nyuuzyou/wikis huggingface dataset:
Set of Order 3 Vectors under Cross Product does not form Ring

== Theorem ==
Let $S$ be the set of all vectors in a vector space of dimension $3$.
Let $\times$ denote the cross product operation.
Then the algebraic structure $\struct {S, +, \times}$ is not a ring.
== Proof ==
For $\struct {S, +, \times}$ to be a ring, it is a necessary condition that $\struct {S, \times}$ is a semigroup.
For $\struct {S, \times}$ to be a semigroup, it is a necessary condition that $\times$ is associative on $S$.
However, from Vector Cross Product is not Associative, this is not the case here.
The result follows.
== Sources ==
* : Chapter $1$: Rings - Definitions and Examples: $2$: Some examples of rings: Some 'non-examples': $\text {(f)}$
Category:Vectors
Category:Examples of Rings
Set of Order m times n Matrices does not form Ring

== Theorem ==
Let $m, n \in \N_{>0}$ be non-zero natural numbers such that $m > n$.
Let $S$ be the set of all matrices of order $m \times n$.
Then the algebraic structure $\struct {S, +, \times}$ is not a ring.
Note that $\times$ denotes conventional matrix multiplication.
== Proof ==
For $\struct {S, +, \times}$ to be a ring, it is a necessary condition that $\struct {S, \times}$ is a semigroup.
For $\struct {S, \times}$ to be a semigroup, it is a necessary condition that $\struct {S, \times}$ is closed.
That is:
:$\forall x, y \in S: x \times y \in S$
Let $\mathbf A$ and $\mathbf B$ be elements of $S$.
The matrix multiplication operation is defined on $\mathbf A$ and $\mathbf B$ $\mathbf A$ is of order $m \times p$ and $\mathbf A$ is of order $p \times n$, for some $m, n, p \in \N_{>0}$.
That is, the second dimension of $\mathbf A$ must be equal to the first dimension of $\mathbf B$.
But in this case, the second dimension of $\mathbf A$ is $n$, and the first dimension of $\mathbf B$ is $m$.
As we have been given that $m \ne n$, it follows that matrix multiplication operation is not defined on $S$.
Hence the result.
== Sources ==
* : Chapter $1$: Rings - Definitions and Examples: $2$: Some examples of rings: Some 'non-examples': $\text {(e)}$
Category:Matrices
Category:Examples of Rings
Set of Ordered Pairs of Integers is Countable Infinite

== Theorem ==
The set of all ordered pairs of integers $\Z$ is countably infinite.
== Proof ==
The set of all ordered pairs of a set $S$ is by definition the Cartesian product $S \times S$.
In this context we are determining the cardinality of $\Z \times \Z$.
From Integers are Countably Infinite, we have that $\Z$ is a countably infinite set.
The result then follows from Cartesian Product of Countable Sets is Countable.
== Sources ==
* : Chapter $1$: Preliminaries: Exercises: $1.11$
Category:Countable Sets
Category:Integers
Set of Pairwise Disjoint Intervals is Countable

== Theorem ==
Let $X$ be a subset of $\powerset \R$ such that:
:$(1): \quad X$ is pairwise disjoint:
::::$\forall A, B \in X: A \ne B \implies A \cap B = \O$.
:$(2): \quad$ every element of $X$ contains an open interval:
::::$\forall A \in X: \exists x, y \in \R: x < y \land \openint x y \subseteq A$.
Then $X$ is countable.
== Proof ==
By Between two Real Numbers exists Rational Number:
:$\forall A \in X: \exists x, y \in \R, q \in \Q: x < y \land q \in \openint x y \subseteq A$
By the Axiom of Choice define a mapping $f: X \to \Q$:
:$\forall A \in X: \map f A \in A$
First it needs to be shown that $f$ is an injection by definition.
Let $A, B \in X$ such that:
:$\map f A = \map f B$
By definition of $f$:
:$\map f A \in A$ and $\map f B \in B$
By definition of intersection:
:$\map f A \in A \cap B$
Then by definition of empty set:
:$A \cap B \ne \O$
Thus by definition of pairwise disjoint:
:$A = B$
Hence $f$ is an injection.
By Set is Subset of Itself, $X$ is a subset of $X$.
Thus by Cardinality of Image of Injection:
:$\card X = \card {f^\to \sqbrk X}$
By definition of image:
:$f^\to \sqbrk X \subseteq \Q$
By Rational Numbers are Countably Infinite:
:$\Q$ is countable.
Hence by Subset of Countable Set is Countable:
:$f^\to \sqbrk X$ is countable.
Thus by Set is Countable if Cardinality equals Cardinality of Countable Set the result:
:$X$ is countable.
== Sources ==
* 
Category:Countable Sets
Set of Palindromes over 0, 1 forms Formal Language

