hcju/proofwikiqa huggingface dataset:
Let $U_1, U_2 \in \tau$ be open in $T$. Let $\relcomp \R {U_1} = V_1$ and $\relcomp \R {U_2} = V_2$. By definition of compact complement topology, $V_1, V_2 \subseteq \R$ such that $V_1$ and $V_2$ are both compact. $V_1$ and $V_2$ are both bounded by definition of compact in $\R$, so their union is likewise bounded: above by $\max \set {\sup V_1, \sup V_2}$ and below by $\min \set {\inf V_1, \inf V_2}$. So $V_1 \cup V_2$ can not equal $\R$ as $\R$ is not bounded. So: {{begin-eqn}} {{eqn | l = \relcomp \R {V_1 \cup V_2}       | o = \ne       | r = \O       | c =  }} {{eqn | ll= \leadsto       | l = \relcomp \R {V_1} \cap \relcomp \R {V_2}       | o = \ne       | r = \O       | c = De Morgan's Laws: Complement of Union }} {{eqn | ll= \leadsto       | l = U_1 \cap U_2       | o = \ne       | r = \O       | c =  }} {{end-eqn}} So any two open sets in $T$ are not disjoint and so $T$ is irreducible space by definition. {{qed}}
We have that a Compact Complement Space is a $T_1$ space. From $T_1$ Space is $T_0$ Space, $T$ is a $T_0$ space. We have: :Compact Complement Topology is Irreducible :Irreducible Hausdorff Space is Singleton From $T_{2 \frac 1 2}$ (completely Hausdorff) space is $T_2$ (Hausdorff) space, $T$ is not a $T_{2 \frac 1 2}$ (completely Hausdorff) space. From Regular Space is $T_2$ Space, $T$ is not a regular space. By definition, a regular space is a space that is both a $T_0$ space and a $T_3$ space. But $T$ is a $T_0$ space and not a regular space. So it follows that $T$ can not be a $T_3$ space. Next we have that a Normal Space is a $T_3$ Space. But as $T$ is not a $T_3$ space, $T$ can not be a normal space. By definition, a normal space is a space that is both a $T_1$ space and a $T_4$ space. But $T$ is a $T_1$ space and not a normal space. So it follows that $T$ can not be a $T_4$ space. Finally we have that a $T_5$ Space is a $T_4$ Space. But as $T$ is not a $T_4$ space, $T$ can not be a $T_5$ space. {{qed}}
{{WLOG}}, suppose $a = 0$ and $r = 1$. {{AimForCont}} $\exists s < 1$ such that: :$f \sqbrk {\map B {0, s} \setminus \set 0}$ is not a dense subset of $\C$. Then, by definition of dense subset: :$\exists z_0 \in \C: \exists r_0 > 0: \map B {z_0, r_0} \cap f \sqbrk {\map B {0, s} \setminus \set 0} = \O$ {{explain|This definition of dense still needs to be added to {{ProofWiki}}. Whichever definition of denseness (either everywhere dense or dense-in-itself) will need to be expanded for the Complex case so as to make it clear that the above definition follows.}} Hence, the function $\varphi$ defined on $\map B {z_0, r_0}$ by: :$\map \varphi z = \dfrac 1 {\map f z - z_0}$ is analytic on $\map B {0, s} \setminus \set 0$ and bounded near to $0$, because: :$\forall z \in \map B {0, s} \setminus \set 0: \cmod {\map f z - z_0} > r_0 \implies \cmod {\map \varphi z} < \dfrac 1 {r_0}$ Therefore, we can extend the domain of $\varphi$ (using the Analytic Continuation Principle). Let $\map \varphi 0 \ne 0$. Then: :$\map f 0 = z_0 + \dfrac 1 {\map \varphi 0}$ and the singularity of $f$ was removable. Otherwise, let the power series of $\varphi$ be written: :$\displaystyle \map \varphi z = \sum_{n \mathop = 1}^{+\infty} a_n z^n$ Then as $\varphi \ne 0$: :$E = \set {k \in \N: a_k \ne 0} \ne \O$ Let $p = \min E$. Then $0$ is a pole of order $p$ of $f$. In each case, the assumption that: :$\exists s < 1: f \sqbrk {\map B {0, s} \setminus \set 0}$ is not a dense subset of $\C$ contradicts the fact that $0$ is an essential singularity of $f$. Hence the result, by Proof by Contradiction. {{qed}}
