nyuuzyou/wikis huggingface dataset:
Real Number Line with Point Removed is Not Path-Connected

== Theorem ==
Let $\R$ be the real number line considered as an Euclidean space.
Let $x \in \R$ be a real number.
Then $\R \setminus \set x$, where $\setminus$ denotes set difference, is not path-connected.
== Proof ==
We have that $x - 1$ and $x + 1$ are both real numbers, so:
:$x - 1 \in \R \setminus \set x$
:$x + 1 \in \R \setminus \set x$
Let $\mathbb I := \closedint 0 1$ be the closed unit interval.
Suppose there exists a path $f: \mathbb I \to \R \setminus \set x$ from $x - 1$ to $x + 1$.
Then by Image of Interval by Continuous Function is Interval, it follows that $x = \map f x$ for some $s \in \mathbb I$.
But $x \notin \R \setminus \set x$ by definition of set difference.
Hence such an $f$ can not exist.
The result follows by definition of path-connected.
== Sources ==
* : Chapter $\text{III}$: Metric Spaces: Path-Connectedness
Category:Path-Connected Spaces
Category:Analysis
Real Number Multiplied by Complex Number

== Theorem ==
Let $a \in \R$ be a real number.
Let $c + d i \in \C$ be a complex number.
Then:
:$a \times \paren {c + d i} = \paren {c + d i} \times a = a c + i a d$
== Proof ==
$a$ can be expressed as a wholly real complex number $a + 0 i$.
Then we have:
The result for $\paren {c + d i} \times a$ follows from Complex Multiplication is Commutative.
== Sources ==
* : $\S 1.2$. The Algebraic Theory
Category:Complex Multiplication
Real Number Ordering is Compatible with Addition

== Theorem ==
:$\forall a, b, c \in \R: a < b \implies a + c < b + c$
where $\R$ is the set of real numbers.
== Proof ==
From Real Numbers form Ordered Integral Domain, $\struct {\R, +, \times, \le}$ forms an ordered integral domain.
By definition of ordered integral domain, the usual ordering $\le$ is compatible with ring addition.
== Examples ==
== Sources ==
* : Introduction: Review of Algebra, Geometry, and Trigonometry: $\text{0-2}$: Inequalities
* : Chapter $1$: Inequalities
* : $\S 1$: Real Numbers: $\S 1.4$: Inequalities: $\text{(II)}$
* 
* : $(3)$
* 
* : $(3)$
Category:Real Number Ordering is Compatible with Addition
Category:Real Addition
Category:Inequalities
Real Number Ordering is Compatible with Addition/Examples/15+3 Greater than 12+3

== Examples of Real Number Ordering is Compatible with Addition ==
We have that:
:$15 > 12$
so by Real Number Ordering is Compatible with Addition:
:$15 + 3 > 12 + 3$
That is:
:$18 > 15$
== Sources ==
* : Chapter $1$: Inequalities
Category:Real Number Ordering is Compatible with Addition
Real Number Ordering is Compatible with Addition/Examples/15-3 Greater than 12-3

== Examples of Real Number Ordering is Compatible with Addition ==
We have that:
:$15 > 12$
so by Real Number Ordering is Compatible with Addition:
:$15 - 3 > 12 - 3$
That is:
:$12 > 9$
== Sources ==
* : Chapter $1$: Inequalities
Category:Real Number Ordering is Compatible with Addition
Real Number Ordering is Compatible with Multiplication/Examples

== Examples of Real Number Ordering is Compatible with Multiplication ==
=== $15 \times 3$ Greater than $12 \times 3$ ===
=== $\dfrac {15} 3$ Greater than $\dfrac {12} 3$ ===
=== $15 \times \paren {-3}$ Less than $12 \times \paren {-3}$ ===
=== $\dfrac {15} {-3}$ Less than $\dfrac {12} {-3}$ ===
Category:Real Number Ordering is Compatible with Multiplication
Real Number Ordering is Compatible with Multiplication/Examples/15 over -3 Less than 12 over -3

== Examples of Real Number Ordering is Compatible with Multiplication ==
We have that:
:$15 > 12$
so by Real Number Ordering is Compatible with Multiplication:
:$\dfrac {15} {-3} < \dfrac {12} {-3}$
That is:
:$-5 < -4$
== Sources ==
* : Chapter $1$: Inequalities
Category:Real Number Ordering is Compatible with Multiplication
Real Number Ordering is Compatible with Multiplication/Examples/15 over 3 Greater than 12 over 3

