hcju/proofwikiqa huggingface dataset:
Let $a, b \in S$ such that $a \ne b$. Let us assume the truth of the Continuum Hypothesis. Then $a$ and $b$ are contained in a subset $X \subseteq S$ whose cardinality is the same as that of $\closedint 0 1$. So by definition of cardinality we can set up a bijection $f: \closedint 0 1 \leftrightarrow X$ such that $\map f 0 = a$ and $\map f 1 = b$. Each element of $S$ is closed from Finite Complement Space is $T_1$. Hence $f$ is continuous and therefore an arc in $T$. As $a$ and $b$ are arbitrary, $T$ is arc-connected. Now let $\BB$ be a basis for $T$. Let $B \in \BB$ and let $a, b \in B$. By definition of a finite complement topology on a uncountable set, $B$ is uncountable. So, by the same argument as above, we can set up an arc $f$ between any points of $B$. So $T$ is locally arc-connected. {{qed}}
By definition, we have that $\O \in \tau$. We also have that $S \in \tau$ as $\complement_S \left({S}\right) = \O$ which is trivially countable. Now suppose $A, B \in \tau$. Let $H = A \cap B$. 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}} But as $A, B \in \tau$ it follows that $\relcomp S A$ and $\relcomp S B$ are both countable. Hence from Countable Union of Countable Sets is Countable their union is also countable and so $\relcomp S H$ is countable. So $H = A \cap B \in \tau$ as its complement is countable. Now let $\UU \subseteq \tau$. Then: :$\displaystyle \relcomp S {\bigcup \UU} = \bigcap_{U \mathop \in \UU} \relcomp S U$ by De Morgan's Laws: Complement of Union. But as: :$\forall U \in \UU: \relcomp S U \in \tau$ each of the $\relcomp S U$ is countable. Hence so is their intersection. So $\displaystyle \relcomp S {\bigcup \UU}$ is countable which means: :$\displaystyle \bigcup \UU \in \tau$ So $\tau$ is a topology on $T$. {{qed}}
Let $U \in \tau', U \ne \O$. Then $\relcomp S U$ is finite by definition of finite complement topology. Then by definition of countable complement topology, we have that $U \in \tau$. So $\tau' \subseteq \tau$ and hence the result by definition of expansion. {{qed}}
We have that the Countable Complement Topology is Expansion of Finite Complement Topology.  We also have that a Finite Complement Space is $T_1$. Then from Separation Properties Preserved by Expansion, we have that $T$ is a $T_1$ (Fréchet) space. {{qed}}
Let $U \in \tau$ be any non-empty open set of $T$. Then $\relcomp S U$ is countable. Suppose $H \cap U = \O$. Then from Intersection with Complement is Empty iff Subset it follows that $H \subseteq \relcomp S U$ and so $H$ is countable. So if $H$ is uncountable it is bound to have a non-empty intersection with every open set in $T$. {{finish|It still needs to be shown that $U \cap H$ is uncountable {{iff}} $H$ is uncountable.}}
We have: :Countable Complement Space is Irreducible :Irreducible Hausdorff Space is Singleton {{qed}}
Let $U_1, U_2 \in \tau$ be non-empty open sets of $T$. We have that both $U_1$ and $U_2$ are both uncountable if $S$ is. From Uncountable Subset of Countable Complement Space Intersects Open Sets, they intersect each other. Hence the result from definition of irreducible space. {{qed}}
We have that a Countable Complement Space is a $T_1$ space. From $T_1$ Space is $T_0$ Space, $T$ is a $T_0$ space. We then have that a Countable Complement Space is not $T_2$. 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 $T$ cannot 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, $T$ cannot 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 $T$ cannot 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, $T$ cannot be a $T_5$ space. {{qed}}