Follows from: :Compact Complement Topology is Irreducible :Irreducible Space is Connected {{qed}}
Follows from: :Compact Complement Topology is Irreducible :Irreducible Space is Locally Connected {{qed}}
By definition, closed sets of $T$ are compact sets of $T$. So, for example, $\left[{0 \,.\,.\, 1}\right]$ and $\left[{2 \,.\,.\, 3}\right]$ are disjoint compact sets and therefore closed sets of $T$. Hence the result by definition of ultraconnected. {{qed}}
We have: : Compact Complement Topology is $T_1$ : Non-Trivial Ultraconnected Space is not $T_1$ Hence the result. {{qed}}
Let $\CC$ be an open cover of $\R$. Let $U \in \CC$. Then by definition $\R \setminus U$ is compact in the usual (Euclidean) topology. Since each $U$ is open in the Euclidean topology, a finite number must cover $\R \setminus U$. {{qed}}
Let $p \in \R$. Let: :$\BB_p = \set {\openint {-\infty} {-n} \cup \openint {p - \dfrac 1 n} {p + \dfrac 1 n} \cup \openint n \infty: n \in \N}$ Let $n \in \N$ and $P_n \in \BB$, so that: :$P_n = \openint {-\infty} {-n} \cup \openint {p - \dfrac 1 n} {p + \dfrac 1 n} \cup \openint n \infty$ Then: :$\R \setminus P_n = \closedint {-n} {p - \dfrac 1 n} \cup \closedint {p + \dfrac 1 n} n$ which is the union of two compact sets in $\R$ and therefore itself compact. Clearly $p \in P_n$. So $\BB$ is a set of open neighborhoods of $p$ in $T$ Now let $U \in \tau$ such that $p \in U$. Then $V := \R \setminus U$ is compact and is therefore bounded. Suppose $\sup V = M, \inf V = m$. Then $\exists n_B \in \N: n_T > M, -n_B > m$. But as $p \notin V$ we have: :$\exists \epsilon \in \R: p + \epsilon \notin V, p - \epsilon \notin V$ So $\exists n \in \N: \epsilon > \dfrac 1 n$ and so by making $n$ large enough you can fix it so that: :$\openint {-\infty} {-n} \cup \openint {p - \dfrac 1 n} {p + \dfrac 1 n} \cup \openint n \infty \subseteq U$ so fulfilling the conditions for $\BB$ to be a local basis for $p$ which is countable. {{qed}}
Let $\CC$ be an open cover of $\R$. Let $p \in \R$. Consider the set: :$\BB_p = \set {\openint {-\infty} {-n} \cup \openint {p - \dfrac 1 n} {p + \dfrac 1 n} \cup \openint n \infty: n \in \N}$ From Countable Local Basis in Compact Complement Topology, $\BB_p$ is a countable local basis for $T$. Hence the result, by definition of first-countable space. {{qed}}
We have: :Compact Complement Topology is Compact :Compact Space is Countably Compact :Compact Complement Topology is First-Countable :First-Countable Space is Sequentially Compact iff Countably Compact {{qed}}
Let $U \in \tau$. Then $V = \R \setminus U$ is compact. So by definition $V$ is closed in the Euclidean topology. That is, $U = \R \setminus V$ is open in the Euclidean topology. That is, every open set in the compact complement topology is also open in the Euclidean topology. Hence the result, by definition of coarser topology. {{qed}}
