avewright/proofwiki-math huggingface dataset:
Theorem: Spacing Limit Theorem

Let $X_{\paren i}$ be the $i$th ordered statistic of $N$ samples from a continuous random distribution with density function $\map {f_X} x$.
Then the spacing between the ordered statistics given $X_{\paren i}$ converges in distribution to exponential for sufficiently large sampling according to:
:$N \paren {X_{\paren {i + 1} } - X_{\paren i} } \xrightarrow D \map \exp {\dfrac 1 {\map f { X_{\paren i} } } }$
as $N \to \infty$ for $i = 1, 2, 3, \dotsc, N - 1$.

Proof:
Given $i$ and $N$, the ordered statistic $X_{\paren i}$ has the probability density function:
:$\map {f_{X_{\paren i} } } {x \mid i, N} = \dfrac {N!} {\paren {i - 1}! \paren {N - i}!} \map {F_X} x^{i - 1} \paren {1 - \map {F_X} x}^{N - i} \map {f_X} x$
where $\map {F_X} x$ 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 $\tuple {x \mid i, N}$ in the context of a PDF}}
Let $Y_i = N \paren {X_{\paren {i + 1} } - X_{\paren i} }$ be an independent spacing variable valid for $i = 1, 2, 3, \dotsc, N - 1$ which is always positive.
The joint density function of both $X_{\paren i}$ and $Y_{i}$ is then:
:$\map {f_{X_{\paren i}, Y_i} } {x, y \mid i, N} = \dfrac {\paren {N - 1}!} {\paren {i - 1}! \paren {N - i - 1}!} \map {F_X} x^{i - 1} \paren {1 - \map {F_X} {x + \dfrac y N} }^{N - i - 1} \map {f_X} x \map {f_X} {x + \dfrac y N}$
The conditional density function of $Y_i$ given $X_{\paren i}$ is:
:$f_{Y_i} = \dfrac {f_{X_{\paren i}, Y_i} } {f_{X_{\paren i} } }$
which turns into:
:$\map {f_{Y_i} } {y \mid x = X_{\paren i}, i, N} = \dfrac {N - i} N \dfrac {\paren {1 - \map {F_X} {x + \dfrac y N} }^{N - i - 1} } {\paren {1 - \map {F_X} x}^{N - i} } \map {f_X} {x + \dfrac y N}$
The conditional cumulative function of $Y_i$ given $X_{\paren i}$ is:
:$\map {F_{Y_i} } {y \mid x = X_{\paren i}, i, N} = 1 - \paren {\dfrac {1 - \map {F_X} {x + \dfrac y N} } {1 - \map {F_X} x} }^{N - i}$
The following Taylor expansion in $y$ is an approximation of $\map {F_X} {x + \dfrac y N}$:
:$\map {F_X} {x + \dfrac y N} = \map {F_X} x + \map {f_X} x \dfrac y N + \map \OO {N^{-2} }$
Inserting this produces:
:$\map {F_{Y_i} } {y \mid x = X_{\paren i}, i, N} = 1 - \paren {1 - \dfrac {\map {f_X} x y} {N \paren {1 - \map {F_X} x} } + \map \OO {N^{-2} } }^{N - i}$
The limit as $N$ gets large is the exponential function:
:$\map {F_{Y_i} } {y \mid x = X_{\paren i}, i, N} = 1 -e^{- \map {f_X} x y \dfrac {1 - \dfrac i N} {1 - \map {F_X} x} } + \map \OO {N^{-1} }$
The distribution of $F_X$ is uniform by definition of a random pick:
:$\map {F_X} x \sim \map U {0, 1}$
Therefore $i$ is uniformly distributed as well:
:$\map {F_X} {X_{\paren i} } \approx \dfrac i N$
The limit of $F_{Y_i}$ is then:
:$\ds \lim_{N \mathop \to \infty} \map {F_{Y_i} } {y \mid x = X_{\paren i}, i, N} = 1 - e^{- \map {f_X} x y}$
{{qed}}
Category:Probability Theory
Category:Named Theorems
5d2721amilx7f2ulbir6vfq06ivxb0k
Theorem: Rational Number Space is Topological Space

Let $\struct {\Q, \tau_d}$ be the rational number space formed by the rational numbers $\Q$ under the usual (Euclidean) topology $\tau_d$.
Then $\tau_d$ forms a topology.

Proof:
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}}
Theorem: Rational Number Space is Topological Space

Let $\struct {\Q, \tau_d}$ be the rational number space formed by the rational numbers $\Q$ under the usual (Euclidean) topology $\tau_d$.
Then $\tau_d$ forms a topology.

