nyuuzyou/wikis huggingface dataset:
Open Interval Defined by Absolute Value

== Theorem ==
Let $\xi, \delta \in \R$ be real numbers.
Let $\delta > 0$.
Then:
:$\set {x \in \R: \size {\xi - x} < \delta} = \openint {\xi - \delta} {\xi + \delta}$
where $\openint {\xi - \delta} {\xi + \delta}$ is the open real interval between $\xi - \delta$ and $\xi + \delta$.
== Proof ==
But:
:$\openint {\xi - \delta} {\xi + \delta} = \set {x \in \R: \xi - \delta < x < \xi + \delta}$
== Also presented as ==
:$\set {x \in \R: \size {x - \xi} < \delta} = \openint {\xi - \delta} {\xi + \delta}$
which is immediate from:
:$\size {x - \xi} = \size {\xi - x}$
== Also see ==
* Closed Interval Defined by Absolute Value
* Complement of Open Interval Defined by Absolute Value
* Complement of Closed Interval Defined by Absolute Value
== Sources ==
* : Chapter $1$: Real Numbers and Functions of a Real Variable: $1.2$ Operations with Real Numbers
* : $\S 2$: Continuum Property: Exercise $\S 2.10 \ (2)$
Category:Real Intervals
Category:Absolute Value Function
Open Mapping is not necessarily Closed Mapping

== Theorem ==
Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $f: T_1 \to T_2$ be a mapping which is not a bijection.
Let $f$ be an open mapping.
Then it is not necessarily the case that $f$ is also a closed mapping.
== Proof ==
Note that if $f$ is a bijection, the result Bijection is Open iff Closed applies.
It is to be shown that if $f$ is not a bijection, this is not necessarily the case.
This is achieved by Proof by Counterexample:
Let $\struct {\R^2, d}$ be the real number plane with the usual (Euclidean) topology.
Let $\rho: \R^2 \to \R$ be the first projection on $\R^2$ defined as:
:$\forall \tuple{x, y} \in \R^2: \map \rho {x, y} = x$
From Projection on Real Euclidean Plane is Open Mapping, $\rho$ is an open mapping.
From Projection on Real Euclidean Plane is not Closed Mapping, $\rho$ is not a closed mapping.
The result is apparent.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Functions
Category:Open Mappings
Category:Closed Mappings
Open Neighborhood contains Affine Open Neighborhood

== Theorem ==
Let $\struct {X, \OO_X}$ be a scheme.
Let $U \subset X$ be an open subset.
Let $x \in U$.
Then there exists an open subset $V \subset U$ with $x \in V$, such that the restriction $\struct {V, \OO_X {\restriction_V}}$ of $\struct {X, \OO_X}$ to $V$ is an affine scheme.
== Proof ==
Category:Algebraic Geometry
Category:Schemes
Open Neighborhood of Dilation of Point in Topological Vector Space contains Pointwise Scalar Multiplication of Open Neighborhood of Scalar with Open Neighborhood of Vector

== Theorem ==
Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$.
Let $\lambda \in K$ and $x \in X$. 
Let $U$ be an open neighborhood of $\lambda x$.
Then there exists an open neighborhood $D$ of $\lambda$ in $K$ and an open neighborhood $V$ in $X$ such that:
:$D V \subseteq U$
where:
:$D V = \set {\mu y : \mu \in D, \, y \in V}$
== Proof ==
Equip $K \times X$ with the product topology.
From Box Topology on Finite Product Space is Product Topology, this is precisely the box topology.
Define $m : K \times X \to X$ by: 
:$\map m {\mu, y} = \mu y$
for each $\tuple {\mu, y} \in K \times X$.
From the definition of a topological vector space, $m$ is continuous.
In particular, $m$ is continuous at $\tuple {\lambda, x}$.
Hence, there exists an open neighborhood $W$ of $\tuple {\lambda, x}$ such that: 
:$m \sqbrk W \subseteq U$
From Basis for Box Topology and the definition of a topology generated by a synthetic basis, there exists an open neighborhood $D$ of $\lambda$ in $K$ and an open neighborhood $V$ of $x$ in $X$ such that:
:$D \times V \subseteq W$
Then we have, from Image of Subset under Mapping is Subset of Image:
:$m \sqbrk {D \times V} \subseteq m \sqbrk W \subseteq U$
Since:
:$m \sqbrk {D \times V} = D V$
we have:
:$D V \subseteq U$
Category:Topological Vector Spaces
Open Neighborhood of Origin in Topological Vector Space contains Balanced Open Neighborhood