We have that $p \in \relcomp S \O = S$, so $\O \in \tau_p$. We have that $\relcomp S S = \O$ which is finite, so $S \in \tau_p$. Now consider $A, B \in \tau_p$, and let $H = A \cap B$. If $p \notin A$ or $p \notin B$ then $p \notin A \cap B$ and so $H \in \tau_p$. Now suppose $p \in A$ and $p \in B$. Then: {{begin-eqn}} {{eqn | l = H       | r = A \cap B       | c =  }} {{eqn | ll= \leadsto       | l = \relcomp S H       | r = \relcomp S {A \cap B}       | c =  }} {{eqn | r = \relcomp S A \cup \relcomp S B       | c = De Morgan's Laws: Complement of Union }} {{end-eqn}} In order for $A$ and $B$ to be open sets it follows that $\relcomp S A$ and $\relcomp S B$ are both finite. Hence their union is also finite and so $\relcomp S H$ is finite. So $H = A \cap B \in \tau_p$ as its complement is finite. Now let $\UU \subseteq \tau_p$. Then from De Morgan's Laws: Complement of Union: :$\displaystyle \relcomp S {\bigcup \UU} = \bigcap_{U \mathop \in \UU} \relcomp S U$ We have either of two options: :$(1): \quad \forall U \in \UU: p \in \relcomp S U$ in which case: :$\displaystyle p \in \bigcap_{U \mathop \in \UU} \relcomp S U$ Or: :$(2): \quad \exists U \in \UU: \relcomp S U$ is finite in which case: :$\displaystyle \bigcap_{U \mathop \in \UU} \relcomp S U$ is finite, from Intersection is Subset. So in either case $\displaystyle \bigcup \UU \in \tau_p$. {{qed}}
Let $T_1 = \struct {S, \tau_1}$ be the excluded point space on $S$ from $p$. Let $T_2 = \struct {S, \tau_2}$ be the finite complement space on $S$. By definition: :$\tau_1 = \set {H \subseteq S: p \in \relcomp S H} \cup \set S$  :$\tau_2 = \leftset {H \subseteq S: \relcomp S H}$ is finite$\rightset {} \cup \set \O$ By definition of Fort space, we have: :$U \in \tau_1 \implies U \in \tau_p$ :$U \in \tau_2 \implies U \in \tau_p$ So $\tau_1 \cup \tau_2 \subseteq \tau_p$. Similarly: :$U \in \tau_p \implies U \in \tau_1 \lor U \in \tau_2$ and so $\tau_p \subseteq \tau_1 \cup \tau_2$. So $\tau_p = \tau_1 \cup \tau_2$ and the result follows from Union is Smallest Superset. {{qed}}
From Fort Space is Excluded Point Space with Finite Complement Space, $T$ is an expansion of a finite complement space. Then we have that a Finite Complement Space is $T_1$. Then from Separation Properties Preserved by Expansion we have that as a finite complement space is a $T_1$ space, then so is $T$. {{qed}}
Let $A, B \in \tau_p$ such that $A$ and $B$ are separated. If $p \notin A$ and $p \notin B$ then $A$ and $B$ are both open. Otherwise $p$ must be in exactly one of them, because if $p$ were in both they could not be separated. {{WLOG}}, suppose that $p \in A$. Then $p \notin B$ so $B$ is open. Now suppose $B$ were not closed. Then $B$ is infinite by definition of Fort space. But every open set containing $p$ is cofinite in $S$. So $p$ would be in $B^-$, where $B^-$ is the closure of $B$. But then $p \in A$, by hypothesis. So $A \cap B^- \ne \O$ and so $A$ and $B$ are not separated. So $B$ must be closed. Therefore $\relcomp S B$ is an open set such that $A \subseteq \relcomp S B$, and the $T_5$ separation axiom is satisfied. {{qed}}
We have: :Fort Space is $T_1$ :Fort Space is $T_5$ and so by definition $T$ is completely normal. {{qed}} See Sequence of Implications of Separation Axioms for confirmation of the statement about weaker separation axioms.
