hcju/proofwikiqa huggingface dataset:
This follows immediately from Value of Finite Continued Fraction equals Numerator Divided by Denominator. {{qed}} Category:Continued Fractions h2cpxoogn7dq8uudqsqt7b9v6bzsd37
Let $T$ be the Either-Or topological space. From Either-Or Topology is Locally Path-Connected, $T$ is a locally path-connected space. From Either-Or Topology is not Locally Arc-Connected, $T$ is not a locally arc-connected space. Hence the result. {{qed}}
Let $T$ be a countable finite complement Space. From Finite Complement Space is Locally Connected, $T$ is a locally connected space. From Countable Finite Complement Space is not Locally Path-Connected, $T$ is not a locally path-connected space. Hence the result. {{qed}}
Let $T$ be the extended topologist's sine curve. From Extended Topologist's Sine Curve is Path-Connected, $T$ is a path-connected space. From Extended Topologist's Sine Curve is not Locally Path-Connected, $T$ is not a locally path-connected space. Hence the result. {{qed}}
By Denominators of Simple Continued Fraction are Strictly Increasing, with the possible exception of $1 = q_0 = q_1$, the sequence $\left \langle {q_n}\right \rangle$ is strictly increasing. Now, since $q_2 > q_1 \ge q_0 = 1$, we have $q_2 \geq 2$. Then $q_{k+1} \ge q_k + q_{k-1}$ shows that from $q_3$ onwards, the $q_k$s increase in steps of at least $2$. As $q_3 \ge 3$, it follows that $q_4 \ge 5$. The result follows. {{qed}} Category:Simple Continued Fractions 9daac5tv8wsgntbb249wb510nbsnj3p
Note that by definition, $\Q$ is the field of quotients of $\Z$. === 1 implies 2 === Let $\dfrac a b = \dfrac c d$ in $\Q$. Then $b c = a d$ in $\Q$. By Canonical Mapping to Field of Quotients is Injective, $b c = a d$ in $\Z$. {{qed|lemma}} === 2 implies 1 === Let $bd = ac$ in $\Z$. By definition of ring homomorphism, $b c = a d$ in $\Q$. Thus Let $\dfrac a b = \dfrac c d$ in $\Q$. {{qed}}
Let $T := \struct {\R, \tau_d}$ be the real number line $\R$ under the usual (Euclidean) topology $\tau_d$. Let $a, b, c \in \R$ where $a < b < c$. Let $A$ be the union of the two open intervals: :$A := \openint a b \cup \openint b c$ Let $T' := \struct {A, \tau_A}$ be the subspace composed of $A$ with the subspace topology induced on $\tau_d$ by $A$. From Union of Adjacent Open Intervals is Locally Path-Connected, $T'$ is a locally path-connected space. From Union of Adjacent Open Intervals is not Path-Connected, $T$ is not a path-connected space. Hence the result. {{qed}}
Let $T$ be the extended topologist's sine curve. From Extended Topologist's Sine Curve is Arc-Connected, $T$ is an arc-connected space. From Extended Topologist's Sine Curve is not Locally Arc-Connected, $T$ is not a locally arc-connected space. Hence the result. {{qed}}
Let $T := \struct {\R, \tau_d}$ be the real number line $\R$ under the usual (Euclidean) topology $\tau_d$. Let $a, b, c \in \R$ where $a < b < c$. Let $A$ be the union of the two open intervals: :$A := \openint a b \cup \openint b c$ Let $T' := \struct {A, \tau_A}$ be the subspace composed of $A$ with the subspace topology induced on $\tau_d$ by $A$. From Union of Adjacent Open Intervals is Locally Arc-Connected, $T'$ is a locally arc-connected space. From Union of Adjacent Open Intervals is not Arc-Connected, $T$ is not a arc-connected space. Hence the result. {{qed}}
