avewright/proofwiki-math huggingface dataset:
Theorem: Compact Set of Rational Numbers is Nowhere Dense

Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Let $S \subseteq \Q$ be a compact set of $\Q$.
Then $S$ is nowhere dense in $\Q$.

Proof:
By Compact Subspace of Hausdorff Space is Closed, $S$ is closed in $\Q$.
By Set is Closed iff Equals Topological Closure, $S = S^-$.
{{AimForCont}} $S$ is not nowhere dense in $\Q$.
Then $S^-$ contains some non-empty open set.
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 $\Q$ and open real intervals of $\R$ form a basis for $\struct {\Q, \tau_d}$.
So there exists $V = \Q \cap \openint {a'} {b'} \subseteq S^- = S$ for some $a' < b'$.
By Between two Real Numbers exists Irrational Number we have:
:$\exists r \in \R \setminus \Q: a' < r < b'$
By Between two Real Numbers exists Rational Number we have:
:$\exists a, b \in \Q: a' < a < r < b < b'$
Thus $\Q \cap \closedint a b$ is a closed subset of $S$ in $\Q$.
By Closed Subspace of Compact Space is Compact, $\Q \cap \closedint a b$ is compact.
Now we construct an open cover for $\Q \cap \closedint a b$:
Let $C = \openint r {b + 1} \cap \Q$.
Let $C_q = \openint {a - 1} q \cap \Q$ for each $q \in \openint a r \cap \Q$.
Then $C$ and $C_q$ are open sets.
Let $\CC = \set C \cup \set {C_q: q \in \openint a r \cap \Q}$.
To show that $\CC$ is an open cover for $\Q \cap \closedint a b$:
:Pick any $x \in \Q \cap \closedint a b$.
:Suppose $x > r$. 
:Then $r < x \le b < b + 1$.
:Therefore $x \in \openint r {b + 1} = C \in \CC$.
:Suppose $x < r$.
:By Between two Real Numbers exists Rational Number we have:
::$\exists q \in \Q: a - 1 < a \le x < q < r$
:Then $x \in \openint {a - 1} q$ and $q \in \openint a r$.
:Thus $x \in C_q \in \CC$.
:Therefore $x \in \bigcup \CC$, showing that $\CC$ is an open cover for $\Q \cap \closedint a b$. 
Since $\Q \cap \closedint a b$ is compact, $\CC$ has a finite subcover, $\CC'$.
This finite subcover can only contain a finite number of sets in $\set {C_q: q \in \openint a r \cap \Q}$.
Therefore $\set {q: C_q \in \CC'}$ has a greatest element, $q'$.
We have that $q' \in \openint a r \cap \Q$.
By Between two Real Numbers exists Rational Number we have:
:$\exists q^* \in \Q: a < q' < q^* < r < b$
So $q^* \in \Q \cap \closedint a b$.
But for all $q \in \hointl a {q'} \cap \Q$:
:$q^* \notin C_q = \openint {a - 1} q \cap \Q$
We also have $q^* \notin C = \openint r {b + 1} \cap \Q$.
Hence $q^* \notin \bigcup \CC'$.
So $\CC'$ is not a cover for $\Q \cap \closedint a b$.
This is a contradiction.
Thus $S$ is nowhere dense in $\Q$.
{{qed}}
Theorem: Rational Number Space is not Locally Compact Hausdorff Space

Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Then $\struct {\Q, \tau_d}$ is not a locally compact Hausdorff Space.

Proof:
For $\struct {\Q, \tau_d}$ to be a locally compact Hausdorff Space, it is required that every point of $\Q$ has a compact neighborhood.
Let $x \in \Q$.
Let $N \subseteq \Q$ be a neighborhood of $x$.
Then:
:$\exists U \in \tau: x \in U \subseteq N \subseteq \Q$.
{{AimForCont}} $N$ is compact.
By Compact Set of Rational Numbers is Nowhere Dense, $N$ is nowhere dense.
Thus $N^-$ contains no open set of $\Q$ which is non-empty.
But $U$ is a non-empty open set of $\Q$.
By Set is Subset of its Topological Closure, $N \subseteq N^-$.
Because Subset Relation is Transitive:
:$U \subseteq N^-$
This is a contradiction.
Hence $\struct {\Q, \tau_d}$ is not a locally compact Hausdorff Space.
{{qed}}
Theorem: Irrational Number Space is not Locally Compact Hausdorff Space

Let $\struct {\R \setminus \Q, \tau_d}$ be the irrational number space under the Euclidean topology $\tau_d$.
Then $\struct {\R \setminus \Q, \tau_d}$ is not a locally compact Hausdorff Space.

