nyuuzyou/wikis huggingface dataset:
Real Multiplication Identity is One

== Theorem ==
The identity element of real number multiplication is the real number $1$:
:$\exists 1 \in \R: \forall a \in \R_{\ne 0}: a \times 1 = a = 1 \times a$
=== Corollary ===
== Proof ==
From the definition, the real numbers are the set of all equivalence classes $\eqclass {\sequence {x_n} } {}$ of Cauchy sequences of rational numbers.
Let $x = \eqclass {\sequence {x_n} } {}, y = \eqclass {\sequence {y_n} } {}$, where $\eqclass {\sequence {x_n} } {}$ and $\eqclass {\sequence {y_n} } {}$ are such equivalence classes.
From the definition of real multiplication, $x \times y$ is defined as:
:$\eqclass {\sequence {x_n} } {} \times \eqclass {\sequence {y_n} } {} = \eqclass {\sequence {x_n \times y_n} } {}$
Let $\sequence {1_n}$ be such that $\forall i: 1_n = 1$.
Then we have:
Similarly for $\eqclass {\sequence {x_n} } {} \times \eqclass {\sequence {1_n} } {}$.
So the identity element of $\struct {\R_{\ne 0}, \times}$ is the real number $1$.
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 1$. Rings and Fields
* : Chapter $1$: Real Numbers and Functions of a Real Variable: $1.2$ Operations with Real Numbers: $(7)$
Category:Multiplicative Group of Real Numbers
Category:Real Multiplication
Category:1
Real Multiplication is Associative

== Theorem ==
The operation of multiplication on the set of real numbers $\R$ is associative:
:$\forall x, y, z \in \R: x \times \paren {y \times z} = \paren {x \times y} \times z$
== Proof ==
From the definition, the real numbers are the set of all equivalence classes $\eqclass {\sequence {x_n} } {}$ of Cauchy sequences of rational numbers.
Let $x = \eqclass {\sequence {x_n} } {}, y = \eqclass {\sequence {y_n} } {}, z = \eqclass {\sequence {z_n} } {}$, where $\eqclass {\sequence {x_n} } {}$, $\eqclass {\sequence {y_n} } {}$ and $\eqclass {\sequence {z_n} } {}$ are such equivalence classes.
From the definition of real multiplication, $x \times y$ is defined as $\eqclass {\sequence {x_n} } {} \times \eqclass {\sequence {y_n} } {} = \eqclass {\sequence {x_n \times y_n} } {}$.
Thus we have:
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: $\text{1-2}$ Arithmetical properties of real numbers: Axiom $2$
* : Chapter $\text {I}$: Elementary Definitions: $\S 1$. Rings and Fields
* : Chapter $1$: Real Numbers and Functions of a Real Variable: $1.2$ Operations with Real Numbers: $(4)$
Category:Real Multiplication
Category:Examples of Associative Operations‎
Category:Associative Law of Multiplication
Real Multiplication is Closed

== Theorem ==
The operation of multiplication on the set of real numbers $\R$ is closed:
:$\forall x, y \in \R: x \times y \in \R$
== Proof ==
From the definition, the real numbers are the set of all equivalence classes $\eqclass {\sequence {x_n} } {}$ of Cauchy sequences of rational numbers.
Let $x = \eqclass {\sequence {x_n} } {}, y = \eqclass {\sequence {y_n} } {}$, where $\eqclass {\sequence {x_n} } {}$ and $\eqclass {\sequence {y_n} } {}$ are such equivalence classes.
From the definition of real multiplication, $x \times y$ is defined as:
:$\eqclass {\sequence {x_n} } {} \times \eqclass {\sequence {y_n} } {} = \eqclass {\sequence {x_n \times y_n} } {}$
We have that:
:$\forall i \in \N: x_i \in \Q, y_i \in \Q$
therefore $x_i \times y_i \in \Q$.
So it follows that:
:$\eqclass {\sequence {x_n \times y_n} } {} \in \R$
== Sources ==
* : Chapter $1$: Integral Domains: $\S 2$. Operations: Example $1$
* : Chapter $1$: Axioms for the Real Numbers: $2$. Fields: Example $1$
Category:Real Multiplication
Category:Algebraic Closure
Real Multiplication is Commutative

