hcju/proofwikiqa huggingface dataset:
From Derivative of Cosine Function: :$\dfrac \d {\d x} \cos x = -\sin x$ From Derivative of Sine Function: :$\dfrac \d {\d x} \sin x = \cos x$ Hence: {{begin-eqn}} {{eqn | l = \dfrac {\d^2} {\d x^2} \cos x       | r = -\cos x       | c =  }} {{eqn | l = \dfrac {\d^3} {\d x^3} \cos x       | r = \sin x       | c =  }} {{eqn | l = \dfrac {\d^4} {\d x^4} \cos x       | r = \cos x       | c =  }} {{end-eqn}} and so for all $m \in \N$: {{begin-eqn}} {{eqn | ll = m = 4 k:       | l = \dfrac {\d^m} {\d x^m} \cos x       | r = \cos x       | c =  }} {{eqn | ll = m = 4 k + 1:       | l = \dfrac {\d^m} {\d x^m} \cos x       | r = -\sin x       | c =  }} {{eqn | ll = m = 4 k + 2:       | l = \dfrac {\d^m} {\d x^m} \cos x       | r = -\cos x       | c =  }} {{eqn | ll = m = 4 k + 3:       | l = \dfrac {\d^m} {\d x^m} \cos x       | r = \sin x       | c =  }} {{end-eqn}} where $k \in \Z$. This leads to the Maclaurin series expansion: {{begin-eqn}} {{eqn | l = \sin x       | r = \sum_{k \mathop = 0}^\infty \paren {\frac {x^{4 k} } {\paren {4 k}!} \map \cos 0 - \frac {x^{4 k + 1} } {\paren {4 k + 1}!} \map \sin 0 - \frac {x^{4 k + 2} } {\paren {4 k + 2}!} \map \cos 0 + \frac {x^{4 k + 3} } {\paren {4 k + 3}!} \map \sin 0}       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^\infty \paren {\frac {x^{4 k} } {\paren {4 k}!} - \frac {x^{4 k + 2} } {\paren {4 k + 2}!} }       | c = Sine of Zero is Zero, Cosine of Zero is One }} {{eqn | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!}       | c = setting $n = 2 k$ }} {{end-eqn}} From Series of Power over Factorial Converges, it follows that this series is convergent for all $x$. {{qed}}
Given $i$ and $N$, the ordered statistic $X_{\left({i}\right)}$ has the probability density function: :$f_{X_{\left({i}\right)} } \left({x \mid i, N}\right) = \dfrac {N!} {\left({i - 1}\right)! \left({N - i}\right)!} F_X \left({x}\right)^{i-1} \left({1 - F_X \left({x}\right)}\right)^{N - i} f_X \left({x}\right)$ where $F_X \left({x}\right)$ is the cumulative distribution function of $X$. {{MissingLinks|The page Definition:Conditional Probability needs to be expanded so as to define and explain the notation $\left({x \mid i, N}\right)$ in the context of a PDF}} Let $Y_i = N \left({X_{\left({i + 1}\right)} - X_{\left({i}\right)} }\right)$ be an independent spacing variable valid for $i = 1, 2, 3, \dotsc, N - 1$ which is always positive. The joint density function of both $X_{\left({i}\right)}$ and $Y_{i}$ is then: :$f_{X_{\left({i}\right)}, Y_i} \left({x, y \mid i, N}\right) = \dfrac {\left({N - 1}\right)!} {\left({i - 1}\right)! \left({N - i - 1}\right)!} F_X \left({x}\right)^{i - 1} \left({1 - F_X \left({x + \dfrac y N}\right)}\right)^{N - i - 1} f_X \left({x}\right) f_X \left({x + \dfrac y N}\right)$ The conditional density function of $Y_i$ given $X_{\left({i}\right)}$ is: :$f_{Y_i} = \dfrac {f_{X_{\left({i}\right)}, Y_i} } {f_{X_{\left({i}\right)} } }$ which turns into: :$f_{Y_i} \left({y \mid x = X_{\left({i}\right)}, i, N}\right) = \dfrac {N-i} N \dfrac {\left({1 - F_X \left({x + \dfrac y N}\right)}\right)^{N - i - 1} } {\left({1 - F_X \left({x}\right)}\right)^{N - i} } f_X \left({x + \dfrac y N}\right)$ The conditional cumulative function of $Y_i$ given $X_{\left({i}\right)}$ is: :$F_{Y_i} \left({y \mid x = X_{\left({i}\right)}, i, N}\right) = 1 - \left({\dfrac {1 - F_X \left({x + \dfrac y N}\right)} {1 - F_X \left({x}\right)} }\right)^{N - i}$ The following Taylor expansion in $y$ is an approximation of $F_X \left({x + \dfrac y N}\right)$: :$F_X \left({x + \dfrac y N}\right) = F_X \left({x}\right) + f_X \left({x}\right) \dfrac y N + O \left({N^{-2} }\right)$ Inserting this produces: :$F_{Y_i} \left({y \mid x = X_{\left({i}\right)}, i, N}\right) = 1 - \left({1 - \dfrac {f_X \left({x}\right) y} {N \left({1 - F_X \left({x}\right)}\right)} + O \left({N^{-2} }\right)}\right)^{N - i}$ The limit as $N$ gets large is the exponential function: :$F_{Y_i} \left({y \mid x = X_{\left({i}\right)}, i, N}\right) = 1 -e^{-f_X \left({x}\right) y \dfrac {1 - \dfrac i N} {1 - F_X \left({x}\right)} } + O \left({N^{-1} }\right)$ The distribution of $F_X$ is uniform by definition of a random pick: :$F_X \left({x}\right) \sim U \left({0, 1}\right)$ Therefore $i$ is uniformly distributed as well: :$F_X \left({X_{\left({i}\right)} }\right) \approx \dfrac i N$ The limit of $F_{Y_i}$ is then: :$\displaystyle \lim_{N \to \infty} F_{Y_i} \left({y \mid x = X_{\left({i}\right)}, i, N}\right) = 1 - e^{-f_X \left({x}\right) y}$ {{qed}} Category:Probability Theory Category:Named Theorems 6r1tid9e917j9xc5shsuwecjamz3hkd
From Rational Numbers form Metric Space we have that $\Q$ is a metric space under the Euclidean metric. From Metric Induces Topology, it follows that the Euclidean topology forms a topology on $\Q$. {{qed}}
Let $\left({\R, \tau_d}\right)$ be the real number space $\R$ under the Euclidean topology $\tau_d$. By definition of rational numbers, $\Q \subseteq \R$. From Topological Subspace is Topological Space we have that $\left({\Q, \tau_d}\right)$ is a topology. {{qed}}
Let $\struct {\R, \tau_d}$ be the real number space $\R$ under the Euclidean topology $\tau_d$. By definition of irrational numbers, $\R \setminus \Q \subseteq \R$. From Topological Subspace is Topological Space we have that $\struct {\R \setminus \Q, \tau_d}$ is a topology. {{qed}}
Let $\alpha \in \Q$. By Real Number is Closed in Real Number Line, $B_\alpha = \set \alpha$ is closed in $\R$. From Closed Set Equals its Closure, it follows that: :$B_\alpha = \map \cl {B_\alpha}$ Hence the result. {{qed}}
By definition: :$B_\alpha = \set \alpha$ Thus: :$\displaystyle \bigcup_{\alpha \mathop \in \Q} B_\alpha = \Q$ The result follows from Closure of Rational Numbers is Real Numbers. {{qed}}
From Rationals are Everywhere Dense in Topological Space of Reals, $\Q$ is everywhere dense in $\R$. It follows by definition of everywhere dense that $\Q^- = \R$. {{qed}} Category:Rational Number Space Category:Real Number Line with Euclidean Topology Category:Set Closures h1q0jst1ieie6aoa4ptu2s188m4f779
By definition: :$B_\alpha = \set \alpha$ Thus: :$\displaystyle \bigcup_{\alpha \mathop \in \Q} B_\alpha = \Q$ By definition, the exterior of $\Q$ is the complement of the closure of $\Q$ in $\R$. By Closure of Rational Numbers is Real Numbers: :$\Q^- = \R$ By Relative Complement with Self is Empty Set: :$\relcomp \R \R = \O$ Hence the result. {{qed}}
From Open Sets in Real Number Line, the set: :$S := \openint \gets \alpha \cup \openint \alpha \to$ is open in $\R$. Thus by definition of closed, its complement relative to $\R$: :$\R \setminus S = \set \alpha$ is closed in $\R$. {{qed}} Category:Real Number Line with Euclidean Topology Category:Closed Sets h6rshhni5fod5b84tm211sjyz7u3v09