== Theorem ==
Let $\Bbb F \in \set {\R, \C}$. 
Let $X$ be a topological vector space over $\Bbb F$. 
Let $U$ be an open neighborhood of ${\mathbf 0}_X$.
Then there exists a balanced open neighborhood $W$ of ${\mathbf 0}_X$ with $W \subseteq U$.
== Proof ==
Equip $\Bbb F \times X$ with the product topology.
From Box Topology on Finite Product Space is Product Topology, this is precisely the box topology.
Define $M : \Bbb F \times X \to X$ by: 
:$\map M {\lambda, x} = \lambda x$
for each $\tuple {\lambda, x} \in \Bbb F \times X$.
From the definition of a topological vector space, $M$ is continuous.
In particular, $M$ is continuous at $\tuple {0, {\mathbf 0}_X}$.
So there exists an open neighborhood $A$ of $\tuple {0, {\mathbf 0}_X}$ such that: 
:$M \sqbrk A \subseteq U$
From Basis for Box Topology and the definition of a topology generated by a synthetic basis, there exists an open neighborhood $D$ of $0$ and an open neighborhood $V$ of ${\mathbf 0}_X$ such that: 
:$D \times V \subseteq A$
Using:
:the definition of an open set in $\R$ if $\Bbb F = \R$
:the definition of an open set in $\C$ if $\Bbb F = \C$
there exists $r > 0$ such that if $\alpha \in \Bbb F$ has $\cmod \alpha < r$ we have $\alpha \in D$.
Then for $\alpha \in \Bbb F$ with $\cmod \alpha < r$ we have: 
:$\set \alpha \times V \subseteq D \times V$
so that: 
:$M \sqbrk {\set \alpha \times V} \subseteq U$
from Image of Subset under Mapping is Subset of Image.
We have: 
:$M \sqbrk {\set \alpha \times V} = \alpha V$
so we have: 
:$\alpha V \subseteq U$
for each $\alpha \in \Bbb F$ with $\cmod \alpha < r$. 
Let: 
:$\ds W = \bigcup_{\cmod \alpha < r} \alpha V$
From Dilation of Open Set in Topological Vector Space is Open, $\alpha V$ is open for each $\alpha \in \Bbb F$.
Since $W$ is the union of open sets, $W$ is open.
From Union of Subsets is Subset, we have $W \subseteq U$.
Since $0 \in \alpha V$ for all $\alpha \in \Bbb F$ with $\cmod \alpha < r$, we also have $0 \in W$.
So $W$ is an open neighborhood of $0$.
It remains to show that $W$ is balanced.
Let $x \in W$ and $\lambda \in \Bbb F$ have $\cmod \lambda \le 1$. 
Then there exists $\alpha \in \Bbb F$ with $\cmod \alpha < r$ such that $x \in \alpha V$. 
Then we have $\lambda x \in \paren {\alpha \lambda} V$.
We have $\cmod {\alpha \lambda} < r$, so $\paren {\alpha \lambda} V \subseteq W$.
So we have $\lambda x \in W$.
So we have $\lambda W \subseteq W$ for all $\lambda \in \Bbb F$ with $\cmod \lambda \le 1$, so $W$ is balanced.
== Sources ==
* : $1.14$: Theorem
Category:Balanced Sets
Category:Topological Vector Spaces
Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods

== Theorem ==
Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$. 
Let $a, b \in X$.
Let $W$ be an open neighborhood of $a + b$.
Then there exists an open neighborhood $W_1$ of $a$ and an open neighborhood $W_2$ of $b$ such that: 
:$W_1 + W_2 \subseteq W$
where $W_1 + W_2$ denotes the sum of $W_1$ and $W_2$.
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
Equip $X \times X$ with the product topology.
From Box Topology on Finite Product Space is Product Topology, this is precisely the box topology.
Define $S : X \times X \to X$ by: 
:$\map S {x, y} = x + y$
for each $\tuple {x, y} \in X \times X$. 
From the definition of a topological vector space, $S$ is continuous.
In particular, $S$ is continuous at $\tuple {a, b}$.
Hence, there exists an open neighborhood $V$ of $\tuple {a, b}$ such that: 
:$S \sqbrk V \subseteq W$ 
From Basis for Box Topology and the definition of a topology generated by a synthetic basis, there exists an open neighborhood $W_1$ of $a$ and an open neighborhood $W_2$ of $b$ such that:
:$W_1 \times W_2 \subseteq V$
Then we have, from Image of Subset under Mapping is Subset of Image: 
:$S \sqbrk {W_1 \times W_2} \subseteq S \sqbrk V$
Since: 
:$S \sqbrk {W_1 \times W_2} = W_1 + W_2$
this gives: 
:$W_1 + W_2 \subseteq W$
as required.
Category:Topological Vector Spaces
Category:Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods
Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods/Corollary 1

== Theorem ==
Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$. 
Let $W$ be an open neighborhood of ${\mathbf 0}_X$. 
Then there exists a symmetric open neighborhood $U$ of ${\mathbf 0}_X$ such that: 
:$U + U \subseteq W$
== Proof ==
Since ${\mathbf 0}_X + {\mathbf 0}_X = {\mathbf 0}_X$, we can apply Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods to obtain an open neighborhood $V_1$ of ${\mathbf 0}_X$ and an open neighborhood $V_2$ of ${\mathbf 0}_X$ such that: 
:$V_1 + V_2 \subseteq W$
Note at this point we do not necessarily have $V_1 = V_2$.
From Dilation of Open Set in Topological Vector Space is Open, $-V_1$ and $-V_2$ are open neighborhoods of ${\mathbf 0}_X$.
Let:
:$U = V_1 \cap V_2 \cap \paren {-V_1} \cap \paren {-V_2}$
Since $U$ is the finite intersection of open sets, it is itself open.
Since ${\mathbf 0}_X$ is contained in each of the four sets of the intersection, we have ${\mathbf 0}_X \in U$.
So $U$ is an open neighborhood of ${\mathbf 0}_X$.
We show that $U$ is symmetric and that: 
:$U + U \subseteq W$
Let $x \in U$. 
Then $x \in V_1$, $x \in V_2$, $x \in -V_1$ and $x \in -V_2$.
So we have that $-x \in -V_1$, $-x \in -V_2$, $-x \in V_1$ and $-x \in -V_2$.
So:
:$-x \in \paren {-V_1} \cap \paren {-V_2} \cap V_1 \cap V_2 = U$
So we have that $U$ is symmetric.
Now let $x, y \in U$. 
We want to show that $x + y \in W$.
Since $x, y \in U$, we have $x \in V_1$ and $y \in V_2$.
So, we have $x + y \in V_1 + V_2$.
We have $V_1 + V_2 \subseteq W$, so $x + y \in W$.
So, we have $U + U \subseteq W$. 
Hence $U$ is our desired open neighborhood.
Category:Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods
Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods/Corollary 2

== Corollary ==
Let $K \in \set {\R, \C}$.
Let $X$ be a topological vector space over $K$.
Let $W$ be an open neighborhood of ${\mathbf 0}_X$. 
Then there exists a balanced open neighborhood $U$ of ${\mathbf 0}_X$ such that: 
:$U + U \subseteq W$
== Proof ==
From Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods: Corollary, there exists an open neighborhood $V$ of ${\mathbf 0}_X$ such that:
:$V + V \subseteq W$
From Open Neighborhood of Origin in Topological Vector Space contains Balanced Open Neighborhood, there exists a balanced open neighborhood $U$ of ${\mathbf 0}_X$ such that $U \subseteq V$.
Then $U + U \subseteq V + V \subseteq W$.
Category:Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods
Open Neighborhoods of Point form Directed Ordering

