hcju/proofwikiqa huggingface dataset:
From Compact Sets in Fortissimo Space we have that the only compact sets in $T$ are finite. A union of countably many finite sets can not be an uncountable set. Hence the result by definition of $\sigma$-compact space. {{qed}}
Let $U$ be a countable subset of $S$. By the definition of the fortissimo space, $U$ is closed. From Closed Set Equals its Closure, $U^- = U \ne S$. Thus, by definition, $U$ is not everywhere dense in $T$. Thus, there exists no countable subset of $S$ which is everywhere dense in $T$. So, by definition, $T$ is not a separable space. {{qed}}
This proof follows the proof from Countable Complement Space is not First-Countable. {{AimForCont}} that $p \in S$ has a countable local basis. That means: :there exists a countable set of sets $\BB_p \subseteq \tau$ such that: :$\forall B \in \BB_p: p \in B$ and such that: :every open neighborhood of $p$ contains some $B \in \BB_p$. So: {{begin-eqn}} {{eqn | l = \bigcap \BB_p       | r = \set p       | c =  }} {{eqn | ll= \leadsto       | l = S \setminus \set p       | r = S \setminus \bigcap \BB_p       | c =  }} {{eqn | r = \bigcup_{B \mathop \in \BB_p} \paren {S \setminus B}       | c = De Morgan's Laws: Difference with Intersection }} {{end-eqn}} By definition of the fortissimo space, each of $S \setminus B$ is countable. From Countable Union of Countable Sets is Countable it follows that $\displaystyle \bigcup_{B \mathop \in \BB_p} \paren {S \setminus B}$ is also countable. So $S \setminus \set p$ and therefore $S$ is also countable. From this contradiction (as we have specified that $S$ is uncountable) it follows that our assumption that $p \in S$ has a countable local basis must be false. Hence by definition $T$ can not be first-countable. {{qed}}
Let $S = N \cup \set {a, b}$ where $N$ is infinite, $a \ne b$ and $a, b \notin N$. We have that $\O \subseteq N$ so $\O \in \tau_{a, b}$. We have that $a, b \in S, S \setminus S = \O$ and $\O$ is trivially finite, so $S \in \tau_{a, b}$. Now consider $A, B \in \tau_{a, b}$, and let $H = A \cap B$. If $A \subseteq N$ and $B \subseteq N$ then $A \cap B \subseteq N$ from Intersection is Largest Subset. So by definition $A \cap B \in \tau_{a, b}$ Now suppose $A \cap \set {a, b} \ne \O$ and $B \cap \set {a, b} \ne \O$. Then: {{begin-eqn}} {{eqn | l = H       | r = A \cap B       | c =  }} {{eqn | ll= \leadsto       | l = N \setminus H       | r = N \setminus \paren {A \cap B}       | c =  }} {{eqn | r = \paren {N \setminus A} \cup \paren {N \setminus B}       | c = De Morgan's Laws: Difference with Intersection }} {{end-eqn}} In order for $A$ and $B$ to be open sets we have that $N \setminus A$ and $N \setminus B$ are both finite. Hence their union is also finite and so $N \setminus \paren {A \cap B}$ is finite. So $H = A \cap B \in \tau_{a, b}$ as its complement is finite. Now let $\UU \subseteq \tau_{a, b}$. Then from De Morgan's Laws: Difference with Union: :$\displaystyle N \setminus \paren {\bigcup \UU} = \bigcap_{U \mathop \in \UU} \paren {N \setminus U}$ We have either of two options: :$(1): \quad \forall U \in \UU: U \subseteq N$ in which case: :$\displaystyle \bigcup \UU \subseteq N$ Or: :$(2): \quad \exists U \in \UU: N \setminus U$ is finite in which case: :$\displaystyle \bigcap_{U \mathop \in \UU} \paren {N \setminus U}$ is finite, from Intersection is Subset. So in either case: :$\displaystyle \bigcup \UU \in \tau_{a, b}$ {{qed}}
A subset of the natural numbers is also a subset of the real numbers $\R$. By definition, a bounded subset of $\R$ is bounded below in $\R$ and bounded above in $\R$. By the Well-Ordering Principle, $X$ is bounded below Thus $X$ is bounded {{iff}} $X$ is bounded above. That is, {{iff}}: :$\exists p \in \N: \forall x \in X: x \le p$ === Necessary Condition === Let $X$ be finite. Then $X = \set {x_1, x_2, \ldots, x_n}$ for some $n \in \N$. Let $p = x_1 + x_2 + \ldots + x_n$. Thus $x \in X \implies x \le p$. Thus $X$ is bounded above by $p$. So by definition $X$ is bounded. {{qed|lemma}} === Sufficient Condition === Let $X \subseteq \N$ be bounded. Then for some $p \in \N$, it is contained in the initial segment $\N_p = \set {0, 1, \ldots, p}$ of $\N$. It follows from Subset of Finite Set is Finite that $X$ is finite. {{qed}}
