hcju/proofwikiqa huggingface dataset:
Let $x \in \R \setminus \Q$. Let $U \subseteq \R$ be an open set of $\struct {\R \setminus \Q, \tau_d}$ such that $x \in U$. From Basis for Euclidean Topology on Real Number Line, the set of all open real intervals of $\R$ form a basis for $\struct {\R, \tau_d}$. By Basis for Topological Subspace, the set of all intersections of $\R \setminus \Q$ and open real intervals of $\R$ form a basis for $\struct {\R \setminus \Q, \tau_d}$. So there exists $V = \paren {\R \setminus \Q} \cap \openint {x - \epsilon} {x + \epsilon} \subseteq U$ for some $\epsilon > 0$ such that $x \in V$. By Between two Real Numbers exists Irrational Number we have: :$\exists r \in \R \setminus \Q: x < r < x + \epsilon$. Then $r \in \paren {\R \setminus \Q} \cap \openint {x - \epsilon} {x + \epsilon}$. Therefore $x$ is not isolated. Hence $\struct {\R \setminus \Q, \tau_d}$ is dense-in-itself. {{qed}}
For a space to be scattered, it needs by definition to have no subset which is dense-in-itself. From Irrational Number Space is Dense-in-itself, $\struct {\R \setminus \Q, \tau_d}$ is dense-in-itself. Hence the result. {{qed}}
We first show that $\openint 0 1 \cap \Q$ is totally bounded. Let $\epsilon \in \R_{> 0}$. By the Archimedean Property of $\R$: :$\exists n \in \N: \dfrac 1 n < \epsilon$ We pick the numbers $\dfrac i n \in \openint 0 1 \cap \Q$, where $i \in \N$ and $0 < i < n$. Then for all $x \in \openint 0 1 \cap \Q$ and $x \ge \dfrac 1 n$: {{begin-eqn}} {{eqn | l = \inf_{0 \mathop < i \mathop < n} \map d {x, \frac i n}       | r = \inf_{0 \mathop < i \mathop < n} \size {x - \frac i n}       | c = {{Defof|Euclidean Metric on Real Number Line}} }} {{eqn | o = \le       | r = \size {\frac {n x} n - \frac {\floor {n x} } n }       | c = $\floor {n x}$ is an integer strictly between $0$ and $n$ }} {{eqn | o = <       | r = \frac 1 n       | c = Real Number minus Floor }} {{eqn | o = <       | r = \epsilon }} {{end-eqn}} For $x \le \dfrac 1 n$: {{begin-eqn}} {{eqn | l = \inf_{0 \mathop < i \mathop < n} \map d {x, \frac i n}       | r = \inf_{0 \mathop < i \mathop < n} \size {x - \frac i n}       | c = {{Defof|Euclidean Metric on Real Number Line}} }} {{eqn | o = \le       | r = \frac 1 n - x }} {{eqn | o = \le       | r = \frac 1 n }} {{eqn | o = <       | r = \epsilon }} {{end-eqn}} This shows that $\openint 0 1 \cap \Q$ totally bounded. From the Heine-Borel Theorem on a metric space, $\openint 0 1 \cap \Q$ is compact {{iff}} it is both totally bounded and complete. From Rational Number Space is not Complete Metric Space it follows that $\openint 0 1 \cap \Q$ is not compact. {{qed}}
We have that $A \subseteq \R$. By definition, $\struct {A, \tau_d}$ is a subspace of $\struct {\R, \tau_d}$. Hence the result from Topological Subspace is Topological Space. {{qed}}
By definition, the closure of $A$ is: :$A \cup A'$ where $A'$ is the derived set of $A$. By definition of derived set, $A'$ consists of all the limit points of $A$ in $\R$. From Zero is Limit Point of Integer Reciprocal Space, the only limit point of $A$ is $0$. Hence the result: :$A^- = A \cup \set 0$ {{qed}}