From Clopen Points in Fort Space, $\set p$ is closed in $T$. Consider a countable intersection of open sets of $T$ which contain $p$. By definition, all these are cofinite in $S$ and so uncountable. So this intersection must itself contain all but a countable number of points of $S$. So $\set p$ is not a $G_\delta$ set. Hence $T$ is not a perfectly normal space as not all its closed sets is a $G_\delta$ set. {{qed}}
We have from Closed Set of Countable Fort Space is $G_\delta$ that every closed set in $T$ is a $G_\delta$ set. From Fort Space is $T_5$ and $T_5$ Space is $T_4$ Space, we have that a Fort space is a $T_4$ space. From Fort Space is $T_1$ it follows by definition that $T$ is a perfectly normal space. {{qed}}
Let $\CC$ be an open cover of $T$. Then $\exists U \in \CC$ such that $p \in U$ and so $\relcomp S U$ is finite. For each $x \in \relcomp S U$ there exists some $C_x \in \CC$ such that $x \in C$. So $U$, together with each of those $C_x \in \CC$, is a finite subcover of $\CC$. Hence the result by definition of compact space. {{qed}}
Let $\sequence {x_n}_{n \mathop \in \N}$ be an infinite sequence in $T$. Suppose $\sequence {x_n}$ takes an infinite number of distinct values in $S$. Then there is an infinite subsequence $\sequence {x_{n_r} }_{r \mathop \in \N}$ with distinct terms. Let $U$ be a neighborhood of $p$. Then $S \setminus U$ is a finite set by definition. Thus there exists $N \in \N$ such that $\forall r > N: x_{n_r} \in U$. Thus $\sequence {x_{n_r} }$ converges to $p$. Otherwise $\sequence {x_n}$ only takes a finite number of distinct values. Then, since $\sequence {x_n}$ is infinite, there exists $x \in S$ such that: :$\forall N \in \N: \exists n > N: x = x_n$ This implies that we can take a subsequence of $\sequence {x_n}$ which is constant, and which converges to that constant. We can conclude then that, by definition, $T$ is a sequentially compact space. {{qed}}
Let $\UU$ be a countable set of open neighborhoods of $p$. Let $U \in \UU$. Then as $p \in U$, $p \notin \relcomp S U$ and so for $U$ to be open it must follow that $\relcomp S U$ is finite. From De Morgan's Laws: Complement of Intersection we have: :$\displaystyle H := \bigcup_{U \mathop \in \UU} \relcomp S U = \relcomp S {\bigcap \UU}$ From Countable Union of Countable Sets is Countable it follows that $H$ can be no more than countable. So $\displaystyle \bigcap \UU = \relcomp S H$ must be uncountable. So (trivially): :$H \ne \set p$ So: :$\exists q \ne p: q \in \displaystyle \bigcap \UU$ So $\relcomp S {\set q}$ is an open neighborhood of $P$ which does not contain any of the elements of $\UU$. So $\UU$ is not a countable local basis of $p$. Hence by definition $T$ is not a first-countable space. {{qed}}
From the definition of the Boubaker polynomials, we have:     :$ \begin{cases} B_0 \left({x}\right) = 1 \\     B_1 \left({x}\right) = x \\     \end{cases}$ We have also, for $T_n$, the Chebyshev polynomials of the first kind and $U_n$, the Chebyshev polynomials of the second kind:         :$ \begin{cases} T_0 \left({x}\right) = 1 \\     T_1 \left({x}\right) = x \\     \end{cases}$ and : :$ \begin{cases} U_{-1} \left({x}\right) = 0 \\     U_0 \left({x}\right) = 1 \\     U_1 \left({x}\right) = 2x \\     \end{cases}$ Then, from the definition of the exponential generating functions of the Chebyshev polynomials of the first kind: : $\displaystyle f_{T_n, \operatorname{EXP}} \left({x, t}\right) = e^{t \frac x 2} \cos \left({t \sqrt {1 - \left({\frac x 2}\right)^2}}\right) $ and the Chebyshev polynomials of the second kind:  : $\displaystyle f_{U_n, \operatorname{EXP}} \left({x, t}\right) = e^{t \frac x 2} \frac {\sin \left({t \sqrt {1 - \left({\frac x 2}\right)^2}}\right)} {\sqrt {1 - \left({\frac x 2}\right)^2}} $     and from the relation: : $B_n \left({x}\right) = 4 x U_{n-1}\left({\frac x 2}\right) -2 T_n \left({\frac x 2}\right)$ which is valid only for $n>1$, we have: : $\displaystyle B_0 \left({x}\right) \frac{t^0}{0!