== Theorem ==
The operation of multiplication on the set of real numbers $\R$ is commutative:
:$\forall x, y \in \R: x \times y = y \times x$
== Proof ==
From the definition, the real numbers are the set of all equivalence classes $\eqclass {\sequence {x_n} } {}$ of Cauchy sequences of rational numbers.
Let $x = \eqclass {\sequence {x_n} } {}, y = \eqclass {\sequence {y_n} } {}$, where $\eqclass {\sequence {x_n} } {}$ and $\eqclass {\sequence {y_n} } {}$ are such equivalence classes.
From the definition of real multiplication, $x \times y$ is defined as $\eqclass {\sequence {x_n} } {} \times \eqclass {\sequence {y_n} } {} = \eqclass {\sequence {x_n \times y_n} } {}$.
Thus we have:
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: $\text{1-2}$ Arithmetical properties of real numbers: Axiom $1$
* : Chapter $\text {I}$: Elementary Definitions: $\S 1$. Rings and Fields
* : Chapter $1$: Real Numbers and Functions of a Real Variable: $1.2$ Operations with Real Numbers: $(3)$
Category:Real Multiplication
Category:Examples of Commutative Operations
Category:Commutative Law of Multiplication
Real Multiplication is Well-Defined

== Theorem ==
The operation of multiplication on the set of real numbers $\R$ is well-defined.
== Proof ==
From the definition, the real numbers are the set of all equivalence classes $\eqclass {\sequence {x_n} } {}$ of Cauchy sequences of rational numbers.
Let:
:$x = \eqclass {\sequence {x_n} } {}$
:$y = \eqclass {\sequence {y_n} } {}$
where $\eqclass {\sequence {x_n} } {}$ and $\eqclass {\sequence {y_n} } {}$ are such equivalence classes.
From the definition of real multiplication, $x \times y$ is defined as:
:$\eqclass {\sequence {x_n} } {} \eqclass {\sequence {y_n} } {} = \eqclass {\sequence {x_n \times y_n} } {}$
We need to show that:
:$\sequence {x_n}, \sequence {x'_n} \in \eqclass {\sequence {x_n} } {}, \sequence {y_n}, \sequence {y'_n} \in \eqclass {\sequence {y_n} } {} \implies \sequence {x_n \times y_n} = \sequence {x'_n \times y'_n}$
That is:
:$\forall \epsilon > 0: \exists N: \forall i, j > N: \size {\paren {x_i \times y_i} - \paren {x'_j \times y'_j} } < \epsilon$
As $\eqclass {\sequence {x_n} } {}$ and $\eqclass {\sequence {y_n} } {}$ are Cauchy sequences, from Cauchy Sequence in Metric Space is Bounded they are bounded.
Let:
:$B_x = 2 \map \sup {\sequence {x_n} }$
:$B_y = 2 \map \sup {\sequence {y_n} }$
Let:
:$B = \max \set {B_x, B_y}$
where $\max \set {B_x, B_y}$ denotes the max operation on $B_x$ and $B_y$.
Now let $\epsilon > 0$.
Then:
:$\exists N_1: \forall i, j > N_1: \size B \size {x_i - x'_j} < \epsilon / 2$
:$\exists N_2: \forall i, j > N_2: \size B \size {y_i - y'_j} < \epsilon / 2$
Now let $N = \max \set {N_1, N_2}$.
Then we have:
:$\forall i, j \ge N: \size B \size {x_i - x'_j} + \size B \size {y_i - y'_j} < \epsilon$
So:
Hence the result.
Category:Real Multiplication
Real Natural Logarithm Function is Continuous/Proof 2

== Theorem ==
== Proof ==
From Bounds of Natural Logarithm:
:$\dfrac 1 2 < \map \ln 2 < 1$
Fix $x \in \R$.
Consider $\dfrac x {\map \ln 2}$.
From Rationals are Everywhere Dense in Topological Space of Reals:
:$\forall \epsilon \in \R_{>0} \exists r \in \Q : \size {r - \dfrac x {\map \ln 2} } < \epsilon$
Thus:
Thus:
:$\forall \epsilon \in \R_{>0}: \exists t \in \R_{>0}: \size {\map \ln t - x} < \epsilon$
Thus, the image of $\R_{>0}$ under $\ln$ is everywhere dense in $\R$.
From Monotone Real Function with Everywhere Dense Image is Continuous, $\ln$ is continuous on $\R_{>0}$.
Hence the result.
Category:Real Natural Logarithm Function is Continuous
Real Natural Logarithm is Restriction of Complex Natural Logarithm