{{begin-eqn}} {{eqn | l = \bigcap_{\alpha \mathop \in \Q} B_\alpha^e       | r = \bigcap_{\alpha \mathop \in \Q} \R \setminus B_\alpha^-       | c = {{Defof|Exterior (Topology)|index = 1|Exterior}} }} {{eqn | r = \bigcap_{\alpha \mathop \in \Q} \R \setminus B_\alpha       | c = Real Number is Closed in Real Number Line }} {{eqn | r = \R \setminus \bigcup_{\alpha \mathop \in \Q} B_\alpha       | c = De Morgan's Laws: Difference with Union }} {{eqn | r = \R \setminus \Q       | c = {{Defof|Rational Number|Set of Rational Numbers}} }} {{end-eqn}} {{qed}}
Let $\alpha \in \Q$ be a rational number. From Real Number is Closed in Real Number Line, $\set \alpha$ is a closed set of $\R$. From Rational Numbers are Countably Infinite, $\displaystyle \bigcup_{\alpha \mathop \in \Q} \set \alpha$ is a countable union. Thus $\Q = \displaystyle \bigcup_{\alpha \mathop \in \Q} \set \alpha$ is a countable union of closed sets of $\R$. Hence the result by definition of $F_\sigma$ set. {{qed}}
Let $\alpha \in \R \setminus \Q$. Let $I := \openint a b$ be an open interval in $\R$ such that $\alpha \in I$. By Between two Real Numbers exists Rational Number: :$\exists \beta \in \Q: \beta \in I$. Thus $I$ contains elements of $\Q$ and so $\R \setminus \Q$ is not open in $\R$. Thus by definition, $\Q$ is not closed in $\R$. {{qed}}
{{AimForCont}} $\Q$ is a $G_\delta$ set in $\R$. Let $\Q = \displaystyle \bigcap_{i \mathop \in \N} V_i$. Since Rational Numbers are Countably Infinite, there exists an enumeration of $\Q$. Write $\Q = \set {q_i}_{i \mathop \in \N}$. Define $U_i = V_i \setminus \set {q_i}$. We show that $U_i$ is dense in $\R$. :Let $A \subseteq \R$ be an open set of $\struct {\R, \tau}$. :From Basis for Euclidean Topology on Real Number Line, the set of all open real intervals of $\R$ form a basis for $\struct {\R, \tau}$. :So there exists $\openint a b \subseteq A$ for some $a < b$. :One of $q_i \le a$, $q_i > a$ must hold. :Suppose $q_i \le a$. :By Between two Real Numbers exists Rational Number we have: ::$\exists r \in Q: a < r < b$. :Since $r > a \ge q_i$: ::$r \in \Q \setminus \set {q_i} \subseteq V_i \setminus \set {q_i} = U_i$. :Suppose $q_i > a$. :By Between two Real Numbers exists Rational Number we have: ::$\exists r \in Q: a < r < \min \set {q_i, b}$. :Since $r < q_i$: ::$r \in \Q \setminus \set {q_i} \subseteq V_i \setminus \set {q_i} = U_i$. :In both cases we see that $r \in \openint a b \cap U_i \subseteq A \cap U_i$. :Therefore $A \cap U_i \ne \O$. :Thus $U_i$ is dense in $\R$. Also we have: {{begin-eqn}} {{eqn | l = \bigcap_{i \mathop \in \N} U_i       | r = \bigcap_{i \mathop \in \N} \paren {V_i \setminus \set {q_i} } }} {{eqn | r = \paren {\bigcap_{i \mathop \in \N} V_i} \setminus \paren {\bigcup_{i \mathop \in \N} \set {q_i} }       | c = Difference with Union }} {{eqn | r = \Q \setminus \Q }} {{eqn | r = \O       | c = Set Difference with Self is Empty Set }} {{end-eqn}} By Real Number Line is Complete Metric Space and Baire Category Theorem, $\struct {\R, \tau}$ is a Baire space. By definition of Baire Space, $\displaystyle \bigcap_{i \mathop \in \N} U_i$ is dense in $\R$. But Empty Set is Nowhere Dense. This is a contradiction. Hence $\Q$ is not a $G_\delta$ set in $\R$. {{qed}}