== Theorem ==
Let $\struct{ S, \tau }$ be a topological space.
Let $x \in S$.
Let $\NN \subseteq \tau$ be the set of open neighborhoods of $x$.
Then $\supseteq$, the ordering of $\NN$ by reverse inclusion, is a directed ordering on $\NN$.
== Proof ==
By Subset Relation is Ordering and Dual Ordering is Ordering $\supseteq$ is an ordering on $\NN$.
To show that $\supseteq$ is directed, let $U, V \in \NN$.
Then $x \in U, V$, so that $x \in U \cap V$.
Hence $U \cap V \in \NN$ is an open neighborhood of $x$.
Moreover by Intersection is Subset:
:$U, V \supseteq U \cap V$
Hence $\supseteq$ is directed.
== Sources ==
* : Appendix $\text{A}$ Preliminaries: $\S 2.$ Topology
Category:Topology
Category:Directed Orderings
Category:Subset Relation
Open Ordinal Space is not Compact in Closed Ordinal Space

== Theorem ==
Let $\Gamma$ be a limit ordinal.
Let $\hointr 0 \Gamma$ denote the open ordinal space on $\Gamma$.
Consider the compact subspace $\hointr 0 \Gamma$.
Then $\hointr 0 \Gamma$ is not compact in $\closedint 0 \Gamma$.
== Proof ==
Consider the set:
:$\set {\hointr 0 \Gamma: \alpha < \Gamma}$
This is an open cover of $\hointr 0 \Gamma$.
But because $\Gamma$ is a limit ordinal, it has no finite subcover.
Hence the result by definition of compact.
== Sources ==
* : Part $\text {II}$: Counterexamples: $41$. Closed Ordinal Space $[0, \Gamma] \ (\Gamma < \Omega)$: $7$
Category:Ordinal Spaces
Category:Compact Spaces
Open Projection and Closed Graph Implies Quotient is Hausdorff

== Theorem ==
Let $\RR \subseteq X \times X$ be an equivalence relation on a topological space $\struct {X, \tau}$.
Let $X / \RR$ be the quotient space.
Let $p$ denote the quotient mapping.
Let:
:$\RR$ be closed in $X \times X$
:$p$ be an open mapping.
Then $X / \RR$ is Hausdorff.
== Proof ==
Let $\eqclass x \RR, \eqclass y \RR \in X / \RR$ such that $\eqclass x \RR \ne \eqclass y \RR$.
Then $\tuple {x, y} \notin \RR$.
Since $\RR$ is closed, $\paren {X \times X} \setminus \RR$ is open.
Let:
:$\BB = \set { U \times V : U, V \in \tau}$
By Natural Basis of Product Topology, $\BB$ is a basis for the product topology on $X \times X$.
Then choose $U, V \in \tau$ such that $\tuple {x, y} \in U \times V \subseteq \paren {X \times X} \setminus \RR$.
Thus $x \in U$, so:
:$\eqclass x \RR = \map p x \in p \sqbrk U$
Similarly, $y \in V$, so:
:$\eqclass y \RR = \map p y \in p \sqbrk V$
Since $p$ is open, $p \sqbrk U$ and $p \sqbrk V$ are open.
 that $p \sqbrk U \cap p \sqbrk V \ne \O$.
Then there exist $z \in U$ and $w \in V$ such that $\eqclass z \RR = \eqclass w \RR$.
Then $z \RR w$, and so $\tuple {z, w} \in \RR$.
However, $\tuple {z, w} \in U \times V$, contradicting $U \times V \subseteq \paren {X \times X} \setminus \RR$.
Thus:
:$p \sqbrk U \cap p \sqbrk V = \O$
Hence the result.
== Also see ==
* Hausdorff Space iff Diagonal Set on Product is Closed
Category:Quotient Spaces
Open Rational-Number Balls form Neighborhood Basis in Real Number Line