Let $ k \ge 3$ be an odd integer. From Difference between Adjacent Convergents But One of Simple Continued Fraction: :$C_k - C_{k - 2} = \dfrac {\paren {-1}^k a_k} {q_k q_{k - 2} } = \dfrac {-a_k} {q_k q_{k - 2} }$ By definition of simple continued fraction, $a_k > 0$. By Convergents of Simple Continued Fraction are Rationals in Canonical Form, $q_k > 0$ and $q_{k - 2} > 0$. Thus $C_k < C_{k - 2}$. {{qed}}
Let $k \ge 2$ be an even integer. From Difference between Adjacent Convergents But One of Simple Continued Fraction: :$C_k - C_{k - 2} = \dfrac {\paren {-1}^k a_k} {q_k q_{k - 2} } = \dfrac {a_k} {q_k q_{k - 2} }$ By definition of simple continued fraction, $a_k > 0$. By Convergents of Simple Continued Fraction are Rationals in Canonical Form, $q_k > 0$ and $q_{k - 2} > 0$. Thus $C_k > C_{k - 2}$. {{qed}}
Follows from: * $q_0 = 1$ by definition * Denominators of Simple Continued Fraction are Strictly Increasing {{qed}} Category:Continued Fractions 9lh3u0v2mya7widc2v8a5e10h4p62lb
By definition, the subgroup generated by $a$ is the intersection of all subgroups containing $a$. By Powers of Element form Subgroup, the set $H = \set {a^n : n \in \Z}$ is a subgroup. Thus $\gen a \subseteq H$. By Power of Element in Subgroup, $H \subseteq \gen a$. By definition of set equality, $\gen a = H$. {{qed}}
The proof proceeds by induction. For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition: :$T$ is connected {{iff}} each of $T_1, T_2, \ldots, T_n$ are connected. === Basis for the Induction === $\map P 2$ is the case: :The product space $T_1 \times T_2$ is connected {{iff}} $T_1$ and $T_2$ are connected. {{:Finite Product Space is Connected iff Factors are Connected/Basis for the Induction}} This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$\displaystyle \prod_{i \mathop = 1}^k T_i$ is connected {{iff}} each of $T_1, T_2, \ldots, T_k$ are connected. from which it is to be shown that: :$\displaystyle \prod_{i \mathop = 1}^{k + 1} T_i$ is connected {{iff}} each of $T_1, T_2, \ldots, T_{k + 1}$ are connected. === Induction Step === This is the induction step: By definition of product space: :$\displaystyle \prod_{i \mathop = 1}^{k + 1} T_i = \paren {\prod_{i \mathop = 1}^k T_i} \times T_{k + 1}$ But from the basis for the induction: :$\displaystyle \prod_{i \mathop = 1}^{k + 1} T_i$ is connected {{iff}} $\displaystyle \prod_{i \mathop = 1}^k T_i$ is connected and $T_{k + 1}$ is connected and from the induction hypothesis: :$\displaystyle \prod_{i \mathop = 1}^k T_i$ is connected {{iff}} each of $T_1, T_2, \ldots, T_k$ are connected. Hence: :$\displaystyle \prod_{i \mathop = 1}^{k + 1} T_i$ is connected {{iff}} each of $T_1, T_2, \ldots, T_{k + 1}$ are connected. So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction: :For all $n \in \Z_{\ge 2}$, $T$ is connected {{iff}} each of $T_1, T_2, \ldots, T_n$ are connected. {{qed}}
By definition of binary biproduct, it remains to verify that: :$i_1\circ p_1 + i_2 \circ p_2 = 1 : a_1 \times a_2 \to a_1 \times a_2$. By definition of binary product and identity morphism, $1 : a_1 \times a_2 \to a_1 \times a_2$ is the unique morphism with: :$p_1 \circ 1 = p_1$ :$p_2 \circ 1 = p_2$ Thus it remains to verify that $i_1\circ p_1 + i_2 \circ p_2$ satisfies the same condition: {{begin-eqn}} {{eqn | l = p_1 \circ (i_1\circ p_1 + i_2 \circ p_2)       | r = (p_1 \circ i_1) \circ p_1 + (p_1 \circ i_2) \circ p_2       | c =  }} {{eqn | r = 1 \circ p_1 + 0 \circ p_2       | c =  }} {{eqn | r = p_1       | c = {{Defof|Identity Morphism}}, {{Defof|Bilinear Mapping}} }} {{end-eqn}} {{begin-eqn}} {{eqn | l = p_2 \circ (i_1\circ p_1 + i_2 \circ p_2)       | r = (p_2 \circ i_1) \circ p_1 + (p_2 \circ i_2) \circ p_2       | c =  }} {{eqn | r = 0 \circ p_1 + 1 \circ p_2       | c =  }} {{eqn | r = p_2       | c = {{Defof|Identity Morphism}}, {{Defof|Bilinear Mapping}} }} {{end-eqn}} {{qed}}