== Example of Formal Language ==
The set of palindromes over $\set {0, 1}$ forms a formal language.
== Proof ==
$\set {0, 1}$ forms a set of symbols.
== Sources ==
* : Chapter $1$: Preliminaries: $1.1$ Strings, Alphabets and Languages
Category:Examples of Formal Languages
Set of Palindromes over Infinite Set does not form Formal Language

== Example of Formal Language ==
The set of palindromes over an infinite set of symbols does not form a formal language.
== Proof ==
According to the alternative definition, formal language consists of a finite set of symbols.
As the set of symbols specified here is infinite, this does not fit the criteria.
== Sources ==
* : Chapter $1$: Preliminaries: $1.1$ Strings, Alphabets and Languages
Category:Examples of Formal Languages
Set of Points at which Sequence of Measurable Functions does not Converge to Given Measurable Function is Measurable

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space. 
Let $f : X \to \R$ be a $\Sigma$-measurable function.
For each $n \in \N$, let $f_n : X \to \R$ be a $\Sigma$-measurable function.
Then:
:$\ds \set {x \in X : \sequence {\map {f_n} x}_{n \in \N} \text { does not converge to } \map f x}$ is $\Sigma$-measurable.
== Proof ==
From Expression for Set of Points at which Sequence of Functions does not Converge to Given Function, we have: 
:$\ds \set {x \in X : \sequence {\map {f_n} x}_{n \in \N} \text { does not converge to } \map f x} = \bigcup_{k \mathop = 1}^\infty \bigcap_{N \mathop = 1}^\infty \bigcup_{n \mathop = N}^\infty \set {x \in X : \size {\map {f_n} x - \map f x} \ge \frac 1 k}$
From Pointwise Difference of Measurable Functions is Measurable, we have: 
:$f_n - f$ is $\Sigma$-measurable for each $n \in \N$.
From Absolute Value of Measurable Function is Measurable, we have: 
:$\size {f_n - f}$ is $\Sigma$-measurable for each $n \in \N$.
From Characterization of Measurable Functions, we have that: 
:$\ds \set {x \in X : \size {\map {f_n} x - \map f x} \ge \frac 1 k}$ is $\Sigma$-measurable for each $k, n \in \N$.
Since $\sigma$-Algebras are closed under countable union, we have: 
:$\ds \bigcup_{n \mathop = N}^\infty \set {x \in X : \size {\map {f_n} x - \map f x} \ge \frac 1 k}$ is $\Sigma$-measurable for each $k, N \in \N$.
Then, from Sigma-Algebra Closed under Countable Intersection: 
:$\ds \bigcap_{N \mathop = 1}^\infty \bigcup_{n \mathop = N}^\infty \set {x \in X : \size {\map {f_n} x - \map f x} \ge \frac 1 k}$ is $\Sigma$-measurable for each $k \in \N$.
Again, since Since $\sigma$-Algebras are closed under countable union, we have: 
:$\ds \bigcup_{k \mathop = 1}^\infty \bigcap_{N \mathop = 1}^\infty \bigcup_{n \mathop = N}^\infty \set {x \in X : \size {\map {f_n} x - \map f x} \ge \frac 1 k}$ is $\Sigma$-measurable.
So:
:$\ds \set {x \in X : \sequence {\map {f_n} x}_{n \in \N} \text { does not converge to } \map f x}$ is $\Sigma$-measurable.
Category:Measurable Functions
Category:Convergence
Set of Points for which Measurable Function is Real-Valued is Measurable