== Examples of Real Number Ordering is Compatible with Multiplication ==
We have that:
:$15 > 12$
so by Real Number Ordering is Compatible with Multiplication:
:$\dfrac {15} 3 > \dfrac {12} 3$
That is:
:$5 > 4$
== Sources ==
* : Chapter $1$: Inequalities
Category:Real Number Ordering is Compatible with Multiplication
Real Number Ordering is Compatible with Multiplication/Examples/15 x -3 Less than 12 x -3

== Examples of Real Number Ordering is Compatible with Multiplication ==
We have that:
:$15 > 12$
so by Real Number Ordering is Compatible with Multiplication:
:$15 \times \paren {-3} < 12 \times \paren {-3}$
That is:
:$-45 < -36$
== Sources ==
* : Chapter $1$: Inequalities
Category:Real Number Ordering is Compatible with Multiplication
Real Number Ordering is Compatible with Multiplication/Examples/15 x 3 Greater than 12 x 3

== Examples of Real Number Ordering is Compatible with Multiplication ==
We have that:
:$15 > 12$
so by Real Number Ordering is Compatible with Multiplication:
:$15 \times 3 > 12 \times 3$
That is:
:$45 > 36$
== Sources ==
* : Chapter $1$: Inequalities
Category:Real Number Ordering is Compatible with Multiplication
Real Number Ordering is Compatible with Multiplication/Negative Factor

== Theorem ==
:$\forall a, b, c \in \R: a < b \land c < 0 \implies a c > b c$
where $\R$ is the set of real numbers.
== Proof ==
From Real Numbers form Ordered Integral Domain, $\struct {\R, +, \times, \le}$ forms an ordered integral domain.
Thus:
== Examples ==
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: $\text{1-3}$ Order properties of real numbers
* : Introduction: Review of Algebra, Geometry, and Trigonometry: $\text{0-2}$: Inequalities
* : Chapter $1$: Inequalities
* : $\S 1$: Real Numbers: $\S 1.4$: Inequalities: $\text{(IV)}$
* 
* 
* : $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $2 \ \text{(e)}$
* 
* 
Category:Real Number Ordering is Compatible with Multiplication
Real Number Ordering is Compatible with Multiplication/Positive Factor

== Theorem ==
:$\forall a, b, c \in \R: a < b \land c > 0 \implies a c < b c$
where $\R$ is the set of real numbers.
=== Corollary ===
== Proof ==
From Real Numbers form Ordered Integral Domain, $\struct {\R, +, \times, \le}$ forms an ordered integral domain.
Thus:
== Examples ==
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: $\text{1-3}$ Order properties of real numbers
* : Introduction: Review of Algebra, Geometry, and Trigonometry: $\text{0-2}$: Inequalities
* : Chapter $1$: Inequalities
* : $\S 1$: Real Numbers: $\S 1.4$: Inequalities: $\text{(III)}$
* 
* : $(4)$
* 
* : $(4)$
Category:Real Number Ordering is Compatible with Multiplication
Real Number Ordering is Compatible with Multiplication/Positive Factor/Corollary

== Theorem ==
:$\forall a, b, c, d \in \R: 0 < a < b \land 0 < c < d \implies a c < b d$
where $\R$ is the set of real numbers.
== Proof ==
The result follows by the Transitive Law.
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: $\text{1-3}$ Order properties of real numbers
Category:Real Number Ordering is Compatible with Multiplication
Real Number Subtracted from Itself leaves Zero

== Theorem ==
Let $x \in \R$ be a real number.
Then:
:$x - x = 0$
where $x - x$ denotes the operation of real subtraction.
== Proof ==
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: $\text{1-2}$ Arithmetical properties of real numbers: Axiom $4$
Category:Subtraction
Real Number at Distance Zero from Closed Real Interval is In Interval