By definition of binary biproduct, it remains to verify that: :$i_1\circ p_1 + i_2 \circ p_2 = 1 : a_1 \sqcup a_2 \to a_1 \sqcup a_2$. By definition of binary coproduct and identity morphism, $1 : a_1 \sqcup a_2 \to a_1 \sqcup a_2$ is the unique morphism with: :$1 \circ i_1 = i_1$ :$1 \circ i_2 = i_2$ Thus it remains to verify that $i_1\circ p_1 + i_2 \circ p_2$ satisfies the same condition: {{begin-eqn}} {{eqn | l = (i_1\circ p_1 + i_2 \circ p_2) \circ i_1       | r = i_1 \circ (p_1 \circ i_1) + i_2 \circ (p_2 \circ i_1)       | c =  }} {{eqn | r = i_1 \circ 1 + i_2 \circ 0       | c =  }} {{eqn | r = i_1       | c = {{Defof|Identity Morphism}}, {{Defof|Bilinear Mapping}} }} {{end-eqn}} {{begin-eqn}} {{eqn | l = (i_1\circ p_1 + i_2 \circ p_2) \circ i_2       | r = i_1 \circ (p_1 \circ i_2) + i_2 \circ (p_2 \circ i_2)       | c =  }} {{eqn | r = i_1 \circ 0 + i_2 \circ 1       | c =  }} {{eqn | r = i_2       | c = {{Defof|Identity Morphism}}, {{Defof|Bilinear Mapping}} }} {{end-eqn}} {{qed}}
Let the Axiom of Choice be assumed. Let $I$ be well-ordered. Let $x = \family {x_\alpha} \in T$ be some arbitrary fixed element of $T$. Let $S_\alpha = \set {\family {y_\beta} \in T: y_\beta = x \beta \text { for all } \beta \ge \alpha}$. We have that $S_\alpha$ is homeomorphic to $S_{\alpha - 1} \times T_\alpha$. Then from Finite Product Space is Connected iff Factors are Connected, $S_\alpha$ is connected {{iff}} $S_{\alpha - 1}$ is. Let $\alpha$ be a limit ordinal. Then: :$S_\alpha = \displaystyle \paren {\bigcup_{\beta \mathop < \alpha} S_\beta}^-$ where $X^-$ denotes the closure of $X$. So if each $S_\beta$ is connected for $\beta < \alpha$, it follows that $S_\alpha$ must likewise be connected, as $\family \gamma$ is nested. Thus: :$T = \displaystyle \overline {\bigcup_{\alpha \mathop \in I} S_\alpha}$ is connected. {{qed}} {{AoC}}
Let $H$ be a component of an open set $U$ of $T_2$. By definition of continuous mapping, $f^{-1} \sqbrk U$ is an open set of $T_1$. Let $G$ be a component of $f^{-1} \sqbrk U$. Thus by Continuous Image of Connected Space is Connected, $f \sqbrk G$ is connected in $T_2$. Thus either: :$f \sqbrk G \subseteq H$ or: :$f \sqbrk G \cap H = \O$ Thus every component of $f^{-1} \sqbrk H$ is a component of $f^{-1} \sqbrk U$. From Component of Locally Connected Space is Open, $f^{-1} \sqbrk H$ is open. Let $S_1 \setminus G$ denote the complement of $f^{-1} \sqbrk H$ relative to $S_1$. By definition of closed set, $S_1 \setminus f^{-1} \sqbrk H$ is closed in $T_1$. From Closed Subspace of Compact Space is Compact, $S_1 \setminus f^{-1} \sqbrk H$ is also compact in $T_1$. From Continuous Image of Compact Space is Compact, $f \sqbrk {S_1 \setminus f^{-1} \sqbrk H} = S_2 \setminus H$ is compact in $T_2$. From Compact Subspace of Hausdorff Space is Closed, $S_2 \setminus H$ is closed in $T_2$. Thus by definition of closed set, $H$ is open in $T_2$. {{Proofread}} {{qed}}