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f: X \to \overline \R$ be a $\Sigma$-measurable.
Then: 
:$\set {x \in X : \map f x \in \R}$ is $\Sigma$-measurable.
=== Corollary ===
== Proof ==
Since $f$ is $\Sigma$-measurable, we have that: 
:for all $n \in \N$ the set $\set {x \in X : \map f x \le n}$ is $\Sigma$-measurable
and:
:for all $n \in \N$ the set $\set {x \in X : -n \le \map f x}$ is $\Sigma$-measurable.
From $\sigma$-Algebra Closed under Countable Intersection, we have: 
:$\set {x \in X : -n \le \map f x \le n} = \set {x \in X : \map f x \le n} \cap \set {x \in X : -n \le \map f x}$ is $\Sigma$-measurable.
Since $\sigma$-algebras are closed under countable union, we also have: 
:$\ds \bigcup_{n \mathop = 1}^\infty \set {x \in X : -n \le \map f x \le n}$ is $\Sigma$-measurable.
We will finally show that: 
:$\ds \set {x \in X : \map f x \in \R} = \bigcup_{n \mathop = 1}^\infty \set {x \in X : -n \le \map f x \le n}$
which gives the claim.
If $x \in X$ has $\map f x \in \R$, then we have: 
:$-\paren {\floor {\size {\map f x} } + 1} \le \map f x \le \floor {\size {\map f x} } + 1$
with:
:$\floor {\size {\map f x} } + 1 \in \N$
So:
:$\ds x \in \bigcup_{n \mathop = 1}^\infty \set {x \in X : -n \le \map f x \le n}$
Now, let: 
:$\ds x \in \bigcup_{n \mathop = 1}^\infty \set {x \in X : -n \le \map f x \le n}$
Then: 
:$-n \le \map f x \le n$
for some $n \in \N$, so certainly: 
:$\map f x \in \R$
So we have: 
:$\ds \set {x \in X : \map f x \in \R} = \bigcup_{n \mathop = 1}^\infty \set {x \in X : -n \le \map f x \le n}$
and hence the claim.
Category:Measurable Sets
Category:Measurable Functions
Category:Set of Points for which Measurable Function is Real-Valued is Measurable
Set of Points for which Measurable Function is Real-Valued is Measurable/Corollary

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f: X \to \overline \R$ be a $\Sigma$-measurable.
Then: 
:$\set {x \in X : \size {\map f x} = +\infty}$ is $\Sigma$-measurable. 
== Proof ==
We have: 
:$\set {x \in X : \size {\map f x} = +\infty} = X \setminus \set {x \in X : \map f x \in \R}$
From Set of Points for which Measurable Function is Real-Valued is Measurable, we have: 
:$\set {x \in X : \map f x \in \R}$ is $\Sigma$-measurable.
Since $\sigma$-algebras are closed under complementation, we have that: 
:$\set {x \in X : \size {\map f x} = +\infty}$ is $\Sigma$-measurable.
Category:Set of Points for which Measurable Function is Real-Valued is Measurable
Set of Points for which Seminorm is Zero is Vector Subspace

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $X$ be a vector space over $\GF$.
Let $p$ be a seminorm on $X$.
Let:
:$U = \set {x \in X : \map p x = 0}$
Then $U$ is a vector subspace of $X$. 
== Proof ==
From Seminorm Maps Zero Vector to Zero, $\map p {\mathbf 0_X} = 0$.
So $\mathbf 0_X \in U$ and in particular $U \ne \O$.
So we look to apply One-Step Vector Subspace Test.
Let $x, y \in U$ and $\lambda, \mu \in \GF$.
Then we have: 
Since $\map p {\lambda x + \mu y} \ge 0$, so:
:$\map p {\lambda x + \mu y} = 0$
so $\lambda x + \mu y \in U$.
By One-Step Vector Subspace Test, we have that $U$ is a vector subspace of $X$. 
== Sources ==
* : $1.34$: Theorem
Category:Seminorms
Category:Vector Subspaces
Set of Points on Line Segment is Infinite

== Theorem ==
The set of points on a line segment is infinite.
== Proof ==
Let $S$ denote the set of points on a line segment.
 $S$ is finite.
Then there exists $n \in \N$ such that $S$ has $n$ elements.
Let $s_1$ and $s_2$ be two arbitrary adjacent points in $S$.
That is, such that there are no points in $S$ between $s_1$ and $s_2$.
But there exists (at least) one point on the line segment between $s_1$ and $s_2$ which is not in $S$.
Hence there must be more than $n$ elements of $S$.
From that contradiction it follows by Proof by Contradiction that $S$ is not finite.
== Sources ==
* : Chapter $1$: General Background: $\S 1$ What is infinity?
Category:Lines
Category:Infinite Sets
Set of Polynomials over Infinite Set has Same Cardinality

