nyuuzyou/wikis huggingface dataset:
Hahn-Banach Separation Theorem/Normed Vector Space/Complex Case/Open Convex Set and Convex Set

== Theorem ==
Let $\struct {X, \norm \cdot}$ be a normed vector space over $\C$.
Let $\struct {X^\ast, \norm \cdot_{X^\ast} }$ be the normed dual space of $\struct {X, \norm \cdot}$.
Let $A \subseteq X$ be an open convex set.
Let $B \subseteq X$ be a convex set disjoint from $A$. 
Then there exists $f \in X^\ast$ and $c \in \R$ such that: 
:$A \subseteq \set {x \in X : \map \Re {\map f x} < c}$
and:
:$B \subseteq \set {x \in X : \map \Re {\map f x} \ge c}$
That is: 
:there exists $f \in X^\ast$ and $c \in \R$ such that $\map \Re {\map f a} < c \le \map \Re {\map f b}$ for each $a \in A$ and $b \in B$.
== Proof ==
Let $\struct {X_\R, \norm {\, \cdot \,}_\R}$ be the realification of $X$ equipped with the restricted norm. 
Applying Hahn-Banach Separation Theorem: Real Case: Open Convex Set and Convex Set to $\struct {X_\R, \norm {\, \cdot \,}_\R}$, there exists a bounded linear functional $g : X_\R \to \R$ and $c \in \R$ such that: 
:$A \subseteq \set {x \in X : \map g x < c}$
and:
:$B \subseteq \set {x \in X : \map g x \ge c}$
From Bounded Real-Valued Linear Functional is Real Part of Unique Bounded Complex-Valued Linear Functional, there exists $f \in X^\ast$ such that: 
:$\map g x = \map \Re {\map f x}$ 
for each $x \in X$. 
Then $f \in X^\ast$ satisfies: 
:$A \subseteq \set {x \in X : \map \Re {\map f x} < c}$
and:
:$B \subseteq \set {x \in X : \map \Re {\map f x} \ge c}$
as desired.
Category:Hahn-Banach Separation Theorem
Hahn-Banach Separation Theorem/Normed Vector Space/Real Case

== Theorem ==
Let $\struct {X, \norm \cdot}$ be a normed vector space over $\R$.
Let $\struct {X^\ast, \norm \cdot_{X^\ast} }$ be the normed dual space of $\struct {X, \norm \cdot}$.
=== Open Convex Set and Convex Set ===
=== Compact Convex Set and Closed Convex Set ===
Category:Hahn-Banach Separation Theorem
Hahn-Banach Separation Theorem/Normed Vector Space/Real Case/Open Convex Set and Convex Set/Lemma

== Lemma ==
Let $\struct {X, \norm \cdot}$ be a normed vector space over $\R$.
Let $\struct {X^\ast, \norm \cdot_{X^\ast} }$ be the normed dual space of $\struct {X, \norm \cdot}$.
Let $A \subseteq X$ be an open convex set.
Let $B \subseteq X$ be a convex set disjoint from $A$. 
Let $a_0 \in A$ and $b_0 \in B$.
Let:
:$v_0 = b_0 - a_0$
Let:
:$C = v_0 + A - B = \set {v_0 + a - b : a \in A, \, b \in B}$
Then: 
:$C$ is open, convex and contains $0$.
== Proof ==
We first show that $C$ is open. 
Note that we can write: 
:$\ds A - B = \bigcup_{b \mathop \in B} \set {a - b : a \in A} = \bigcup_{b \mathop \in B} \paren {A - b}$
Since $A$ is open, from Translation of Open Set in Normed Vector Space is Open, we have: 
:$A - b$ is open.
From Union of Open Sets of Normed Vector Space is Open, we therefore have: 
:$A - B$ is open.
From Translation of Open Set in Normed Vector Space is Open, we then obtain: 
:$C$ is open.
We now show that $C$ is convex.
From , we have: 
:$A - B$ is convex.
From Translation of Convex Set in Vector Space is Convex, we have: 
:$v_0 + A - B = C$ is convex.
We finally show that $0 \in C$.
Since:
:$-v_0 = a_0 - b_0 \in A - B$
We have:
:$b_0 - a_0 + a_0 - b_0 = 0 \in C$
Category:Hahn-Banach Separation Theorem
Hahn-Banach Theorem/Complex Vector Space