}+ B_1 \left({x}\right) \frac{t^1}{1!} + \sum_{n=2}^\infty B_n \left({x}\right) \frac{t^n}{n!} = 4x \sum_{n=2}^\infty U_{n-1} \left({x}\right) \frac{t^n}{n!} - 2 \sum_{n=2}^\infty T_n \left({x}\right) \frac{t^n}{n!}+B_0 \left({x}\right) \frac{t^0}{0!}+ B_1 \left({x}\right) \frac{t^1}{1!} -4x \sum_{n=0}^1 U_{n-1} \left({x}\right) \frac{t^n}{n!} - 2 \sum_{n=0}^1 T_n \left({x}\right) \frac{t^n}{n!} $      Finally, by simplifying the two sides and considering the exponential generating functions and the values of the first terms of $T_n$ and $U_n$, we obtain:       : $\displaystyle \sum_{n=0}^\infty B_n \left({x}\right) \frac{t^n}{n!} =    4xe^{t \frac x 2} \frac {\sin \left({t \sqrt {1 - \left({\frac x 2}\right)^2}}\right)} {\sqrt {1 - \left({\frac x 2}\right)^2}} - 2 e^{t \frac x 2} \cos \left({t \sqrt {1 - \left({\frac x 2}\right)^2}}\right) - 3$ Hence the result.      {{qed}} {{MissingLinks}}   Category:Boubaker Polynomials Category:Generating Functions m9fsgzrtn86mth7uyu6jjnvjf7o36dp
We have that $\set q$ is finite so $\relcomp S {\set q}$ is cofinite. So $\relcomp S {\set q}$ is open and so $\set q$ is closed. Then we have that $p \notin \set q$ so $\set q$ is open. However, $p \notin \relcomp S {\set p}$ and $\relcomp S {\set p}$ is infinite. So $\set p$ is not open in $S$. But $\set p$ is finite, so $\relcomp S {\set p}$ is cofinite. Hence, as for $\set q$, we have that $\set p$ is closed. {{qed}}
Let $a, b \in S$ such that $a \ne b$. From Clopen Points in Fort Space, all points in $S$ apart from $p$ are both open and closed in $T$. {{WLOG}}, suppose that $a \ne p$. Then as $\set a$ is closed, $\relcomp S {\set a}$ is open in $T$. So we have a partition $\set a \mid \relcomp S {\set a}$ of $S$ such that $a \in \set a, b \in \relcomp S {\set a}$. So, by definition  $T$ is a totally separated space. {{qed}}
We note from Fort Space is Completely Normal that $T$ is a $T_2$ (Hausdorff) space. Let $H \subseteq S$ be an infinite subset of $S$ such that $p \in \relcomp S H$ and $\relcomp S H$ is infinite. Then $H$ is open as $p \in \relcomp S H$. From Limit Points in Fort Space, as $H$ is infinite, $p$ is the only limit point of $H$. So $H^- = H \cup \set p$ where $H^-$ is the closure of $H$. But as $\relcomp S {H^-}$ is infinite and does not contain $p$, it follows that $H^-$ is not open in $T$. So by definition $T$ is not an extremally disconnected space. {{qed}}
Let $q \in S$ such that $q \ne p$. Then from Clopen Points in Fort Space, $\set q$ is clopen. So $\forall q \in S, q \ne p: \set {\set q}$ is a local basis for $q$. If we take the open neighborhoods of $p$ we get a local basis $\UU_p$ with the following property: :Since $p \in U \in \UU_p$, its complement does not contain $p$ and so it is open. This implies that $U$ is also closed. The union of the local basis forms a basis for the topology. This basis is formed with clopen sets. So, by definition, $T$ is a zero dimensional space. {{qed}}
