Limit Point of $F$. Since $F$ is closed, then $y \in F$. Therefore: :$\displaystyle F \supseteq \bigcap \limits_{n \mathop = 1}^\infty F_{\frac 1 n}$ We conclude that $\displaystyle F = \bigcap \limits_{n \mathop = 1}^\infty F_{\frac 1 n}$, a countable intersection of open sets. Therefore $F$ is a $G_\delta$ set by definition.  {{qed}} Category:G-Delta Sets Category:Metric Spaces 6w7rbg0x01ukufe29lnccbhs0910lxd
Let $T = \struct {S, \tau}$ be a topological space which is totally disconnected. Then by definition $T$ contains no non-degenerate connected sets. {{AimForCont}} $T$ is not a totally pathwise disconnected space. That is, there exist two points $x, y \in S$ such that there exists a path between $x$ and $y$. That is, $x$ and $y$ are path-connected. Thus they are in the same path component. From Path-Connected Space is Connected, $x$ and $y$ are therefore in the same component. But that contradicts the definition of $T$ having no non-degenerate connected sets. Hence, by Proof by Contradiction, $T$ is a totally pathwise disconnected space. {{qed}}
From Closure of Open Set of Particular Point Space, we have that $A^- = S$. The result follows from Infinite Particular Point Space is not Compact. {{qed}}