{{AimForCont}} $G$ is a simple connected planar graph with $\map \delta G \ge 6$.  Let $m$ and $n$ denote the number of edges and vertices respectively in $G$.  Then by the Handshake Lemma:  {{begin-eqn}} {{eqn | l = 2 m       | r = \sum_{i \mathop = 0}^n \map {\deg_G} {v_i} }} {{eqn | o = \ge       | r = \sum_{i \mathop = 0}^n 6 }} {{eqn | r = 6 n }} {{eqn | ll = \leadsto       | l = m       | o = \ge       | r = 3 n }} {{end-eqn}} This contradicts the Linear Bound Lemma: :$m \le 3 n - 6$ Hence $\map \delta G \le 5$. {{qed}} Category:Graph Theory r3rkr0y9ywj66c04mx411opbi5bmv0s
Let $f$ denote the number of faces of $G_n$.  Let the sequence $\left\langle{s_i}\right\rangle_{i \mathop = 1}^f$ be the regions of a planar embedding of $G_n$.  Consider the sequence $\left\langle{r_i}\right\rangle_{i \mathop = 1}^f$ where $r_i$ denotes the number of boundary edges for $s_i$.  Since $G$ is simple, then (by the definition of planar embedding):  * every region has at least $3$ boundary edges * every edge is a boundary edge of at most two regions in the planar embedding. Using this two facts, we can find the boundary for $\displaystyle \sum \limits_{i + 1}^f r_i$ as: :$3 f \le \displaystyle \sum \limits_{i + 1}^f r_i \le 2m$ Now calculating the Euler Polyhedron Formula with $f \le 2 m /3$, we will arrive to $m \le 3 n − 6$. {{qed}} Category:Graph Theory Category:Named Theorems p01vh7i0djr1tgy0m7un1cqk319n6i2
Let $T$ be Gustin's sequence space. From Gustin's Sequence Space is Connected, $T$ is a connected space. From Gustin's Sequence Space is Totally Pathwise Disconnected, $T$ is a totally pathwise disconnected space. Hence the result. {{qed}}
=== $(1)$ iff $(2)$ === ;Extremally Disconnected by Interior of Closed Sets: {{:Extremally Disconnected by Interior of Closed Sets}} {{qed|lemma}} === $(1)$ iff $(3)$ === ;Extremally Disconnected by Disjoint Open Sets: {{:Extremally Disconnected by Disjoint Open Sets}} {{qed}} Category:Extremally Disconnected Spaces Category:Equivalence of Definitions of Extremally Disconnected Space ljgspy0eqptnaazv9mmrizibtjqrilo
Let $T = \struct {S, \tau}$ be a partition space. From Partition Topology is Zero Dimensional, $T$ is a zero dimensional topological space. From Partition Topology is not $T_0$, $T$ is not a $T_0$ (Kolmogorov) space. {{qed}}
Let $T = \struct {S, \tau}$ be a non-trivial particular point space. From Particular Point Space is Scattered, $T$ is a scattered space. From Non-Trivial Particular Point Topology is not $T_1$, $T$ is not a $T_1$ (Fréchet) space. {{qed}}
Let $T$ be a divisor space. From Divisor Space is Connected, $T$ is a connected space. From Divisor Space is not $T_1$, $T$ is not a $T_1$ (Fréchet) space. From Divisor Space is Scattered, $T$ is a scattered space. Hence the result. {{qed}}
Suppose $\BB$ is a basis for $\tau$. By definition of basis: :$\forall U \in \tau: \forall x \in U: \exists B \in \BB: x \in B \subseteq U$ For all $x \in \R$, pick $U = \hointr x {x + \epsilon} \in \tau$ for some $\epsilon > 0$. Now: :$\forall x \in \R: \exists B_x \in \BB: x \in B_x \subseteq \hointr x {x + \epsilon}$ This $\BB_x$ has an infimum equal to $x$. So for different $x$, the corresponding $\BB_x$ is different. So the cardinality of $\BB$ is at least $\size \R$, which is uncountable. {{qed}}
Let $T$ be a Miller's biconnected set. From Miller's Biconnected Set is Biconnected, $T$ is a biconnected set. From Miller's Biconnected Set is has no Dispersion Point, $T$  does not have a dispersion point. Hence the result. {{qed}}
Let $T$ be Cantor's leaky tent. From Cantor's Leaky Tent is Connected, $T$ is a connected topological space. From Cantor's Leaky Tent is Punctiform, $T$ is a punctiform space. Hence the result. {{qed}}