{{begin-eqn}} {{eqn | l = \Q       | r = \bigcup_{\alpha \mathop \in \Q} \set \alpha       | c = Rational Numbers form F-Sigma Set in Reals }} {{eqn | ll= \leadsto       | l = \R \setminus \Q       | r = \R \setminus \bigcup_{\alpha \mathop \in \Q} \set \alpha       | c =  }} {{eqn | r = \bigcap_{\alpha \mathop \in \Q} \paren {\R \setminus \set \alpha}       | c = De Morgan's Laws: Difference with Union }} {{end-eqn}} The result follows from Rational Numbers are Countably Infinite. {{qed}}
Proof by induction: For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition: :$A^n \mathbf v = \lambda^n \mathbf v$ === Basis for the Induction === $P \left({1}\right)$ is true, as this just says: :$A \mathbf v = \lambda \mathbf v$ which follows by definition of eigenvector. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true. So this is our induction hypothesis: :$A^k \mathbf v = \lambda^k \mathbf v$ Then we need to show: :$A^{k+1} \mathbf v = \lambda^{k+1} \mathbf v$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \lambda^{k+1} \mathbf v       | r = \lambda \cdot \lambda^k \mathbf v }} {{eqn | r = \lambda A^k \mathbf v       | c = Induction Hypothesis }} {{eqn | r = A^k \lambda \mathbf v }} {{eqn | r = A^k A \mathbf v       | c = Basis for the Induction }} {{eqn | r = A^{k+1} {\mathbf v} }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \N_{> 0}: A^n \mathbf v = \lambda^n \mathbf v$ {{Qed}} Category:Matrix Algebra 888f16pnpzouxje4om8ew0eanyi2lvl
From Set Difference Intersection with Second Set is Empty Set: :$\Q \cap \paren {\R \setminus \Q} = \O$ By Empty Set is Closed in Topological Space, $\O$ is closed in $\R$. From Closed Set Equals its Closure: :$\O^- = \O$ Hence the result. {{qed}}
From Closure of Rational Numbers is Real Numbers: :$\Q^- = \R$ From Closure of Irrational Numbers is Real Numbers: :$\paren {\R \setminus \Q}^- = \R$ The result follows from Intersection is Idempotent. {{qed}}
From the definition of absolute value: :$\size {x_1 - x_2} = \sqrt {\paren {x_1 - x_2}^2}$ This is the Euclidean metric. in Euclidean Metric on Real Vector Space is Metric this is shown to be a metric. {{qed}}
From Euclidean Space is Complete Metric Space, a Euclidean space is a metric space. From Metric Space fulfils all Separation Axioms it follows that $\struct {\Q, \tau_d}$ is a completely normal space. {{qed}}
From Euclidean Space is Complete Metric Space, a Euclidean space is a metric space. From Metric Space fulfils all Separation Axioms it follows that $\struct {\R \setminus \Q, \tau_d}$ is a completely normal space. {{qed}}
From Euclidean Space is Complete Metric Space, a Euclidean space is a metric space. The result follows from Metric Space is Paracompact. {{qed}}
Let $d: \R \times \R \to \R$ be the Euclidean plus metric: :$\map d {x, y} := \size {x - y} + \displaystyle \sum_{i \mathop = 1}^\infty 2^{-i} \map \inf {1, \size {\max_{j \mathop \le i} \frac 1 {\size {x - r_j} } - \max_{j \mathop \le i} \frac 1 {\size {y - r_j} } } }$ Let $\sequence {x_n}$ converge in $\struct {\R \setminus \Q, \tau_d}$ to a rational number. For each $x_i$ in $\sequence {x_n}$, there exists $j > i$ such that $\map d {x_i, x_j} \ge \size {x_i - x_j} + 2^{-k}$ for some $k$. Hence $\sequence {x_n}$ is not a Cauchy sequence. Thus if $\sequence {x_n}$ is a Cauchy sequence it converges to an irrational number. {{finish|Tighten up which metrics and which metric spaces are being converged in or not, and establish the direct relevance of the metric $d$. This has been lifted more or less verbatim from Steen and Seebach and isn't fully understood.}}