== Theorem ==
Let $S$ be a set of infinite cardinality $\kappa$.
Let $S \sqbrk x$ be the set of polynomial forms over $S$ in the indeterminate $x$.
Then $S \sqbrk x$ has cardinality $\kappa$.
== Proof ==
Since $S \sqbrk x$ contains a copy of $S$ as constant polynomials, we have an injection $S \to S \sqbrk x$.
We define an injection from $S \sqbrk x$ to the set $\FF$ of finite sequences over $S$ as follows:
Each polynomial in $f \in S \sqbrk x$ is of the form:
:$f = a_0 + a_1 x + a_2 x^2 + \dotsb + a_n x^n$
where $a_n$ is non-zero and each $a_i$ is in $S$.
We send each polynomial $f$ to the sequence of its coefficients $\sequence {a_0, \dotsc, a_n}$.
By the definition of equality of polynomials, this is injective.
Now the set of finite sequences over $S$ is a countable union of sets of cardinality $\kappa$.
From Cardinality of Infinite Union of Infinite Sets, $\FF$ has cardinality $\kappa$.
Therefore there is a bijection $\FF \leftrightarrow S$.
Composing this with the injection $S \sqbrk x \to \FF$, we have an injection $S \sqbrk x \to S$.
So by the Cantor-Bernstein-Schröder Theorem there is a bijection $S \sqbrk x \leftrightarrow S$.
Hence, we have:
:$\card {S \sqbrk x} = \card S = \kappa$
Category:Polynomial Theory
Set of Polynomials over Integral Domain is Subring

== Theorem ==
Let $\struct {R, +, \circ}$ be a commutative ring.
Let $\struct {D, +, \circ}$ be an integral subdomain of $R$.
Then $\forall x \in R$, the set $D \sqbrk x$ of polynomials in $x$ over $D$ is a subring of $R$.
== Proof ==
By application of the Subring Test:
As $D$ is an integral domain, it has a unity $1_D$ and so $x = 1_D x$.
Hence $x \in D \sqbrk x$ and so $D \sqbrk x \ne \O$.
Let $p, q \in D \sqbrk x$.
Then let:
:$\ds p = \sum_{k \mathop = 0}^m a_k \circ x^k, q = \sum_{k \mathop = 0}^n b_k \circ x^k$
Thus:
:$\ds -q = -\sum_{k \mathop = 0}^n b_k \circ x^k = \sum_{k \mathop = 0}^n \paren {-b_k} \circ x^k$
and so:
:$q \in D \sqbrk x$
Thus as Polynomials Closed under Addition‎, it follows that:
:$p + \paren {-q} \in D \sqbrk x$
Finally, from Polynomials Closed under Ring Product, we have that $p \circ q \in D \sqbrk x$.
All the criteria of the Subring Test are satisfied.
Hence the result.
== Sources ==
* : $\S 64.1$ Polynomial rings over an integral domain
Category:Polynomial Theory
Category:Subrings
Set of Positive Elements of C*-Algebra is Closed

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra.
Let $A^+$ be the set of positive elements of $A$.
Then $A^+$ is closed.
== Proof ==
Let $\struct {A_+, \ast, \norm {\, \cdot \,}_\ast}$ be the unitization of $A$. 
Let ${\mathbf 1}_+$ be the identity element of $A_+$. 
Let $\iota : A \to A_+$ be the mapping defined by:
:$\map \iota a = \tuple {a, 0}$
for each $a \in A$. 
Let $\sequence {a_n}_{n \mathop \in \N}$ be a sequence in $A^+$ such that $a_n \to a$ in $\struct {A, \norm {\, \cdot \,} }$.
From Convergence in Direct Product Norm, $\sequence {\tuple {a_n, 0} }_{n \mathop \in \N}$ converges to $\struct {a, 0}$ in $\struct {A_+, \norm {\, \cdot \,}_\ast}$. 
From Convergent Sequence in Normed Vector Space is Bounded, there exists $t \ge 0$ such that:
:$\norm {\tuple {a_n, 0} } \le t$
From Element of C*-Algebra is Positive iff Positive in Unitization, $\tuple {a_n, 0}$ is positive for each $n \in \N$.
Hence from Characterization of Positive Element of Unital C*-Algebra, we must have:
:$\norm {\tuple {a_n, 0} - t {\mathbf 1}_+} \le t$ 
Taking $n \to \infty$ we obtain:
:$\norm {\tuple {a, 0} - t {\mathbf 1}_+} \le t$ 
from Norm is Continuous.
Hence by Characterization of Positive Element of Unital C*-Algebra, $\tuple {a, 0}$ is positive in $A_+$.
Hence by Element of C*-Algebra is Positive iff Positive in Unitization, $a \in A^+$. 
Hence by the definition of a closed set in a normed vector space, $A^+$ is closed.
== Sources ==
* : $2.2$: Positive Elements of $C^\ast$-Algebras
Category:Positive Elements of C*-Algebras
Set of Positive Elements of C*-Algebra is Convex Cone