Let $M = \struct {A, d}$ be a metric space whose metric is $d$. === $(1)$ implies $(2)$ === Let $T = \struct {A, \tau_d}$ be the topological space of which $\tau_d$ is the topology induced on $M$ by $d$ by definition 1. Then by definition: :$\tau_d$ is the set of all open sets of $M$. Let $U \in \tau_d$. From definition: :$\forall y \in U: \exists \epsilon_y \in \R_{>0}: \map {B_{\epsilon_y} } y \subseteq U$ where $\map {B_{\epsilon_y}} y$ is the open $\epsilon$-ball of $y$. We find these $\epsilon$ for every $y \in U$. Consider the set $\displaystyle \bigcup_{y \mathop \in U} \map {B_{\epsilon_y} } y$. From $y \in \map {B_{\epsilon_y} } y \subseteq U$ for each $y$: :$\displaystyle \bigcup_{y \mathop \in U} \map {B_{\epsilon_y} } y = U$. This shows that any $U \in \tau_d$ is a union of open $\epsilon$-balls in $M$. So open $\epsilon$-balls in $M$ form a basis for $\tau_d$. Thus $\tau_d$ is the topology induced on $M$ by $d$ by definition 2. {{qed|lemma}} === $(2)$ implies $(1)$ === Let $T = \struct {A, \tau_d}$ be the topological space of which $\tau_d$ is the topology induced on $M$ by $d$ by definition 2. Then by definition: :$\tau$ is the topology generated by the basis consisting of the set of all open $\epsilon$-balls in $M$. Let $U \in \tau_d$. Since open $\epsilon$-balls in $M$ form a basis for $\tau_d$, $U$ is a union of open $\epsilon$-balls in $M$. Hence for each $x \in U$: :$\exists y \in U: \exists \epsilon \in \R_0: x \in \map {B_\epsilon} y$ By Open Ball is Subset of Open Ball: :$x \in \map {B_{\epsilon - \map d {x, y} } } x \subseteq \map {B_\epsilon} y \subseteq U$ Therefore $U$ is an open set of $M$. Thus $\tau_d$ is the topology induced on $M$ by $d$ by definition 1. {{qed}} Category:Metric Spaces hvhyt9rf3o3blid1z6lb1alwm8hs7gp
By definition, a topological space is completely normal space {{iff}} it is: :a $T_5$ space :a $T_1$ (Fréchet) space. We have that: :a Metric Space is $T_5$ :a Metric Space is $T_2$ (Hausdorff) :a $T_2$ (Hausdorff) Space is a $T_1$ (Fréchet) Space. Hence the result. {{qed}}
By definition, a topological space is perfectly normal space {{iff}} it is: :a perfectly $T_4$ space :a $T_1$ (Fréchet) space. We have that: :a Metric Space is Perfectly $T_4$ :a Metric Space is $T_2$ (Hausdorff) :a $T_2$ (Hausdorff) Space is a $T_1$ (Fréchet) Space. {{qed}}
Let $M = \struct {\R, d}$ be the real number line with the Euclidean metric. Let $M' = \struct {\R, \delta}$ be the unity-bounded metric space on $M$ where $\delta$ is defined as: :$\delta = \dfrac d {1 + d}$ From Unity-Bounded Metric Space is Bounded, $M'$ is a bounded metric space. From Unity-Bounded Metric Space on Real Number Line is not Totally Bounded, $M'$ is not a totally bounded metric space. {{qed}}
{{ProofWanted|According to S&S (citation below), this is proved somehow using the metric $\delta {{=}} \dfrac d {1 + d}$, given some metric space $M {{=}} \struct {A, d}$, but the derivation of this is obscure.}}