== Theorem ==
Let $X$ be a vector space over $\C$. 
Let $p : X \to \R$ be a seminorm on $X$. 
Let $X_0$ be a linear subspace of $X$. 
Let $f_0 : X_0 \to \C$ be a linear functional such that: 
:$\cmod {\map {f_0} x} \le \map p x$ for each $x \in X_0$.
Then there exists a linear functional $f$ defined on the whole space $X$ which extends $f_0$ and satisfies: 
:$\cmod {\map f x} \le \map p x$ for each $x \in X$.
That is, there exists a linear functional $f : X \to \C$ such that: 
:$\cmod {\map f x} \le \map p x$ for each $x \in X$
and:
:$\map f x = \map {f_0} x$ for each $x \in X_0$.
=== Corollary ===
== Proof ==
Define $g_0 : X_0 \to \R$ by: 
:$\map {g_0} x = \map \Re {\map {f_0} x}$
for each $x \in X_0$.
By Real Part of Linear Functional is Linear Functional, $g_0$ is an $\R$-linear functional.
Also, define $h_0 : X_0 \to \R$ by: 
:$\map {h_0} x = \map \Im {\map {f_0} x}$ 
for each $x \in X$. 
By Imaginary Part of Linear Functional is Linear Functional, $h_0$ is an $\R$-linear functional.
Now, for each $x \in X$, we have: 
Let $X_\R$ be the realification of $X$.
By Hahn-Banach Theorem: Real Vector Space on $X_\R$, there exists a $\R$-linear functional $g : X \to \R$ extending $g_0$ and satisfying: 
:$\cmod {\map g x} \le \map p x$ 
for each $x \in X$. 
Define $f : X \to \C$ by: 
:$\map f x = \map g x - i \map g {i x}$
for each $x \in X$. 
From Real Linear Functional is Real Part of Unique Linear Functional, we have that: 
:$f : X \to \C$ is a linear functional
and: 
:$\map g x = \map \Re {\map f x}$
for each $x \in X$. 
We now want to show that $f$ extends $f_0$.
Let $F_0$ be the restriction of $f$ to $X_0$.
We want to show that:
:$f_0 = F_0$
We have:
:$\map \Re {\map f x} = \map g x$
for each $x \in X$.
Hence: 
:$\map \Re {\map f x} = \map {g_0} x = \map \Re {\map {f_0} x}$ 
for each $x \in X_0$. 
So: 
:$\map \Re {\map {F_0} x} = \map \Re {\map {f_0} x}$
for each $x \in X_0$.
From Linear Functional on Complex Vector Space is Uniquely Determined by Real Part, it follows that: 
:$\map {F_0} x = \map {f_0} x$
for each $x \in X_0$.
So $f$ indeed extends $f_0$.
Now take $x \in X$. 
Pick $\lambda \in \C$ such that:
:$\cmod \lambda = 1$
and: 
:$\lambda \map f x = \cmod {\map f x}$ 
Then, since $f$ is $\C$-linear: 
Since: 
:$\cmod {\map f x} \in \R$
we have that:
:$\map g {i \lambda x} = 0$ 
Finally, we have: 
So $f$ is a linear functional satisfying our requirements.
== Sources ==
Category:Hahn-Banach Theorem
Hahn-Banach Theorem/Complex Vector Space/Corollary

== Corollary ==
Let $\struct {X, \norm \cdot}$ be a normed vector space over $\C$.
Let $X_0$ be a linear subspace of $X$. 
Let $f_0 : X_0 \to \C$ be a bounded linear functional.
Then $f_0$ can be extended to a bounded linear functional $f : X \to \C$ with: 
:$\norm f_{X^\ast} = \norm {f_0}_{\paren {X_0}^\ast}$
where $\norm \cdot_{X^\ast}$ and $\norm \cdot_{\paren {X_0}^\ast}$ are the norms of the normed dual spaces $X^\ast$ and $\paren {X_0}^\ast$.
== Proof ==
Category:Hahn-Banach Theorem
Hahn-Banach Theorem/Real Vector Space

