nyuuzyou/wikis huggingface dataset:
Interesting Number Paradox

== Paradox ==
Some numbers have interesting properties.
For example: $F_{12} = 144$ is a square Fibonacci number which also happens to be the square of its index.
Some numbers, on the other hand, have no such interesting properties.
Suppose all the positive integers were partitioned into $2$ sets: the interesting numbers $I$ and the uninteresting numbers $U$.
By the Well-Ordering Principle, $U$ has a smallest element.
Let $s \in U$ be that smallest element.
Then $s$ has an interesting property: it is the smallest uninteresting number.
Thus $s$ is not an element of $U$, but of $I$.
It follows that the smallest element of $U$ is not $s$ but $s'$ where $s' > s$.
But then $s'$ is now the smallest uninteresting number, and therefore interesting.
And so on.
It follows that there are no uninteresting numbers.
== Resolution ==
This is an antinomy.
The integers in $U$ are defined as uninteresting.
Once $U$ has been defined, redefining $s$ as interesting cannot be done.
Saying that $s$ is interesting because it is the smallest element of $U$ is to say that $U$ is improperly defined.
Ultimately, the point is that $U$ cannot be well-defined.
If $s$ is moved from $U$ to $I$, that which makes it interesting (it being the smallest element of $U$) no longer holds.
$s$ is no longer interesting and so no longer belongs to $I$.
Thus $s$ cannot belong either to $U$ or to $I$ and hence neither $U$ nor $I$ can be defined.
== Sources ==
* : $39$
* : $51$
Category:Antinomies
Category:Semantic Paradoxes
Interior (Topology)/Examples/Closed Real Interval in Closed Unbounded Real Interval

== Examples of Interiors in the context of Topology ==
Let $\R$ be the real number line under the usual (Euclidean) metric.
Let $M$ be the subspace of $\R$ defined as:
:$M = \hointl \gets b$
Let $S$ be the closed real interval defined as:
:$S = \closedint a b$
Then the interior of $S$ in $M$ is given by:
:$S^\circ = \hointl a b$
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Examples $3.7.25$
Category:Examples of Set Interiors
Interior Equals Closure of Subset of Discrete Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a discrete topological space.
Let $A \subseteq S$.
Then:
:$A = A^\circ = A^-$
where:
:$A^\circ$ is the interior of $A$
:$A^-$ is the closure of $A$.
== Proof ==
Let $A \subseteq S$.
Then from Set in Discrete Topology is Clopen it follows that $A$ is both open and closed in $T$.
From Closed Set Equals its Closure we have that $A = A^-$.
From Set Interior is Largest Open Set, we have that $A^\circ = A$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $1 \text { - } 3$. Discrete Topology: $4$
Category:Discrete Topology
Interior equals Complement of Closure of Complement

== Theorem ==
Let $T$ be a topological space.
Let $H \subseteq T$.
Let $H^-$ denote the closure of $H$ and $H^\circ$ denote the interior of $H$.
Let $H^\prime$ denote the complement of $H$ in $T$:
:$H^\prime = T \setminus H$
Then:
:$H^\circ = H^{\prime \, - \, \prime}$
== Proof ==
Category:Set Closures
Category:Set Interiors
Interior is Subset of Exterior of Exterior

== Theorem ==
Let $T$ be a topological space.
Let $H \subseteq T$.
Let $H^e$ denote the exterior of $H$, and let $H^\circ$ denote the interior of $H$.
Then:
:$H^\circ \subseteq \paren {H^e}^e$
== Proof ==
== Also see ==
* Interior may not equal Exterior of Exterior
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Closures and Interiors
Category:Set Interiors
Category:Set Exteriors
Interior is Subset of Interior of Closure