== Theorem ==
Let $\ln: \C_{\ne 0} \to \C$ be the complex natural logarithm.
Let $\ln': \R_{>0} \to \R$ be the real natural logarithm.
Then:
:$\ln' = \ln \restriction_{\R_{>0} \times \R}$
That is, the real natural logarithm is the restriction of the complex natural logarithm.
== Proof ==
From Domain of Real Natural Logarithm:
: $\operatorname{Dom} \left({\ln'}\right) = \R_{>0}$
From Image of Real Natural Logarithm:
: $\operatorname{Im} \left({\ln'}\right) = \R$
Let $z \in \C$ such that $z = x + i y$.
Let $z$ be expressed in exponential form as $z = r e^{i \theta}$.
Let $x > 0$ and $y = 0$.
Thus $z \in \R_{>0}$.
Then:
In order for $\ln z \in \R$ it is necessary that $k = 0$.
Thus $\ln z = \ln' x$.
Hence the result.
Category:Natural Logarithms
Real Null Sequence/Examples/n^alpha over y^n

== Example of Real Null Sequence ==
Let $\alpha \in \R$ be a (strictly) positive real number.
Let $y \in \R$ be a real number such that $\size y > 1$.
Let $\sequence {a_n}_{n \mathop \ge 1}$ be the real sequence defined as:
:$\forall n \in \Z_{>0}: a_n = \dfrac {n^\alpha} {y^n}$
Then $\sequence {a_n}$ is a null sequence:
:$\ds \lim_{n \mathop \to \infty} \dfrac {n^\alpha} {y^n} = 0$
== Proof ==
Let $y = 1 + x$.
Then:
== Sources ==
* : $1$: Review of some real analysis: Exercise $1.5: 9$
Category:Examples of Null Sequences
Real Null Sequence/Examples/n^alpha x^n

== Example of Real Null Sequence ==
Let $\alpha \in \Q$ be a (strictly) positive rational number.
Let $x \in \R$ be a real number such that $\size x < 1$.
Let $\sequence {a_n}_{n \mathop \ge 1}$ be the real sequence defined as:
:$\forall n \in \Z_{>0}: a_n = n^\alpha x^n$
Then $\sequence {a_n}$ is a null sequence:
:$\ds \lim_{n \mathop \to \infty} n^\alpha x^n = 0$
== Proof ==
First a lemma:
=== Lemma ===
Let $x \ne 0$.
Consider the expression:
:$\size {\dfrac {\paren {n + 1}^{\alpha + 1} x^{n + 1} } {n^{\alpha + 1} x^n} } = \paren {1 - \dfrac 1 n}^{\alpha + 1} \size x$
For all $n \ge N$, we have:
:$\size {\paren {n + 1}^{\alpha + 1} x^{n + 1} } \le \size {n^{\alpha + 1} x^n}$
Hence for all $n \ge N$:
:$\size {n^{\alpha + 1} x^n} \le \size {N^{\alpha + 1} x^n}$
So for $n \ge N$:
:$\size {n^\alpha x^n} \le \dfrac 1 N \size {N^{\alpha + 1} x^N}$
The result follows from the Squeeze Theorem for Real Sequences.
== Sources ==
* : $\S 4$: Convergent Sequences: Exercise $\S 4.20 \ (5)$
Category:Examples of Null Sequences
Category:Real Null Sequence/Examples/n^alpha x^n
Real Null Sequence/Examples/n^alpha x^n/Lemma

== Lemma for Real Null Sequence: $n^\alpha x^n$ ==
Let $\alpha \in \Q$ be a (strictly) positive rational number.
Let $x \in \R$ be a real number such that $\size x < 1$.
There exists $N \in \N$ such that:
:$\paren {1 + \dfrac 1 N}^{\alpha + 1} \, \size x \le 1$
== Proof ==
:
:$\forall n \in \N: \paren {1 + \dfrac 1 n}^{\alpha + 1} \, \size x > 1$
Then:
But this contradicts Sequence of Powers of Reciprocals is Null Sequence.
Hence by Proof by Contradiction:
:$\exists N \in \N: \paren {1 + \dfrac 1 N}^{\alpha + 1} \, \size x \le 1$
== Sources ==
* : $\S 4$: Convergent Sequences: Exercise $\S 4.20 \ (5)$
Category:Real Null Sequence/Examples/n^alpha x^n
Real Number Greater than One is Less than Square