== Theorem ==
Let $S$ be a subset of the set of real numbers $\R$.
Let $x \in \R$ be a real number.
Let $\map d {x, S}$ be the distance between $x$ and $S$.
Let $I \subseteq \R$ be a closed real interval.
Then:
:$\map d {x, I} = 0 \implies x \in I$
== Proof ==
From the definition of distance:
:$\forall x, y \in \R: \map d {x, y} = \size {x - y}$
Thus:
:$\ds \map d {x, S} = \map {\inf_{y \mathop \in S} } {\size {x - y} }$
Because $I$ is an interval, if $x \notin I$ then $x$ is either an upper bound or a lower bound for $I$.
Suppose $x$ is an upper bound for $I$.
Let $B$ be the supremum of $I$.
Then because $I$ is closed, $B \in I$.
So:
Now from Infimum Plus Constant:
:$\inf_{y \mathop \in S} \size {x - y} = x - B + \inf_{y \mathop \in S} \size {B - y}$
But we also have:
:$x - B \ge 0$
:$\map d {B, S} \ge 0$
:$\map d {x, S} = 0$
So it follows that $x = B$ and so $x \in I$.
A similar argument applies if $x$ is a lower bound for $I$.
== Sources ==
* : $\S 2$: Continuum Property: Exercise $\S 2.13 \ (5) \ \text {(iii)}$
Category:Distance Function
Real Number between Zero and One is Greater than Power/Natural Number

== Theorem ==
Let $x \in \R$.
Let $0 < x < 1$.
Let $n$ be a natural number.
Then:
: $0 < x^n \le x$
== Proof 1 ==
== Proof 2 ==
Category:Powers
Category:Real Numbers
Category:Inequalities
Category:Real Number between Zero and One is Greater than Power
Real Number between Zero and One is Greater than Power/Natural Number/Proof 1

== Theorem ==
== Proof ==
For all $n \in \N$, let $\map P n$ be the proposition:
:$0 < x < 1 \implies 0 < x^n \le x$
=== Basis for the Induction ===
$\map P 1$ is true, since $0 < x < 1 \implies 0 < x^1 \le x$ by definition of exponent of $1$.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$0 < x < 1 \implies 0 < x^k \le x$
Then we need to show:
:$0 < x < 1 \implies 0 < x^{k + 1} \le x$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \N: 0 < x < 1 \implies 0 < x^n \le x$
Hence the result.
Category:Real Number between Zero and One is Greater than Power
Real Number between Zero and One is Greater than Square

== Theorem ==
Let $x \in \R$.
Let $0 < x < 1$.
Then:
:$0 < x^2 < x$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 1$: Real Numbers: Exercise $\S 1.8 \ (1) \ \text{(i)}$
Category:Real Number between Zero and One is Greater than Square
Category:Square Function
Category:Real Analysis
Category:Inequalities
Real Number is Ceiling minus Difference

== Theorem ==
Let $x \in \R$ be a real number.
Let $\ceiling x$ be the ceiling of $x$.
Let $n$ be a integer.
:$(1): \quad$ There exists $t \in \hointr 0 1$ such that $x = n - t$
:$(2): \quad n = \ceiling x$
== Proof ==
=== 1 implies 2 ===
Let $x = n - t$, where $t \in \hointr 0 1$.
Because $0 \le t < 1$, we have:
:$0 \leq n - x < 1$
Thus:
:$n - 1 < x \le n$
That is, $n$ is the ceiling of $x$.
=== 2 implies 1 ===
Now let $n = \ceiling x$.
Let $t = \ceiling x - x$.
Then $x = n - t$.
From Ceiling minus Real Number, $t = \ceiling x - x \in \hointr 0 1$
== Also see ==
* Real Number is Floor plus Difference
* Definition:Fractional Part
Category:Ceiling Function
Real Number is Closed in Real Number Line

== Theorem ==
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 $\R$.
Category:Real Number Line with Euclidean Topology
Category:Closed Sets
Real Number is Floor plus Difference

== Theorem ==
Let $x \in \R$ be a real number.
Let $\floor x$ be the floor of $x$.
Then:
:There exists an integer $n \in \Z$ such that for some $t \in \hointr 0 1$:
::$x = n + t$
:
:$n = \floor x$
where:
: $\hointr 0 1$ is the real interval half open on the right from $0$ to $1$
: $\floor x$ is the floor of $x$
== Proof ==
=== Sufficient Condition ===
Let there exist $n \in \Z$ such that $x = n + t$, where $t \in \hointr 0 1$.
We have that $1 - t > 0$.
Thus:
:$0 \le x - n < 1$
Thus:
: $n \le x < n + 1$
That is, $n$ is the floor of $x$.
=== Necessary Condition ===
Let $n = \floor x$.
Let $t = x - \floor x$.
Then $x = n + t$.
From Real Number minus Floor:
:$t = x - \floor x \in \hointr 0 1$
and so:
:$x = n + t: t \in \hointr 0 1$
== Also see ==
* Real Number is Ceiling minus Difference
* Definition:Fractional Part
== Sources ==
* : $\S 10.4 \ \text{(iii)}$: The well-ordering principle
Category:Floor Function
Real Number is Greater than Zero iff its Negative is Less than Zero