We prove the contrapositive. Suppose both $\pi \cup \{\phi\}$ and $\pi \cup \{\neg\phi\}$ fork over $A$. By definition of forking, :$\pi\cup\{\phi\}$ implies $\phi_1 (\bar x, \bar c_1) \vee \cdots \vee \phi_k (\bar x, \bar c_k)$ and :$\pi\cup\{\neg\phi\}$ implies $\psi_1 (\bar x, \bar d_1) \vee \cdots \vee \psi_k (\bar x, \bar d_h)$, where each $\phi_i$ and $\psi_j$ divide over $A$. Then $\pi$ implies the disjunction :$\bigvee_i \phi(\bar x, \bar c_i) \vee \bigvee_j \psi(\bar x, \bar d_i)$ with each component formula dividing over $A$. By definition, this means that $\pi$ forks over $A$. {{qed}} Category:Model Theory fuwj2wm9i715zb7ksuzsg8h7w6n2czx
Suppose $\pi$ does not fork over $A$. We will use Zorn's Lemma to find a candidate for the needed complete type. Consider the collection $\Pi$ of all non-forking sets $\pi'$ of $\mathcal L$-formulas with parameters from $B$ such that $\pi'$ contains $\pi$. Order $\Pi$ by subset inclusion. Since a set forks iff a finite subset forks, the union over any chain is still a non-forking set, and hence is an upper bound for the chain. Thus, by Zorn's Lemma, there is a maximal (with respect to subset inclusion) $p$ in $\Pi$. Suppose $p$ is not complete. :By definition, for some $\phi(\bar x, \bar b)$, $p$ contains neither $\phi(\bar x, \bar b)$ nor $\neg\phi(\bar x, \bar b)$. :Since $p$ is non-forking, by Formula and its Negation Cannot Both Cause Forking, at least one of $p\cup\phi(\bar x, \bar b)$ or $p\cup\neg\phi(\bar x, \bar b)$ is non-forking as well. :Hence $p$ is not maximal in $\Pi$, contradicting the choice of $p$. Thus $p$ is complete. {{qed}} {{AoC|Zorn's Lemma}} Category:Model Theory ppskkx54i644iki7i51u3hpdka104hw
The proof is straightforward using the definition of forking and the fact that proofs in first-order logic are finite. Suppose $\pi$ forks over $A$. :By definition, $\pi$ implies a disjunction of formulas which each divide over $A$. :Since proofs are finite, this means that there is a finite subset of $\pi$ which implies this disjunction, completing this direction of the proof. Suppose a finite subset of $\pi$ forks over $A$. :By definition, the finite subset implies a disjunction of formulas which each divide over $A$. :But then $\pi$ clearly implies this disjunction as well, completing this direction of the proof. {{qed}} Category:Model Theory 5xn75ly0rbxveqozd29d7zr8ixy1slr
From the definition, $X$ is finite {{iff}} $\exists n \in \N$ such that there exists a bijection: :$f: X \leftrightarrow \N_n$ where $\N_n$ is the set of all elements of $\N$ less than $n$, that is: :$\N_n = \set {0, 1, 2, \ldots, n - 1}$ The case in which $X$ is empty is trivial. We begin proving the following particular case: :''If $X$ is finite and $a \in X$, then $X \setminus \set a$ is also finite.'' From Bijection between Specific Elements there exists a bijection $f: \N_n \to X$, which satisfies $\map f n = a$. Next we prove the general case by induction. === Basis for the Induction === If $n = 1$ then $X \setminus \set a = \O$ is finite. If $n > 1$, the restriction of $f$ to $\set {k \in \N: k \le n - 1}$ yields a bijection into $X \setminus \set a$. Hence $X \setminus \set a$ is finite and has $n - 1$ elements. So, we have that if $n = 1$, then its subsets ($\O$ and $X$) are finite. This is the basis for the induction. === Induction Hypothesis === Now we need to show that, if our Theorem is valid for sets with $n$ elements, it is also true for sets with $n + 1$ elements. This is our induction hypothesis. === Induction Step === This is our induction step: Let $X$ have $n + 1$ elements, and let $Y \subseteq X$. If $Y = X$, there is nothing to prove. Otherwise, $\exists a \in X \setminus Y$. This means that $Y$ is actually a subset of $X \setminus \set a$. Since $X \setminus \set a$ has $n$ elements, it follows that $Y$ is finite. {{qed}}