By definition, we have that $\varnothing \in \tau$. We also have that $S \in \tau$ as $\complement_S \left({S}\right) = \varnothing$ which is trivially finite. Now suppose $A, B \in \tau$. Let $H = A \cap B$. Then: {{begin-eqn}} {{eqn | l = H       | r = A \cap B       | c =  }} {{eqn | ll= \implies       | l = \complement_S \left({H}\right)       | r = \complement_S \left({A \cap B}\right)       | c =  }} {{eqn | r = \complement_S \left({A}\right) \cup \complement_S \left({B}\right)       | c = De Morgan's Laws: Complement of Intersection }} {{end-eqn}} But as $A, B \in \tau$ it follows that $\complement_S \left({A}\right)$ and $\complement_S \left({B}\right)$ both have a cardinality strictly smaller than that of $S$. Hence their union also has a cardinality strictly smaller than that of $S$. So $\complement_S \left({H}\right)$ has a cardinality strictly smaller than that of $S$. So $H = A \cap B \in \tau$ as its complement has a cardinality strictly smaller than that of $S$. Now let $\mathcal U \subseteq \tau$. Then from De Morgan's laws: Complement of Union: : $\displaystyle \complement_S \left({\bigcup \mathcal U}\right) = \bigcap_{U \in \mathcal U} \complement_S \left({U}\right)$ But as: :$\forall U \in \mathcal U: \complement_S \left({U}\right) \in \tau$ each of the $\complement_S \left({U}\right)$ has a cardinality strictly smaller than that of $S$. Hence so has their intersection. So $\displaystyle \complement_S \left({\bigcup \mathcal U}\right)$ has a cardinality strictly smaller than that of $S$, which means: : $\displaystyle \bigcup \mathcal U \in \tau$ So $\tau$ is a topology on $T$. {{qed}}
{{AimForCont}} $\displaystyle \mathbb I = \bigcup_{i \mathop = 1}^\infty C_i$ where $\set {C_i}$ is a set of pairwise disjoint closed sets. Let: :$\displaystyle B = \bigcup \partial C_i = \mathbb I \setminus \bigcup C_i^\circ$ where $\partial C_i$ is the boundary of $C_i$ and $C_i^\circ$ is the interior of $C_i$. Let $J \subseteq \mathbb I$ be a subinterval of $\mathbb I$. Then $J$ is non-meager, and so some $C_k$ is dense in some open interval $L \subseteq J$. Since $C_k$ is closed we have that $L \subseteq C_k^\circ$ and so $L \cap B = \O$. So $J$ contains an open subset $L$ disjoint from $B$. So $B$ is nowhere dense in $\mathbb I$. So every open interval $U$ containing some $x \in \partial C_j$ must intersect $B \setminus \partial C_j$. This is because $U$ is an open neighborhood of $x$ and therefore contains some point $u \in \mathbb I \setminus C_j$, say $u \in C_m$. Then if $U \cap B \cap \partial C_m = \varnothing$ it follows that $C_m^\circ \cap U$ is a non-empty open set in $U$. which is also relatively closed. {{explain|Closed relative to what?}} Now $B$ is itself non-meager, since it is a closed subset of $\mathbb I$. Thus some $\partial C_k$ is dense in some non-empty open set $U \cap B$, for some open interval $U$ in $\mathbb I$. Since $\partial C_k$ is closed, this means $\partial C_k \cap U = B \cap U$. But this is impossible, since if $U \cap C_k \ne \O$ then $U \cap \paren {B \setminus \partial C_k} \ne \O$. This contradiction shows that $\mathbb I$ cannot be expressed as $\displaystyle \bigcup_{i \mathop = 1}^\infty C_i$. {{qed}} {{proofread}}
From Finite Topological Space is Compact, any finite subset of $S$ is compact. {{AimForCont}} $H \subseteq S$ is an infinite compact set. Take a (countably) infinite sequence $\sequence {a_n}_{n \mathop \ge 0}$ of distinct elements of $H$. Consider the open sets: :$V_m := S \setminus \set {a_{m + n} }_{n \mathop \ge 0}$ for $m \ge 0$, which satisfy $V_{m_1} \subset V_{m_2}$ if $m_1 < m_2$. Then: :$\displaystyle H \subseteq \bigcup_{m \mathop \ge 0} V_m$ is an open cover of $H$. Since $H$ is compact, it has a finite open subcover, say: :$\displaystyle H \subseteq \bigcup_{i \mathop = 0}^N V_{m_i}$ with $m_0 < m_1 < \cdots < m_N$. But then: :$\displaystyle \bigcup_{i \mathop = 0}^N V_{m_i} = V_{m_N}$  and so: :$\displaystyle a_{m_N + 1}, a_{m_N + 2}, \ldots \notin V_{m_N} = \bigcup_{i \mathop = 0}^N V_{m_i}$ which implies $a_{m_N + 1}, a_{m_N + 2}, \ldots \notin H$, a contradiction. Hence the result by Proof by Contradiction. {{qed}}