== Theorem ==
:$\forall x \in \R: x > 0 \iff \paren {-x} < 0$
== Proof ==
Let $x > 0$.
Let $\paren {-x} < 0$.
== Sources ==
* : $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $2 \ \text{(c)}$
Category:Real Numbers
Real Number is Integer iff equals Ceiling

== Theorem ==
Let $x \in \R$.
Then:
:$x = \ceiling x \iff x \in \Z$
where $\ceiling x$ is the ceiling of $x$.
== Proof ==
Let $x = \ceiling x$.
As $\ceiling x \in \Z$, then so must $x$ be.
Now let $x \in \Z$.
We have:
:$\ceiling x = \inf \set {m \in \Z: m \ge x}$
As $x \in \inf \set {m \in \Z: m \ge x}$, and there can be no lesser $n \in \Z$ such that $n \in \inf \set {m \in \Z: m \ge x}$, it follows that:
:$x = \ceiling x$
Category:Ceiling Function
Real Number is Integer iff equals Floor

== Theorem ==
Let $x \in \R$.
Then:
:$x = \floor x \iff x \in \Z$
where $\floor x$ is the floor of $x$.
== Proof ==
Let $x = \floor x$.
As $\floor x \in \Z$, then so must $x$ be.
Now let $x \in \Z$.
We have:
:$\floor x = \sup \set {m \in \Z: m \le x}$
As $x \in \sup \set {m \in \Z: m \le x}$, and there can be no greater $n \in \Z$ such that $n \in \sup \set {m \in \Z: m \le x}$, it follows that:
:$x = \floor x$
== Also see ==
* Real Number is Integer iff equals Ceiling
Category:Floor Function
Real Number is Limit Point of Rational Numbers in Real Numbers

== Theorem ==
Let $\R$ be the set of real numbers.
Let $\Q$ be the set of rational numbers.
Let $x \in \R$.
Then $x$ is a limit point of $\Q$.
Hence the interesting case of a $\Q$ is countable set $\Q$ whose set of limit points is uncountable.
== Proof ==
Let $\epsilon \in \R_{>0}$ be arbitrary.
Consider the open interval $\openint x {x + \epsilon}$.
Then by Rationals are Everywhere Dense in Topological Space of Reals, there exists a rational number in $\openint x {x + \epsilon}$.
Hence the result by definition of limit point.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Definition $3.7.10 \ \text {(c)}$
Category:Examples of Limit Points
Real Number is between Ceiling Functions

== Theorem ==
:$\forall x \in \R: \ceiling {x - 1} \le x < \ceiling x$
where $\ceiling x$ is the ceiling of $x$.
== Proof ==
$\ceiling x$ is defined as:
:$\ceiling x = \map \inf {\set {m \in \Z: m \ge x} }$
So $\ceiling x \ge x$ by definition.
Now $\ceiling {x - 1} < \ceiling x$, so by the definition of the infimum:
:$\ceiling {x - 1} > x$
The result follows.
Category:Ceiling Function
Real Number is between Floor Functions

== Theorem ==
:$\forall x \in \R: \floor x \le x < \floor {x + 1}$
where $\floor x$ is the floor of $x$.
== Proof ==
$\floor x$ is defined as:
:$\floor x = \sup \set {m \in \Z: m \le x}$
So $\floor x \le x$ by definition.
From Floor plus One:
:$\floor {x + 1} > \floor x$
Hence by the definition of the supremum:
:$\floor {x + 1} > x$
The result follows.
Category:Floor Function
Real Number is not necessarily Rational Number

== Theorem ==
Let $x$ be a real number.
Then it is not necessarily the case that $x$ is also a rational number.
== Proof ==
By Proof by Counterexample:
Let $x = \sqrt 2$.
From Square Root of 2 is Irrational:
:$\sqrt 2$ is an irrational number.
By definition:
:$x \in \R \setminus \Q$
where:
:$\R$ is the set of real numbers
:$\Q$ is the set of rational numbers
:$\setminus$ denotes the set difference.
Thus $x$, while being a real number, is not also a rational number.
Hence the result.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 1$: The Language of Set Theory
Category:Real Numbers
Category:Rational Numbers
Real Number lies between Unique Pair of Consecutive Integers