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra. 
Let $A^+$ be the set of positive elements of $A$.
Then $A^+$ is a convex cone.
== Proof ==
From Non-Negative Multiple of Positive Element of C*-Algebra is Positive, if $x \in A^+$ and $\alpha \in \R_{\ge 0}$, we have $\alpha x \in A^+$. 
Hence $A^+$ is a cone.
From Sum of Two Positive Elements of C*-Algebra is Positive, we have $x + y \in A^+$ for $x, y \in A^+$.
Hence $A^+$ is a convex cone.
Category:Positive Elements of C*-Algebras
Set of Positive Elements of C*-Algebra is Set of Products of Element with its Star

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra.
Let $a \in A$. 
Let $A^+$ be the set of positive elements of $A$.
We have:
:$A^+ = \set {a^\ast a : a \in A}$
== Proof ==
From Product of Element of C*-Algebra with its Star is Positive, we have:
:$\set {a^\ast a : a \in A} \subseteq A^+$
Conversely suppose that $x \in A^+$.
From Existence and Uniqueness of Positive Nth Root of Positive Element of C*-Algebra, there exists $y \in A^+$ such that $x = y^2$.
Since $y$ is Hermitian, we have $x = y^\ast y$.
So $x \in \set {a^\ast a : a \in A}$.
Hence:
:$A^+ = \set {a^\ast a : a \in A}$
== Sources ==
* : $2.2$: Positive Elements of $C^\ast$-Algebras
Category:Positive Elements of C*-Algebras
Set of Positive Elements of Everywhere Dense Ideal is Dense in Set of Positive Elements of C*-Algebra

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra.
Let $I$ be an everywhere dense ideal of $A$.
Let $A^+$ be the set of positive elements of $A$.
Let $I^+ = I \cap A^+$.
Then $I^+$ is everywhere dense in $A^+$. 
== Proof ==
Let $x \in A^+$. 
From Existence and Uniqueness of Positive Nth Root of Positive Element of C*-Algebra, there exists a positive $y \in A$ such that $x = y^2$.
Since $I$ is everywhere dense in $A$, there exists a sequence $\sequence {y_n}_{n \mathop \in \N}$ in $I$ such that $y_n \to y$. 
From Product of Element of C*-Algebra with its Star is Positive, we have that $y_n^\ast y_n$ is positive for each $n \in \N$. 
Further since $I$ is an ideal, we have $y_n^\ast y_n \in I$ for each $n \in \N$. 
Hence $y_n^\ast y_n \in I^+$ for each $n \in \N$. 
From Multiplication on Normed Algebra is Continuous and since $\ast$ is an isometry, we have $y_n^\ast y_n \to y^\ast y = y^2 = x$ as $n \to \infty$. 
Hence $\sequence {y_n^\ast y_n}_{n \mathop \in \N}$ is a sequence in $I^+$ converging to $x$. 
Hence $I_+$ is everywhere dense in $A^+$.
Category:Positive Elements of C*-Algebras
Set of Positive Integers does not form Ring

== Theorem ==
Let $\Z_{\ge 0}$ denote the set of positive integers.
Then the algebraic structure $\struct {\Z_{\ge 0}, +, \times}$ does not form a ring.
== Proof ==
For $\struct {\Z_{\ge 0}, +, \times}$ to be a ring, it is necessary for the algebraic structure $\struct {\Z_{\ge 0}, +}$ to form a group.
But from the corollary to Natural Numbers under Addition do not form Group:
:$\struct {\Z_{\ge 0}, +}$ is not a group.
== Sources ==
* : Chapter $1$: Rings - Definitions and Examples: $2$: Some examples of rings: Some 'non-examples': $\text {(a)}$
Category:Integers
Category:Examples of Rings
Set of Prime Numbers is Primitive Recursive