=== '''A''' $\implies$ '''B''' === Directly follows from Tutte's Wheel Theorem/Lemma. === '''B''' $\implies$ '''A''' === If $G_i$ is 3-connected then so is $G_{i+1}$ for every $i < n$.  Suppose not. Then $G_i = G_{i+1} \thinspace / \thinspace e$ where $G_{i+1}$ is 2-connected.  Let $S, \: |S| \leq 2$ be a vertex cut, and $C_1, C_2$ be two components of $G_{i+1}-S$. Since $x, y$ are connected, we can assume $V\left({C_1}\right) \cap \{x, y\} = \emptyset$.  Then, $C_2$ cannot contain both $x$ and $y$ nor it can contain any $v \notin \{x,y\}$ (otherwise $v$ or $u_{xy}$  would be separated from $C_1$ in $G_i$ by at most two vertices.) But now $C_2$ contains only one vertex: either $x$ or $y$.  This contradicts $\deg\left({x}\right),  \deg\left({y}\right) \geq 3$.  {{qed}} {{Namedfor|William Thomas Tutte|cat = Tutte}} Category:Named Theorems Category:Graph Theory k7i6esxj62w9ffch8t3va1w1vtv29ia
Let $\struct {\R, d}$ denote the real number line under the Euclidean metric. Let $\struct {\Q, d}$ denote the rational number space under the Euclidean metric. We have that $\Q \subset \R$ by definition. Let $T = \struct {\R, \tau_d}$ be the topological space induced on $\R$ by $d$. By Real Number Line is Complete Metric Space, $\struct {\R, d}$ is a complete metric space. Thus, by definition, $T$ is topologically complete. Let $T' = \struct {\Q, \tau_d}$ be the topological space induced on $\Q$ by $d$. By Rational Number Space is not Complete Metric Space, $\struct {\Q, d}$ is not a complete metric space. Thus, by definition, $T'$ is not topologically complete. Hence the result. {{qed}}
Let $M_1 = \struct {\tilde{A_1}, \tilde{d_1}, \phi_1}$ and $M_2 = \struct {\tilde{A_2}, \tilde{d_2}, \phi_2}$ be two completions of $\struct {A, d}$. Here, $\phi_1: A \to A_1$ and $\phi_2: A \to A_2$ are isometries By Composite of Isometries is Isometry, $\psi = \phi_1^{-1} \circ \phi_2$ gives an isometry from $\phi_1 \sqbrk A$ to $\phi_2 \sqbrk A$. The sets $\phi_1 \sqbrk A$ and $\phi_2 \sqbrk A$ are dense in $A_1$ and $A_2$ respectively. Thus we can extend $\psi$ continuously to a map $\psi: A_1 \to A_2$. {{explain|That needs a result to be linked to}} That is, for $x \in A_1$, we can find a Cauchy sequence $\sequence {w_n}$ in $A_1$ with limit $x$. Then we define: :$\displaystyle \map \psi x := \lim_{n \mathop \to \infty} \map \psi {w_n}$ which converges as $A_2$ is complete. By Metric Space is Hausdorff, $A$ is a Hausdorff space. By $T_2$ (Hausdorff) Space is Preserved under Homeomorphism, $A_1$ and $A_2$ are also Hausdorff. Therefore, by Convergent Sequence in Hausdorff Space has Unique Limit, $\psi$ is well-defined. {{qed|lemma}} It is to be shown that $\psi$ is surjective: For $y \in \tilde {A_2}$, let $\sequence {w_n'}$ be a Cauchy sequence in $\phi_2 \sqbrk A$ with limit $y$ in $\tilde {A_2}$. Let $z_n$ be the preimage of the $w_n'$ under $\psi$. Then, as $A_2$ is Hausdorff: :$\displaystyle \lim_{n \mathop \to \infty} \map \psi {z_n} = y$ as required. {{qed|lemma}} injectivity of $\psi$ holds because $A_1$ is Hausdorff, as follows: Suppose that: :$\displaystyle \lim_{n \mathop \to \infty} \map \psi {w_n} = \lim_{n \mathop \to \infty} \map \psi {w_n'}$ :$\displaystyle \lim_{n \mathop \to \infty} w_n = w$ and: :$\displaystyle \lim_{n \mathop \to \infty} w_n' = w'$. For $\epsilon > 0$, pick $M \in \N$ such that $\map \psi {w_n}$, $\map \psi {w_n'}$ lie in the open $\epsilon$-ball $\map {B_{\epsilon / 2} } {\map \psi w}$ for all $n \ge M$. Then we have: :$\map {\tilde {d_1} } {w_n, w_n'} = \map {\tilde {d_2} } {\map \psi {w_n}, \map \psi {w_n'} } \le \epsilon$ As $A_1$ is Hausdorff, we conclude $w = w'$, and the task is complete. {{qed|lemma}} Finally, from Distance Function of Metric Space is Continuous, it follows that $\psi$ is an isometry on all of $A_1$, and the proof is complete. {{qed}} Category:Completion Theorem qcfvzbwbat9dp4c4x1ch1qd83h2hhsq