== Theorem ==
Let $x \in \R$.
Let $x > 1$.
Then:
:$x^2 > x$
== Proof ==
As $x > 1$ it follows that $x > 0$.
Thus by Real Number Ordering is Compatible with Multiplication:
: $x \times x > 1 \times x$
and the result follows.
== Sources ==
* : $\S 1$: Real Numbers: Exercise $\S 1.8 \ (1) \ \text{(ii)}$
Category:Square Function
Category:Real Analysis
Real Number Inequalities can be Added

== Theorem ==
Let $a, b, c, d \in \R$ such that $a > b$ and $c > d$.
Then:
:$a + c > b + d$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $1$ / Section $1.4$: "The Ordering of the Real Numbers."
* : $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $2 \ \text{(a)}$
Category:Real Number Inequalities can be Added
Category:Real Numbers
Category:Addition
Category:Inequalities
Real Number Line is Banach Space

== Theorem ==
The real number line, along with the absolute value function, forms a Banach space over $\R$.
== Proof ==
From Real Numbers form Vector Space we have the real number line $\R$ is a vector space over $\R$.
That the norm axioms are satisfied by the absolute value function is proven in Absolute Value is Norm.
Then we have Real Number Line is Complete Metric Space.
Hence the result.
Category:Banach Spaces
Category:Real Analysis
Real Number Line is Complete Metric Space

== Theorem ==
The real number line $\R$ with the usual (Euclidean) metric forms a complete metric space.
== Proof ==
From Real Number Line is Metric Space, the distance function defined as $\map d {x, y} = \size {x - y}$ is a metric on $\R$.
It remains to be shown that the metric space $\struct {\R, d}$ is complete.
By definition, this is done by demonstrating that every Cauchy sequence of real numbers has a limit.
This is demonstrated in Cauchy's Convergence Criterion.
Hence the result.
== Sources ==
* : Part $\text {II}$: Counterexamples: $28$. Euclidean Topology: $1$
* 
* 
Category:Complete Metric Spaces
Category:Real Number Line with Euclidean Metric
Real Number Line is First-Countable

== Theorem ==
Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
Then $\struct {\R, \tau_d}$ is first-countable.
== Proof ==
From Real Number Line is Second-Countable we have that $\struct {\R, \tau_d}$ is a second-countable space.
The result follows from Second-Countable Space is First-Countable.
== Sources ==
* : Part $\text {II}$: Counterexamples: $28$. Euclidean Topology: $2$
Category:Real Number Line with Euclidean Topology
Category:First-Countable Spaces
Real Number Line is Lindelöf

== Theorem ==
Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
Then $\struct {\R, \tau_d}$ is Lindelöf.
== Proof ==
From Real Number Line is Second-Countable we have that $\struct {\R, \tau_d}$ is a second-countable space.
The result follows from Second-Countable Space is Lindelöf.
== Sources ==
* : Part $\text {II}$: Counterexamples: $28$. Euclidean Topology: $2$
* 
* 
Category:Lindelöf Spaces
Category:Real Number Line with Euclidean Topology
Real Number Line is Locally Compact Hausdorff Space

== Theorem ==
Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
Then $\struct {\R, \tau_d}$ is a locally compact Hausdorff space.
== Proof ==
We have that a Real Number Line satisfies all Separation Axioms.
Specifically, $\struct {\R, \tau_d}$ is a Hausdorff space.
Consider $\CC$ the set of subsets of $\R$ defined as:
:$\CC = \set {\closedint n {n + 1}: n \in \Z}$
where $\closedint n {n + 1}$ is the closed real interval between successive integers.
By the Heine-Borel Theorem, each element of $\CC$ is compact.
Every element of $\R$ is contained in at least one of the elements of $\CC$.
Hence, by definition, $\R$ is locally compact.
== Sources ==
* : Part $\text {II}$: Counterexamples: $28$. Euclidean Topology: $4$
Category:Real Number Line with Euclidean Topology
Category:Locally Compact Hausdorff Spaces
Real Number Line is Metric Space

== Theorem ==
Let $\R$ be the real number line.
Let $d: \R \times \R \to \R$ 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 $\R$ and so $\struct {\R, d}$ is a metric space.
== Proof ==
=== Proof of ===
So holds for $d$.
=== Proof of ===
So holds for $d$.
=== Proof of ===
So holds for $d$.
=== Proof of ===
So holds for $d$.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 2$: Metric Spaces: Theorem $2.2$
* : Part $\text {II}$: Counterexamples: $28$. Euclidean Topology: $1$
Category:Real Numbers
Category:Real Number Line with Euclidean Metric
Real Number Line is Non-Meager