There are three cases to consider: === Points in $A$ === Consider $x \in \R$ such that $x \in A$. That is, $x = \dfrac 1 n$ for some $n \in \N$. Let: :$d = \dfrac 1 n - \dfrac 1 {n + 1}$ Consider the open real interval: :$I := \openint {\dfrac 1 n - d} {\dfrac 1 n + d} \subseteq \R$ By definition, $I$ is an open set of $\R$. Thus $I$ is an open set of $\R$ which contains no element of $A$ distinct from $x$. Thus $x$ is not a limit point of $A$ in $\R$. {{qed|lemma}} === Non-Zero Points not in $A$ === Let $x \in \R$ such that $x \ne 0$ and $x \notin A$. Let: :$d := \min \set {\size {x - m}: m \in A}$ that is, the smallest distance from $x$ to an element of $A$. Consider the open real interval: :$I := \openint {x - d} {x + d} \subseteq \R$ By definition, $I$ is an open set of $\R$. Thus $I$ is an open set of $\R$ which contains no element of $A$ (distinct from $x$ or not). Thus $x$ is not a limit point of $A$ in $\R$. {{qed|lemma}} === Zero === Finally, consider the point $0$. Let $U$ be an open set of $\R$ which contains $0$. From Open Sets in Real Number Line, there exists an open interval $I$ of the form: :$I := \openint {-a} b \subseteq U$ By the Archimedean Principle: :$\exists n \in \N: n > \dfrac 1 b$ and so: :$\exists n \in \N: \dfrac 1 n < b$ But $\dfrac 1 n \in A$. Thus an open set $U$ which contains $0$ contains at least one element of $A$ (distinct from $0$). Thus, by definition, $0$ is a limit point of $A$ in $\R$. {{qed|lemma}} Thus the '''only''' limit point of $A$ in $\R$ is $0$. {{qed}} Category:Integer Reciprocal Space Category:Limit Points kkrn344ko442q6qhfupxqkb5uf7hryz
Let $U$ be an open set of $\R$ which contains $0$. From Open Sets in Real Number Line, there exists an open interval $I$ of the form: :$I := \openint {-a} b \subseteq U$ By the Archimedean Principle: :$\exists n \in \N: n > \dfrac 1 b$ and so: :$\exists n \in \N: \dfrac 1 n < b$ But $\dfrac 1 n \in B$. Thus an open set $U$ which contains $0$ contains at least one element of $B$ (distinct from $0$). Thus, by definition, $0$ is a limit point of $B$ in $\R$. {{qed}}
Let $U$ be an open set of $\R$ which contains $0$. From Open Sets in Real Number Line, there exists an open interval $I$ of the form: :$I := \openint {-a} b \subseteq U$ By the Archimedean Principle: :$\exists n \in \N: n > \dfrac 1 b$ and so: :$\exists n \in \N: \dfrac 1 n < b$ Let: :$M := \set {m \in \N: m \ge n}$ Then: :$\forall m \in M: 0 < \dfrac 1 m < b$ Thus: :$\forall m \in \N, m \ge n: \dfrac 1 m \in I \cap B$ Thus an open set $U$ which contains $0$ contains a countably infinite number of elements of $B$ (distinct from $0$). Thus, by definition, $0$ is an $\omega$-accumulation point of $B$ in $\R$. {{qed}}
Let $U$ be an open set of $\R$ which contains $0$. From Open Sets in Real Number Line, there exists an open interval $I$ of the form: :$I := \openint {-a} b \subseteq U$ From Zero is Omega-Accumulation Point of Integer Reciprocal Space Union with Closed Interval, there is a countably infinite number of points of $B$ in $U$. However, when $b < 2$ there is not an uncountable number of points of $B$ in $I$. {{qed}}
Let $\sequence {x_n}$ be the sequence $1, \dfrac 1 2, \dfrac 1 3, \ldots$ Let $\epsilon \in \R_{>0}$ be a (strictly) positive real number. By the Archimedean Principle: :$\exists \N \in n: n > \dfrac 1 \epsilon$ and so: :$\exists \N \in n: \dfrac 1 n < \epsilon$ As: :$0 < \dfrac 1 {n + 1} < \dfrac 1 n$ it follows that: :$\size {\dfrac 1 n - \dfrac 1 {n + 1} } < \epsilon$ and so $\sequence {x_n}$ is a Cauchy sequence. From Sequence of Powers of Reciprocals is Null Sequence, $\sequence {x_n}$ is a basic null sequence. That is: :$\displaystyle \lim_{n \mathop \to \infty} \frac 1 n = 0$ As $0 \notin A$, the result follows. {{qed}}
We have: :Integer Reciprocal Space with Zero is Totally Separated :Totally Separated Space is Totally Disconnected :Totally Disconnected and Locally Connected Space is Discrete We also have: :Topological Space is Discrete iff All Points are Isolated :Zero is Limit Point of Integer Reciprocal Space From definition of limit points: :$0$ is not an isolated point of $A$ Hence integer reciprocal space with zero is not the discrete space, and the result follows. {{qed}}