From Irrational Number Space is Complete Metric Space, $\struct {\R \setminus \Q, d}$ is a complete metric space. From the Baire Category Theorem, a complete metric space is also a Baire space. The result follows from Baire Space is Non-Meager. {{qed}}
From Rational Numbers are Countably Infinite, $\Q$ is a countable union of singleton subsets. From Singleton Set is Nowhere Dense in Rational Space, each of those singleton subsets is nowhere dense in $\struct {\Q, \tau_d}$. The result follows from definition of meager. {{qed}}
From Rational Numbers are Countably Infinite, $\Q$ is itself countable. The result follows by Countable Space is Separable. {{qed}}
From Underlying Set of Topological Space is Closed, $S$ is closed in $T$. From Closed Set Equals its Closure, $S = S^-$. The result follows from definition of everywhere dense. {{qed}} Category:Denseness 9a0db4iczcdfb57dc5v1qkct58ghh81
Let $S$ be the set defined as: :$S = \set {\pi + q: q \in \Q}$ From Rational Numbers are Countably Infinite, $\Q$ is countable. Therefore $S$ is also countable. From $\pi$ is Irrational: :$\pi \in \R \setminus \Q$ It follows from Rational Addition is Closed that: :$\forall q \in \Q: \pi + q \in \R \setminus \Q$ and so: :$S \subseteq \R \setminus \Q$ From Rationals plus Irrational are Everywhere Dense in Irrationals, it follows that $S$ is everywhere dense in $\R \setminus \Q$. Thus we have constructed a countable subset $S$ of $\R \setminus \Q$ which is everywhere dense in $\R \setminus \Q$. The result follows by definition of separable space. {{qed}}
From Rational Numbers form Metric Space, $\struct {\Q, \tau_d}$ is a metric space. From Rational Number Space is Separable, $\struct {\Q, \tau_d}$ is a separable space. The result follows from Separable Metric Space is Second-Countable. {{qed}}
From Irrational Numbers form Metric Space, $\struct {\R \setminus \Q, \tau_d}$ is a metric space. From Irrational Number Space is Separable, $\struct {\R \setminus \Q, \tau_d}$ is a separable space. The result follows from Separable Metric Space is Second-Countable. {{qed}}
By Compact Subspace of Hausdorff Space is Closed, $S$ is closed in $\Q$. By Set is Closed iff Equals Topological Closure, $S = S^-$. {{AimForCont}} $S$ is not nowhere dense in $\Q$. Then $S^-$ contains some non-empty open set. From Basis for Euclidean Topology on Real Number Line, the set of all open real intervals of $\R$ form a basis for $\struct {\R, \tau_d}$. By Basis for Topological Subspace, the set of all intersections of $\Q$ and open real intervals of $\R$ form a basis for $\struct {\Q, \tau_d}$. So there exists $V = \Q \cap \openint {a'} {b'} \subseteq S^- = S$ for some $a' < b'$. By Between two Real Numbers exists Irrational Number we have: :$\exists r \in \R \setminus \Q: a' < r < b'$ By Between two Real Numbers exists Rational Number we have: :$\exists a, b \in \Q: a' < a < r < b < b'$ Thus $\Q \cap \closedint a b$ is a closed subset of $S$ in $\Q$. By Closed Subspace of Compact Space is Compact, $\Q \cap \closedint a b$ is compact. Now we construct an open cover for $\Q \cap \closedint a b$: Let $C = \openint r {b + 1} \cap \Q$. Let $C_q = \openint {a - 1} q \cap \Q$ for each $q \in \openint a r \cap \Q$. Then $C$ and $C_q$ are open sets. Let $\CC = \set C \cup \set {C_q: q \in \openint a r \cap \Q}$. To show that $\CC$ is an open cover for $\Q \cap \closedint a b$: :Pick any $x \in \Q \cap \closedint a b$. :Suppose $x > r$.  :Then $r < x \le b < b + 1$. :Therefore $x \in \openint r {b + 1} = C \in \CC$. :Suppose $x < r$. :By Between two Real Numbers exists Rational Number we have: ::$\exists q \in \Q: a - 1 < a \le x < q < r$ :Then $x \in \openint {a - 1} q$ and $q \in \openint a r$. :Thus $x \in C_q \in \CC$. :Therefore $x \in \bigcup \CC$, showing that $\CC$ is an open cover for $\Q \cap \closedint a b$.  Since $\Q \cap \closedint a b$ is compact, $\CC$ has a finite subcover, $\CC'$. This finite subcover can only contain a finite number of sets in $\set {C_q: q \in \openint a r \cap \Q}$. Therefore $\set {q: C_q \in \CC'}$ has a greatest element, $q'$. We have that $q' \in \openint a r \cap \Q$. By Between two Real Numbers exists Rational Number we have: :$\exists q^* \in \Q: a < q' < q^* < r < b$ So $q^* \in \Q \cap \closedint a b$. But for all $q \in \hointl a {q'} \cap \Q$: :$q^* \notin C_q = \openint {a - 1} q \cap \Q$ We also have $q^* \notin C = \openint r {b + 1} \cap \Q$. Hence $q^* \notin \bigcup \CC'$. So $\CC'$ is not a cover for $\Q \cap \closedint a b$. This is a contradiction. Thus $S$ is nowhere dense in $\Q$. {{qed}}
For $\struct {\Q, \tau_d}$ to be a locally compact Hausdorff Space, it is required that every point of $\Q$ has a compact neighborhood. Let $x \in \Q$. Let $N \subseteq \Q$ be a neighborhood of $x$. Then: :$\exists U \in \tau: x \in U \subseteq N \subseteq \Q$. {{AimForCont}} $N$ is compact. By Compact Set of Rational Numbers is Nowhere Dense, $N$ is nowhere dense. Thus $N^-$ contains no open set of $\Q$ which is non-empty. But $U$ is a non-empty open set of $\Q$. By Set is Subset of its Topological Closure, $N \subseteq N^-$. Because Subset Relation is Transitive: :$U \subseteq N^-$ This is a contradiction. Hence $\struct {\Q, \tau_d}$ is not a locally compact Hausdorff Space. {{qed}}
For $\struct {\R \setminus \Q, \tau_d}$ to be a locally compact Hausdorff Space, it is required that every point of $\R \setminus \Q$ has a compact neighborhood. Let $x \in \R \setminus \Q$. Let $N \subseteq \R \setminus \Q$ be a neighborhood of $x$. Then: :$\exists U \in \tau: x \in U \subseteq N \subseteq \R \setminus \Q$. {{AimForCont}} $N$ is compact. By Compact Set of Irrational Numbers is Nowhere Dense, $N$ is nowhere dense. Thus $N^-$ contains no open set of $\R \setminus \Q$ which is non-empty. But $U$ is a non-empty open set of $\R \setminus \Q$. By Set is Subset of its Topological Closure, $N \subseteq N^-$. Because Subset Relation is Transitive: :$U \subseteq N^-$ This is a contradiction. Hence $\struct {\R \setminus \Q, \tau_d}$ is not a locally compact Hausdorff Space. {{qed}}
From Rational Number Space is not Locally Compact Hausdorff Space, $\struct {\Q, \tau_d}$ is not weakly locally compact. Hence the result from definition of weakly $\sigma$-locally compact. {{qed}}
From Irrational Number Space is not Locally Compact Hausdorff Space, $\struct {\R \setminus \Q, \tau_d}$ is not weakly locally compact. Hence the result from definition of weakly $\sigma$-locally compact. {{qed}}
From Rational Numbers are Countably Infinite, $\Q$ is countable. Hence the result from definition of Countable Space is $\sigma$-Compact. {{qed}}