== Theorem ==
Let $T$ be a topological space.
Let $H \subseteq T$.
Let $H^\circ$ denote the interior of $H$.
Let $H^-$ denote the closure of $H$.
Then:
:$H^\circ \subseteq \left({H^-}\right)^\circ$
== Proof ==
From Set is Subset of its Topological Closure, we have $H \subseteq H^-$.
The result follows directly from Interior of Subset.
Category:Set Closures
Category:Set Interiors
Interior is Union of Elements of Basis

== Theorem ==
Let $T = \left({S, \tau}\right)$ be a topological space.
Let $B$ be a basis of $T$.
Let $V$ be a subset of $S$.
Then $V^\circ = \bigcup \left\{ {G \in B: G \subseteq V}\right\}$
where $V^\circ$ denotes the interior of $V$.
== Proof ==
By definition of interior:
:$\left\{ {G \in B: G \subseteq V}\right\} = \left\{ {G \in B: G \subseteq V^\circ}\right\}$
and
:$V^\circ$ is open.
Thus by Open Set is Union of Elements of Basis:
:$V^\circ = \bigcup \left\{ {G \in B: G \subseteq V}\right\}$
== Sources ==
* 
Category:Topological Bases
Interior is Union of Way Above Closures

== Theorem ==
Let $\left({S, \preceq, \tau}\right)$ be a complete continuous topological lattice with Scott topology.
Let $X \subseteq S$.
Then $X^\circ = \bigcup \left\{ {x^\gg: x \in S \land x^\gg \subseteq X}\right\}$
where
:$X^\circ$ denotes the interior of $X$,
:$x^\gg$ denotes the way above closure of $x$.
== Proof ==
We have:
:$\left\{ {G \in \left\{ {g^\gg: g \in S}\right\}: G \subseteq X}\right\} = \left\{ {x^\gg: x \in S \land x^\gg \subseteq X}\right\}$
By Way Above Closures Form Basis:
:$\left\{ {x^\gg: x \in S}\right\}$ is basis of $\left({S, \tau}\right)$.
By Interior is Union of Elements of Basis:
:$X^\circ = \bigcup \left\{ {x^\gg: x \in S \land x^\gg \subseteq X}\right\}$
== Sources ==
* 
* 
Category:Topological Order Theory
Category:Continuous Lattices
Interior may not equal Exterior of Exterior

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $A \subseteq S$ be a subset of the underlying set $S$ of $T$.
Let $A^e$ be the exterior of $A$.
Let $A^\circ$ be the interior of $A$.
Then it is not necessarily the case that:
:$A^{ee} = A^\circ$
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Interior is Subset of Exterior of Exterior
Category:Set Exteriors
Category:Set Interiors
Category:Interior may not equal Exterior of Exterior
Interior may not equal Exterior of Exterior/Proof 1

== Theorem ==
== Proof ==
We have from Interior is Subset of Exterior of Exterior:
:$A^\circ \subseteq A^{ee}$
It remains to be shown that there exist $A \subseteq S$ such that:
:$A^\circ \ne A^{ee}$
Let $a, b, c \in R$ where $a < b < c$.
Let $A$ be the union of the two adjacent open intervals:
:$A := \openint a b \cup \openint b c$
From Exterior of Exterior of Union of Adjacent Open Intervals:
:$A^{ee} = \openint a c$
From Interior of Union of Adjacent Open Intervals:
:$A^\circ := \openint a b \cup \openint b c$
Thus:
:$b \in A^{ee}$
but:
:$b \notin A^\circ$
and so:
:$A^{ee} \subsetneq A^\circ$
== Sources ==
* : Part $\text {II}$: Counterexamples: $32$. Special Subsets of the Real Line: $5 \ \text{(b)}$
Category:Union of Adjacent Open Intervals
Category:Interior may not equal Exterior of Exterior
Interior may not equal Exterior of Exterior/Proof 2