== Theorem ==
Let $X$ be a vector space over $\R$. 
Let $p : X \to \R$ be a sublinear functional.
Let $X_0$ be a linear subspace of $X$. 
Let $f_0 : X_0 \to \R$ be a linear functional such that: 
:$\map {f_0} x \le \map p x$ for each $x \in X_0$. 
Then there exists a linear functional $f$ defined on the whole space $X$ which extends $f_0$ and satisfies: 
:$\map f x \le \map p x$ for each $x \in X$.
That is, there exists a linear functional $f : X \to \R$ such that: 
:$\map f x \le \map p x$ for each $x \in X$
and:
:$\map f x = \map {f_0} x$ for each $x \in X_0$.
=== Corollary 1 ===
 
=== Corollary 2 ===
 
== Proof ==
We first prove a lemma:
=== Lemma 1 ===
Let $P$ be the set of pairs $\tuple {G, g}$ such that: 
:$(1): \quad$ $G$ is a linear subspace of $X$ with $X_0 \subseteq G$
:$(2): \quad$ $g : G \to \R$ is a linear functional extending $f_0$
:$(3): \quad$ $\map g x \le \map p x$ for each $x \in G$.
This set is non-empty from Lemma 1.
Define the relation $\preceq$ on $P$ by:
:$\tuple {G, g} \preceq \tuple {H, h}$ :
::$(1): \quad$ $G \subseteq H$
::$(2): \quad$ $h$ extends $g$. 
=== Lemma 2 ===
We show that every non-empty chain in $\struct {P, \preceq}$ has an upper bound. 
We will then invoke Zorn's Lemma.
=== Lemma 3 ===
From Zorn's Lemma, we have that $P$ has a maximal element.
Let $\struct {G, g}$ be a maximal element of $P$. 
Then: 
:$G$ is a linear subspace of $X$ with $X_0 \subseteq G$
and:
:$g : G \to \R$ is a linear functional extending $f_0$ with: 
::$\map g x \le \map p x$ for each $x \in G$.
 suppose that $G \ne X$.
Then, by Lemma 1, there exists a proper superset $G'$ of $G$ and a linear functional $g^* : G' \to \R$ such that:
:$(1): \quad$ $G'$ is a linear subspace of $X$
:$(2): \quad$ $g^\ast$ extends $f_0$ with $\map {g^\ast} x \le \map p x$ for each $x \in G'$.
So:
:$\tuple {G, g} \preceq \tuple {G', g^\ast}$
Since $\tuple {G, g}$ is maximal, we have: 
:$\tuple {G, g} = \tuple {G', g^\ast}$
So $G = G'$. 
But $G'$ is a proper superset of $G$, which implies $G \ne G'$.
So we have a contradiction, giving $G = X$.
Setting $f = g$, we have therefore found a linear functional $f : X \to \R$ extending $f_0$ with: 
:$\map f x \le \map p x$ for each $x \in X$.
== Sources ==
* : $1.2$: Linear mappings
Category:Hahn-Banach Theorem
Hahn-Banach Theorem/Real Vector Space/Corollary 1

== Corollary ==
Let $X$ be a vector space over $\R$.
Let $p : X \to \R$ be a seminorm on $X$. 
Let $X_0$ be a linear subspace of $X$. 
Let $f_0 : X_0 \to \R$ be a linear functional such that: 
:$\size {\map {f_0} x} \le \map p x$ for each $x \in X_0$. 
Then there exists a linear functional $f$ defined on the whole space $X$ which extends $f_0$.
That is: 
:$\size {\map f x} \le \map p x$ for each $x \in X$
and:
:$\map f x = \map {f_0} x$ for each $x \in X_0$.
== Proof ==
From Seminorm is Sublinear Functional, we have: 
:$p$ is a sublinear functional.
So, from Hahn-Banach Theorem: Real Vector Space, there exists an extension $f$ with: 
:$\forall x \in X: \map f x \le \map p x$
Then, we have: 
So we also have: 
:$\forall x \in X: -\map f x \le \map p x$
So:
:$\forall x \in X: \size {\map f x} \le \map p x$
and so $f$ is the desired extension.
Category:Hahn-Banach Theorem
Hahn-Banach Theorem/Real Vector Space/Corollary 2

