hcju/proofwikiqa huggingface dataset:
Let $\BB$ be the set defined as: :$\BB = \set {\set A: A \in S / \PP}$ From Basis for Partition Topology, $\BB$ forms a basis for a partition space on $S$. From Basis for Discrete Topology, $\BB$ forms a basis for the discrete topology on $S / \PP$. Hence the result, by definition of quotient topology. {{qed}}
By definition, the partition topology on $\PP$ is the set of all unions from $\PP$. This is (trivially, and from Union of Empty Set) $\set {\O, S}$ which is the indiscrete topology on $S$ by definition. {{qed}}
From Basis for Discrete Topology it is shown that $\PP$ as defined here forms the basis of the discrete topology. {{qed}}
As $\PP$ is not the partition of singletons, there exists some $H \in \PP$ such that $a, b \in H: a \ne b$. Any union of sets from $\PP$ which includes $H$ will therefore contain both $a$ and $b$. Therefore, any element of $\tau$ containing $a$ will also contain $b$, and similarly, any element of $\tau$ containing $b$ will also contain $a$. So there is no open set in $T$ containing $a$ and not $b$, or $b$ and not $a$. Hence the result, by definition of $T_0$ (Kolmogorov) space. {{qed}} Note the reason for the exception in the statement as given. If $\PP$ is the partition of singletons: :$\PP = \set {\set x: x \in S}$ then $T$ as constructed is the discrete topology on $S$. From Discrete Space satisfies all Separation Properties, in that case $T$ ''does'' satisfy the $T_0$ property.
Let $A$ and $B$ be subsets of $S$ such that $A^- \cap B = A \cap B^- = \O$. From Open Set in Partition Topology is also Closed, we get that $A^-$ and $B^-$ are both clopen.  From Set is Subset of its Topological Closure: :$A \subseteq A^-$ From $A^- \cap B = \O$ it follows from Intersection with Complement is Empty iff Subset that $B \subseteq S \setminus \paren {A^-}$. Since $A^-$ is clopen, it follows that $A^-$ and $S \setminus \paren {A^-}$ are both open. Also, they are disjoint by definition. Thus we have $A^-$, an open set containing $A$, and $S \setminus \paren {A^-}$, an open set containing $B$. Thus, by definition, $T$ is a $T_5$ space. {{qed}}
Let $F \subseteq S$ be closed, and let $x \in S$, $x \notin F$.  Denote by $S \setminus F$ the relative complement of $F$ in $S$. Define a mapping $f: S \to \closedint 0 1$ as: :$\map f s := \begin{cases}  1 & : \text { if $s \in F$} \\  0 & : \text { if $s \in S \setminus F$} \end{cases}$ Then $f$ is identically $1$ on $F$, and identically $0$ on $\set x$. Now if $f$ is continuous, it will be a Urysohn function for $F$ and $\set y$, and $T$ will be a $T_{3 \frac 1 2}$ space. Now for any $V \subseteq \closedint 0 1$, we have: :$f^{-1} \sqbrk V = \begin{cases}  \O            & : \text{ if $0,1 \notin V$} \\  F             & : \text{ if $0 \notin V$ and $1 \in V$} \\  S \setminus F & : \text{ if $0 \in V$ and $1 \notin V$} \\  S             & : \text{ if $0,1 \in V$} \end{cases}$ By definition of $\tau$, $F$ is open in $T$. By Open Set in Partition Topology is also Closed, $F$ is also closed, and so $S \setminus F$ is open in $T$. Thus, the preimage of any subset $V$ of $\closedint 0 1$ is open in $T$. In particular, this holds for the open sets of $\closedint 0 1$. It follows that $f$ is a continuous mapping, and so a Urysohn function. Hence $T$ is $T_{3 \frac 1 2}$ space. {{qed}}