== Theorem ==
== Proof ==
Proof by Counterexample:
Let $\struct {S, \preccurlyeq}$ be a totally ordered set.
Let $T = \struct {S, \tau}$ be the Sorgenfrey line on $\struct {S, \preccurlyeq}$.
Let $A \subseteq S$ denote the subset of $S$ defined as:
:$A = \openint a {+\infty}$
By Exterior of Exterior in Sorgenfrey Line is not necessarily Interior:
:$A^{ee} = \hointr a {+\infty}$
while:
:$A^\circ = \openint a {+\infty}$
The result is apparent.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Closures and Interiors
Category:Interior may not equal Exterior of Exterior
Interior of Balanced Set containing Origin in Topological Vector Space is Balanced

== Theorem ==
Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a topological vector space over $\Bbb F$.
Let $B \subseteq X$ be a balanced set such that: 
:${\mathbf 0}_X \in B^\circ$
where $B^\circ$ is the interior of $B$.
Then $B^\circ$ is balanced.
== Proof ==
Let $\lambda \in \Bbb F$ have $0 < \cmod \lambda \le 1$. 
Then, we have: 
:$\lambda B \subseteq B$ 
From Interior of Subset, we have: 
:$\paren {\lambda B}^\circ \subseteq B^\circ$
Then from Dilation of Interior of Set in Topological Vector Space is Interior of Dilation we have: 
:$\lambda B^\circ \subseteq B^\circ$
for $\lambda \in \Bbb F$ with $0 < \cmod \lambda \le 1$.
By hypothesis, we have: 
:${\mathbf 0}_X \in B^\circ$
and so:
:$\set { {\mathbf 0}_X} \subseteq B^\circ$
So we have: 
:$\lambda B^\circ \subseteq B^\circ$
if $\lambda = 0$. 
So:
:$\lambda B^\circ \subseteq B^\circ$
for all $\lambda \in \Bbb F$ with $\cmod \lambda \le 1$, so $B^\circ$ is balanced.
== Sources ==
* : $1.13$: Theorem
Category:Balanced Sets
Category:Set Interiors
Category:Topological Vector Spaces
Interior of Cartesian Product is Product of Interiors

== Theorem ==
Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $T = \struct {T_1 \times T_2, \tau}$ be the product space of $T_1$ and $T_2$, where $\tau$ is the product topology on $S$.
Let $H \subseteq T_1$ and $K \subseteq T_2$.
Then:
:$\Int {H \times K} = \Int H \times \Int K$
where $\Int H$, for example, denotes the interior of $H$.
== Proof ==
By definition of interior, both $\Int H$ and $\Int K$ are open in $T_1$ and $T_2$ respectively.
By Projection from Product Topology is Continuous, it follows that $\Int {H \times K}$ is an open set of $T$.
It remains to be shown that $\Int {H \times K}$ is the largest open subset of $H \times K$.
Let $H' \times K'$ be an open set of $T$ such that $H' \times K' \subseteq H \times K$.
Then from Projection from Product Topology is Open, $H'$ and $K'$ are open sets of $T_1$ and $T_2$ respectively such that $H' \subseteq H$ and $K' \subseteq K$.
By definition of interior, $\Int H$ and $\Int K$ are the largest open sets of $T_1$ and $T_2$ respectively such that $\Int H \subseteq H$ and $\Int K \subseteq K$.
Hence $H' \subseteq \Int H$ and $K' \subseteq \Int K$.
Hence by Cartesian Product of Subsets:
:$H' \times K' \subseteq \Int H \times \Int K$
That is, any open subset of $H \times K$ is also a subset of $\Int {H \times K}$.
Hence $\Int {H \times K}$ is the largest open subset of $H \times K$.
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 35 \ \text {(i)}$
Category:Set Interiors
Category:Product Topology
Interior of Closed Real Interval is Open Real Interval