== Corollary ==
Let $\struct {X, \norm \cdot}$ be a normed vector space over $\R$.
Let $X_0$ be a linear subspace of $X$. 
Let $f_0 : X_0 \to \R$ be a bounded linear functional.
Then $f_0$ can be extended to a bounded linear functional $f : X \to \R$ with: 
:$\norm f_{X^\ast} = \norm {f_0}_{\paren {X_0}^\ast}$
where $\norm \cdot_{X^\ast}$ and $\norm \cdot_{\paren {X_0}^\ast}$ are the norms of the normed dual spaces $X^\ast$ and $\paren {X_0}^\ast$.
== Proof ==
First suppose that: 
:$\norm {f_0}_{\paren {X_0}^\ast} = 0$
Then, from positive definiteness of $\norm \cdot_{\paren {X_0}^\ast}$, we have: 
:$f_0 = 0$
That is: 
:$\map {f_0} x = 0$
for each $x \in X_0$.
Then the bounded linear functional $f : X \to \R$ with: 
:$\map f x = 0$
for all $x \in X$ clearly extends $f_0$ with: 
:$\norm f_{X^\ast} = 0 = \norm {f_0}_{\paren {X_0}^\ast}$.
Now take: 
:$\norm {f_0}_{\paren {X_0}^\ast} \ne 0$
From Fundamental Property of Norm on Bounded Linear Functionals, we have: 
:$\size {\map {f_0} x} \le \norm {f_0}_{\paren {X_0}^\ast} \norm x$
for each $x \in X_0$. 
Now define $p : X \to \R$ by: 
:$\map p x = \norm {f_0}_{\paren {X_0}^\ast} \norm x$
for each $x \in X$, so that: 
:$\size {\map {f_0} x} \le \map p x$
for each $x \in X_0$. 
From Positive Scalar Multiple of Norm on Vector Space is Norm, we have: 
:$p$ is a norm. 
In particular, from Norm on Vector Space is Seminorm, we have: 
:$p$ is a seminorm.
So, applying Hahn-Banach Theorem for Real Vector Spaces: Corollary 1, we have that: 
:there exists an extension $f : X \to \R$ of $f_0$ such that: 
::$\size {\map f x} \le \map p x$ for each $x \in X$.
We now only need to check that: 
:$\norm f_{X^\ast} = \norm {f_0}_{\paren {X_0}^\ast}$
Since: 
:$\size {\map f x} \le \norm {f_0}_{\paren {X_0}^\ast} \norm x$
From the definition of the norm on bounded linear functionals, we then have: 
:$\norm f_{X^\ast} \le \norm {f_0}_{\paren {X_0}^\ast}$
We have: 
:$\set {C > 0 : \size {\map f x} \le C \norm x \text { for all } x \in X} \subseteq \set {C > 0 : \size {\map f x} \le C \norm x \text { for all } x \in X_0}$
Since $f$ extends $f_0$ we have: 
:$\set {C > 0 : \size {\map f x} \le C \norm x \text { for all } x \in X_0} = \set {C > 0 : \size {\map {f_0} x} \le C \norm x \text { for all } x \in X_0}$
So, from Infimum of Subset, we have: 
:$\inf \set {C > 0 : \size {\map {f_0} x} \le C \norm x \text { for all } x \in X_0} \le \inf \set {C > 0 : \size {\map f x} \le C \norm x \text { for all } x \in X}$
From the definition of the norm on bounded linear functionals, we then have: 
:$\norm {f_0}_{\paren {X_0}^\ast} \le \norm f_{X^\ast}$
So we obtain the desired: 
:$\norm f_{X^\ast} = \norm {f_0}_{\paren {X_0}^\ast}$
Category:Hahn-Banach Theorem
Hahn-Banach Theorem/Real Vector Space/Lemma 2