Let $\BB_x = \set {\hointr x {x + \dfrac 1 n} : n \in \N_{>0} }$. We will show that: :$(1): \quad \BB_x$ is countable :$(2): \quad \BB_x$ is a local basis at $x$ $(1)$ follows from the fact that $\BB_x$ is a bijection from the set of natural numbers. $(2)$ is demonstrated as follows: By definition of local basis, it suffices to show that $\forall U \in \tau: x \in U: \exists B \in \BB_x: B \subseteq U$. Pick any $U$ in $\tau$. By definition of $\tau$, there exists $\hointr x {x + \epsilon} \subseteq U$ for some $\epsilon \in \R_{>0}$. By the Archimedean Principle there exists $n \in \N$ such that $n > \dfrac 1 \epsilon$ (that is, $\dfrac 1 n < \epsilon$). So: :$x \in \hointr x {x + \dfrac 1 n} \subseteq \hointr x {x + \epsilon} \subseteq U$ Therefore $\BB_x$ is a local basis at $x$. We have seen that the Sorgenfrey line has a countable local basis at every point. By definition of first-countability, $\tau$ is first-countable. {{qed}}
Let $T$ be an uncountable discrete space. From Standard Discrete Metric induces Discrete Topology, $T$ is metrizable. From Uncountable Discrete Space is not Second-Countable, $T$ is not second-countable. {{qed}}
Let $T$ be the Sorgenfrey line. From Sorgenfrey Line satisfies all Separation Axioms, $T$ is a regular space. From Sorgenfrey Line is Paracompact, $T$ is a paracompact space. From Sorgenfrey Line is not Metrizable, $T$ is not a metrizable space. {{qed}}
Let $T = \struct {S, \tau}$ be an uncountable discrete ordinal space. From Uncountable Discrete Ordinal Space is Metrizable, $T$ is a metrizable space. However, from Uncountable Discrete Ordinal Space has Unmetrizable Uniformity, there exists a uniformity $\UU$ which yields the uniformizable space $T = \struct {S, \tau}$ which is not itself a metrizable uniformity. {{qed}}
A set $U$ is open in a topology $\tau$ if $U \in \tau$. In $\tau = \set {\O, S}$, the only open sets are $\O$ and $S$. {{qed}}
A set $U$ is closed in a topology $\tau$ {{iff}}: :$\relcomp S U \in \tau$ where $\relcomp S U$ denotes the complement of $U$ in $S$. That is, the closed sets are the complements of the open sets. From Open Sets in Indiscrete Topology, in $\tau = \set {\O, S}$, the only open sets are $\O$ and $S$. Hence the only closed sets in the indiscrete topology on $S$ are: :$\relcomp S \O = S$ from Relative Complement of Empty Set and: :$\relcomp S S = \O$ from Relative Complement with Self is Empty Set as stated. {{qed}}
An $F_\sigma$ set is a set which can be written as a countable union of closed sets of $S$. Hence the only $F_\sigma$ sets of $T$ are made from unions of $T$ and $\O$. So $T$ and $\O$ are the only $F_\sigma$ sets of $T$. {{qed}}
A $G_\delta$ set is a set which can be written as a countable intersection of open sets of $S$. Hence the only $G_\delta$ sets of $T$ are made from intersections of $T$ and $\O$. So $T$ and $\O$ are the only $G_\delta$ sets of $T$. {{qed}}
The subspace $T_H = \struct {H, \set {\O, S \cap H} }$ is trivially also an indiscrete space. The only open cover of $T_H$ is $\set H$ itself. The only subcover of $H$ is, trivially, also $\set H$, which is finite. So $H$ is (trivially) compact in $T$. {{qed}}
From Sequence in Indiscrete Space converges to Every Point, every sequence in $T$ converges to every point of $S$. So every infinite sequence has a subsequence which converges to every point in $S$. Hence $H$ is (trivially) sequentially compact in $T$. {{qed}}