From Basis for Partition Topology, the set: :$\BB := \set {\set {2 k - 1, 2 k}: k \in \Z, k > 0}$ is a basis for $T$. There is an obvious one-to-one correspondence $\phi: \Z_{>0} \leftrightarrow \BB$ between $\Z_{>0}$ and $\BB$: :$\forall x \in \Z_{>0}: \map \phi x = \set {2 x - 1, 2 x}$ But $\Z_{>0} \subseteq \Z$, and Integers are Countably Infinite. So from Subset of Countably Infinite Set is Countable, $\Z_{>0}$ is countable. Thus $\BB$ is also countable by definition of countability. So we have that $T$ has a countable basis, and so is second-countable by definition. {{qed}}
Let $H \subseteq \Z_{>0}$ such that $H$ is infinite. Let $x \in H$. By definition, the odd-even topology is a partition topology. So $U$ is a union of sets of the form $\set {2 k - 1, 2 k}$. Now if $x \in U$, it will be of the form $2 k - 1$ or $2 k$. So there will exist $y \in U$ of the form $2 k$ or $2 k - 1$. So, by definition, $x$ is a limit point of $H$. So, by definition, $T$ is weakly countably compact. {{qed}}
By definition, the odd-even topology is a partition topology. Let $\PP$ be the partition which is the basis for $T$: :$\PP = \set {\set {2 k - 1, 2 k}: k \in \Z_{>0} }$ Then $\PP$ is a countable open cover of $S$ which has no finite subcover. Hence the result. {{qed}}
Let $A \subseteq S$ be an infinite subset of $S$. Then as Set in Discrete Topology is Clopen, $A$ is closed in $T$. From Closed Set Equals its Closure, $A = A^-$ where $A^-$ is the closure of $A$. From the definition, $x$ is a limit point of $A$ if it belongs to the closure of $A$ but is not an isolated point of $A$. But then we have Topological Space is Discrete iff All Points are Isolated. So $A$ has no limit points in $T$. So $T$ is not weakly countably compact. {{qed}}
Let $\Z_{>0}$ be the strictly positive integers: :$\Z_{>0} = \set {1, 2, 3, \ldots}$ Let $T_A = \struct {\Z_{>0}, \tau_A}$ be the odd-even topology. Let $T_B = \struct {\Z_{>0}, \tau_B}$ be the discrete topology on $\Z_{>0}$. Let $\phi: T_A \to T_B$ be the mapping: :$\map \phi {2 k} = k, \map \phi {2 k - 1} = k$ Then: :$\map {\phi^{-1} } k = \set {2 k, 2 k - 1} \in \tau_A$ demonstrating that $\phi$ is continuous. Now we have that the Odd-Even Topology is Weakly Countably Compact. But we also have that a Countable Discrete Space is not Weakly Countably Compact. Hence the result. {{qed}}
By definition, the deleted integer topology is a partition topology. Let $\PP$ be the partition which is the basis for $T$, that is: :$\PP = \set {\openint {n - 1} n: n \in \Z_{> 0} }$ Then $\PP$ is a countable open cover of $S$ which has no finite subcover. Hence the result. {{qed}}
Let $\Z_{>0}$ be understood as the set of strictly positive integers: :$\Z_{>0} = \set {x \in \Z: x > 0} = \set {1, 2, 3, \ldots}$ From Basis for Partition Topology, the set: :$\BB = \set {\openint {n - 1} n: n \in Z_{>0} }$ is a basis for $T$. There is an obvious one-to-one correspondence $\phi: \Z_{> 0} \leftrightarrow \BB$ between $\Z_{> 0}$ and $\BB$: :$\forall x \in \Z_{>0}: \map \phi x = \openint {x - 1} x$ But $\Z_{>0} \subseteq \Z$ and Integers are Countably Infinite. So from Subset of Countably Infinite Set is Countable, $\Z_{>0}$ is countable. Thus $\BB$ is also countable by definition of countability. So we have that $T$ has a countable basis, and so is second-countable by definition. {{qed}}