== Theorem ==
The set $\Bbb P$ of prime numbers is primitive recursive.
== Proof ==
A prime number is defined as an element of $\N$ with exactly two positive divisors.
So, we have that $n > 0$ is prime $\map \tau n = 2$, where $\tau: \N \to \N$ is the divisor count function.
Thus we can define the characteristic function of the set of prime numbers $\Bbb P$ as:
:$\forall n > 0: \map {\chi_\Bbb P} n := \map {\chi_{\operatorname {eq} } } {\map \tau n, 2}$
Now we let $g: \N^2 \to \N$ be the function given by:
:$\map g {n, z} = \begin{cases}
0 & : z = 0 \\
\ds \sum_{y \mathop = 1}^z \map {\operatorname {div} } {n, y} & : z > 0
\end{cases}$
As:
:$\operatorname {div}$ is primitive recursive
:$\ds \sum_{y \mathop = 1}^z$ is primitive recursive
it follows that $g$ is primitive recursive.
Then for $n > 0$:
:$\ds \map g {n, n} = \sum_{y \mathop = 1}^n \map {\operatorname {div} } {n, y} = \map \tau n$
and from Divisor Count Function is Primitive Recursive we have that $g$ is primitive recursive.
Then let $h: \N \to \N$ be the function defined as:
:$\map h n = \map g {n, n}$
which is also primitive recursive.
So we have, for all $n \in \N$:
:$\map {\chi_\Bbb P} n = \map {\chi_{\operatorname {eq} } } {\map h n, 2}$
Hence the result.
Category:Primitive Recursive Functions
Set of Rational Cuts forms Ordered Field

== Theorem ==
Let $\RR$ denote the set of rational cuts.
Let $\struct {\RR, +, \times, \le}$ denote the ordered structure formed from $\RR$ and:
:the operation $+$ of addition of cuts
:the operation $\times$ of multiplication of cuts
:the ordering $\le$ of cuts.
Then $\struct {\RR, + \times, \le}$ is an ordered field.
== Proof ==
We demonstrate that $\struct {\RR, +, \times}$ is a field by showing it is a subfield of the structure $\struct {\CC, +, \times}$, where $\CC$ denotes the set of all cuts.
We do this by establishing that all $4$ criteria of the Subfield Test are satisfied.
We note that $0^* \in \RR$, where $0^*$ is the rational cut associated with the (rational) number $0$:
:$0^* = \set {r \in \Q: r < 0}$
So $\RR \ne \O$.
Thus criterion $(1)$ is satisfied.
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Real Numbers
Category:Cuts
Category:Totally Ordered Fields
Set of Rational Numbers Strictly between Zero and One has no Greatest or Least Element

== Theorem ==
Let $S \subseteq \Q$ be the subset of the set of rational numbers defined as:
:$S = \set {r \in \Q: 0 < r < 1}$
Then $S$ has no greatest or smallest element.
However, $S$ has a supremum $1$ and an infimum $0$.
== Proof ==
We have that:
:$\forall r \in S: 0 < r$
and:
:$\forall r \in S: r < 1$
Hence $0$ and $1$ are lower and upper bounds of $S$ respectively.
Let $s \in S$.
Then $s \in \Q: 0 < s < 1$.
 $s$ is the greatest element of $S$.
But then we have:
:$0 < s < \dfrac {s + 1} 2 < 1$
and so $\dfrac {s + 1} 2 \in S$ but $s < \dfrac {s + 1} 2$.
This contradicts our assertion that $s$ is the greatest element of $S$.
 $s$ is the smallest element of $S$.
But then we have:
:$0 < \dfrac s 2 < s < 1$
and so $\dfrac s 2 \in S$ but $\dfrac s 2 < s$.
This contradicts our assertion that $s$ is the smallest element of $S$.
So $S$ cannot have either a greatest or smallest element.
Let $H$ be such that $0 < H < 1$.
Suppose $H$ is a smaller upper bound of $S$ than $1$.
Then from Between two Real Numbers exists Rational Number there exists a rational number $r$ such that $H < r < 1$.
But then $r$ is a rational number between $0$ and $1$ which is greater than $H$.
Hence $H$ cannot be an upper bound of $S$.
Thus, by definition, $1$ is the supremum of $S$
Suppose $H$ is a greater lower bound of $S$ than $0$.
Then from Between two Real Numbers exists Rational Number there exists a rational number $r$ such that $0 < r < H$.
But then $r$ is a rational number between $0$ and $1$ which is smaller than $H$.
Hence $H$ cannot be a lower bound of $S$.
Thus, by definition, $0$ is the infimum of $S$.
== Sources ==
* : $\S 3$: Natural Numbers: Exercise $\S 3.6 \ (5)$
Category:Rational Numbers
Category:Infima
Category:Suprema
Set of Rational Numbers is not Closed in Reals

== Theorem ==
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 $\R$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $30$. The Rational Numbers: $2$
Category:Rational Numbers
Category:Real Number Line with Euclidean Topology