Let $T = \struct {S, \tau}$ be a discrete space. From Discrete Space is Locally Path-Connected, $T$ is locally path-connected. From Locally Path-Connected Space is Locally Connected, it follows that $T$ is locally connected. {{qed}}
Proof by Counterexample: Let $\struct {A, \tau_1}$ be any countable discrete space. Let $B \subseteq \R$ be the set of all points on $\R$ defined as: :$B := \set 0 \cup \set {\dfrac 1 n: n \in \Z_{>0} }$ Let $\struct {B, \tau_2}$ be the integer reciprocal space with zero under the usual (Euclidean) topology. From Discrete Space is Locally Connected, $\struct {A, \tau_1}$ is locally connected. Let $f: A \to B$ be a bijection. From Mapping from Discrete Space is Continuous, $f: A \to B$ is a continuous mapping. From Integer Reciprocal Space with Zero is not Locally Connected, $\struct {B, \tau_2} = \struct {\Img f, \tau_3}$ is not locally connected. {{qed}}
Let $a, b \in A$ such that $a < b$. From Between two Rational Numbers exists Irrational Number: :$\exists \alpha \in \R \setminus \Q: a < \alpha < b$ Because $\forall x \in A: x \in \Q$ it follows that $\alpha \notin A$. Consider the half-open intervals $S = \hointr 0 \alpha$ and $T = \hointl \alpha 1$ Let: :$U := S \cap A, V := T \cap A$ Let $\beta \in A$. Then either: :$(1): \quad \beta < \alpha$ in which case: :$\beta \in U$ or: :$(2): \quad \beta > \alpha$ in which case: :$\beta \in V$ Thus $U \cup V = A$. Let $a \in U$. Then $a < \alpha$ and so $a \notin V$. Similarly, let $b \in V$. Then $b > \alpha$ and so $b \notin U$. Then note that $x \in U$ and $y \in V$. Thus $U \ne \O$ and $V \ne \O$. Thus, by definition, $U$ and $V$ constitute a separation of $A$ such that $x \in U$ and $y \in V$. Hence the result by definition of totally separated. {{qed}}
From Integer Reciprocal Space with Zero is Totally Separated, $A$ is totally separated. From Totally Separated Space is Totally Disconnected, $A$ is totally disconnected.  The result follows by definition of totally disconnected. {{qed}}
From Integer Reciprocal Space with Zero is Totally Separated, $A$ is totally separated. The result follows from Totally Separated iff Quasicomponents are Singletons. {{qed}}
$\struct {A, \tau_d}$ is a metric space. We have: :Extremally Disconnected Metric Space is Discrete We also have: :Topological Space is Discrete iff All Points are Isolated :Zero is Limit Point of Integer Reciprocal Space From definition of limit points: :$0$ is not an isolated point of $A$ Hence integer reciprocal space with zero is not the discrete space, and the result follows. {{qed}}
From Intersection of Interiors contains Interior of Intersection it is seen that it is always the case that: :$\displaystyle \paren {\bigcap_{H \mathop \in \mathbb H} H}^\circ \subseteq \bigcap_{H \mathop \in \mathbb H} H^\circ$ From Interior of Finite Intersection equals Intersection of Interiors it is seen that if $\mathbb H$ is finite, then: :$\displaystyle \paren {\bigcap_{H \mathop \in \mathbb H} H}^\circ = \bigcap_{H \mathop \in \mathbb H} H^\circ$ It remains to be shown that in the case where $\mathbb H$ is infinite, then it is possible that: :$\displaystyle \paren {\bigcap_{H \mathop \in \mathbb H} H}^\circ \ne \bigcap_{H \mathop \in \mathbb H} H^\circ$ 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: :$\displaystyle \bigcap A_n = \bigcap A_n^\circ = \set 1$ However: :$\displaystyle \paren {\bigcap A_n}^\circ = \set 1^\circ = \O$ Hence the result. {{qed}}
From Open Sets in Real Number Line, $A$ is open in $\R$. The result follows from Interior of Open Set. {{qed}}