From Compact Sets in Countable Complement Space, the only compact sets in $T$ are finite. A countable union of finite sets can not be an uncountable set. Hence the result by definition of $\sigma$-compact space. {{qed}}
Consider $U \in \tau$. By definition of countable complement topology, $\relcomp S U$ is countably infinite. Then for each $x \in \relcomp S U$, $\relcomp S U \setminus \set x$ is countably infinite. Then for each $x \in \relcomp S U$, $\relcomp S {\relcomp S U \setminus \set x} = U \cup \set x$ is open. Hence $\set {U \cup \set x: x \in \relcomp S U}$ is an open cover. This cover is countable because it is equivalent to $\relcomp S U$. {{AimForCont}} $T$ is a countably compact space. Then by definition there exists a finite subcover $\set {U \cup \set {x_1}, \ldots, U \cup \set {x_n} }$ of $T$. Then: :$\paren {U \cup \set {x_1} } \cup \dotsb \cup \paren {U \cup \set {x_n} } = U \cup \set {x_1, \ldots, x_n}$ Since $\relcomp S U$ is not finite, it follows that $U \cup \set {x_1, \ldots, x_n} \ne S$. Thus there is no finite subcover of $T$. It follows by Proof by Contradiction that $T$ cannot be a countably compact space. {{qed}}
By definition, $T$ is a Lindelöf space {{iff}} every open cover of $S$ has a countable subcover. Let $\CC$ be an open cover of $T$. Let $U \in \CC$ be any set in $C$. It covers all but a countable number of points of $T$. So for each of those points we pick an element of $\CC$ which covers each of those points. Hence we have a countable subcover of $T$. So by definition $T$ is a Lindelöf space. {{qed}}
{{AimForCont}} some $x \in S$ has a countable local basis. That means: :there exists a countable set of sets $\BB_x \subseteq \tau$  such that: :$\forall B \in \BB_x: x \in B$ and such that: :every open neighborhood of $x$ contains some $B \in \BB_x$. So: {{begin-eqn}} {{eqn | l = \bigcap \BB_x       | r = \set x       | c =  }} {{eqn | ll= \leadsto       | l = S \setminus \set x       | r = S \setminus \bigcap \BB_x       | c =  }} {{eqn | r = \bigcup_{B \mathop \in \BB_x} \paren {S \setminus B}       | c = De Morgan's Laws: Difference with Intersection }} {{end-eqn}} By definition, each of $S \setminus B$ is countable. From Countable Union of Countable Sets is Countable it follows that $\displaystyle \bigcup_{B \mathop \in \BB_x} \paren {S \setminus B}$ is also countable. So $S \setminus \set x$ and therefore $S$ is also countable. From this contradiction (as we have specified that $S$ is uncountable) it follows that our assumption that $x \in S$ has a countable local basis must be false. Hence by definition $T$ can not be first-countable. {{qed}}
Let $s_i$ be the $i$th element of set $S$. Begin by defining set $A_m$, which is all of the permutations of $S$ which fixes $S_m$. Then the number of permutations, $W$, with ''at least'' one element fixed, $m$, is: :$\displaystyle W = \size {\bigcup_{m \mathop = 1}^n A_m}$ Applying the Inclusion-Exclusion Principle: {{begin-eqn}} {{eqn | l = W       | r = \sum_{m_1 \mathop = 1}^n \size {A_{m_1} }       | c =  }} {{eqn | o =        | ro= -       | r = \sum_{m_1, m_2 : 1 \mathop \le m_1 \mathop < m_2 \mathop \le n} \size {A_{m_1} \cap A_{m_2} }       | c =  }} {{eqn | o =        | ro= +       | r = \sum_{m_1, m_2, m_3 : 1 \mathop \le m_1 \mathop < m_2 \mathop < m_3 \mathop \le n} \size {A_{m_1} \cap A_{m_2} \cap A_{m_3} }       | c =  }} {{eqn | o =        | ro= -       | r = \cdots       | c =  }} {{end-eqn}} Each value $A_{m_1} \cap \cdots \cap A_{m_p}$ represents the set of permutations which fix $p$ values $m_1, \ldots, m_p$. Note that the number of permutations which fix $p$ values only depends on $p$, not on the particular values of $m$. Thus from Cardinality of Set of Subsets there are $\dbinom n p$ terms in each summation.  