We have that $\tuple {0, 0} \notin \O$ so $\O \in \tau$. We have that $\forall m: \set {n: \tuple {m, n} \notin S} = \O$ which is finite, so $S \in \tau$. Now consider $A, B \in \tau$, and let $H = A \cap B$. If $\tuple {0, 0} \notin A$ or $\tuple {0, 0} \notin B$ then $\tuple {0, 0} \notin A \cap B$ and so $H \in \tau_p$. Now suppose $\tuple {0, 0} \in A$ and $\tuple {0, 0} \in B$. 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: :At most a finite number $m_A$ of sets $S_{m_A} = \set {n: \tuple {m_A, n} \notin A}$ are infinite :At most a finite number $m_B$ of sets $S_{m_B} = \set {n: \tuple {m_B, n} \notin B}$ are infinite. Let $r \in \Z_{\ge 0}$ and let: :$S_{r_A} = \set {n: \tuple {r, n} \notin A}$ :$S_{r_B} = \set {n: \tuple {r, n} \notin B}$ We have that:  {{begin-eqn}} {{eqn | o =       | r = S_{r_{A \mathop \cap B} }       | c =  }} {{eqn | r = \set {n: \tuple {r, n} \notin A \cap B}       | c =  }} {{eqn | r = \set {n: \tuple {r, n} \notin A} \cup \set {n: \tuple {r, n} \notin B}       | c = De Morgan's Laws: Difference with Intersection }} {{eqn | r = S_{r_A} \cup S_{r_B}       | c =  }} {{end-eqn}} If $S_{r_A}$ and $S_{r_B}$ are finite then so is $S_{r_A} \cup S_{r_B}$. If only a finite number of $S_{m_A}$ and $S_{m_B}$ are infinite then it follows that only a finite number of $S_{m_{A \cap B}}$ are infinite. So $A \cap B \in \tau$. 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. We have either of two options: :$(1): \quad \forall U \in \UU: \tuple {0, 0} \notin U$ in which case: :$\displaystyle \tuple {0, 0} \notin \bigcap \UU$ Or: :$(2): \quad$ At most a finite number $m$ of sets $S_m = \set {n: \tuple {m, n} \notin U}$ are infinite in which case: :At most a finite number $m$ of sets $S_m = \set {n: \tuple {m, n} \notin \bigcap \UU}$ are infinite. So in either case $\displaystyle \bigcup \UU \in \tau$. {{qed}}
Let $H \in \tau_p$ where $p = \tuple {0, 0}$. Then either: :$(1): \quad \tuple {0, 0} \in \relcomp S H$ or: :$(2): \quad H$ is cofinite in $S$, that is $\relcomp S H$ is finite. Case $(1)$ means that $\tuple {0, 0} \notin H$ and so $H \in \tau$ by definition of the Arens-Fort space. Suppose case $(2)$ applies. Then for all $m \in \Z_{\ge 0}$, the sets $S_m$ defined as $S_m = \set {n: \tuple {m, n} \notin H}$ are finite. Thus in a finite number of these (that is, none of them) the set $S_m$ is infinite. So $H \in \tau$. So, if $H \in \tau_p$ it follows that $H \in \tau$ and so $\tau_p \subseteq \tau$. Hence the result by definition of expansion. Finally, from Arens-Fort Space is Countable we have that $S$ is countably infinite. {{qed}}
We have that the Arens-Fort Space is Completely Hausdorff. Then from Sequence of Implications of Separation Axioms, a $T_{2 \frac 1 2}$ (completely Hausdorff) space is a $T_1$ space. {{qed}}
We have: :Fort Space is Completely Normal :Arens-Fort Space is Expansion of Fort Space. From Sequence of Implications of Separation Axioms we have that a Fort space is a $T_{2 \frac 1 2}$ space. Then in Separation Properties Preserved by Expansion, it follows that as a Fort space is a $T_{2 \frac 1 2}$ space, then so is the Arens-Fort space. {{qed}}
Let $A, B \subseteq S$ such that $A$ and $B$ are separated. Let $p = \tuple {0, 0}$. If $p \notin A$ and $p \notin B$ then $A$ and $B$ are both open and the problem is solved. 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. $B \cup \set p$ is closed by the definition of the Arens-Fort space. From Set is Subset of its Topological Closure and Set Closure is Smallest Closed Set in Topological Space: :$B \subseteq B^- \subseteq B \cup \set p$ where $B^-$ is the closure of $B$. So $B$ is closed or $p$ is in $B^-$. 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: :Arens-Fort Space is $T_1$ :Arens-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.