== Lemma ==
Let $X$ be a vector space over $\R$. 
Let $p : X \to \R$ be a sublinear functional.
Let $X_0$ be a linear subspace of $X$. 
Let $f_0 : X_0 \to \R$ be a linear functional such that: 
:$\map {f_0} x \le \map p x$ for each $x \in X_0$. 
Let $P$ be the set of pairs $\tuple {G, g}$ such that: 
:$(1): \quad$ $G$ is a linear subspace of $X$ with $X_0 \subseteq G$
:$(2): \quad$ $g : G \to \R$ is a linear functional extending $f_0$
:$(3): \quad$ $\map g x \le \map p x$ for each $x \in G$.
Define the relation $\preceq$ on $P$ by:
:$\tuple {G, g} \preceq \tuple {H, h}$ :
::$(1): \quad$ $G \subseteq H$
::$(2): \quad$ $h$ extends $g$. 
Then: 
:$\struct {P, \preceq}$ is an ordered set.
== Proof ==
We check each of the conditions of an ordering for $\preceq$. 
=== Proof of $(1)$ ===
We show that for each $\tuple {G, g} \in P$, we have: 
:$\tuple {G, g} \preceq \tuple {G, g}$
We have, from Set is Subset of Itself: 
:$G \subseteq G$ 
We also have:
:$g$ extends $g$.
So, we have: 
:$\tuple {G, g} \preceq \tuple {G, g}$
=== Proof of $(2)$ ===
Suppose that $\tuple {G, g}, \tuple {H, h}, \tuple {I, i} \in P$ have: 
:$\tuple {G, g} \preceq \tuple {H, h}$
and:
:$\tuple {H, h} \preceq \tuple {I, i}$
We show that: 
:$\tuple {G, g} \preceq \tuple {I, i}$
We have: 
:$G \subseteq H$
and:
:$H \subseteq I$
From Subset Relation is Transitive, we have: 
:$G \subseteq I$
Since $h$ extends $g$, we have: 
:$\map g x = \map h x$ for each $x \in G$.
Since $i$ extends $h$, we have: 
:$\map h x = \map i x$ for each $x \in H$.
Since $G \subseteq H$, we in particular have: 
:$\map h x = \map i x$ for each $x \in G$.
So, we have: 
:$\map g x = \map i x$ for each $x \in G$. 
Since $G \subseteq I$, we therefore have: 
:$i$ extends $g$.
So, we have: 
:$\tuple {G, g} \preceq \tuple {I, i}$
=== Proof of $(3)$ ===
Suppose that $\tuple {G, g}$ and $\tuple {H, h}$ have: 
:$\tuple {G, g} \preceq \tuple {H, h}$
and:
:$\tuple {H, h} \preceq \tuple {G, g}$
We show that: 
:$\tuple {G, g} = \tuple {H, h}$
That is:
:$G = H$ and $g = h$.
Since: 
:$\tuple {G, g} \preceq \tuple {H, h}$
we have that: 
:$G \subseteq H$
and:
:$g$ extends $h$.
Since: 
:$\tuple {H, h} \preceq \tuple {G, g}$
we have: 
:$H \subseteq G$
and:
:$h$ extends $g$.
So, from the definition of set equality, we have: 
:$G = H$ 
So:
:$h$ is actually a linear functional $G \to \R$.
Since $h$ extends $g$, we have: 
:$\map h x = \map g x$ for each $g \in G$.
That is: 
:$g = h$ 
So, we have: 
:$\tuple {G, g} = \tuple {H, h}$
Category:Hahn-Banach Theorem
Hahn-Banach Theorem for Continuous Linear Functional on Locally Convex Space