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. {{qed}}
{{begin-eqn}} {{eqn | l = A^{\circ -}       | r = A^-       | c = Interior of Union of Adjacent Open Intervals: $A^\circ = A$ }} {{eqn | r = \closedint a c       | c = Closure of Union of Adjacent Open Intervals }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = A^{- \circ -}       | r = \openint a c^-       | c = Interior of Closure of Interior of Union of Adjacent Open Intervals }} {{eqn | r = \closedint a c       | c = Closure of Open Ball in Metric Space }} {{end-eqn}} Hence the result. {{qed}}
By definition of exterior, $A^e$ is the complement relative to $\R$ of the closure of $A$ in $\R$. Thus: {{begin-eqn}} {{eqn | l = A^{ee}       | r = \paren {\relcomp \R {A^-} }^e       | c = {{Defof|Exterior (Topology)|Exterior}} }} {{eqn | r = \paren {\relcomp \R {\closedint a c} }^e       | c = Closure of Interior of Closure of Union of Adjacent Open Intervals }} {{eqn | r = \paren {\openint \gets a \cup \openint c \to}^e       | c = {{Defof|Relative Complement}} }} {{eqn | r = \relcomp \R {\paren {\openint \gets a \cup \openint c \to}^-}       | c = {{Defof|Exterior (Topology)|Exterior}} }} {{eqn | r = \relcomp \R {\openint \gets a^- \cup \openint c \to^-}       | c = Closure of Finite Union equals Union of Closures }} {{eqn | r = \relcomp \R {\hointl \gets a \cup \hointr c \to}       | c = Closure of Open Ball in Metric Space }} {{eqn | r = \openint a c       | c = {{Defof|Relative Complement}} }} {{end-eqn}} {{qed}}
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: :$A^{ee} \subsetneq A^\circ$ {{qed}}
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. {{qed}}
From Open Sets in Real Number Line, $\openint a b$ is open in $\struct {\R, \tau_d}$. From Closure of Open Real Interval is Closed Real Interval: :$\openint a b^- = \closedint a b$ where $\openint a b^-$ denotes the closure of $\openint a b$. From Interior of Closed Real Interval is Open Real Interval: :$\closedint a b^\circ = \openint a b$ where $\closedint a b^\circ$ denotes the interior of $\closedint a b$. Hence the result, by definition of regular open. {{qed}}
From Limit Points of Open Real Interval, the limit points of $\openint a b$ consist of: :the points $\openint a b$ itself and :the points $a$ and $b$. By definition, the closure of $\openint a b$ is the union of $\openint a b$ and its limit points. Hence the result, by definition of the closed interval $\closedint a b$. {{qed}} Category:Open Sets Category:Set Closures Category:Real Number Line with Euclidean Topology 78lv3ewzobt6bmitc1gg8gsylcs6u2u
Let $x \in \openint a b$. Then by definition $a < x < b$. Let $N_x$ be an open neighborhood of $x$ in $\R$. So by Open Superset is Open Neighborhood, $N_x$ is an open set of $\R$ containing $x$. By Open Sets in Real Number Line, $N_x$ is a countable union of pairwise disjoint open intervals. As $x \in N_x$, it follows that $\exists U \subseteq N_x$ such that $x \in U$ and: :$\exists p, q \in \R: U = \openint p q$ Let $r = \max \set {a, p}$ and $s = \min \set {b, q}$. Then as $x > a, x > p$ and $x < b, x < q$ it follows that $x > \max \set {a, p}$ and $x < \min \set {b, q}$. That is: :$x \in \openint r s$ From Real Numbers are Close Packed: :$\exists y \in \R: r < y < x$ and so: :$y \in \openint r s$ By construction, it follows that: :$y \in \openint p q$ That is, $y$ is an element of $\R$ different from $x$ which is also in $N_x$. As $N_x$ was arbitrary, it follows that every such open neighborhood of $x$ has the same property. That is, $x$ is a limit point of $\openint a