{{AimForCont}} $T$ is first-countable. Then there exists a countable local base $B_0 = \sequence {U_i}_{i \mathop = 1}^\infty$ for $\tuple {0, 0}$. Suppose there does not exist a point $\tuple {n_i, m_i} \in U_i$ such that $n_i > i$ and $m_i > i$. Then: :$\forall \tuple {n, m} \in U_i: n \le i$ or $m \le i$ Now suppose that there does not exist some $\tuple {n, m} \in U_i$ with $n > i$. Let $S_k$ be defined as: :$S_k = \set {n: \tuple {k, n} \notin U_i}$ The definition of the Arens-Fort space clearly stipulates that $S_k$ can only be a infinite set for a finite number of $k \in \Z_{\ge 0}$. But: :$k > i \implies S_k = \Z_{\ge 0}$ So we have shown that: :$\exists \tuple {n, m} \in U_i: n > i$ So, let $n > i$, and let $m \le i$. Then $\set {m: m > i} \subseteq S_n$.  Thus $S_n = \set {m: \tuple {n, m} \notin U_i}$ is infinite for all $n > i$. Again, this contradicts the definition of the Arens-Fort space. So, from the definition of the neighborhood of $\tuple {0, 0}$, for each $U_i \in B_0$ there is a point $\tuple {n_i, m_i} \in U_i$ such that both $n_i > i$ and $m_i > i$. Now let the set $E$ be constructed as: :$E := X \setminus \sequence {\tuple {n_i, m_i} }_{i \mathop = 1}^\infty$ We now prove that $E$ is a neighborhood of $\tuple {0, 0}$. By the definition of Arens-Fort space, it is enough to show that: :$\forall i \in \N: S_i$ is finite. We have that: {{begin-eqn}} {{eqn | l = S_i       | r = \set {m: \tuple {i, m} \notin E}       | c =  }} {{eqn | r = \set {m: \tuple {i, m} \in \sequence {\tuple {n_j, m_j} }_{j \mathop = 1}^\infty}       | c =  }} {{end-eqn}} But we defined $n_j > j$. So if $j > i \implies n_j > j > i$. Thus $S_i \subseteq \sequence {m_j}_{j \mathop = 1}^i$ is finite.  However, $\tuple {n_i, m_i} \notin E$. Therefore there exists no $U_i \subseteq E$. From this contradiction it follows that $T$ cannot be first-countable. {{qed}} {{proofread}}
By definition, a topological space $T = \struct {S, \tau}$ is separable if there exists a countable subset of $S$ which is everywhere dense in $T$. The closure of $S$ in $S$ is trivially $S$. So, by definition, $S$ is everywhere dense in $S$. As $S$ is countable by definition, the result follows. {{qed}} Category:Separable Spaces Category:Countable Sets 5sagq1qbpwtck3uym0x314bh0a672vu
We have that the Arens-Fort space is an expansion of a countable Fort space. So $S$ is countable. The result follows from Countable Space is Separable. {{qed}}
From Arens-Fort Space is Countable we have that $S$ is countably infinite. Let $\tau_d$ be the discrete topology on $S$. Consider the identity function $I_S: \left({S, \tau_d}\right) \to \left({S, \tau}\right)$. From Mapping from Discrete Space is Continuous, $I_S$ is continuous. From Identity Mapping is Surjection, $I_S$ is a surjection. From Countable Discrete Space is Sigma-Compact and Sigma-Compactness is Preserved under Continuous Surjection we conclude that $T$ is a $\sigma$-compact space. {{qed}}
The result follows from Arens-Fort Space is Countable and Countable Space is Sigma-Compact. {{qed}}
We have: :The Arens-Fort Space is Sigma-Compact. :A $\sigma$-compact space is Lindelöf. {{qed}}
We have that Neighborhood of Origin in Arens-Fort Space is not Compact. So $\tuple {0, 0}$ is a point in $S$ which is not contained in a compact neighborhood. Hence, by definition, $T$ is not weakly locally compact. {{qed}}
We have: : The Arens-Fort Space is not Weakly Locally Compact. : A Compact Space is Weakly Locally Compact. {{qed}}