Let $A \subseteq S$ such that $A$ is infinite. Let $x \in A$. Let $n \in \Z$ such that $n < x < n + 1$. By the definition of the deleted integer topology, $\openint n {n + 1}$ is open in $T$. We have that $\openint n {n + 1}$ is infinite. Take some $y \in \openint n {n + 1}$ such that $x \ne y$. Now we claim that $y$ is a limit point of $A$. Let $N_y$ be an open neighbourhood of $y$. To be open in $T$, $N_y$ needs to be the union of open intervals of the form $\openint k {k + 1}$ for some $k \in \Z_{\ge 0}$. Therefore to contain $y$ and be open at the same time, $\openint n {n + 1} \subseteq N_y$. Thus $x \in N_y$ and $y \ne x \in A \cap N_y \ne \O$. This is the definition of a limit point. So, by definition, $T$ is weakly countably compact. {{qed}}
Let $\tau_d$ be the set of all $X \subseteq S$ which are open in the sense that: :$\forall y \in X: \exists \epsilon > 0: \map {B_\epsilon} y \subseteq X$ where $\map {B_\epsilon} y$ is the open $\epsilon$-ball of $y$. Equivalently: :$\forall x \in X: \exists \epsilon \in \R_{>0}: \forall y \in S: \map d {x, y} < \epsilon \implies y \in X$ We need to show that $\tau_d$ forms a topology on $S$. We examine each of the open set axioms in turn. === $(\text O 1)$: Union of Open Sets === Let $\left \langle{U_i}\right \rangle_{i \in I}$ be an indexed family of open sets of $S$. Let $\displaystyle V = \bigcup_{i \mathop \in I} U_i$ be the union of $\family {U_i}_{i \mathop \in I}$. Then by the definition of union: :$\forall x \in V: \exists i \in I: x \in U_i$ and so by the definition of open set: :$\exists \epsilon \in \R_{>0}: \map {B_\epsilon} x \subseteq U_i \subseteq V$ Hence $V$ is open by definition. {{qed|lemma}} === $(\text O 2)$: Intersection of Open Sets === Let $U$ and $V$ be open sets of $S$. Let $x \in U \cap V$. Then: :$\exists \epsilon_U \in \R_{>0}: \map {B_{\epsilon_U} } x \subseteq U$ :$\exists \epsilon_V \in \R_{>0}: \map {B_{\epsilon_V} } x \subseteq V$ Let $\epsilon := \min \set {\epsilon_U, \epsilon_V}$. Then: :$\map {B_\epsilon} x \subseteq U \cap V$ Hence $U \cap V$ is open by definition. {{qed|lemma}} === $(\text O 3)$: Empty Set and Set Itself === From Open Sets in Pseudometric Space: :$\O \in \tau_d$ and: :$S \in \tau_d$ {{qed|lemma}} All the open set axioms are fulfilled, and the result follows. {{qed}} {{proofread}}
From the definition of open set: : $\forall y \in U: \exists \epsilon \in \R_{>0}: B_\epsilon \left({y}\right) \subseteq U$ where $U$ is an open set of $P$. That is: :''One cannot get out of $U$ by moving an arbitrarily small distance from any point in $U$.'' Take the case where $U = \varnothing$. The empty set $\varnothing$ is open by dint of the fact that there are no points $y$ in $\varnothing$ for which the condition is false. Thus for $\varnothing$: : $\forall y \in U: \exists \epsilon \in \R_{>0}: B_\epsilon \left({y}\right) \subseteq U$ is vacuously true. When $U = A$, there are no points in $A$ to which one could get to by leaving $U$, an arbitrarily short distance or no, because there are no points in $A$ that ''are'' outside of $U$. {{qed}} Category:Pseudometric Spaces hpnso1ai2e9gfv4bwkqj43qqfxx0g9r
Let $\PP$ be the partition which is the basis for $T$. Let us define $d: S^2 \to \R$ by: :$\forall x, y \in S: \map d {x, y} = \begin{cases} 0 & : \exists U \in \PP: x, y \in U \\ 1 & : \text{otherwise} \end{cases}$ That is, $\map d {x, y} = 0$ when $x$ and $y$ are in the same set in the partition $\PP$, and $1$ otherwise. This is easily established as being a pseudometric on $S$. Hence the result. {{qed}}