Proof:
For $\struct {\R \setminus \Q, \tau_d}$ to be a locally compact Hausdorff Space, it is required that every point of $\R \setminus \Q$ has a compact neighborhood.
Let $x \in \R \setminus \Q$.
Let $N \subseteq \R \setminus \Q$ be a neighborhood of $x$.
Then:
:$\exists U \in \tau: x \in U \subseteq N \subseteq \R \setminus \Q$
{{AimForCont}} $N$ is compact.
By Compact Set of Irrational Numbers is Nowhere Dense, $N$ is nowhere dense.
Thus $N^-$ contains no open set of $\R \setminus \Q$ which is non-empty.
But $U$ is a non-empty open set of $\R \setminus \Q$.
By Set is Subset of its Topological Closure, $N \subseteq N^-$.
Because Subset Relation is Transitive:
:$U \subseteq N^-$
This is a contradiction.
Hence $\struct {\R \setminus \Q, \tau_d}$ is not a locally compact Hausdorff Space.
{{qed}}
Theorem: Rational Number Space is not Weakly Sigma-Locally Compact

Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Then $\struct {\Q, \tau_d}$ is not weakly $\sigma$-locally compact.

Proof:
From Rational Number Space is not Locally Compact Hausdorff Space, $\struct {\Q, \tau_d}$ is not weakly locally compact.
Hence the result from definition of weakly $\sigma$-locally compact.
{{qed}}
Theorem: Irrational Number Space is not Weakly Sigma-Locally Compact

Let $\struct {\R \setminus \Q, \tau_d}$ be the irrational number space under the Euclidean topology $\tau_d$.
Then $\struct {\R \setminus \Q, \tau_d}$ is not weakly $\sigma$-locally compact.

Proof:
From Irrational Number Space is not Locally Compact Hausdorff Space, $\struct {\R \setminus \Q, \tau_d}$ is not weakly locally compact.
Hence the result from definition of weakly $\sigma$-locally compact.
{{qed}}
Theorem: Rational Number Space is Sigma-Compact

Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Then $\struct {\Q, \tau_d}$ is $\sigma$-compact.

Proof:
From Rational Numbers are Countably Infinite, $\Q$ is countable.
Hence the result from definition of Countable Space is $\sigma$-Compact.
{{qed}}
Theorem: Rationals plus Irrational are Everywhere Dense in Irrationals

Let $\struct {\R \setminus \Q, \tau_d}$ be the irrational number space under the Euclidean topology $\tau_d$.
Let $x \in \R \setminus \Q$ be an arbitrary irrational number.
Let $S_x$ be the set defined as:
:$S_x := \set {x + q: q \in \Q}$
Then $S_x$ is everywhere dense in $\struct {\R \setminus \Q, \tau_d}$.

Proof:
Let $y \in \R \setminus \Q$.
Let $U \subseteq \R \setminus \Q$ 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 $V = \paren {\R \setminus \Q} \cap \openint {y - \epsilon} {y + \epsilon} \subseteq U$ for some $\epsilon > 0$ such that $y \in V$.
By Between two Real Numbers exists Rational Number we have:
:$\exists r \in Q: y - \epsilon - x < r < y + \epsilon - x$.
Then:
:$y - \epsilon < x + r < y + \epsilon$
Then $x + r \in S_x \cap V \subseteq S_x \cap U$.
Since $y$ and $U$ are arbitrary:
$S_x$ is everywhere dense in $\struct {\R \setminus \Q, \tau_d}$.
{{qed}}
Theorem: Rational Number Space is Totally Separated

Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Then $\struct {\Q, \tau_d}$ is totally separated.

Proof:
Let $x, y \in \Q$.
From Between two Rational Numbers exists Irrational Number:
:$\exists \alpha \in \R \setminus \Q: x < \alpha < y$
Consider the unbounded open real intervals:
:$A := \openint \gets \alpha$, $B := \openint \alpha \to$
Let:
:$U := A \cap \Q, V := B \cap \Q$
Let $\beta \in \Q$.
Then either:
:$(1): \quad \beta < \alpha$
in which case:
:$\beta \in U$
or:
:$(2): \quad \beta > \alpha$
in which case:
:$\beta \in V$
Thus $U \cup V = \Q$.
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 $\Q$ such that $x \in U$ and $y \in V$.
Hence the result by definition of totally separated.
{{qed}}
Theorem: Irrational Number Space is Totally Separated

Let $\struct {\R \setminus \Q, \tau_d}$ be the irrational number space under the Euclidean topology $\tau_d$.
Then $\struct {\R \setminus \Q, \tau_d}$ is totally separated.