{{AimForCont}} the Arens-Fort space is countably compact. From Arens-Fort Space is Lindelöf, it is also Lindelöf. From Countably Compact Lindelöf Space is Compact, $T$ is compact. But this contradicts  Arens-Fort Space is not Compact. Hence the result from Proof by Contradiction. {{qed}}
Let $\CC$ be any open cover of $T$. Let $H \in \CC$ be any open set which contains $\tuple {0, 0}$. For all $s \in S$ such that $s \ne \tuple {0, 0}$, we have that $\set s$ is open in $T$ by definition of the Arens-Fort space. So the open cover of $T$ which consists of $H$ together with all the open sets $\set s$, where $s \in S \setminus H$ is a refinement of $T$ which is locally finite. Hence the result, by definition of paracompact space. {{qed}}
We have that: :The Arens-Fort Space is Completely Normal. :The Arens-Fort Space is Lindelöf. From Sequence of Implications of Separation Axioms, it follows that $T$ is a $T_3$ space. The result follows from Lindelöf $T_3$ Space is Paracompact. {{qed}}
Let $T_S = \struct {S, \tau_S}$ be the Arens-Fort space. Let $T_D = \struct {S, \tau_D}$ be the discrete space, also on $S$. Let $I_S: S \to S$ be the identity mapping on $S$. From Mapping from Discrete Space is Continuous, we have that $I_S$ is a continuous mapping. Then we have: :Discrete Space is First-Countable :Arens-Fort Space is not First-Countable Thus we have demonstrated a continuous mapping from a first-countable space to a space which is not first-countable space. {{qed}}
Let $T_S = \struct {S, \tau_S}$ be the Arens-Fort space. Let $T_D = \struct {S, \tau_D}$ be the discrete space, also on $S$. As $S$ is countable, from Arens-Fort Space is Expansion of Countable Fort Space, it follows that $T_D = \struct {S, \tau_D}$ is a countable discrete space. Let $I_S: S \to S$ be the identity mapping on $S$. From Mapping from Discrete Space is Continuous, we have that $I_S$ is a continuous mapping. Then we have that a Countable Discrete Space is Second-Countable. We have that the Arens-Fort Space is not First-Countable. It follows from Second-Countable Space is First-Countable that the Arens-Fort space is not second-countable either. Thus we have demonstrated a continuous mapping from a second-countable space to a space which is not second-countable. {{qed}}
Consider $p \in S$ such that $p \ne \tuple {0, 0}$. From Clopen Points in Arens-Fort Space, we have that $\set p$ is both open and closed in $T$. So by definition of closed set, $\relcomp S {\set p}$ is also both open and closed in $T$. So, by definition, $\set p \mid \relcomp S {\set p}$ is a separation of $T$ Hence the result, by definition of connected space. {{qed}}
Let $\UU_0$ be a local basis for $\tuple {0, 0}$. Let $U \in \UU_0$. By definition of local basis, $U$ is open in $T$. From Clopen Points in Arens-Fort Space, $\set {\tuple {0, 0} }$ is not open in $T$. So $U \ne \set {\tuple {0, 0} }$.  Therefore: :$\exists p \in U: p \ne \tuple {0, 0}$ From Singleton of Element is Subset: :$\set p \subseteq U$ From Clopen Points in Arens-Fort Space it follows that $\set p$ is clopen. As $U \in U_0$ it follows by definition of local basis that: :$\tuple {0, 0} \in U$ and so: :$U \ne \set p$ That is: :$\set p \subsetneq U$ From Connected iff no Proper Clopen Sets, the set $U$ is not connected. {{explain|Re-establish the definition of Definition:Connected Topological Space as the current flow does not make it obvious that a connected subset has no proper clopen sets, only that this condition applies to the full space.}} It is deduced that any local basis is formed with disconnected sets. Thus, by definition, $T$ is not locally connected. {{qed}}