b$. {{qed|lemma}} Let $N_a$ be an open neighborhood of $a$ in $\R$. So by Open Superset is Open Neighborhood, $N_a$ is an open set of $\R$ containing $a$. By Open Sets in Real Number Line, $N_a$ is a countable union of pairwise disjoint open intervals. As $a \in N_a$, it follows that $\exists U \subseteq N_a$ such that $a \in U$ and: :$\exists p, q \in \R: U = \openint p q$ Let $s = \min \set {b, q}$. As $a < b, a < q$ it follows that $a < \min \set {b, q}$. That is: :$a \in \openint p s$ From Real Numbers are Close Packed: :$\exists y \in \R: a < y < s$ and so: :$y \in \openint p s$ By construction, it follows that: :$y \in \openint p q$ That is, $y$ is an element of $\R$ different from $a$ which is also in $N_a$. As $N_a$ was arbitrary, it follows that every such open neighborhood of $a$ has the same property. That is, $a$ is a limit point of $\openint a b$. By a similar argument it is shown that $a$ is a limit point of $\openint a b$. {{qed|lemma}} Now let: :$x \notin \openint a b \cup \set {a, b}$ Then either: :$x < a$ or :$x > b$ Let $x < a$. Then by Real Numbers are Close Packed: :$\exists q \in \R: x < q < a$ and so $\exists p \in \R$ such that: :$x \in \openint p q$ such that: :$\openint p q \cap \openint a b = \O$ That is, $\openint p q$ is an open set of $\R$ containing $x$ which contains no elements of $\openint a b$. Hence by definition, $x$ is not a limit point of $\openint a b$. By a similar argument it is shown that if $x > b$ then $x$ is not a limit point of $\openint a b$. {{qed}} Category:Limit Points Category:Real Intervals sfvwu84r4v7iqtpm2z6siwo4ma4q62d
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 Close Packed: :$\exists r \in \R: p < r < a$ 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. {{qed}} Category:Closed Sets Category:Set Interiors Category:Real Number Line with Euclidean Topology gwdsvnq3skrpye0p3bup2a3cyjxy7sd
Proof by Counterexample: By Open Real Interval is Regular Open, the open real intervals: :$\openint 0 {\dfrac 1 2}, \openint {\dfrac 1 2} 1$ are both regular open sets of $\R$. Consider $A$, the union of the adjacent open intervals: :$A := \openint 0 {\dfrac 1 2} \cup \openint {\dfrac 1 2} 1$ From Interior of Closure of Interior of Union of Adjacent Open Intervals: :$A^{- \circ} = \openint 0 1$ Thus $A$ is not a regular open set of $\R$. {{qed}}
Let $\struct {\R, \tau_d}$ be the real number line under the usual (Euclidean) topology. Let $H_1 = \openint 0 {\dfrac 1 2}$ and $H_2 = \openint {\dfrac 1 2} 1$. By inspection it can be seen that: :$H_1 \cap H_2 = \O$ Thus from Closure of Empty Set is Empty Set: :$\paren {H_1 \cap H_2}^- = \O$ From Closure of Open Real Interval is Closed Real Interval: :$H_1 = \closedint 0 {\dfrac 1 2}, H_2 = \closedint {\dfrac 1 2} 1$ Thus: :${H_1}^- \cap {H_2}^- = \set {\dfrac 1 2}$ So $\paren {H_1 \cap H_2}^- \ne {H_1}^- \cap {H_2}^-$ Hence the result. {{qed}}
Let $\struct {\R, \tau}$ be the real number line under the usual (Euclidean) topology. Let $\Q$ denote the set of rational numbers. Let $\R \setminus \Q$ denote the set of irrational numbers. From Closure of Intersection of Rationals and Irrationals is Empty Set: :$\paren {\Q \cap \paren {\R \setminus \Q} }^- = \O$ From Intersection of Closures of Rationals and Irrationals is Reals: :$\Q^- \cap \paren {\R \setminus \Q}^- = \R$ The result follows. {{qed}}
From Empty Set is Closed in Topological Space, $\varnothing$ is closed in $T$. The result follows from Closed Set equals its Closure. {{qed}} Category:Set Closures Category:Empty Set husyxvcdkwhmlykudfkl9ehdaf894hf