So: {{begin-eqn}} {{eqn | l = W       | r = \binom n 1 \size {A_1}       | c =  }} {{eqn | o =        | ro= -       | r = \binom n 2 \size {A_1 \cap A_2}       | c =  }} {{eqn | o =        | ro= +       | r = \binom n 3 \size {A_1 \cap A_2 \cap A_3}       | c =  }} {{eqn | o =        | ro= -       | r = \cdots       | c =  }} {{eqn | o =        | ro= +       | r = \paren {-1}^{p - 1} \binom n p \size {A_1 \cap \cdots \cap A_p}       | c =  }} {{eqn | o =        | r = \cdots       | c =  }} {{end-eqn}} $\size {A_1 \cap \cdots \cap A_p}$ is the number of permutations fixing $p$ elements in position. This is equal to the number of permutations which rearrange the remaining $n - p$ elements, which is $\paren {n - p}!$. Thus we finally get:   :$W = \dbinom n 1 \paren {n - 1}! - \dbinom n 2 \paren {n - 2}! + \dbinom n 3 \paren {n - 3}! - \cdots + \paren {-1}^{p - 1} \dbinom n p \paren {n - p}! \cdots$ That is: :$\displaystyle W = \sum_{p \mathop = 1}^n \paren {-1}^{p - 1} \binom n p \paren {n - p}!$ Noting that $\dbinom n p = \dfrac {n!} {p! \paren {n - p}!}$, this reduces to: :$\displaystyle W = \sum_{p \mathop = 1}^n \paren {-1}^{p - 1} \dfrac {n!} {p!}$  {{qed}}
A derangement is a permutation that fixes no elements. So, to compute the total number of derangements, we compute the number of permutations that fix elements, that is leave elements in the same place. Let $\pi$ be a permutation of $S$. Let $x, y \in S$. Let $i$ be a position in a given permutation of set $S$, where $1 \le i \le n$. Let $\map {P_i} \pi$ be the property that $\map \pi x = y$ at position $i$. Let $\map N {P_i, \dots, P_j}$ be the number of permutations where $\map {P_i} \pi, \dots, \map {P_j} \pi$ hold. If we consider all the permutations of a set where the property $P_i$ holds, we have fixed a single element of the set. So there are $\paren {n - 1}!$ such possible permutations. Furthermore, we could have chosen $i$ as any element in the set so there are $\dbinom n 1$ terms to choose from. So $\map N {P_i} = \paren {n - 1}!$ and there are $\dbinom n 1$ terms. Similarly, $\map N {P_i, P_j} = \paren {n - 2}!$ and there are $\dbinom n 2$ terms. So now we can compute the total number of derangements: {{begin-eqn}} {{eqn | l = D_n       | r = n! - \dbinom n 1 \paren {n - 1}! + \dbinom n 2 \paren {n - 2}! - \ldots + \paren {-1}^n \dbinom n n \paren {n - n}!       | c =  }} {{eqn | r = n! - \frac {n!} {1! \paren {n - 1}!} \paren {n - 1}! + \frac {n!} {2! \paren {n - 2}!} \paren {n - 2}! - \cdots + \paren {-1}^n \frac {n!} {n! \paren {n - n}!} \paren {n - n}!       | c = {{Defof|Binomial Coefficient}} }} {{eqn | r = n! \paren {1 - \frac 1 {1!} + \frac 1 {2!} - \frac 1 {3!} + \cdots + \paren {-1}^n \frac 1 {n!} }       | c =  }} {{end-eqn}} {{qed}}
Let $U \in \tau$ be any open set of $T$. Let $H \subseteq S$ be an uncountable. From Uncountable Subset of Countable Complement Space Intersects Open Sets, we have that $U \cap H$ is uncountable {{iff}} $H$ is uncountable. Let $x \in S$. Then every open set $U$ in $T$ such that $x \in U$ also contains an uncountable number of points of $H$ other than $x$. Thus, by definition, $x$ is a limit point of $H$. Now let $H$ be a countable subset of $S$. By definition $\relcomp S H$ is open in $T$ and so $H$ is closed. So by Closed Set Equals its Closure we have by definition that if $x$ is a limit point of $H$, then $x \in H$. {{qed}}
Let $U$ be a countable subset of $S$. By the definition of $T$, $U$ is closed. As $U$ is countable but $S$ is uncountable: :$U \subsetneq S$ and so: :$U \ne S$ From Closed Set Equals its Closure: :$U^- = U$ where $U^-$ is the closure of $U$. Thus: :$U^- \ne S$ So by definition, $U$ is not everywhere dense in $T$. As $U$ is arbitrary, there is no countable set which is everywhere dense in $T$. The result follows by definition of separable space. {{qed}}