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 $\tuple {0, 0} \notin \set q$ so $\set q$ is open. However, $\tuple {0, 0} \notin \relcomp S {\set {\tuple {0, 0} } }$ and $\relcomp S {\set {\tuple {0, 0} } }$ is clearly not open in $T$. But as $\set {\tuple {0, 0} }$ is finite so $\relcomp S {\set {\tuple {0, 0} } }$ is cofinite. Hence, as for $\set q$, we have that $\set {\tuple {0, 0} }$ is closed. {{qed}}
Let $H \subseteq T$ such that: :$\exists U \in \tau: \tuple {0, 0} \in U \subseteq H \subseteq S$ that is: such that $H$ is a neighborhood of $\tuple {0, 0}$ in $T$. As $\tuple {0, 0} \in H$ it follows that $\tuple {0, 0} \notin \relcomp S H$. So, by definition of the Arens-Fort space, $\relcomp S H$ is open in $T$. So by definition, we have that $H$ is closed. {{qed}}
Let $q \in S$ such that $q \ne \tuple {0, 0}$. Then from Clopen Points in Arens-Fort Space, $\set q$ is clopen. So $\forall q \in S, q \ne \tuple {0, 0}: \set {\set q}$ is a local basis for $q$. If we take the neighborhoods of $\tuple {0, 0}$ that are open we get that they are also closed from Neighborhood of Origin of Arens-Fort Space is 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 have that: :The Arens-Fort Space is Zero Dimensional. :The Arens-Fort Space is $T_1$ and so by $T_1$ Space is $T_0$ Space is a $T_0$ (Kolmogorov) space. Then we have that a Zero Dimensional $T_0$ Space is Totally Separated. {{qed}}
If $q \ne \tuple {0, 0}$ then from Clopen Points in Arens-Fort Space we have that $\set q$ is both closed and open in $T$. In particular, $\set q$ is open in $T$. The result follows from Point in Topological Space is Open iff Isolated. {{qed}}
Let $\left\{{x}\right\}$ be open in $T$. Then we have that: :$\exists \left\{{x}\right\} \in \tau: x \in \left\{{x}\right\}\subseteq S$ This is precisely the condition which ensures that $x$ is an isolated point of $T$. Now suppose that $x$ is an isolated point of $T$. Then by definition there exists a open set of $T$ containing no points other than $x$: :$\exists U \in \tau: U = \left\{{x}\right\}$ That is, if $x$ is an isolated point of $T$ then $\left\{{x}\right\}$ is open in $T$. {{qed}} Category:Open Sets Category:Isolated Points luapupy565wng4ac9qxl0b6ulox69fo
Let $H \subseteq T$ such that $H \ne \set {\tuple {0, 0} }$. Then $\exists x \in H: x \ne \tuple {0, 0}$. From Clopen Points in Arens-Fort Space, every point of $T$ apart from $\tuple {0, 0}$ is open in $T$. So $\set x$ is an open set of $T$. So $H \cap \set x = \set x$ and so $x$ is isolated in $H$. Thus $H$ contains at least one point which is isolated in $H$. On the other hand, suppose $H = \set {\tuple {0, 0} }$. From Singleton Point is Isolated, $\tuple {0, 0}$ is an isolated point in $\set {\tuple {0, 0} } = H$. So again $H$ contains at least one point which is isolated in $H$. So for all $H \subseteq S$ we have that $H$ contains at least one point which is isolated in $H$. Hence the result, by definition of scattered space. {{qed}}
From Metric Space is First-Countable and Fortissimo Space is not First-Countable, it is deduced that $T$ is not a metrizable topology. {{qed}}
Let $\CC$ be an open cover of $T$. Let $U_p \in \CC$ be an open set in that cover which contains $p$. All the points of $S \setminus U_p$ are open points by definition of Fortissimo space. Then $\DD = \set {\set s: s \in S \setminus U_p} \cup \set {U_p}$ is an open refinement of $\CC$. Take $k \in S$: :If $k = p$, then $U_p$ is a neighbourhood of $k$ which intersects only with $U_p$ from all the subsets in $\DD$ :If $k \ne p$, then $\set k$ is a neighbourhood of $k$ which intersects only with $\set k$ from all the subsets in $\DD$ Thus $\DD$ is by definition locally finite. Hence, by definition, $T$ is a paracompact space. {{qed}}