== Theorem ==
Let $x$ be a real number.
=== Right-open Interval ===
There exists a unique integer $n\in\Z$ such that:
:$n \leq x < n + 1$
=== Left-open Interval ===
There exists a unique integer $n\in\Z$ such that:
:$n - 1 < x \leq n$
== Proof ==
=== Existence ===
By Set of Integers Bounded Above by Real Number has Greatest Element, the set:
:$S = \left\{{m \in \Z: m \le x}\right\}$
has a greatest element, say $n$.
Because $n+1>n$, $n+1\notin S$.
Thus $n+1> x$.
Thus $n\leq x < n+1$.
=== Uniqueness ===
Let $n\in\Z$ be such that:
:$n \leq x < n + 1$
We show that $n$ is a greatest element of the set:
:$S = \left\{{m \in \Z: m \le x}\right\}$
so that the uniqueness follows from Greatest Element is Unique.
Because $n\leq x$, we have $n \in S$.
Let $m \in S$.
Because $m \leq x < n+1$, $n+1 > m$.
By Weak Inequality of Integers iff Strict Inequality with Integer plus One:
:$n \geq m$.
Because $m$ was arbitrary, $n$ is a greatest element of $S$.
== Also see ==
* Definition:Floor Function
* Definition:Ceiling Function
Category:Integers
Category:Floor Function
Real Number minus Floor

== Theorem ==
Let $x \in \R$ be any real number.
Then:
:$x - \floor x \in \hointr 0 1$
where $\floor x$ is the floor of $x$.
That is:
: $0 \le x - \floor x < 1$
== Proof ==
== Also denoted as ==
The expression $x - \floor x$ is sometimes denoted $\fractpart x$ and called the fractional part of $x$.
== Also see ==
* Definition:Fractional Part
* Definition:Modulo 1: $x \bmod 1 = x - \floor x$
== Sources ==
* : $\S 10.4 \ \text{(ii)}$: The well-ordering principle
Category:Floor Function
Real Number to Negative Power/Integer

== Theorem ==
Let $r \in \R_{> 0}$ be a positive real number.
Let $n \in \Z$ be an integer.
Let $r^n$ be defined as $r$ to the power of $n$.
Then:
:$r^{-n} = \dfrac 1 {r^n}$
== Proof ==
Let $n \in \Z_{\ge 0}$.
Then from Real Number to Negative Power: Positive Integer:
:$r^{-n} = \dfrac 1 {r^n}$
It remains to show that this holds when $n < 0$.
Let $n \in \Z_{<0}$.
Then $n = - m$ for some $m \in \Z_{> 0}$.
Thus:
Category:Real Number to Negative Power
Real Number to Negative Power/Positive Integer

== Theorem ==
Let $r \in \R_{> 0}$ be a (strictly) positive real number.
Let $n \in \Z_{\ge 0}$ be a positive integer.
Let $r^n$ be defined as $r$ to the power of $n$.
Then:
:$r^{-n} = \dfrac 1 {r^n}$
== Proof ==
Proof by induction on $m$:
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$r^{-n} = \dfrac 1 {r^n}$
$\map P 0$ is the case:
=== Basis for the Induction ===
$\map P 1$ is the case:
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$r^{-k} = \dfrac 1 {r^k}$
Then we need to show:
:$r^{-\paren {k + 1} } = \dfrac 1 {r^{k + 1} }$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 0}: r^{-n} = \dfrac 1 {r^n}$
Category:Real Number to Negative Power
Real Numbers Between Epsilons

== Theorem ==
Let $a, b \in \R$ such that $\forall \epsilon \in \R_{>0}: a - \epsilon < b < a + \epsilon$.
Then $a = b$.
== Proof ==
From Real Plus Epsilon:
:$b < a + \epsilon \implies b \le a$
From Real Number Ordering is Compatible with Addition:
:$a - \epsilon < b \implies a < b + \epsilon$
Then from Real Plus Epsilon:
:$a < b + \epsilon \implies a \le b$
The result follows.
== Sources ==
* : $\S 1$: Real Numbers: Exercise $\S 1.8 \ (5)$
Category:Real Analysis
Real Numbers are Densely Ordered