By definition, the double pointed topology $\tau$ on $T_1$ is the product topology on $T_1 \times D$. Let $x \in S$, and consider the point $\tuple {x, a} \in S \times A$. Then: :$\forall U \in \tau: \tuple {x, a} \in U \implies \tuple {x, b} \in U$ :$\forall U \in \tau: \tuple {x, b} \in U \implies \tuple {x, a} \in U$ as $D$ is an indiscrete space. Hence the result, by definition of $T_0$ (Kolmogorov) space. {{qed}}
By definition, the double pointed topology $\tau$ on $T_1$ is the product topology on $T_1 \times D$. By definition, $D$ is the indiscrete space on a doubleton. {{AimForCont}} $T$ is a $T_0$ (Kolmogorov) space. Then from Product Space is $T_0$ iff Factor Spaces are $T_0$ it follows that $D$ is also a $T_0$ (Kolmogorov) space. But from Indiscrete Non-Singleton Space is not $T_0$, $D$ is not a $T_0$ (Kolmogorov) space. The result follows by Proof by Contradiction. {{qed}}
We have that $T$ is a partition topology, whose basis $\PP$ is defined as: :$\PP = \set {\set {\tuple {s, a}, \tuple {s, b} }: s \in \R}$ Let $A \subseteq \R \times D$ such that $A$ is infinite. Let $x \in A$. Let $U$ be the union of sets of the form $\set {\tuple {s, a}, \tuple {s, b} }$, and hence open in $T$. Now if $x \in U$, it will be an element in some $\set {\tuple {s, a}, \tuple {s, b} }$. So there will exist $y \in U$ which will also be an element in that $\set {\tuple {s, a}, \tuple {s, b} }$. So, by definition, $x$ is a limit point of $A$. So, by definition, $T$ is weakly countably compact. {{qed}}
We have that $T$ is a partition topology, whose basis $\PP$ is defined as: :$\PP = \set {\set {\paren {s, a}, \paren {s, b} }: s \in \R}$ We have that $\PP$ is an open cover of $T$. But $\PP$ has no countable subcover. Hence the result, by definition of Lindelöf space. {{qed}}
We have by definition that $\O \in \tau_p$, and as $p \in S$ we have that $S \in \tau_p$. Now let $U_1, U_2 \in \tau_p$. By definition $p \in U_1$ and $p \in U_2$, and so $p \in U_1 \cap U_2$ by definition of set intersection. So $U_1 \cap U_2 \in \tau_p$. Now let $\UU \subseteq \tau_p$. We have that $\forall U \in \UU: p \in U$. Hence from Subset of Union $p \in \bigcup \UU$. So all the properties are fulfilled for $\tau_p$ to be a topology on $S$. {{qed}}
We note from Limit Points in Particular Point Space that every $x \in U$ such that $x \ne p$ is a limit point of $U$. Let $\sequence {a_i}$ be the (infinite) sequence in $T$ whose terms are the elements of $U$. {{AimForCont}} $\beta \in S$ is an accumulation point of $\sequence {a_i}$. Then from Accumulation Points for Sequence in Particular Point Space, $\beta$ is such that an infinite number of terms of $\sequence {a_i}$ are equal either to $\beta$ or to $p$. But this contradicts the assertion that $\sequence {a_i}$ is a sequence of distinct terms. Hence no element of $U$ is an accumulation point of $\sequence {a_i}$. {{qed}}
Follows directly from: :Particular Point Topology is Closed Extension Topology of Discrete Topology :Closure of Open Set of Closed Extension Space {{qed}}
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. {{qed}}
Let $x \in S, x \ne p$. By Limit Points in Particular Point Space, $x$ is a limit point of $H$. Consider the set $U = \set {x, p} \subseteq S$. By definition of the particular point topology, $U$ is open in $T$. But as $U$ contains only $x$ and $p$, it is clear that $U$ does not contain an infinite number of points of $H$. Hence, by definition, $x$ can not be an $\omega$-accumulation point of $H$. {{qed}}