Proof:
Let $x, y \in \R \setminus \Q$.
From Between two Real Numbers exists Rational Number:
:$\exists \alpha \in \Q: x < \alpha < y$
Consider the unbounded open real intervals:
:$A := \openint \gets \alpha$, $B := \openint \alpha \to$
Let:
:$U := A \cap \paren {\R \setminus \Q}$, $V := B \cap \paren {\R \setminus \Q}$
Let $\beta \in \R \setminus \Q$.
Then either:
:$(1): \quad \beta < \alpha$
in which case:
:$\beta \in U$
or:
:$(2): \quad \beta > \alpha$
in which case:
:$\beta \in V$
Thus $U \cup V = \R \setminus \Q$.
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 $\R \setminus \Q$ such that $x \in U$ and $y \in V$.
Hence the result by definition of totally separated.
{{qed}}
Theorem: Rational Number Space is not Scattered

Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Then $\struct {\Q, \tau_d}$ is not scattered.

Proof:
For a space to be scattered, it needs by definition to have no subset which is dense-in-itself.
From Rational Number Space is Dense-in-itself, $\struct {\Q, \tau_d}$ is dense-in-itself.
Hence the result.
{{qed}}
Theorem: Rational Number Space is Dense-in-itself

Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Then $\struct {\Q, \tau_d}$ is dense-in-itself.

Proof:
Let $x \in \Q$.
Let $U \subseteq \R$ be an open set of $\struct {\Q, \tau_d}$ 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 $\Q$ and open real intervals of $\R$ form a basis for $\struct {\Q, \tau_d}$.
So there exists $V = \Q \cap \openint {x - \epsilon} {x + \epsilon} \subseteq U$ for some $\epsilon > 0$ such that $x \in V$.
By Between two Real Numbers exists Rational Number we have:
:$\exists r \in \Q: x < r < x + \epsilon$.
Then $r \in \Q \cap \openint {x - \epsilon} {x + \epsilon}$.
Therefore $x$ is not isolated.
Hence $\struct {\Q, \tau_d}$ is dense-in-itself.
{{qed}}
Theorem: Irrational Number Space is Dense-in-itself

Let $\struct {\R \setminus \Q, \tau_d}$ be the irrational number space under the Euclidean topology $\tau_d$.
Then $\struct {\R \setminus \Q, \tau_d}$ is dense-in-itself.

Proof:
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}}
Theorem: Irrational Number Space is not Scattered

Let $\struct {\R \setminus \Q, \tau_d}$ be the irrational number space under the Euclidean topology $\tau_d$.
Then $\struct {\R \setminus \Q, \tau_d}$ is not scattered.

Proof:
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}}
Theorem: Open Unit Interval on Rational Number Space is Totally Bounded

Let $T = \struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Then:
:$\openint 0 1 \cap \Q$ is totally bounded in $T$
where $\openint 0 1$ is the open unit interval.

Proof:
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.
{{qed}}
Theorem: Integer Reciprocal Space is Topological Space

Let $\struct {\R, \tau_d}$ be the real number line $\R$ under the usual (Euclidean) topology $\tau_d$.
Let $A \subseteq \R$ be the set of all points on $\R$ defined as:
:$A := \set {\dfrac 1 n: n \in \Z_{>0} }$
Then the integer reciprocal space $\struct {A, \tau_d}$ is a topological space.

Proof:
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}}
Theorem: Closure of Integer Reciprocal Space

Let $A \subseteq \R$ be the set of all points on $\R$ defined as:
:$A := \set {\dfrac 1 n: n \in \Z_{>0} }$
Let $\struct {A, \tau_d}$ be the integer reciprocal space under the usual (Euclidean) topology.
Then:
:$A^- = A \cup \set 0$
where $A^-$ denotes the closure of $A$ in $\R$.

Proof:
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}}
Theorem: Zero is Limit Point of Integer Reciprocal Space

Let $A \subseteq \R$ be the set of all points on $\R$ defined as:
:$A := \set {\dfrac 1 n : n \in \Z_{>0} }$
Let $\struct {A, \tau_d}$ be the integer reciprocal space under the usual (Euclidean) topology.
Then $0$ is the only limit point of $A$ in $\R$.

Proof:
There are three cases to consider:
Theorem: Zero is Limit Point of Integer Reciprocal Space Union with Closed Interval

Let $A \subseteq \R$ be the set of all points on $\R$ defined as:
:$A := \set {\dfrac 1 n : n \in \Z_{>0} }$
Let $T = \struct {A, \tau_d}$ be the integer reciprocal space under the usual (Euclidean) topology.
Let $B$ be the uncountable set:
:$B := A \cup \closedint 2 3$
where $\closedint 2 3$ is a closed interval of $\R$.
$2$ and $3$ are to all intents arbitrary, but convenient.
Then $0$ is a limit point of $B$ in $\R$.