== Corollary to Between two Real Numbers exists Rational Number ==
:$\forall a, b \in \R: a < b \implies \paren {\exists c \in \R: a < c < b}$
That is, the set of real numbers is densely ordered.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $1$: Sets and mappings: $\S 1.1$: The need for logic: Exercise $(8)$
Category:Real Numbers
Category:Densely Ordered
Category:Real Numbers are Densely Ordered
Real Numbers are Densely Ordered/Proof 1

== Theorem ==
== Proof ==
We can express $a$ and $b$ as:
:$a = \dfrac a 1, b = \dfrac b 1$
Thus from Mediant is Between:
:$\dfrac a 1 < \dfrac {a + b} {1 + 1} < \dfrac b 1$
Hence $c = \dfrac {a + b} 2$ is an element of $\R$ between $a$ and $b$.
== Sources ==
* : $\S 1$: Real Numbers: Exercise $\S 1.8 \ (6)$
Category:Real Numbers are Densely Ordered
Real Numbers are Uncountably Infinite

== Theorem ==
The set of real numbers $\R$ is uncountably infinite.
== Cantor's First Proof ==
== Cantor's Second Proof ==
== Cantor's Diagonal Argument ==
== Set-Theoretical Approach: Proof 1 ==
== Set-Theoretical Approach: Proof 2 ==
== Proof 1 using Ternary Notation ==
== Proof 2 using Ternary Notation ==
== Historical Note ==
== Sources ==
* : $\S 1.1$. Sets: Example $6$
* 
* 
Category:Real Numbers are Uncountably Infinite
Category:Real Numbers
Category:Uncountable Sets
Real Numbers are Uncountably Infinite/Cantor's Diagonal Argument

== Theorem ==
== Proof ==
We show that the unit interval $\hointr 0 1$ is uncountable, from which uncountability of $\R$ follows immediately.
 that $\hointr 0 1$ is countable.
First we note that $\hointr 0 1$ is not finite because $\dfrac 1 n \in \hointr 0 1$ are distinct for all $n \in \N$.
Therefore an injection $\hointr 0 1 \hookrightarrow \N$ enumerates $\hointr 0 1$ with a subset of the natural numbers.
By renaming, we can associate each $x \in \hointr 0 1$ to exactly one natural number to obtain a bijection.
(We have that such a mapping can be constructed).
Let $g$ be such a correspondence:
where juxtaposition of digits describes the decimal expansion of a number.
By Existence of Base-$N$ Representation, any decimal expansion of a real number exists and is unique, or else has exactly two representative strings.
In this case, if there are exactly two representations, one will have an infinite trail of $9$s, and one will terminate.
Restrict $g$ such that there exists no infinite strings of $9$s in the decimal expansions in the question, that is, to the set:
:$S := \set {f: \N \to \set {0, 1, 2, \ldots, 9} \middle\vert \, \forall M \in \N: \exists m \ge M: \map f m \ne 9}$
of sequences in $\set {0, 1, 2, \ldots, 9}$ not ending in infinitely many $9$s.
Define $g'$ as the restriction of $g$ to $S$:
:$g' := g \restriction S$
That is, construct $g'$ such that there exist no infinite strings of $9$s in the decimal expansions in question.
From Injection to Image is Bijection, $g'$ is a bijection
Hence, , $g'$ is a surjection.
For every $k \in \N$, define $f_k = d_{k k} + 1$ taken modulo $10$. 
That is, $f: 0 \mapsto 1, 1 \mapsto 2, \dots, 8 \mapsto 9, 9 \mapsto 0$.
Let $y$ be defined by the decimal expansion:
:$y = 0.f_1 f_2 f_3 \ldots$
Now:
:$y$ differs from $\map {g'} 1$ in the first digit of the decimal expansion
:$y$ differs from $\map {g'} 2$ the second digit of the decimal expansion
and generally the $n$th digit of the decimal expansion of $\map {g'} n$ and $y$ is different.
So $y$ can be none of the numbers $\map {g'} n$ for $n \in \N$.
But this contradicts our $g'$ is a surjection.
From this contradiction it is deduced that $\hointr 0 1$ is not countable.
== Historical Note ==
== Sources ==
* : $\S 2.4$: Uncountability of the real numbers: Theorem $5$
* : Chapter $1$: Preliminaries: $1.4$ Set Notation: Infinite sets
* 
* : $2.8$
* : Appendix $\text{A}$: Set Theory: The Continuum
* 
* : Appendix $\text{A}.5$: Theorem $\text{A}.27$
Category:Real Numbers are Uncountably Infinite