== Theorem ==
Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
 
Let $\closedint a b$ be a closed interval of $\R$.
Then the interior of $\closedint a b$ is the open interval $\openint a b$.
== Proof ==
By definition, the interior of $\closedint a b$ is the largest open set contained in $\closedint a b$.
From Open Sets in Real Number Line it follows that $\openint a b$ is an open set of $\struct {\R, \tau_d}$.
By definition of open interval, $\openint a b$ is contained in $\closedint a b$.
Suppose $U$ is an open set of $\struct {\R, \tau_d}$ which is contained in $\closedint a b$.
The only way $U$ could be bigger than $\openint a b$ is if either $a \in U$ or $b \in U$ or both.
Suppose $a \in U$.
Then by Open Sets in Real Number Line it follows that $a \in \openint p q$ for some $p, q \in \R$ such that $\openint p q \subseteq U$.
From Real Numbers are Densely Ordered:
:$\exists r \in \R: p < r < a$
and so:
:$\exists r \in U: r < a$
which means:
:$r \notin \openint a b$
That is, $a$ is not in an open set of $\struct {\R, \tau_d}$ which is contained in $\closedint a b$.
Thus $a$ is not in the interior of $\closedint a b$.
By a similar argument it is shown that neither is $b$ in the interior of $\closedint a b$.
Hence the result.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Examples $3.7.25$
Category:Closed Sets
Category:Examples of Set Interiors
Category:Real Number Line with Euclidean Topology
Interior of Closed Set of Particular Point Space

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a particular point space.
Let $V \subseteq S$ be closed in $T$ such that $V \ne S$.
Then:
:$V^\circ = \O$
where $V^\circ$ denotes the interior of $V$.
== Proof ==
By definition:
:$\forall U \in \tau_p, U \ne \O: p \in U$
Thus if $V$ is closed in $T$:
:$\exists U \subseteq T: V = \relcomp S U$
So $p \notin V$.
Hence no open set of $T$ can be a subset of $V$ unless it is $\O$.
Hence the result, by definition of interior.
== Sources ==
* : Part $\text {II}$: Counterexamples: $8 \text { - } 10$. Particular Point Topology: $2$
Category:Closed Sets
Category:Examples of Set Interiors
Category:Particular Point Topology
Interior of Closure is Regular Open

== Theorem ==
Let $T = \left({S, \tau}\right)$ be a topological space.
Let $H \subseteq S$.
Then $H^{- \circ}$ is regular open.
== Proof ==
We must show that $H^{- \circ - \circ} = H^{- \circ}$.
First we show that $H^{- \circ - \circ} \subseteq H^{- \circ}$.
By the definition of interior:
:$H^{- \circ} \subseteq H^-$
By Topological Closure of Subset is Subset of Topological Closure:
:$H^{- \circ -} \subseteq H^{--} = H^-$
By Interior of Subset:
:$H^{- \circ - \circ } \subseteq H^{- \circ}$
Next we show $H^{- \circ - \circ} \supseteq H^{- \circ}$.
Again by the definition of interior:
: $H^{- \circ -} \supseteq H^{- \circ}$
By Interior of Subset:
: $H^{- \circ - \circ} \supseteq H^{- \circ \circ} = H^{- \circ}$
Thus by definition of set equality:
: $H^{- \circ - \circ} = H^{- \circ}$
== Sources ==
* : $\text{I}: \ \S 2$: Interior, Exterior, Boundary and Closure
Category:Regular Open Sets
Interior of Closure of Interior of Union of Adjacent Open Intervals

== Theorem ==
Let $a, b, c \in R$ where $a < b < c$.
Let $A$ be the union of the two adjacent open intervals:
:$A := \openint a b \cup \openint b c$
Then:
:$A^{\circ - \circ} = A^{- \circ} = \openint a c$
where:
:$A^\circ$ is the interior of $A$
:$A^-$ is the closure of $A$.
== Proof ==
From Interior of Union of Adjacent Open Intervals:
:$A^\circ = A$
From Closure of Union of Adjacent Open Intervals:
:$A^- = \closedint a c$
From Interior of Closed Real Interval is Open Real Interval:
:$\closedint a c^\circ = \openint a c$
whence the result.
== Sources ==
* : Part $\text {II}$: Counterexamples: $32$. Special Subsets of the Real Line: $5 \ \text{(a)}$
Category:Examples of Set Interiors
Category:Set Closures
Category:Union of Adjacent Open Intervals
Interior of Convex Angle is Convex Set