Let $y \in \R \setminus \Q$. Let $U \subseteq \R \setminus \Q$ be an open set of $\struct {\R \setminus \Q, \tau_d}$ such that $x \in U$. From Basis for Euclidean Topology on Real Number Line, the set of all open real intervals of $\R$ form a basis for $\struct {\R, \tau_d}$. By Basis for Topological Subspace, the set of all intersections of $\R \setminus \Q$ and open real intervals of $\R$ form a basis for $\struct {\R \setminus \Q, \tau_d}$. So there exists $V = \paren {\R \setminus \Q} \cap \openint {y - \epsilon} {y + \epsilon} \subseteq U$ for some $\epsilon > 0$ such that $y \in V$. By Between two Real Numbers exists Rational Number we have: :$\exists r \in Q: y - \epsilon - x < r < y + \epsilon - x$. Then: :$y - \epsilon < x + r < y + \epsilon$ Then $x + r \in S_x \cap V \subseteq S_x \cap U$. Since $y$ and $U$ are arbitrary: $S_x$ is everywhere dense in $\struct {\R \setminus \Q, \tau_d}$. {{qed}}
Let $x, y \in \Q$. From Between two Rational Numbers exists Irrational Number: :$\exists \alpha \in \R \setminus \Q: x < \alpha < y$ Consider the unbounded open real intervals: :$A := \openint \gets \alpha$, $B := \openint \alpha \to$ Let: :$U := A \cap \Q, V := B \cap \Q$ Let $\beta \in \Q$. Then either: :$(1): \quad \beta < \alpha$ in which case: :$\beta \in U$ or: :$(2): \quad \beta > \alpha$ in which case: :$\beta \in V$ Thus $U \cup V = \Q$. Let $a \in U$. Then $a < \alpha$ and so $a \notin V$. Similarly, let $b \in V$. Then $b > \alpha$ and so $b \notin U$. Then note that $x \in U$ and $y \in V$. Thus $U \ne \O$ and $V \ne \O$. Thus, by definition, $U$ and $V$ constitute a separation of $\Q$ such that $x \in U$ and $y \in V$. Hence the result by definition of totally separated. {{qed}}
Let $x, y \in \R \setminus \Q$. From Between two Real Numbers exists Rational Number: :$\exists \alpha \in \Q: x < \alpha < y$ Consider the unbounded open real intervals: :$A := \openint \gets \alpha$, $B := \openint \alpha \to$ Let: :$U := A \cap \paren {\R \setminus \Q}$, $V := B \cap \paren {\R \setminus \Q}$ Let $\beta \in \R \setminus \Q$. Then either: :$(1): \quad \beta < \alpha$ in which case: :$\beta \in U$ or: :$(2): \quad \beta > \alpha$ in which case: :$\beta \in V$ Thus $U \cup V = \R \setminus \Q$. Let $a \in U$. Then $a < \alpha$ and so $a \notin V$. Similarly, let $b \in V$. Then $b > \alpha$ and so $b \notin U$. Then note that $x \in U$ and $y \in V$. Thus $U \ne \O$ and $V \ne \O$. Thus, by definition, $U$ and $V$ constitute a separation of $\R \setminus \Q$ such that $x \in U$ and $y \in V$. Hence the result by definition of totally separated. {{qed}}
For a space to be scattered, it needs by definition to have no subset which is dense-in-itself. From Rational Number Space is Dense-in-itself, $\struct {\Q, \tau_d}$ is dense-in-itself. Hence the result. {{qed}}
Let $x \in \Q$. Let $U \subseteq \R$ be an open set of $\struct {\Q, \tau_d}$ such that $x \in U$. From Basis for Euclidean Topology on Real Number Line, the set of all open real intervals of $\R$ form a basis for $\struct {\R, \tau_d}$. By Basis for Topological Subspace, the set of all intersections of $\Q$ and open real intervals of $\R$ form a basis for $\struct {\Q, \tau_d}$. So there exists $V = \Q \cap \openint {x - \epsilon} {x + \epsilon} \subseteq U$ for some $\epsilon > 0$ such that $x \in V$. By Between two Real Numbers exists Rational Number we have: :$\exists r \in \Q: x < r < x + \epsilon$. Then $r \in \Q \cap \openint {x - \epsilon} {x + \epsilon}$. Therefore $x$ is not isolated. Hence $\struct {\Q, \tau_d}$ is dense-in-itself. {{qed}}