We have that a Countable Complement Space is Irreducible. Therefore there is no disjoint set of open sets of $T$. So the result holds vacuously. {{qed}}
We have that a Countable Complement Space is Irreducible. The result follows from Irreducible Space is Connected. {{qed}}
We have that a Countable Complement Space is Irreducible. The result follows from Irreducible Space is Locally Connected. {{qed}}
We have that a Countable Complement Space is Irreducible. The result follows from Irreducible Space is Pseudocompact. {{qed}}
From Uncountable Subset of Countable Complement Space Intersects Open Sets, the intersection of any open sets is uncountable. So for any open cover $\CC$ of $T$, every point is in an infinite number of open sets of $T$. So no refinement of any open cover of $T$ can be point finite. Hence the result by definition of countably metacompact. {{qed}}
By definition, $T$ is weakly countably compact {{iff}} every infinite subset of $S$ has a limit point in $S$. Let $H \subseteq S$ be a countable set. From Limit Points of Countable Complement Space, it contains all its limit points. Let $x \in H$. $\relcomp S {\relcomp S H \cup \set x} = H \setminus \set x$ is countable. Then $\relcomp S H \cup \set x$ is an open neighborhood of $x$.  However $H \cap \paren {\paren {\relcomp S H \cup \set x} \setminus \set x} = H \cap \relcomp S H = \O$. So $x$ is not a limit point of $H$. Since $x$ is arbitrary, there are no limit points in $H$.  This shows that $T$ is not weakly countably compact. {{qed}}
The idea of the proof is that $\mathcal{M}$ being saturated means that when we want to define an elementary map $\mathcal{N}\to\mathcal{M}$, we can find an image $y\in \mathcal{M}$ for an element $x\in \mathcal{N}$ by realizing the type made up of the formulas that such a $y$ would need to satisfy. Let $\mathcal{N}$ be a model of $T$ with universe of cardinality strictly less than $\kappa$. We will construct an elementary embedding of $\mathcal{N}$ into $\mathcal{M}$ by transfinite recursion. Since $|\mathcal{N}|<\kappa$, we can write its elements as $n_\alpha$ for ordinals $\alpha < \kappa$. For each ordinal $\alpha < \kappa$, let $A_\alpha$ be the subset $\{n_\beta : \beta < \alpha\}$ of the universe of $\mathcal{N}$. Note for clarity that $n_\alpha \in A_{\alpha+1}$ but $n_\alpha \notin A_\alpha$. *Base case $\alpha = 0$: Define $f_0 = \emptyset$. Note that $f_0$ is trivially an elementary embedding from $A_0 = \emptyset$ into $\mathcal{M}$ *Limit ordinals $\alpha$, assuming $f_\beta$ is defined and elementary $A_\beta \to \mathcal{M}$ for all $\beta < \alpha$: Let $\displaystyle f_\alpha = \bigcup_{\beta < \alpha} f_\beta$. If $\phi$ is an $\mathcal{L}$-sentence with parameters from $A_\alpha$, then since it involves only finitely many such parameters, they must all be contained in some $A_\beta$ for $\beta < \alpha$.  But $f_\alpha \restriction A_\beta = f_\beta$ is elementary, so $f_\alpha$ must be as well. * Successor ordinals $\alpha = \beta + 1$, assuming $f_\beta$ is defined and elementary $A_\beta \to \mathcal{M}$: We need to extend $f_\beta$ to $A_\alpha = A_\beta \cup \{n_\beta\}$ so that truth of $\mathcal{L}$-sentences with parameters from $A_\alpha$ is preserved. Consider the subset $p = \{\phi(v, f_\beta(\bar{a})):\bar{a}\text{ is a tuple from }A_\beta \text{ and }\mathcal{N}\models\phi(n_\beta,\bar{a})\}$ of the set of $\mathcal{L}$-formulas with one free variable and parameters from the image $f_\beta (A_\beta)$ of $A_\beta$ under $f_\beta$. The set $p$ is a $1$-type over the image $f_\beta (A_\beta)$ in $\mathcal{M}$. Since $|A_\beta| < \kappa$ and by assumption $\mathcal{M}$ is $\kappa$-saturated, this means that $p$ is realized in $\mathcal{M}$ by some element $b$. Thus, defining $f_\alpha$ to be $f_\beta \cup \{(n_\beta, b)\}$ makes it an elementary embedding $A_\alpha \to \mathcal{M}$. Now, define $\displaystyle f = \bigcup_{\alpha < \kappa} f_\alpha$. Then $f$ is an elementary embedding from $\mathcal{N}$ to $\mathcal{M}$ since $\displaystyle \bigcup_{\alpha < \kappa} A_\alpha = \mathcal{N}$, any finite set of parameters from $\mathcal{N}$ must belong to one $A_\alpha$, and $f \restriction A_\alpha = f_\alpha$ is elementary. {{qed}} {{MissingLinks}} o0chgtswe94jl7ke5tvpt0un4m8c9fg