Proof:
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 Axiom of Archimedes:
:$\exists n \in \N: n > \dfrac 1 b$
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}}
Theorem: Zero is Omega-Accumulation Point of Integer Reciprocal Space Union with Closed Interval

Let $A \subseteq \R$ be the set of all points on $\R$ defined as:
:$A := \set {\dfrac 1 n : n \in \Z_{>0} }$
Let $\struct {A, \tau_d}$ be the integer reciprocal space under the usual (Euclidean) topology.
Let $B$ be the uncountable set:
:$B := A \cup \closedint 2 3$
where $\closedint 2 3$ is a closed interval of $\R$.
$2$ and $3$ are to all intents arbitrary, but convenient.
Then $0$ is an $\omega$-accumulation point of $B$ in $\R$.

Proof:
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 Axiom of Archimedes:
:$\exists n \in \N: n > \dfrac 1 b$
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}}
Theorem: Zero is not Condensation Point of Integer Reciprocal Space Union with Closed Interval

Let $A \subseteq \R$ be the set of all points on $\R$ defined as:
:$A := \set {\dfrac 1 n : n \in \Z_{>0} }$
Let $\struct {A, \tau_d}$ be the integer reciprocal space under the usual (Euclidean) topology.
Let $B$ be the uncountable set:
:$B := A \cup \closedint 2 3$
where $\closedint 2 3$ is a closed interval of $\R$.
$2$ and $3$ are to all intents arbitrary, but convenient.
Then $0$ is not a condensation point of $B$ in $\R$.

Proof:
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}}
Theorem: Integer Reciprocal Space contains Cauchy Sequence with no Limit

Let $A \subseteq \R$ be the set of all points on $\R$ defined as:
:$A := \set {\dfrac 1 n : n \in \Z_{>0} }$
Let $\struct {A, \tau_d}$ be the integer reciprocal space under the usual (Euclidean) topology.
Then $A$ has a Cauchy sequence which has no limit in $A$.

Proof:
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 Axiom of Archimedes:
:$\exists \N \in n: n > \dfrac 1 \epsilon$
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:
:$\ds \lim_{n \mathop \to \infty} \frac 1 n = 0$
As $0 \notin A$, the result follows.
{{qed}}
Theorem: Integer Reciprocal Space with Zero is not Locally Connected

Let $A \subseteq \R$ be the set of all points on $\R$ defined as:
:$A := \set 0 \cup \set {\dfrac 1 n : n \in \Z_{>0} }$
Let $\struct {A, \tau_d}$ be the integer reciprocal space with zero under the usual (Euclidean) topology.
Then $A$ is not locally connected.

Proof:
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 point:
:$0$ is not an isolated point of $A$
Hence integer reciprocal space with zero is not the discrete space, and the result follows.
{{qed}}
Theorem: Discrete Space is Locally Connected

Let $T = \struct {S, \tau}$ be a discrete topological space.
Then $T$ is locally connected.

Proof:
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}}
Theorem: Local Connectedness is not Preserved under Continuous Mapping

Let $\struct {A, \tau_1}$ and $\struct {B, \tau_2}$ be topological spaces.
Let $f: A \to B$ be a continuous mapping.
Let $\struct {\Img f, \tau_3}$ be the image of $f$ with the subspace topology of $B$.
Let $\struct {A, \tau_1}$ be locally connected.
Then it is not necessarily the case that $\struct {\Img f, \tau_3}$ is also locally connected.

Proof:
;Proof by Counterexample
Let $\struct {A, \tau_1}$ be an arbitrary countable discrete space.
Let $B \subseteq \R$ be the set of all points of $\R$ 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}}
Theorem: Integer Reciprocal Space with Zero is Totally Separated

Let $A \subseteq \R$ be the set of all points on $\R$ defined as:
:$A := \set 0 \cup \set {\dfrac 1 n : n \in \Z_{>0} }$
Let $\struct {A, \tau_d}$ be the integer reciprocal space with zero under the usual (Euclidean) topology.
Then $A$ is totally separated.

Proof:
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}}
Theorem: Components of Integer Reciprocal Space with Zero are Single Points

Let $A \subseteq \R$ be the set of all points on $\R$ defined as:
:$A := \set 0 \cup \set {\dfrac 1 n : n \in \Z_{>0} }$
Let $\struct {A, \tau_d}$ be the integer reciprocal space with zero under the usual (Euclidean) topology.
Then the components of $A$ are singletons.

Proof:
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}}
Theorem: Quasicomponents of Integer Reciprocal Space with Zero are Single Points

Let $A \subseteq \R$ be the set of all points on $\R$ defined as:
:$A := \set 0 \cup \set {\dfrac 1 n : n \in \Z_{>0} }$
Let $\struct {A, \tau_d}$ be the integer reciprocal space with zero under the usual (Euclidean) topology.
Then the quasicomponents of $A$ are singletons.

Proof:
From Integer Reciprocal Space with Zero is Totally Separated, $A$ is totally separated.
The result follows from Totally Separated iff Quasicomponents are Singletons.
{{qed}}