== Theorem ==
Let $\R$ be the real number line with the usual (Euclidean) metric.
Let $a \in \R$ be a point in $\R$.
Let $\BB_a$ be defined as:
:$\BB_a := \set {\map {B_\epsilon} a: \epsilon \in \Q_{>0} }$
that is, the set of all open $\epsilon$-balls of $a$ for rational $\epsilon$.
Then the $\BB_a$ is a basis for the neighborhood system of $a$.
== Proof ==
Let $N$ be a neighborhood of $a$ in $M$.
Then by definition:
:$\exists \epsilon' \in \R_{>0}: \map {B_{\epsilon'} } a \subseteq N$
where $\map {B_{\epsilon'} } a$ is the open $\epsilon'$-ball at $a$.
From Open Ball in Real Number Line is Open Interval:
:$\map {B_{\epsilon'} } a = \openint {a - \epsilon'} {a + \epsilon'}$
From Between two Real Numbers exists Rational Number:
:$\exists \epsilon \in \Q: 0 < \epsilon < \epsilon'$
and so:
:$\openint {a - \epsilon} {a + \epsilon} \subseteq \openint {a - \epsilon'} {a + \epsilon'}$
From Open Real Interval is Open Ball
:$\map {B_\epsilon} a = \openint {a - \epsilon} {a + \epsilon}$
is the open $\epsilon$-ball at $a$.
From Subset Relation is Transitive:
:$\openint {a - \epsilon} {a + \epsilon} \subseteq N$
From Open Ball is Neighborhood of all Points Inside, $\openint {a - \epsilon} {a + \epsilon}$ is a neighborhood of $a$ in $M$.
Hence there exists a neighborhood $\openint {a - \epsilon} {a + \epsilon}$ of $a$ which is a subset of $N$.
Hence the result by definition of basis for the neighborhood system of $a$.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 4$: Open Balls and Neighborhoods: Exercise $4 \ \text{ii)}$
Category:Real Intervals
Category:Real Number Line with Euclidean Metric
Open Ray is Dual to Open Ray

== Theorem ==
Let $\struct {S, \preceq}$ be a totally ordered set.
Let $R$ be an open ray in $\struct {S, \preceq}$.
Then $R$ is an open ray in $\struct {S, \succeq}$, where $\succeq$ is the dual ordering of $\preceq$.
== Proof ==
By the definition of open ray, there is some $p \in S$ such that:
:$R$ is the strict upper or strict lower closure of $p$ with respect to $\preceq$.
By Strict Lower Closure is Dual to Strict Upper Closure, the dual statement is:
:$R$ is the strict upper or strict lower closure of $p$ with respect to $\succeq$.
Thus $R$ is an open ray in $\struct {S, \succeq}$.
Category:Total Orderings
Open Ray is Open in GO-Space

== Theorem ==
Let $\left({S, \preceq, \tau}\right)$ be a generalized ordered space.
Let $p \in S$.
Then:
: $p^\prec$ and $p^\succ$ are $\tau$-open
where:
: $p^\prec$ is the strict lower closure of $p$
: $p^\succ$ is the strict upper closure of $p$.
== Proof for Definition 1 of GO-Space ==
== Proof for Definition 2 of GO-Space ==
Category:Generalized Ordered Spaces
Open Ray is Open in GO-Space/Definition 1

== Theorem ==
Let $\struct {S, \preceq, \tau}$ be a generalized ordered space.
Let $p \in S$.
Then:
:$p^\prec$ and $p^\succ$ are $\tau$-open
where:
:$p^\prec$ is the strict lower closure of $p$
:$p^\succ$ is the strict upper closure of $p$.
== Proof ==
We will prove that $U = p^\succ$ is $\tau$-open.
That $p^\prec$ is $\tau$-open will follow by duality.
Let $u \in U$.
Since $p \notin U$, $p \ne u$.
By definition of GO-space, $\tau$ is Hausdorff.
From $T_2$ Space is $T_1$ Space, $\tau$ is $T_1$.
Thus by definition of GO-space, there is an open, convex set $M$ such that $u \in M$ and $p \notin M$.
Next we will show that $M \subseteq U$:
Let $x \in S \setminus U$.
Then $x \preceq p \preceq u$.
 $x \in M$.
Since $x, u \in M$, $p \in M$ because $M$ is convex, contradicting the choice of $M$.
Thus $x \notin M$.
Since this hold for all $x \in S \setminus U$, $M \subseteq U$.
Thus $U$ contains a neighborhood of each of its points.
From Set is Open iff Neighborhood of all its Points, $U$ is open.
Category:Generalized Ordered Spaces
Open Ray is Open in GO-Space/Definition 2