== Theorem ==
Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
Then $\struct {\R, \tau_d}$ is non-meager.
== Proof ==
We have that the Real Number Line is 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.
== Sources ==
* : Part $\text {II}$: Counterexamples: $28$. Euclidean Topology: $1$
Category:Real Number Line with Euclidean Topology
Category:Non-Meager Spaces
Real Number Line is Paracompact

== Theorem ==
Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
Then $\struct {\R, \tau_d}$ is paracompact.
== Proof ==
Let $\CC$ be an open cover for $\R$.
Then $\CC$ covers each of the closed real intervals $\closedint n {n + 1}$ for all $n \in \Z$.
By the Heine-Borel Theorem, each of $\closedint n {n + 1}$ is compact.
So, for each of these intervals $\closedint n {n + 1}$, it follows that $\CC$ can be reduced to a sequence $\sequence {G_i^{\paren n} }$ of finite subcovers.
Then each of $G_i^{\paren n} \cap \openint {n - 1} {n + 2}$ forms a refinement of $\CC$ which is locally finite.
Hence, by definition, $\R$ is paracompact.
== Sources ==
* : Part $\text {II}$: Counterexamples: $28$. Euclidean Topology: $6$
Category:Real Number Line with Euclidean Topology
Category:Paracompact Spaces
Real Number Line is Second-Countable

== Theorem ==
Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
Then $\struct {\R, \tau_d}$ is second-countable.
== Proof ==
From Countable Basis of Real Number Line we have that $\struct {\R, \tau_d}$ has a countable basis.
The result follows directly from the definition of a second-countable space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $28$. Euclidean Topology: $2$
Category:Real Number Line with Euclidean Topology
Category:Second-Countable Spaces
Real Number Line is Separable

== Theorem ==
Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
Then $\struct {\R, \tau_d}$ is separable.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* 
Category:Real Number Line with Euclidean Topology
Category:Separable Spaces
Category:Real Number Line is Separable
Real Number Line is Separable/Proof 1

== Theorem ==
== Proof ==
The rational numbers $\Q$ form a metric space.
We have that the Rationals are Everywhere Dense in Topological Space of Reals.
We also have that the Rational Numbers are Countably Infinite.
The result follows from the definition of separable space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $28$. Euclidean Topology: $2$
Category:Real Number Line is Separable
Real Number Line is Sigma-Compact

== Theorem ==
Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
Then $\struct {\R, \tau_d}$ is $\sigma$- compact.
== Proof ==
We have that a Real Number Line satisfies all Separation Axioms.
Specifically, $\struct {\R, \tau_d}$ is a Hausdorff space.
Consider $\CC$ the set of subsets of $\R$ defined as:
:$\CC = \set {\closedint n {n + 1}: n \in \Z}$
where $\closedint n {n + 1}$ is the closed real interval between successive integers.
By the Heine-Borel Theorem, each element of $\CC$ is compact.
$\CC$ itself is countable, as there is a (trivial) one-to-one-correspondence between $\CC$ and $\Z$.
Every element of $\R$ is contained in at least one of the elements of $\CC$.
Thus $\R$ is the union of $\CC$.
Hence, by definition, $\R$ is $\sigma$-compact.
== Sources ==
* : Part $\text {II}$: Counterexamples: $28$. Euclidean Topology: $4$
Category:Real Number Line with Euclidean Topology
Category:Sigma-Compact Spaces
Real Number Line is not Compact

== Theorem ==
Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
Then $\struct {\R, \tau_d}$ is not compact.
== Proof ==
We have:
:Compact Space is Countably Compact
:Real Number Line is not Countably Compact
Hence, as $\struct {\R, \tau_d}$ is not countably compact, it follows that it is not compact.
Category:Real Number Line with Euclidean Topology
Category:Compact Spaces
Real Number Line is not Countably Compact

== Theorem ==
Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
Then $\struct {\R, \tau_d}$ is not countably compact.
== Proof ==
Let $\CC$ be the set of subsets of $\R$ defined as:
:$\CC = \set {\openint n {n + 2}: n \in \Z}$
Then $\CC$ is an open cover of $\R$ which is countable.
However, there is no finite subcover for $\R$ of $\CC$.
Hence the result, by definition of countably compact.
== Sources ==
* : Part $\text {II}$: Counterexamples: $28$. Euclidean Topology: $4$
Category:Real Number Line with Euclidean Topology
Category:Countably Compact Spaces
Real Number Line is not Topological Continuum