Proof:
Let $\struct {\R, \tau_d}$ 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 $\struct {\Q, \tau_d}$ is a topological space.
{{qed}}
Theorem: Irrational Number Space is Topological Space

Let $\struct {\R \setminus \Q, \tau_d}$ be the irrational number space formed by the irrational numbers $\R \setminus \Q$ under the usual (Euclidean) topology $\tau_d$.
Then $\tau_d$ forms a topology.

Proof:
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}}
Theorem: Union of Closures of Singleton Rationals is Rational Space

Let $\struct {\Q, \tau_d}$ be the rational number space under the usual (Euclidean) topology $\tau_d$.
Let $B_\alpha$ denote the singleton containing the rational number $\alpha$.
Then the union of the closures in the set of real numbers $\R$ of all $B_\alpha$ is $\Q$:
:$\ds \bigcup_{\alpha \mathop \in \Q} \map \cl {B_\alpha} = \Q$

Proof:
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}}
Theorem: Closure of Union of Singleton Rationals is Real Number Line

Let $\struct {\Q, \tau_d}$ be the rational number space under the usual (Euclidean) topology $\tau_d$.
Let $B_\alpha$ be the singleton containing the rational number $\alpha$.
Then the closure in the set of real numbers $\R$ of the union of all $B_\alpha$ is $\R$ itself:
:$\ds \map \cl {\bigcup_{\alpha \mathop \in \Q} B_\alpha} = \R$

Proof:
By definition:
:$B_\alpha = \set \alpha$
Thus:
:$\ds \bigcup_{\alpha \mathop \in \Q} B_\alpha = \Q$
The result follows from Closure of Rational Numbers is Real Numbers.
{{qed}}
Theorem: Closure of Rational Numbers is Real Numbers

Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
Let $\struct {\Q, \tau_d}$ be the rational number space under the same topology.
Then:
:$\Q^- = \R$
where $\Q^-$ denotes the closure of $\Q$.

Proof:
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
Theorem: Exterior of Union of Singleton Rationals is Empty

Let $B_\alpha$ be the singleton containing the rational number $\alpha$.
Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology $\tau_d$.
Then the exterior in $\struct {\R, \tau_d}$ of the union of all $B_\alpha$ is the empty set:
:$\ds \paren {\bigcup_{\alpha \mathop \in \Q} B_\alpha}^e = \O$

Proof:
By definition:
:$B_\alpha = \set \alpha$
Thus:
:$\ds \bigcup_{\alpha \mathop \in \Q} B_\alpha = \Q$
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}}
Theorem: Real Number is Closed in Real Number Line

Let $\struct {\R, \tau}$ be the real number line with the usual (Euclidean) topology.
Let $\alpha \in \R$ be a real number.
Then $\set \alpha$ is closed in $\struct {\R, \tau}$.

Proof:
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
Theorem: Intersection of Exteriors of Singleton Rationals is Irrationals

Let $\struct {\Q, \tau_d}$ be the rational number space under the usual (Euclidean) topology $\tau_d$.
Let $B_\alpha$ be the singleton containing the rational number $\alpha$.
Then:
:$\ds \bigcap_{\alpha \mathop \in \Q} B_\alpha^e = \R \setminus \Q$
where $B_\alpha^e$ denotes the exterior of $B_\alpha$ in $\R$.

Proof:
{{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}}
Theorem: Rational Numbers form F-Sigma Set in Reals

Let $\Q$ be the set of rational numbers.
Let $\struct {\R, \tau}$ be the real number line with the usual (Euclidean) topology.
Then $\Q$ is a $F_\sigma$ set in $\R$.

Proof:
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, $\ds \bigcup_{\alpha \mathop \in \Q} \set \alpha$ is a countable union.
Thus $\Q = \ds \bigcup_{\alpha \mathop \in \Q} \set \alpha$ is a countable union of closed sets of $\R$.
Hence the result by definition of $F_\sigma$ set.
{{qed}}
Theorem: Set of Rational Numbers is not Closed in Reals

Let $\Q$ be the set of rational numbers.
Let $\struct {\R, \tau}$ denote the real number line with the usual (Euclidean) topology.
Then $\Q$ is not closed in $\R$.

Proof:
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}}
Theorem: Set of Rational Numbers is not G-Delta Set in Reals

Let $\Q$ be the set of rational numbers.
Let $\struct {\R, \tau}$ denote the real number line with the usual (Euclidean) topology.
Then $\Q$ is not a $G_\delta$ set in $\R$.

Proof:
{{AimForCont}} $\Q$ is a $G_\delta$ set in $\R$.
Let $\Q = \ds \bigcap_{i \mathop \in \N} V_i$.
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, $\ds \bigcap_{i \mathop \in \N} U_i$ 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}}
Theorem: Irrational Numbers form G-Delta Set in Reals

Let $\R \setminus \Q$ denote the set of irrational numbers.
Let $\struct {\R, \tau}$ denote the real number line with the usual (Euclidean) topology.
Then $\R \setminus \Q$ forms a $G_\delta$ set in $\R$.