== Theorem ==
Let $\struct {S, \preceq, \tau}$ be a generalized ordered space by Definition 2.
That is:
:Let $\struct {S, \preceq}$ be a totally ordered set.
:Let $\struct {S, \tau}$ be a topological space.
Let there be:
:a linearly ordered space $\struct {S', \preceq', \tau'}$
and:
:a mapping $\phi: S \to S'$ which is both:
::a $\preceq$-$\preceq'$ order embedding
:and:
::a $\tau$-$\tau'$ topological embedding.
Let $p \in S$.
Then:
:$p^\prec$ and $p^\succ$ are $\tau$-open
where:
:$p^\prec$ is the strict lower closure of $p$
:$p^\succ$ is the strict upper closure of $p$.
== Proof ==
We will prove that $p^\succ$ is open.
That $p^\prec$ is open will follow by duality.
By Inverse Image under Order Embedding of Strict Upper Closure of Image of Point:
:$\map {\phi^{-1} } {\map \phi p^\succ} = p^\succ$
:$\map \phi p^\succ$ is an open ray in $S'$
Therefore $\tau'$-open by the definition of the order topology.
Since $\phi$ is a topological embedding, it is continuous.
Thus $p^\succ$ is $\tau$-open.
Category:Generalized Ordered Spaces
Open Real Interval/Examples/Example 1

== Example of Open Real Interval ==
Let $I$ be the open real interval defined as:
:$I := \openint 1 2$
Then $3 \notin I$.
== Proof ==
By definition of open real interval:
:$I = \set {x \in \R: 1 < x < 2}$
As $3 > 2$ it follows that $3 \notin I$.
== Sources ==
* : $\S 2$: Continuum Property: Exercise: $\S 2.10 \ (1) \ \text{(i)}$
Category:Examples of Intervals
Open Real Interval/Examples/Example 2

== Example of Open Real Interval ==
Let $I$ be the open real interval defined as:
:$I := \openint 0 2$
Then $2 \notin I$.
== Proof ==
By definition of open real interval:
:$I = \set {x \in \R: 0 < x < 2}$
As it is not the case that $2 < 2$ it follows that $2 \notin I$.
== Sources ==
* : $\S 2$: Continuum Property: Exercise: $\S 2.10 \ (1) \ \text{(iv)}$
Category:Examples of Intervals
Open Real Interval is Homeomorphic to Real Number Line

== Theorem ==
Let $\R$ be the real number line with the Euclidean topology.
Let $I := \openint a b$ be a non-empty open real interval.
Then $I$ and $\R$ are homeomorphic.
== Proof 1 ==
== Proof 2 ==
Category:Examples of Homeomorphisms
Category:Open Real Interval is Homeomorphic to Real Number Line
Open Real Interval is Homeomorphic to Real Number Line/Proof 1

== Theorem ==
== Proof ==
By definition of open real interval, for $I$ to be non-empty it must be the case that $a < b$.
In particular it is noted that $a \ne b$.
Thus $a - b \ne 0$.
Let $I' := \openint {-1} 1$ denote the open real interval from $-1$ to $1$.
From Open Real Intervals are Homeomorphic, $I$ and $I'$ are homeomorphic.
Consider the real function $f: I' \to \R$ defined as:
:$\forall x \in I': \map f x = \dfrac x {1 - \size x}$
Then after some algebra:
:$\forall x \in \R: \map {f^{-1} } x = \dfrac x {1 + \size x}$
Both of these are defined, as $\size x < 1$.
By the Combination Theorem for Continuous Real Functions, both $f$ and $f^{-1}$ are continuous on the open real intervals on which they are defined.
Thus by definition $I'$ and $\R$ are homeomorphic.
From Homeomorphism Relation is Equivalence it follows that $I$ and $\R$ are homeomorphic.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.6$: Homeomorphisms: Examples $3.6.2 \ \text{(b)}$
Category:Open Real Interval is Homeomorphic to Real Number Line
Open Real Interval is Homeomorphic to Real Number Line/Proof 2