== Theorem ==
The real number line is not a continuum in the topological sense.
== Proof ==
Recall the definition of continuum:
:A topological space $T$ is a continuum $T$ is both compact and connected.
However, we have the result Real Number Line is not Compact.
Hence the result.
== Sources ==
* 
* 
Category:Continua (Topology)
Category:Real Numbers
Real Number Line less Zero is Disconnected Space

== Theorem ==
Let $S := \R \setminus \set 0$ be the real number line with $0$ excluded.
Let $\struct {S, \tau_d}$ be $S$ with the usual (Euclidean) topology.
Then $\struct {S, \tau_d}$ is disconnected.
== Proof ==
We note that:
:$S = \openint \gets 0 \cup \openint 0 \to$
:$\openint \gets 0 \cap \openint 0 \to = \O$
Hence we have partitioned $S$ into $2$ disjoint open sets whose union is $S$.
Hence by definition $S$ is not connected.
Hence the result by definition of disconnected space.
== Sources ==
* 
Category:Disconnected Spaces
Category:Real Number Line with Euclidean Topology
Real Number Line satisfies all Separation Axioms

== Theorem ==
Let $\struct {\R, \tau_d}$ be the the real number line with the usual (Euclidean) topology.
Then $\struct {\R, \tau_d}$ fulfils all separation axioms:
:$\struct {\R, \tau_d}$ is a $T_0$ (Kolmogorov) space
:$\struct {\R, \tau_d}$ is a $T_1$ (Fréchet) space
:$\struct {\R, \tau_d}$ is a $T_2$ (Hausdorff) space
:$\struct {\R, \tau_d}$ is a semiregular space
:$\struct {\R, \tau_d}$ is a $T_{2 \frac 1 2}$ (completely Hausdorff) space 
:$\struct {\R, \tau_d}$ is a $T_3$ space 
:$\struct {\R, \tau_d}$ is a regular space
:$\struct {\R, \tau_d}$ is an Urysohn space
:$\struct {\R, \tau_d}$ is a $T_{3 \frac 1 2}$ space
:$\struct {\R, \tau_d}$ is a Tychonoff (completely regular) space
:$\struct {\R, \tau_d}$ is a $T_4$ space
:$\struct {\R, \tau_d}$ is a normal space
:$\struct {\R, \tau_d}$ is a $T_5$ space 
:$\struct {\R, \tau_d}$ is a completely normal space
:$\struct {\R, \tau_d}$ is a perfectly $T_4$ space
:$\struct {\R, \tau_d}$ is a perfectly normal space
== Proof ==
* Real Number Line is Metric Space
* Metric Space fulfils all Separation Axioms
== Sources ==
* : Part $\text {II}$: Counterexamples: $28$. Euclidean Topology: $1$
Category:Real Number Line with Euclidean Topology
Category:Separation Axioms
Real Number Line with Off-Center Distance Function is Quasimetric Space

== Theorem ==
Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
Then $\tau_d$ can be given by a quasimetric defined as:
:$\map d {x, y} = \begin {cases}
y - x & : y \ge x \\
2 \paren {x - y} & : y < x \end {cases}$
Thus $\struct {\R, \tau_d}$ is a quasimetric space.
== Proof ==
To show that $\map d {x, y}$ is a quasimetric, we need to show that $d: \R \times \R \to \R$ satisfies the following conditions for all $x, y, z \in \R$:
:$(\text M 1): \quad \map d {x, x} = 0$
:$(\text M 2): \quad \map d {x, y} + \map d {y, z} \ge \map d {x, z}$
:$(\text M 4): \quad x \ne y \implies \map d {x, y} > 0$
=== Proof of $\text M 1$ ===
:$\map d {x, x} = x - x = 0$
So $(\text M 1)$ is shown to hold.
=== Proof of $\text M 2$ ===
Suppose $x \le y \le z$.
Then:
Suppose $x > y > z$.
Then:
Equality holds, hence so does the inequality.
=== Proof of $\text M 4$ ===
Suppose $x \ne y$.
There are two possibilities:
:$(1): \quad y \ge x$, in which case:
::$\map d {x, y} = y - x > 0$
:$(2): \quad y < x$, in which case:
::$\map d {x, y} = 2 \paren {x - y} > 0$
So $(\text M 4)$ is shown to hold.
All criteria $(\text M 1)$, $(\text M 2)$ and $(\text M 4)$ are shown to hold, and so $\struct {\R, \tau_d}$ is shown to be a quasimetric space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $28$. Euclidean Topology: $7$
Category:Real Number Line with Euclidean Topology
Category:Quasimetric Spaces