== Theorem ==
Let $\mathbf v, \mathbf w$ be two non-zero vectors in $\R^2$, and let $p$ be a point in $\R^2$.
Suppose that the angle between $\mathbf v$ and $\mathbf w$ is a convex angle.
Then the set 
:$U = \set {p + s t \mathbf v + \paren {1 - s} t \mathbf w : s \in \openint 0 1, t \in \R_{>0} }$
is a convex set.
== Proof ==
Let $p_1, p_2 \in U$.
Then for $i \in \set {1, 2}$, $p_i = p + s_i t_i \mathbf v + \paren {1 - s_i} t_i \mathbf w$ for some $s_i \in \openint 0 1, t_i \in \R_{>0}$.
:
, assume that $t_1 \le t_2$.
Suppose that $q \in \R^2$ lies on the line segment joining $p_1$ and $p_2$, so:
As $t_1 \le t_2$, it follows that $r \in \R_{>0}$.
We have:
:$\dfrac {\paren {1 - s} s_1 t_1 + s s_2 t_2} r > 0$
and so:
It follows that:
:$\dfrac {\paren {1 - s} s_1 t_1 + s s_2 t_2} r \in \openint 0 1$
Then $q \in U$.
It follows by definition that $U$ is convex.
Category:Vector Spaces
Category:Convex Sets (Vector Spaces)
Interior of Convex Set in Topological Vector Space is Convex

== Theorem ==
Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a topological vector space over $\Bbb F$.
Let $C \subseteq X$ be a convex set.
Then the interior of $C$, $C^\circ$, is convex.
== Proof ==
Let $t \in \closedint 0 1$. 
Since $C$ is convex, we have: 
:$t C + \paren {1 - t} C \subseteq C$
Since $C^\circ \subseteq C$, we have: 
From Set Interior is Largest Open Set we have that $C^\circ$ is open.
From Dilation of Open Set in Topological Vector Space is Open, we have that $t C^\circ$ and $\paren {1 - t} C^\circ$ are open.
From Sum of Set and Open Set in Topological Vector Space is Open, we have that $t C^\circ + \paren {1 - t} C^\circ$ is open.
So $t C^\circ + \paren {1 - t} C^\circ$ is an open set contained in $C$, so we have: 
:$t C^\circ + \paren {1 - t} C^\circ \subseteq C^\circ$
from the definition of interior.
Since this holds for all $t \in \closedint 0 1$, it follows that $C^\circ$ is convex.
== Sources ==
* : $1.13$: Theorem
Category:Set Interiors
Category:Topological Vector Spaces
Category:Convex Sets (Vector Spaces)
Interior of Finite Intersection equals Intersection of Interiors

== Theorem ==
Let $T$ be a topological space.
Let $n \in \N$.
Let:
:$\forall i \in \set {1, 2, \dotsc, n}: H_i \subseteq T$
Then:
:$\ds \paren {\bigcap_{i \mathop = 1}^n H_i}^\circ = \bigcap_{i \mathop = 1}^n H_i^\circ$
where $H_i^\circ$ denotes the interior of $H_i$.
== Proof ==
In the following, $H_i^-$ denotes the closure of the set $H_i$.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Closures and Interiors
Category:Set Interiors
Category:Set Intersection
Interior of Intersection may not equal Intersection of Interiors