== Theorem ==
== Proof ==
Let $I := \openint {-\dfrac \pi 2} {\dfrac \pi 2}$ denote the open real interval from $-\dfrac \pi 2$ to $\dfrac \pi 2$.
Consider the real function $f: I \to \R$ defined as:
:$\forall x \in I: \map f x = \tan x$
Then we have:
:$\forall x \in \R: \map {f^{-1} } x = \arctan x$
From Homeomorphism Relation is Equivalence it follows that $I$ and $\R$ are homeomorphic.
Then by Open Real Intervals are Homeomorphic, $I$ is homeomorphic to every other open real interval.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 7$: Subspaces and Equivalence of Metric Spaces: Exercise $2$
* 
* 
Category:Open Real Interval is Homeomorphic to Real Number Line
Open Real Interval is Open Ball

== Theorem ==
Let $\R$ denote the real number line with the usual (Euclidean) metric.
Let $I := \openint a b \subseteq \R$ be an open real interval.
Then $I$ is the open $\epsilon$-ball $\map {B_\epsilon} \alpha$ of some $\alpha \in \R$.
== Proof ==
Let:
Then:
Thus:
:$\openint a b = \openint {\alpha - \epsilon} {\alpha + \epsilon}$
From Open Ball in Real Number Line is Open Interval:
:$\openint {\alpha - \epsilon} {\alpha + \epsilon} = \map {B_\epsilon} \alpha$
where $\map {B_\epsilon} \alpha$ is the open $\epsilon$-ball of $\alpha$ in $\R$.
Hence the result
== Also see ==
* Open Ball in Real Number Line is Open Interval
Category:Real Intervals
Category:Open Balls
Category:Real Number Line with Euclidean Metric
Open Real Interval is Open Set

== Theorem ==
Let $\R$ be the real number line considered as an Euclidean space.
Let $\openint a b \subset \R$ be an open interval of $\R$.
Then $\openint a b$ is an open set of $\R$.
=== Corollary ===
== Proof ==
Let $c \in \R$ such that $a < c < b$.
Let $\epsilon = \min \set {b - c, c - a}$.
From the definition of positive it follows that $\epsilon \in \R_{>0}$.
Let $\map {B_\epsilon} c = \openint {c - \epsilon} {c + \epsilon}$ be the open $\epsilon$-ball of $c$.
Let $y \in \map {B_\epsilon} c$.
Then:
:$\size {c - y} < \epsilon $
implies:
:$-\epsilon < c - y < \epsilon$
We have that:
In other words:
:$y \in \map {B_\epsilon} c \implies y \in \openint a b$
Thus:
:$\map {B_\epsilon} c \subseteq \openint a b$
It follows that, by definition, $\openint a b$ is a neighborhood of $c$.
The result follows by definition of open set.
== Sources ==
* : Chapter $\text{III}$: Metric Spaces: Compactness
* : $2$: Continuity generalized: metric spaces: $2.3$: Open sets in metric spaces: Example $2.3.9$
* 
* 
* : $\S 1.3$: Normed and Banach spaces. Topology of normed spaces
Category:Real Intervals
Category:Open Sets
Open Real Interval is Open Set/Corollary

== Corollary to Open Real Interval is Open Set ==
Let $\R$ be the real number line considered as an Euclidean space.
Let $A := \openint a \infty \subset \R$ be an open interval of $\R$.
Let $B := \openint {-\infty} b \subset \R$ be an open interval of $\R$.
Then both $A$ and $B$ are open sets of $\R$.
== Proof ==
From Open Real Interval is Open Set we have that for any $c \in \openint a b$ there exists an open $\epsilon$-ball of $c$ lying wholly within $\openint a b$.
When either of $a \to -\infty$ or $b \to \infty$ the result still holds.
The result follows by definition of open set.
Category:Real Intervals
Category:Open Sets