Let $p \in S: p \notin \set {a, b}$. From the definition of modified Fort space, any subset of $S \setminus \set {a, b}$ is open in $T$. It follows directly that as $p \in S \setminus \set {a, b}$ we have that $\set p \subseteq S \setminus \set {a, b}$. Hence $p$ is open in $T$. As for $a$ and $b$, we have that $S \setminus \set a$ and $S \setminus \set b$ are not finite and so $a$ and $b$ are not open in $T$. {{qed|lemma}} For all points $p \in S$ (including $a$ and $b$), we have that $\set p$ is (trivially) finite. It follows that while $S \setminus \set p$ contains either $a$ or $b$, it is cofinite. Thus $S \setminus \set p$ is open in $T$. It follows by definition that $p$ is closed in $T$. {{qed}}
Let $\CC$ be an open cover of $T$. Then $\exists U \in \CC: a \in U$. Then by definition of modified Fort space, $U$ is cofinite. That is, $S \setminus U$ is finite. Then $S \setminus U$ is covered by a finite subcover of $\CC$. Hence, by definition, $T$ is compact. {{qed}}
Let $p, q \in S: p \ne q$. There are two options: :$(1): \quad$ At least one of $p$ and $q$ is not in $\set {a, b}$ :$(2): \quad$ Both $p$ and $q$ are in $\set {a, b}$. $(1): \quad$ First let $p \notin \set {a, b}$ or $q \notin \set {a, b}$. {{WLOG}}, let $p \notin \set {a, b}$. From Clopen Points in Modified Fort Space both $\set p \in \tau_{a, b}$ and $S \setminus \set p \in \tau_{a, b}$. So: :$p \in \set p, q \notin \set p$ :$p \notin S \setminus \set p, q \in S \setminus \set p$ Similarly for $q \notin \set {a, b}$. $(2): \quad$ Next, let $p, q \in \set {a, b}$. Let $V_p \subseteq S$ such that $p \in V_p, q \notin V_p$ and $V_p$ is finite. Then $S \setminus V_p$ is open in $T$ containing $q$ but not $p$. Similarly, let $V_q \subseteq S$ such that $q \in V_q, p \notin V_q$ and $V_q$ is finite. Then $S \setminus V_q$ is open in $T$ containing $p$ but not $q$. Hence it can be seen that $T$ is a $T_1$ (Fréchet) space. {{qed}}
Consider $U, V \in \tau_{a, b}$ such that $a \in U, b \in V$. We have that both $U$ and $V$ are cofinite. So $U$ and $V$ must be infinite. Suppose $U \cap V = \O$. Then from Intersection with Complement is Empty iff Subset it follows that $U \subseteq \relcomp S V$ and so $U$ is finite. But this contradicts the fact that $U$ is infinite. So $U \cap V \ne \O$. Hence the result by definition of $T_2$ (Hausdorff) space. {{qed}}
We have that a Modified Fort Space is $T_1$. From $T_1$ Space is $T_0$ Space, $T$ is a $T_0$ space. We then have that a Modified Fort 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 $T$ 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}}
From Real Number Line is Metric Space, one can define an open interval on the set of real numbers in terms of an $\epsilon$-neighborhood. From Open Real Interval is Open Ball, an open interval $\openint a b$ is the open $\epsilon$-ball $\map {B_\epsilon} \alpha$. Then from Metric Induces Topology we have that: :$\BB = \set {\openint a b: a, b \in \R}$ forms a topology on $\R$. {{qed}}
* Real Number Line is Metric Space * Metric Space fulfils all Separation Axioms {{qed}}
We have that the Real Number Line is Complete Metric Space. From the Baire Category Theorem, a complete metric space is also a Baire space. The result follows from Baire Space is Non-Meager. {{qed}}