== Theorem ==
Let $T$ be a topological space.
Let $\mathbb H$ be a set of subsets of $T$.
That is, let $\mathbb H \subseteq \powerset T$ where $\powerset T$ is the power set of $T$.
Then it is not necessarily the case that:
:$\ds \paren {\bigcap_{H \mathop \in \mathbb H} H}^\circ = \bigcap_{H \mathop \in \mathbb H} H^\circ$
where $H^\circ$ denotes the interior of $H$.
== Proof ==
From Intersection of Interiors contains Interior of Intersection it is seen that it is always the case that:
:$\ds \paren {\bigcap_{H \mathop \in \mathbb H} H}^\circ \subseteq \bigcap_{H \mathop \in \mathbb H} H^\circ$
From Interior of Finite Intersection equals Intersection of Interiors it is seen that if $\mathbb H$ is finite, then:
:$\ds \paren {\bigcap_{H \mathop \in \mathbb H} H}^\circ = \bigcap_{H \mathop \in \mathbb H} H^\circ$
It remains to be shown that in the case where $\mathbb H$ is infinite, then it is possible that:
:$\ds \paren {\bigcap_{H \mathop \in \mathbb H} H}^\circ \ne \bigcap_{H \mathop \in \mathbb H} H^\circ$
Proof by Counterexample:
Let $A_n$ denote the open real interval:
:$A_n := \openint {1 - \dfrac 1 n} {1 + \dfrac 1 n}$
for all $n \in \Z_{>0}$.
From Open Sets in Real Number Line, each of $A_n$ is an open set in $\R$.
From Interior of Open Set:
:$\forall n \in \Z_{>0}: A_n^\circ = A_n$
Thus:
:$\ds \bigcap A_n = \bigcap A_n^\circ = \set 1$
However:
:$\ds \paren {\bigcap A_n}^\circ = \set 1^\circ = \O$
Hence the result.
== Also see ==
* Interior of Finite Intersection equals Intersection of Interiors
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Closures and Interiors
* : Part $\text {II}$: Counterexamples: $32$. Special Subsets of the Real Line: $4$
Category:Set Interiors
Category:Set Intersection
Interior of Jordan Curve is Simply Connected

== Theorem ==
Let $\gamma : \closedint 0 1 \to \R^2$ be a Jordan curve.
Let $\Int \gamma$ denote the interior of $\gamma$.
Let $\tau_0$ denote the subspace topology on $\Int \gamma$, induced by the Euclidean topology on $\R^2$.
Then $\struct {\Int \gamma, \tau_0}$ is simply connected.
== Proof ==
Let $\map {B_1} { \mathbf 0 }$ denote the open ball in $\R^2$ with radius $1$ and center equal to the origin $\bszero$.
We are given $\Int \gamma$ is the interior of $\gamma$.
From the Jordan-Schönflies Theorem, it follows that $\Int \gamma$ and $\map {B_1} { \mathbf 0 }$ are homeomorphic.
From Open Ball is Simply Connected, it follows that $\map {B_1} { \mathbf 0 }$, considered as a subspace of $\R^2$, is simply connected.
From Simple Connectedness is Preserved under Homeomorphism, it follows that $\struct {\Int \gamma, \tau_0}$ is simply connected.
Category:Jordan Curves
Category:Simply Connected Spaces
Interior of Open Set

== Theorem ==
Let $T = \left({S, \tau}\right)$ be a topological space.
Let $U \subseteq T$ be open in $T$.
Then:
: $U^\circ = U \iff U \in \tau$
where $U^\circ$ is the interior of $U$.
That is, a subset of $S$ is open in $T$ it equals its interior.
== Proof ==
=== Necessary Condition ===
Let $U \subseteq T$ be open in $T$.
From Set Interior is Largest Open Set:
: $U \subseteq U^\circ$
But by definition of interior:
: $U^\circ \subseteq U$
Hence $U^\circ = U$.
=== Sufficient Condition ===
Suppose $U^\circ = U$.
By definition $U^\circ$ is the union of all subsets of $U$ which are open in $U$.
By the definition of a topology, that means $U$ itself must also be open.
Category:Set Interiors
Category:Open Sets
Interior of Proper Subspace of Normed Vector Space is Empty

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space over $\GF$. 
Let $U$ be a proper vector subspace of $X$. 
Let $U^\circ$ be the interior of $U$. 
Then $U^\circ = \O$. 
 