Proof:
{{begin-eqn}}
{{eqn | l = \Q
      | r = \bigcup_{\alpha \mathop \in \Q} \set \alpha
      | c = 
}}
{{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}}
Theorem: Eigenvalue of Matrix Power

Let $A$ be a square matrix.
Let $\lambda$ be an eigenvalue of $A$ and $\mathbf v$ be the corresponding eigenvector.
Then:
:$A^n \mathbf v = \lambda^n \mathbf v$
holds for each positive integer $n$.
Here $A^n$ is the $n$th power of $A$.

Proof:
Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$A^n \mathbf v = \lambda^n \mathbf v$
Theorem: Closure of Intersection of Rationals and Irrationals is Empty Set

Let $\struct {\R, \tau}$ denote the real number line with the usual (Euclidean) topology.
Let $\Q$ be the set of rational numbers.
Then:
:$\paren {\Q \cap \paren {\R \setminus \Q} }^- = \O$
where:
:$\R \setminus \Q$ denotes the set of irrational numbers
:$\paren {\Q \cap \paren {\R \setminus \Q} }^-$ denotes the closure of $\Q \cap \paren {\R \setminus \Q}$.

Proof:
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}}
Theorem: Intersection of Closures of Rationals and Irrationals is Reals

Let $\struct {\R, \tau}$ be the real number line with the usual (Euclidean) topology.
Let $\Q$ be the set of rational numbers.
Then:
:$\Q^- \cap \paren {\R \setminus \Q}^- = \R$
where:
:$\R \setminus \Q$ denotes the set of irrational numbers
:$\Q^-$ denotes the closure of $\Q$.

Proof:
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 Set Intersection is Idempotent.
{{qed}}
Theorem: Irrational Numbers form Metric Space

Let $\mathbb I = \R \setminus \Q$ be the set of all irrational numbers.
Let $d: \mathbb I \times \mathbb I \to \R$ be defined as:
:$\map d {x_1, x_2} = \size {x_1 - x_2}$
where $\size x$ is the absolute value of $x$.
Then $d$ is a metric on $\mathbb I$ and so $\struct {\mathbb I, d}$ is a metric space.

Proof:
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}}
Theorem: Rational Number Space is Completely Normal

Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Then $\struct {\Q, \tau_d}$ is a completely normal space.

Proof:
From Euclidean Space is Complete Metric Space, a Euclidean space is a metric space.
From Metric Space is Completely Normal it follows that $\struct {\Q, \tau_d}$ is a completely normal space.
{{qed}}
Theorem: Irrational Number Space is Completely Normal

Let $\struct {\R \setminus \Q, \tau_d}$ be the irrational number space under the Euclidean topology $\tau_d$.
Then $\struct {\R \setminus \Q, \tau_d}$ is a completely normal space.

Proof:
From Euclidean Space is Complete Metric Space, a Euclidean space is a metric space.
Hence in particular $\struct {\R, \tau_d}$ is a metric space.
From Subspace of Metric Space is Metric Space, it follows that $\struct {\R \setminus \Q, \tau_d}$ is likewise a metric space.
From Metric Space is Completely Normal it follows that $\struct {\R \setminus \Q, \tau_d}$ is a completely normal space.
{{qed}}
Theorem: Rational Number Space is Paracompact

Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Then $\struct {\Q, \tau_d}$ is paracompact.

Proof:
From Rational Numbers form Metric Space, $\struct {\Q, \tau_d}$ is a metric space.
The result follows from Metric Space is Paracompact.
{{qed}}
Theorem: Irrational Number Space is Paracompact

Let $\struct {\R \setminus \Q, \tau_d}$ be the irrational number space under the Euclidean topology $\tau_d$.
Then $\struct {\R \setminus \Q, \tau_d}$ is paracompact.

Proof:
From Euclidean Space is Complete Metric Space, a Euclidean space is a metric space.
The result follows from Metric Space is Paracompact.
{{qed}}
Theorem: Euclidean Plus Metric is Metric

Let $\R$ be the set of real numbers.
Let $d: \R \times \R \to \R$ be the Euclidean plus metric:
:$\map d {x, y} := \size {x - y} + \ds \sum_{i \mathop = 1}^\infty 2^{-i} \map \inf {1, \size {\max_{j \mathop \le i} \frac 1 {\size {x - r_j}} - \max_{j \mathop \le i} \frac 1 {\size {y - r_j} } } }$
Then $d$ is indeed a metric.