From Separation Axioms on Double Pointed Topology, we have that: :$T \times D$ is not a $T_0$ space. :$T \times D$ is not a $T_1$ space. :$T \times D$ is not a $T_2$ space. Also from Separation Axioms on Double Pointed Topology, we have that: :$T \times D$ is a $T_3$ space {{iff}} $T$ is a $T_3$ space :$T \times D$ is a $T_4$ space {{iff}} $T$ is a $T_4$ space :$T \times D$ is a $T_5$ space {{iff}} $T$ is a $T_5$ space But from Countable Complement Space is not $T_3$, $T_4$ or $T_5$: :$T$ is not a $T_3$ space :$T$ is not a $T_4$ space :$T$ is not a $T_5$ space Hence the result. {{qed}}
By definition, $T$ is weakly countably compact {{iff}} every infinite subset of $S$ has a limit point in $S$. Let $D = \set {0, 1}$. Let $\tuple {p, 0}$ belong to some infinite $A \subseteq S$. Then its twin $\tuple {p, 1}$ is a limit point of $A$. Hence the result by definition of weakly countably compact. {{qed}}
From Open Extension Topology is T4 we have that $\paren {T \times D}^*_{\bar p}$ is a $T_4$ space. From Double Pointed Countable Complement Topology fulfils no Separation Axioms, we have that $T \times D$ is not a $T_0$ space or a $T_5$ space. From Condition for Open Extension Space to be $T_0$ Space, it follows that $\paren {T \times D}^*_{\bar p}$ is not a $T_0$ space. From Condition for Open Extension Space to be $T_5$ Space, it follows that $\paren {T \times D}^*_{\bar p}$ is not a $T_5$ space. From Open Extension Topology is not $T_1$, we have that $\paren {T \times D}^*_{\bar p}$ is not a $T_1$ space. From Open Extension Topology is not $T_3$ we have that $\paren {T \times D}^*_{\bar p}$ is not a $T_3$ space. Finally, from $T_2$ Space is $T_1$ Space, as $\paren {T \times D}^*_{\bar p}$ is not a $T_1$ space it is not a $T_2$ space. {{qed}}
By definition, we have that $\O \in \tau$. We also have that $\R \in \tau$ as $\relcomp \R \R = \O$ which is trivially compact. Now suppose $A, B \in \tau$. Let $H = A \cap B$. Then: {{begin-eqn}} {{eqn | l = H       | r = A \cap B       | c =  }} {{eqn | ll= \leadsto       | l = \relcomp \R H       | r = \relcomp \R {A \cap B}       | c =  }} {{eqn | r = \relcomp \R A \cup \relcomp \R B       | c = De Morgan's Laws: Complement of Intersection }} {{end-eqn}} But as $A, B \in \tau$ it follows that $\relcomp \R A$ and $\relcomp \R B$ are both compact. Hence from Union of Compact Sets is Compact their union is also compact and so $\relcomp \R H$ is compact. So $H = A \cap B \in \tau$ as its complement is compact. Now let $\UU \subseteq \tau$. Then from De Morgan's Laws: Complement of Union: :$\displaystyle \relcomp \R {\bigcup \UU} = \bigcap_{U \mathop \in \UU} \relcomp \R U$ But as: :$\forall U \in \UU: \relcomp \R U \in \tau$ each of the $\relcomp \R U$ is compact. Hence from Intersection of Compact Sets is Compact, so is their intersection. So $\displaystyle \relcomp \R {\bigcup \UU}$ is compact which means $\displaystyle \bigcup \UU \in \tau$. So $\tau$ is a topology on $T$. {{qed}}
We have that a Finite Topological Space is Compact. So any finite subspace of $T$ is compact. Let $\tau^*$ be the set defined as: :$\tau^* = \leftset {S \subseteq \R: S = \O \text { or } \relcomp \R S}$ is finite $\rightset {}$ where $\relcomp \R S$ denotes the complement of $S$ in $\R$. Then $\tau^*$ is a subset of $\tau$ by definition of the compact complement topology. But $\tau^*$ is the finite complement topology. The result follows from Finite Complement Topology is Minimal $T_1$ Topology. {{qed}}