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \PP}$ be a locally convex space over $\GF$ with its standard topology.
Let $X_0$ be a linear subspace of $X$. 
Let $f_0 : X_0 \to \GF$ be a continuous linear functional.
Then there exists a continuous linear functional $f : X \to \GF$ such that $f$ extends $f_0$. 
== Proof ==
By Normed Vector Space is Locally Convex Space and Norm on Vector Space is Seminorm, we can view the normed vector space $\struct {\GF, \cmod {\, \cdot \,} }$ as the locally convex space $\struct {\GF, \set {\cmod {\, \cdot \,} } }$.
From Characterization of Continuous Linear Transformations between Locally Convex Spaces, there exists $C > 0$, $n \in \N$ and $p_1, \ldots, p_n \in \PP$ such that: 
:$\ds \cmod {\map {f_0} x} \le C \max_{1 \le i \le n} \map {p_i} x$
for each $x \in X_0$.
Define $p : X \to \R_{\ge 0}$ by: 
:$\ds \map p x = C \max_{1 \le i \le n} \map {p_i} x$
for each $x \in X$. 
From Pointwise Maximum of Finite Family of Seminorms is Seminorm and Non-Negative Scalar Multiple of Seminorm on Vector Space is Seminorm, we have that $p$ is a seminorm.
So we have: 
:$\cmod {\map {f_0} x} \le \map p x$
for each $x \in X_0$.
From:
:the Hahn-Banach Theorem: Real Vector Space: Corollary 1 if $\GF = \R$
:the Hahn-Banach Theorem: Complex Vector Space: Corollary if $\GF = \C$
there exists a linear functional $f$ extending $f_0$ such that: 
:$\ds \cmod {\map f x} \le \map p x = C \max_{1 \le i \le n} \map {p_i} x$ 
for each $x \in X_0$. 
From Characterization of Continuous Linear Transformations between Locally Convex Spaces, $f$ is continuous.
So $f$ is a continuous linear functional satisfying the demand.
Category:Functional Analysis
Category:Locally Convex Spaces
Haidao Suanjing/Examples/Example 1

== Example of Problem from by ==
:What is the size of a square inscribed in the corner of a right-angled triangle to touch the hypotenuse?
== Solution ==
Let the lengths of the legs of the given right-angled triangle be $a$ and $b$.
Then the length of the side of the inscribed square is $\dfrac {a b} {a + b}$.
== Proof ==
Let $x$ be the length of the side of the inscribed square be $x$.
, let $a < b$.
Let the right-angled triangle be half of a rectangle whose sides are of length $a$ and $b$.
Let the rectangle be dissected along the straight lines shown.
:
Let the pieces of the dissection be assembled into a rectangle whose sides are of length $a + b$ and $x$.
Then we have:
:$a b = x \paren {a + b}$
and the result follows.
== Sources ==
* 
* 
* : Sun Tsu Suan Ching: $72$
Category:Haidao Suanjing
Half-Integer is Half Odd Integer

== Theorem ==
Let $r$ be a number.
Then $r$ is a half-integer $r = \dfrac n 2$ where $n$ is an odd integer.
== Proof ==
=== Necessary Condition ===
Let $r$ be a half-integer.
Then by definition $r = n + \dfrac 1 2$ for some $n \in \Z$.
Thus:
thus showing that $r$ is half of $2 n + 1$ for some $n \in \Z$.
By Odd Integer 2n + 1 it follows that $r$ is half of an odd integer.
=== Sufficient Condition ===
Let $k$ be an odd integer.
Then by Odd Integer 2n + 1:
:$k = 2 n + 1$
where $n \in \Z$.
Then:
thus showing that $\dfrac k 2$ is a half-integer.
Category:Integers
Category:Odd Integers
Category:Numbers
Half-Life of Radioactive Substance

== Theorem ==
Let a radioactive element $S$ decay with a rate constant $k$.
Then its half-life $T$ is given by:
:$T = \dfrac {\ln 2} k$
:
== Proof ==
Let $x_0$ be the quantity of $S$ at time $t = 0$.
At time $t = T$ the quantity of $S$ has been reduced to $x = \dfrac {x_0} 2$.
This gives:
== Sources ==
* : $1$: The Nature of Differential Equations: $\S 4$: Growth, Decay and Chemical Reactions
Category:Radioactive Decay
Half-Open Real Interval/Examples/Example 1

== Example of Half-Open Real Interval ==
Let $I$ be the unbounded closed real interval defined as:
:$I := \hointr 1 2$
Then $1 \in I$.
== Proof ==
By definition of open real interval:
:$I = \set {x \in \R: 1 \le x < 2}$
As $1 \le 1$ it follows that $1 \in I$.
== Sources ==
* : $\S 2$: Continuum Property: Exercise: $\S 2.10 \ (1) \ \text{(iii)}$
Category:Examples of Intervals
Half-Open Real Interval is Closed in some Open Intervals