Proof:
Recall that $\set {r_j}_{j \mathop \in \N}$ is an enumeration of the rational numbers $\Q$.
Also, we note that:
{{begin-eqn}}
{{eqn | o = 
      | r = \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 {x - r_j} } } }
      | c = 
}}
{{eqn | o = \le
      | r = \sum_{i \mathop = 1}^\infty 2^{-i} 1
      | c = {{Defof|Infimum of Subset of Real Numbers}}
}}
{{eqn | r = 1
      | c = Sum of Infinite Geometric Sequence: Corollary 1 with $z = \dfrac 1 2$
}}
{{end-eqn}}
meaning that this is a convergent series and so the definition is meaningful.
Next the axioms for a metric are checked in turn.
Theorem: Open Ball in Euclidean Plus Metric is Subset of Equivalent Ball in Euclidean Metric

Let $\R$ be the set of real numbers.
Let $d: \R \times \R \to \R$ be the Euclidean plus metric:
:$\map d {x, y} := \size {x - y} + \ds \sum_{i \mathop = 1}^\infty 2^{-i} \map \inf {1, \size {\max_{j \mathop \le i} \frac 1 {\size {x - r_j} } - \max_{j \mathop \le i} \frac 1 {\size {y - r_j} } } }$
Let $d': \R \times \R \to \R$ be the Euclidean metric.
Let $\epsilon \in \R_{>0}$ be a (strictly) positive real number.
Let $\map {B_\epsilon} {p; d}$ be an open $\epsilon$-ball of $p$ in $\R$ on $d$.
Let $\map {B'_\epsilon} {p; d'}$ be an open $\epsilon$-ball of $p$ in $\R$ on $d'$.
Then:
:$\map {B_\epsilon} {p; d} \subseteq \map {B'_\epsilon} {p; d'}$

Proof:
Let $p \in \R$.
Let $\epsilon \in \R_{>0}$.
Let $x \in \map {B_\epsilon} {p; d}$.
Then:
{{begin-eqn}}
{{eqn | l = \map d {x, p}
      | o = <
      | r = \epsilon
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \epsilon
      | o = >
      | r = \size {x - y} + \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} } } }
      | c = 
}}
{{eqn | o = \ge
      | r = \size {x - y}
      | c = 
}}
{{eqn | r = \map {d'} {x, p}
      | c = 
}}
{{end-eqn}}
Thus:
:$x \in \map {B_\epsilon} {p; d} \implies x \in \map {B_\epsilon} {p; d'}$
and the result follows by {{Defof|Subset}}.
{{qed}}
Theorem: Irrational Number Space is Completely Metrizable

Let $\struct {\R \setminus \Q, \tau_d}$ be the irrational number space under the Euclidean topology $\tau_d$.
Then $\struct {\R \setminus \Q, \tau_d}$ is completely metrizable.

Proof:
Let $\set {r_i}$ be an enumeration of rational numbers.
Let $d: \R \times \R \to \R$ be the Euclidean plus metric:
:$\ds \map d {x, y} := \size {x - y} + \sum_{i \mathop = 1}^\infty 2^{-i} \map \inf {1, \size {\max_{j \mathop \le i} \frac 1 {\size {x - r_j} } - \max_{j \mathop \le i} \frac 1 {\size {y - r_j} } } }$
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.}}
Theorem: Irrational Number Space is Non-Meager

Let $\struct {\R \setminus \Q, \tau_d}$ be the irrational number space under the Euclidean topology $\tau_d$.
Then $\struct {\R \setminus \Q, \tau_d}$ is non-meager.

Proof:
From Irrational Number Space is Completely Metrizable, $\struct {\R \setminus \Q, d}$ 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}}
Theorem: Rational Number Space is Meager

Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Then $\struct {\Q, \tau_d}$ is meager.

Proof:
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}}
Theorem: Rational Number Space is Separable

Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Then $\struct {\Q, \tau_d}$ is separable.

Proof:
From Rational Numbers are Countably Infinite, $\Q$ is itself countable.
The result follows by Countable Space is Separable.
{{qed}}
Theorem: Underlying Set of Topological Space is Everywhere Dense

Let $T = \struct {S, \tau}$ be a topological space.
Then the underlying set $S$ of $T$ is everywhere dense in $T$.

Proof:
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
Theorem: Irrational Number Space is Separable

Let $\struct {\R \setminus \Q, \tau_d}$ be the irrational number space under the Euclidean topology $\tau_d$.
Then $\struct {\R \setminus \Q, \tau_d}$ is separable.

Proof:
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}}
Theorem: Rational Number Space is Second-Countable

Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Then $\struct {\Q, \tau_d}$ is second-countable.

Proof:
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}}
Theorem: Irrational Number Space is Second-Countable

Let $\struct {\R \setminus \Q, \tau_d}$ be the irrational number space under the Euclidean topology $\tau_d$.
Then $\struct {\R \setminus \Q, \tau_d}$ is second-countable.

Proof:
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}}