From Closed Real Interval is Closed in Real Number Line, $\closedint a b$ is closed in $\struct {\R, \tau_d}$. From Interior of Closed Real Interval is Open Real Interval: :$\closedint a b^\circ = \openint a b$ where $\closedint a b^\circ$ denotes the interior of $\closedint a b$. From Closure of Open Real Interval is Closed Real Interval: :$\openint a b^- = \closedint a b$ where $\openint a b^-$ denotes the closure of $\openint a b$. Hence the result, by definition of regular closed. {{qed}}
From Open Sets in Real Number Line: :$U := \openint \gets a \cup \openint b \to$ is an open set in $\struct {\R, \tau_d}$. Consider the complement relative to $\R$: :$V := \relcomp \R U = \closedint a b$ By definition, $\relcomp \R U$ is closed (in the topological sense) in $\struct {\R, \tau_d}$. But by construction: :$\relcomp \R U = \closedint a b$ Hence the result. {{qed}} Category:Closed Sets Category:Real Number Line with Euclidean Topology a69kamvjw8cvvdh5n1jb38ihjnidaxf
Proof by Counterexample: By Closed Real Interval is Regular Closed, the closed real intervals: :$\closedint 0 {\dfrac 1 2}, \closedint {\dfrac 1 2} 1$ are both regular closed sets of $\R$. Consider $A$, the intersection of the two half-unit closed intervals: :$A := \closedint 0 {\dfrac 1 2} \cap \closedint {\dfrac 1 2} 1 = \set {\dfrac 1 2} = \closedint {\dfrac 1 2} {\dfrac 1 2}$ From Interior of Closed Real Interval is Open Real Interval: :$A^\circ = \openint {\dfrac 1 2} {\dfrac 1 2} = \O$ From Closure of Empty Set is Empty Set: :$A^{\circ -} = \O \ne A$ Thus $A$ is not a regular closed set of $\R$. {{qed}}
From Union of Interiors is Subset of Interior of Union: :$\paren {H_1 \cup H_2}^\circ \supseteq {H_1}^\circ \cup {H_2}^\circ$ It remains to be shown that it is not necessarily the case that: :$\paren {H_1 \cup H_2}^\circ = {H_1}^\circ \cup {H_2}^\circ$ Proof by Counterexample: Let $\struct {\R, \tau_d}$ be the real number line under the usual (Euclidean) topology. Let $H_1 = \closedint 0 {\dfrac 1 2}$ and $H_2 = \closedint {\dfrac 1 2} 1$. Then: {{begin-eqn}} {{eqn | l = \paren {H_1 \cup H_2}^\circ       | r = \paren {\closedint 0 {\dfrac 1 2} \cup \closedint {\dfrac 1 2} 1}^\circ       | c =  }} {{eqn | r = \closedint 0 1^\circ       | c = by definition }} {{eqn | r = \openint 0 1       | c = Interior of Closed Real Interval is Open Real Interval }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = {H_1}^\circ \cup {H_2}^\circ       | r = \closedint 0 {\dfrac 1 2}^\circ \cup \closedint {\dfrac 1 2} 1^\circ       | c =  }} {{eqn | r = \openint 0 {\dfrac 1 2} \cup \openint {\dfrac 1 2} 1       | c = Interior of Closed Real Interval is Open Real Interval }} {{eqn | o = \ne       | r = \openint 0 1       | c = as $\dfrac 1 2$ is in one but not the other }} {{end-eqn}} Hence the result. {{qed}}
Let $\struct {\R, \tau_d}$ be the real number line under the usual (Euclidean) topology. Let $S_1 := \closedint 0 1$ be the closed unit interval. Let $S_2 := \openint 0 1$ be the open unit interval. Let $H_1 := \openint 0 1$ and $H_2 := \closedint {\dfrac 1 4} {\dfrac 3 4}$. Then: :$(1): \quad \struct {S_1, \tau_d}$ is homeomorphic to $\struct {H_2, \tau_d}$ by the mapping $f: S_1 \to H_2$ defined as $\map f x = \dfrac x 2 + \dfrac 1 4$ :$(2): \quad \struct {S_2, \tau_d}$ is trivially homeomorphic to $\struct {H_1, \tau_d}$ From Continuous Image of Compact Space is Compact, if $S_1$ and $S_2$ are homeomorphic then both must be compact. From Closed Real Interval is Compact in Metric Space, $S_1$ is compact. But from Open Real Interval is not Compact, $S_2$ is not compact. Hence: :$(3): \quad \struct {S_1, \tau_d}$ and $\struct {S_2, \tau_d}$ are not homeomorphic. {{qed}}