== Theorem ==
Let $\R$ be the real number line considered as an Euclidean space.
Let $\hointr a b \subset \R$ be a half-open interval of $\R$.
Let $c < a$.
Then $\hointr a b$ is a closed set of $\openint c b$.
Similarly, let $d > b$.
Then the half-open interval $\hointl a b$ is a closed set of $\openint a d$.
== Proof ==
Consider:
:$A := \openint c b \setminus \hointr a b = \openint c a$
Then $A$ is an open interval.
By Open Real Interval is Open Set, $A$ is open in $\R$.
By the definition of the subspace topology it follows that $A$ is open in $\openint c b$.
Thus by the definition of closed set, $\openint c b \setminus A = \hointr a b$ is closed in $\openint c b$.
Mutatis mutandis, the argument also shows that $\hointl a b \subset \R$ is a closed set of $\openint a d$.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Examples $3.7.2$
Category:Real Intervals
Category:Closed Sets
Half-Open Real Interval is neither Open nor Closed

== Theorem ==
Let $\R$ be the real number line considered as an Euclidean space.
Let $\hointr a b \subset \R$ be a right half-open interval of $\R$.
Then $\hointr a b$ is neither an open set nor a closed set of $\R$.
Similarly, the left half-open interval $\hointl a b \subset \R$ is neither an open set nor a closed set of $\R$.
== Proof ==
From Half-Open Real Interval is not Open Set we have that neither $\hointr a b$ nor $\hointl a b$ is an open set of $\R$.
From Half-Open Real Interval is not Closed in Real Number Line we have that neither $\hointr a b$ nor $\hointl a b$ is a closed set of $\R$.
== Sources ==
* : Chapter $\text{III}$: Metric Spaces: Compactness
* : $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Examples $3.7.3 \ \text{(a)}$
Category:Real Intervals
Category:Open Sets
Category:Closed Sets
Half-Open Real Interval is not Closed in Real Number Line

== Theorem ==
Let $\R$ be the real number line considered as an Euclidean space.
Let $\hointr a b \subset \R$ be a right half-open interval of $\R$.
Then $\hointr a b$ is not a closed set of $\R$.
Similarly, the left half-open interval $\hointl a b \subset \R$ is not a closed set of $\R$.
== Proof ==
Consider:
:$A := \R \setminus \hointr a b = \openint \gets a \cup \hointr b \to$
Let $A:= \openint \gets a$ and let $B := \hointr b \to$.
Let $\map {B_\epsilon} b$ be the open $\epsilon$-ball of $b$.
We have that $b - \epsilon < b$ and so $\map {B_\epsilon} b = \openint {b - \epsilon} {b + \epsilon}$ does not lie entirely in $\hointr b \to$.
Now $b - \epsilon$ may itself lie in $A$.
However:
:$\hointr b \to \cap \map {B_\epsilon} b \ne \O$
and so there are elements of $\map {B_\epsilon} b$ which are not in $\openint \gets a \cup \hointr b \to$.
So $\openint \gets a \cup \hointr b \to$ is not an open set of $\R$.
Thus, by definition, $\hointr a b$ is not a closed set of $\R$.
Hence the result.
Mutatis mutandis, the argument also shows that $\hointl a b \subset \R$ is neither an open set nor a closed set of $\R$.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Examples $3.7.2$
Category:Real Intervals
Category:Closed Sets
Half-Open Real Interval is not Open Set

== Theorem ==
Let $\R$ be the real number line considered as an Euclidean space.
Let $\hointr a b \subset \R$ be a half-open interval of $\R$.
Then $\hointr a b$ is not an open set of $\R$.
Similarly, the half-open interval $\hointl a b \subset \R$ is not an open set of $\R$.
== Proof ==
Let $\epsilon \in \R_{>0}$.
Let $\map {B_\epsilon} a$ be the open $\epsilon$-ball of $a$.
We have that $a - \epsilon < a$ and so $\map {B_\epsilon} a = \openint {a - \epsilon} {a + \epsilon}$ does not lie entirely in $\hointr a b$.
Thus $\hointr a b$ is not a neighborhood $a$.
It follows that $\hointr a b$ is not an open set of $\R$.
the argument also shows, , that $\hointl a b \subset \R$ is not an open set of $\R$.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.3$: Open sets in metric spaces: Example $2.3.9$
Category:Real Intervals
Category:Open Sets