Let $T$ be a trivial space, so that $S = \set p$. Then the result holds vacuously as there are no two distinct points in $T$. Now suppose $T$ is not trivial. Then $\exists x \in S: x \ne p$. Now we have that $\set p \subseteq T$ is open in $T$ such that $p \in \set p$ but $x \notin \set p$. Finally, suppose that $x, y \in S: x \ne y, x \ne p, y \ne p$. Then we have that $\set {p, x} \subseteq T$ is open in $T$ such that $x \in \set {p, x}$ but $y \notin \set {p, x}$. Hence the result. {{qed}}
We have: :Particular Point Topology is Closed Extension Topology of Discrete Topology :Discrete Space satisfies all Separation Properties (including being a $T_0$ space) Then by Condition for Closed Extension Space to be $T_0$ Space, as a discrete space is $T_0$ then so is its closed extension. {{qed}}
Follows directly from: :Particular Point Topology is Closed Extension Topology of Discrete Topology :Closed Extension Topology is not $T_1$. {{qed}}
We have that there are at least three elements of $S$. So, consider $x, y, p \in S: x \ne y, x \ne p, y \ne p$. Then $X = \set x, Y = \set y$ are closed in $T$ and $X \cap Y = \O$. Suppose $U, V \in \tau_p$ are open sets in $T$ such that $X \subseteq U, Y \subseteq V$. But as $p \in U, p \in V$ we have that $U \cap V \ne \O$. So $T$ is not a $T_4$ space. {{qed}}
We have that there are at least two distinct elements of $S$. So, consider $x, p \in S: x \ne p$. Then $X = \set x$ is closed in $T$ and $p \notin X$. Suppose $U \in \tau_p$ is an open set in $T$ such that $X \subseteq U$. We have that $\set p \in \tau_p$ such that $p \in \set p$. But as $p \in U, p \in \set p$ we have that $U \cap \set p \ne \O$. So $T$ is not a $T_3$ space. {{qed}}
Any open cover of $\set p$ has a finite subcover: any single set that contains $p$ is a cover for $\set p$. So $\set p$ is compact in $T$. {{qed}}
As C is a finite abelian group, due to the Fundamental Theorem of Finite Abelian Groups, $C$ is the Internal Group Direct Product of Cyclic Groups $H_1, \ldots, H_r$ with orders $p_1^{e_1}, \ldots, p_r^{e_r}$ respectively. As Internal and External Group Direct Products are Isomorphic, :$C \cong H_1 \times \ldots \times H_r$ where $\cong$ denotes (group) isomorphism. As Powers of Coprime Numbers are Coprime, the orders of $H_1, \ldots, H_r$ are coprime. By Group Direct Product of Cyclic Groups, $H_1 \times \ldots \times H_r$ is cyclic, so $C$ is cyclic. {{qed}} Category:Field Theory Category:Cyclic Groups 88txuy93qmfwfwqcmp2083jevg4kga2
Let $f = a_0 + \cdots + a_n X^n$, $g = b_0 + \cdots + b_m X^m \in R \left[{X}\right]$. We have: {{begin-eqn}} {{eqn|l=\overline{\phi} \left({f}\right) \overline{\phi} \left({g}\right)      |r=\left[{\phi \left({a_0}\right) + \cdots + \phi \left({a_n}\right) X^n}\right] \left[{\phi \left({b_0}\right) + \cdots + \phi \left({b_m}\right) X^m}\right]      |c= }} {{eqn|r=\sum_{r \mathop = 0}^{m+n} \sum_{i \mathop + j \mathop = r} \phi \left({a_i}\right) \phi \left({b_j}\right) X^r      |c=by the definition of multiplication of polynomial forms }} {{eqn|r=\sum_{r \mathop = 0}^{m+n} \phi \left({\sum_{i \mathop + j \mathop = r} a_i b_j}\right) X^r      |c=because $\phi$ is a homomorphism }} {{eqn|r=\overline{\phi} \left({f g}\right)      |c= }} {{end-eqn}} and {{begin-eqn}} {{eqn|l=\overline{\phi} \left({f}\right) + \overline{\phi} \left({g}\right)      |r=\left[{\phi \left({a_0}\right) + \cdots + \phi \left({a_n}\right) X^n}\right] + \left[{\phi \left({b_0}\right) + \cdots + \phi \left({b_m}\right) X^m}\right]      |c= }} {{eqn|r=\sum_{i \mathop = 0}^{\max \left\{ {m, n}\right\} } \left[ \phi \left({a_i}\right) + \phi \left({b_i}\right) \right] X^i      |c=by the definition of addition of polynomial forms }} {{eqn|r=\sum_{i \mathop = 0}^{\max \left\{ {m, n}\right\} } \phi \left({a_i + b_i}\right) X^i      |c=because $\phi$ is a homomorphism }} {{eqn|r=\overline{\phi} \left({f + g}\right)      |c= }} {{end-eqn}} Thus $\overline{\phi}$ is a ring homomorphism. {{qed}}
Define $\phi: K \left[{X}\right] \to K \left[{\alpha}\right]$ by: :$\phi \left({f}\right) = f \left({\alpha}\right)$. We have :$\phi \left({f}\right) = 0 \iff f \left({\alpha}\right) = 0 \iff \mu_\alpha | f$ where the last equivalence is proved in Minimal Polynomial. {{explain|Presumably $\mu_\alpha \vert f$ means divisibility -- clarification needed.}} {{refactor|The above needs to be extracted into a different page, as Minimal Polynomial is to be refactored: merged, probably, into somewhere else.}} Thus: :$\ker \phi = \left\{ {f \in K \left[{X}\right]: \mu_\alpha | f}\right\} =: \left\langle{\mu_\alpha}\right\rangle$ By the corollary to Field Adjoined Set $\phi$ is surjective. So by the First Isomorphism Theorem, :$K \left[{X}\right] / \left\langle{\mu_\alpha}\right\rangle \simeq K \left[{\alpha}\right]$ By Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal: :$\left\langle{\mu_\alpha}\right\rangle$ is maximal. So $K \left[{\alpha}\right]$ is a field by Maximal Ideal iff Quotient Ring is Field. Also $K \left[{\alpha}\right]$ is the smallest ring containing $K \cup \left\{ {\alpha}\right\}$. So because a field is a ring it is also the smallest field containing $K \cup \left\{ {\alpha}\right\}$. This shows that $K \left[{\alpha}\right] = K \left({\alpha}\right)$. By Field Adjoined Set: : $K \left[{\alpha}\right] = \left\{ {f \left({\alpha}\right) : f \in K \left[{X}\right] }\right\}$ where $K \left[{X}\right]$ is the ring of polynomial functions over $K$. From Division Theorem for Polynomial Forms over Field, for each $f \in K \left[{X}\right]$ there are $q, r \in K \left[{X}\right]$ such that: :$f = q \mu_\alpha + r$ and: :$\deg r < d =: \deg \mu_\alpha$ Therefore, since $\mu_\alpha \left({\alpha}\right) = 0$, we have  {{begin-eqn}} {{eqn | l = K \left[{\alpha}\right]       | r = \left\{ {r \left({\alpha}\right): r \in K \left[{X}\right], \ \deg r < \deg \mu_\alpha}\right\} }} {{eqn | r = \left\{ {a_0 + a_1 \alpha + \dotsb + a_{d - 1} \alpha^{d - 1}: \alpha_0, \dotsc, \alpha_d \in K}\right\} }} {{end-eqn}} Therefore $1, \dotsc, \alpha^{d - 1}$ span $K \left[{\alpha}\right] / K$ as a vector space. Moreover if $s \in K \left[{X}\right]$, then $s \left({\alpha}\right) = 0$ and $\deg s < \deg \mu_\alpha$ implies that $s = 0$. Thus $1, \dotsc, \alpha^{d - 1}$ are linearly independent. Therefore no non-zero $K$-combination of $1, \dotsc, \alpha^{d-1}$ is zero. Therefore $\left\{ {1, \dotsc, \alpha^{d-1} }\right\}$ is a basis of $K \left[{\alpha}\right] / K$. The degree of $K \left[{\alpha}\right] / K$ is by definition the number of elements of a basis for $K \left[{\alpha}\right] / K$. Therefore: :$d = \deg \mu_\alpha = \left[{K \left[{\alpha}\right] : K}\right]$ {{qed}} Category:Field Extensions im9j193ofrijralo77nx3lswqxz3h8p