We have that $p \in \relcomp S \O = S$ so $\O \in \tau_p$. We have that $\relcomp S S = \O$ so $\O \in \tau_p$. Now consider $A, B \in \tau_p$, and let $H = A \cap B$. If $p \notin A$ or $p \notin B$ then $p \notin A \cap B$ and so $H \in \tau_p$. Now suppose $p \in A$ and $p \in B$. Then: {{begin-eqn}} {{eqn | l = H       | r = A \cap B       | c =  }} {{eqn | ll= \leadsto       | l = \relcomp S H       | r = \relcomp S {A \cap B}       | c =  }} {{eqn | r = \relcomp S A \cup \relcomp S B       | c = De Morgan's Laws: Complement of Intersection }} {{end-eqn}} In order for $A$ and $B$ to be open sets it follows that $\relcomp S A$ and $\relcomp S B$ are both countable. From Countable Union of Countable Sets is Countable, their union is also countable and so $\relcomp S H$ is countable. So $H = A \cap B \in \tau_p$ as its complement is countable. Now let $\UU \subseteq \tau_p$. Then: :$\displaystyle \relcomp S {\bigcup \UU} = \bigcap_{U \mathop \in \UU} \relcomp S U$ by De Morgan's Laws: Complement of Union. We have either of two options: :$(1): \quad \forall U \in \UU: p \in \relcomp S U$ in which case: :$\displaystyle p \in \bigcap_{U \mathop \in \UU} \relcomp S U$ Or: :$(2): \quad \exists U \in \UU: \relcomp S U$ is countable in which case: :$\displaystyle \bigcap_{U \mathop \in \UU} \relcomp S U$ is countable, from Intersection is Subset. So in either case $\displaystyle \bigcup \UU \in \tau_p$. {{qed}}
We have: :Fortissimo Space is $T_1$ :Fortissimo Space is $T_5$ and so by definition $T$ is completely normal. {{qed}} See Sequence of Implications of Separation Axioms for confirmation of the statement about weaker separation axioms.
From Fortissimo Space is Excluded Point Space with Countable Complement Space, $T$ is an expansion of a countable complement space. Then we have that a Countable Complement Space is $T_1$. Then from Separation Properties Preserved by Expansion we have that as a cuntable  complement space is a $T_1$ space, then so is $T$. {{qed}}
Let $T_1 = \struct {S, \tau_1}$ be the excluded point space on $S$ from $p$. Let $T_2 = \struct {S, \tau_2}$ be the countable complement space on $S$. By definition: :$\tau_1 = \set {H \subseteq S: p \in \relcomp S H} \cup \set S$  :$\tau_2 = \leftset {H \subseteq S: \relcomp S H}$ is countable $\rightset{} \cup \set \O$ By definition of Fortissimo space, we have: :$U \in \tau_1 \implies U \in \tau_p$ :$U \in \tau_2 \implies U \in \tau_p$ So $\tau_1 \cup \tau_2 \subseteq \tau_p$. Similarly: :$U \in \tau_p \implies U \in \tau_1 \lor U \in \tau_2$ and so $\tau_p \subseteq \tau_1 \cup \tau_2$. So $\tau_p = \tau_1 \cup \tau_2$ and the result follows from Union is Smallest Superset. {{qed}}
Let $A, B \in \tau_p$ such that $A$ and $B$ are separated. If $p \notin A$ and $p \notin B$ then $A$ and $B$ are both open. Otherwise $p$ must be in exactly one of them, because if $p$ were in both they could not be separated. {{WLOG}}, suppose $p \in A$. Then $p \notin B$ so $B$ is open. Now suppose $B$ were not closed. Then $B$ is uncountable by definition of Fortissimo space. But every open set not containing $p$ is co-countable in $S$. So $p$ would be in $B^-$, where $B^-$ is the closure of $B$. But then $p \in A$, by hypothesis. So $A \cap B^- \ne \O$ and so $A$ and $B$ are not separated. So $B$ must be closed. Therefore $\relcomp S B$ is an open set such that $A \subseteq \relcomp S B$, and the $T_5$ separation axiom is satisfied. {{qed}}
Let $\CC$ be an open cover of $T$. Then $\exists U \in \CC$ such that $p \in U$ and so $\relcomp S U$ is countable. So $U$, together with an open neighborhood of each of the elements of $\relcomp S U$, is a countable subcover of $\CC$. Hence the result by definition of Lindelöf space. {{qed}}