== Proof ==
 $U^\circ \ne \O$.
Take $x \in U^\circ$.
Then there exists $\epsilon > 0$ such that $\map {B_\epsilon} x \subseteq U$, where $\map {B_\epsilon} x$ is the open ball centered at $x$ with radius $\epsilon$.
Since $U$ is a vector subspace, we have:
:$\map {B_\epsilon} 0 = \map {B_\epsilon} x - x \subseteq U$
So if $x \in X$ has $\norm x < \epsilon$, then $x \in U$.
Take $x \in U \setminus \set {\mathbf 0_X}$. 
Then, we have: 
:$\norm {\dfrac \epsilon {2 \norm x} x} = \dfrac \epsilon 2 < \epsilon$
so that: 
:$\dfrac \epsilon {2 \norm x} x \in U$
Since $U$ is a vector subspace, we have $x \in U$. 
We also have:
:$\mathbf 0_X \in U$
so:
:$U = X$
This contradicts the hypothesis that $U \ne X$.
So by Proof by Contradition we have:
:$U^\circ = \O$
Category:Normed Vector Spaces
Category:Set Interiors
Interior of Set of Rational Numbers in Real Numbers is Empty

== Theorem ==
Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
 
Let $\Q$ be the subspace of rational numbers.
Then the interior of $\Q$ in $\R$ is the empty set $\O$.
== Proof ==
Consider the set of set of irrational numbers $\R \setminus \Q$.
By definition:
:$\R \setminus \Q = \relcomp \R \Q$
where $\relcomp \R \Q$ denotes the relative complement of $\Q$ in $\R$.
We have that Irrationals are Everywhere Dense in Reals.
Hence by definition of everywhere dense, the closure of $\R \setminus \Q$ in $\R$ is $\R$ itself.
By Relative Complement with Self is Empty Set:
:$\relcomp \R \R = \O$
The result follows from Interior equals Complement of Closure of Complement.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Examples $3.7.25$
Category:Examples of Set Interiors
Category:Rational Number Space
Interior of Set of Real Numbers in Complex Numbers is Empty

== Theorem ==
Let $\tuple {\C, d}$ be the complex Euclidean space.
Let $\R$ be the subspace of real numbers.
Then the interior of $\R$ in $\C$ is the empty set $\O$.
=== Corollary ===
== Proof ==
 that: 
:$\R^\circ \ne \O$
Let $x \in \R^\circ$.
From the definition of an open subset of $\C$, there exists $\epsilon > 0$ such that:
:$\set {z \in \C : \cmod {z - x} < \epsilon} \subseteq \R^\circ$
Consider:
:$\ds z = x + \frac \epsilon 2 i \in \C \setminus \R$
Then, we have:
:$\ds \cmod {z - x} = \cmod {x + \frac \epsilon 2 i - x} = \frac \epsilon 2 < \epsilon$ 
Hence $z \in \R^\circ$.
But by the definition of interior, we have $\R^\circ \subseteq \R$. 
This contradicts the fact that $z \in \C \setminus \R$.
Category:Examples of Set Interiors
Category:Complex Numbers
Category:Real Numbers
Category:Interior of Set of Real Numbers in Complex Numbers is Empty
Interior of Set of Real Numbers in Complex Numbers is Empty/Corollary

== Theorem ==
Let $\tuple {\C, d}$ be the complex Euclidean space.
Consider $S \subseteq \R$ as a topological subspace of $\tuple {\C, d}$.
Then the interior of $S$ in $\C$ is the empty set $\O$.
== Proof ==
From Interior of Subset we have:
:$S^\circ \subseteq \R^\circ$
From Interior of Set of Real Numbers in Complex Numbers is Empty, we have:
:$\R^\circ = \O$
Hence $S^\circ \subseteq \O$ and so:
:$S^\circ = \O$
from Subset of Empty Set.
Category:Interior of Set of Real Numbers in Complex Numbers is Empty