Let $H \subseteq S$. There are two cases to consider: :$p \in H$ :$p \notin H$ If $p \in H$ then by definition of a particular point topology, $H$ is open. If $p \notin H$, then $p \in \complement_S \left({H}\right)$, where $\complement_S \left({H}\right)$ is the relative complement of $H$ in $S$. So $\complement_S \left({H}\right)$ is open by definition of a particular point topology. From the definition of a closed set, $\complement_S \left({\complement_S \left({H}\right)}\right)$ is closed in $T$. From Relative Complement of Relative Complement we have that $\complement_S \left({\complement_S \left({H}\right)}\right) = H$ and so $H$ is closed in $T$. Now suppose $H \subseteq T$ is both closed and open in $T$. From Open and Closed Sets in Topological Space, if $H = \varnothing$ or $H = S$ then H is both closed and open in $T$. So, suppose $H \ne \varnothing$ or $H \ne S$. As $H \subseteq T$ is both closed and open in $T$, $\complement_S \left({H}\right)$ is also both closed and open in $T$. From Boundary is Intersection of Closure with Closure of Complement we have: :$\partial H = H^- \cap \left({\complement_S \left({H}\right)}\right)^-$ where $H^-$ is the closure of $H$. By Closure of Open Set of Particular Point Space we have that $H^- = S$, and of course $\left({\complement_S \left({H}\right)}\right)^- = S$. So: :$\partial H = H^- \cap \left({\complement_S \left({H}\right)}\right)^- = S$ However, from Set Clopen iff Boundary is Empty we have that $\partial H = \varnothing$. Hence if $H$ is both closed and open in $T$, then $H = \varnothing$ or $H = S$. {{qed}} Category:Open Sets Category:Closed Sets Category:Clopen Sets Category:Particular Point Topology braurrhwyos5294fdt69nsx7if5n7fv
By Compact Space in Particular Point Space, $\set p$ is compact in $T$. But by definition $\set p$ is open in $T$. So $p$ is contained in a compact neighborhood, that is, $\set p$. Now let $x \in S: x \ne p$. Then $\set {x, p}$ is open in $T$. We have that $\set {x, p}$ trivially has an open cover, that is, $\set {\set {x, p} }$ itself. Hence any open cover of $\set {x, p}$ has a finite subcover: any one set that contains $x$ and $p$ is a cover for $\set {x, p}$. So $\set {x, p}$ is a neighborhood of $x$ which is compact in $T$. So $x$ is contained in a compact neighborhood. Hence the result, by definition of weakly locally compact. {{qed}}
By definition, $T$ is strongly locally compact {{iff}} every point of $S$ is contained in an open set whose closure is compact. Let $x \in S: x \ne p$. Let $x \in U$ where $U$ is open in $T$. From Closure of Open Set of Particular Point Space we have: :$U^- = S$ where $U^-$ is the closure of $U$. But from Infinite Particular Point Space is not Compact we have that $S$ is not compact. Hence the result, by definition of strongly locally compact. {{qed}}
Consider the open cover of $T$: :$\CC = \set {\set {x, p}: x \in S, x \ne p}$ As $S$ is infinite, then so is $\CC$, as we can set up a bijection from $\phi: S \setminus \set p \leftrightarrow \CC$: :$\forall x \in S \setminus \set p: \map \phi x = \set {x, p}$ Hence $\CC$ has no finite subcover. The result follows by definition of compactness. {{qed}}
Consider the open cover of $T$: :$\CC = \set {\set {x, p}: x \in S, x \ne p}$ As $S$ is uncountable, then so is $\CC$, as we can set up a bijection $\phi: S \setminus \set p \leftrightarrow \CC$: :$\forall x \in S \setminus \set p: \map \phi x = \set {x, p}$ Hence $\CC$ has no countable subcover. The result follows by definition of Lindelöf space. {{qed}}