avewright/proofwiki-math huggingface dataset:
Theorem: Image of Set Difference under Mapping/Corollary 2

Let $f: S \to T$ be a mapping.
Let $X$ be a subset of $S$.
Then:
:$\relcomp {\Img f} {f \sqbrk X} \subseteq f \sqbrk {\relcomp S X}$
where:
:$\Img f$ denotes the image of $f$
:$\complement_{\Img f}$ denotes the complement relative to $\Img f$.
This can be expressed in the language and notation of direct image mappings as:
:$\forall X \in \powerset S: \relcomp {\Img f} {\map {f^\to} X} \subseteq \map {f^\to} {\relcomp S X}$
That is:
:$\forall X \in \powerset S: \map {\paren {\complement_{\Img f} \circ f^\to} } X \subseteq \map {\paren {f^\to \circ \complement_S} } X$
where $\circ$ denotes composition of mappings.

Proof:
From {{Corollary|Image of Set Difference under Relation|2}}:
:$\relcomp {\Img \RR} {\RR \sqbrk X} \subseteq \RR \sqbrk {\relcomp S X}$
where $\RR \subseteq S \times T$ is a relation on $S \times T$.
As $f$, being a mapping, is also a relation, it follows directly that:
:$\relcomp {\Img f} {f \sqbrk X} \subseteq f \sqbrk {\relcomp S X}$
{{Qed}}
Theorem: Preimage of Set Difference under Mapping/Corollary 1

Let $f: S \to T$ be a mapping.
Let $T_1 \subseteq T_2 \subseteq T$.
Then:
:$\relcomp {f^{-1} \sqbrk {T_2} } {f^{-1} \sqbrk {T_1} } = f^{-1} \sqbrk {\relcomp {T_2} {T_1} }$
where:
:$\complement$ (in this context) denotes relative complement
:$f^{-1} \sqbrk {T_1}$ denotes preimage.

Proof:
From One-to-Many Image of Set Difference: Corollary 1 we have:
:$\relcomp {\RR \sqbrk {T_2} } {\RR \sqbrk {T_1} } = \RR \sqbrk {\relcomp {T_2} {T_1} }$
where $\RR \subseteq T \times S$ is a one-to-many relation on $T \times S$.
Hence as $f^{-1}: T \to S$ is a one-to-many relation:
:$\relcomp {f^{-1} \sqbrk {T_2} } {f^{-1} \sqbrk {T_1} } = f^{-1} \sqbrk {\relcomp {T_2} {T_1} }$
{{qed}}
Category:Preimages under Mappings
Category:Set Difference
Category:Set Complement
gjhcrxi22fejbbz1x17d9e0jmuouvz9
Theorem: Complement of Preimage equals Preimage of Complement

Let $f: S \to T$ be a mapping.
Let $T_1$ be a subset of $T$.
Then:
:$\relcomp S {f^{-1} \sqbrk {T_1} } = f^{-1} \sqbrk {\relcomp T {T_1} }$
where:
:$\complement_S$ (in this context) denotes relative complement
:$f^{-1} \sqbrk {T_1}$ denotes preimage.

Proof:
From {{Corollary|One-to-Many Image of Set Difference|2}} we have:
:$\relcomp {\Img \RR} {\RR \sqbrk {S_1} } = \RR \sqbrk {\relcomp S {S_1} }$
where:
:$S_1 \subseteq S$
:$\RR \subseteq T \times S$ is a one-to-many relation on $T \times S$.
Hence as $f^{-1}: T \to S$ is a one-to-many relation:
:$\relcomp {\Preimg f} {f^{-1} \sqbrk {T_1} } = f^{-1} \sqbrk {\relcomp T {T_1} }$
But from Preimage of Mapping equals Domain, we have that:
:$\Preimg f = S$
Hence:
:$\relcomp S {f^{-1} \sqbrk {T_1} } = f^{-1} \sqbrk {\relcomp T {T_1} }$
{{Qed}}
Theorem: One-to-Many Image of Set Difference/Corollary 1

Let $\RR \subseteq S \times T$ be a relation which is one-to-many.
Let $A \subseteq B \subseteq S$.
Then:
:$\relcomp {\RR \sqbrk B} {\RR \sqbrk A} = \RR \sqbrk {\relcomp B A}$
where $\complement$ (in this context) denotes relative complement.

Proof:
We have that $A \subseteq B$.
Then by definition of relative complement:
:$\relcomp B A = B \setminus A$
:$\relcomp {\RR \sqbrk B} {\RR \sqbrk A} = \RR \sqbrk B \setminus \RR \sqbrk A$
Hence, when $A \subseteq B$:
:$\relcomp {\RR \sqbrk B} {\RR \sqbrk A} = \RR \sqbrk {\relcomp B A}$
means exactly the same thing as:
:$\RR \sqbrk B \setminus \RR \sqbrk A = \RR \sqbrk {B \setminus A}$
Hence the result from One-to-Many Image of Set Difference.
{{qed}}
Category:Relative Complement
Category:One-to-Many Image of Set Difference
djw2ypmjumv1rqtegj63pbqi71w5ph7
Theorem: One-to-Many Image of Set Difference/Corollary 2

Let $\RR \subseteq S \times T$ be a relation which is one-to-many.
Let $A$ be a subset of $S$.
Then:
:$\relcomp {\Img \RR} {\RR \sqbrk A} = \RR \sqbrk {\relcomp S A}$
where $\complement$ (in this context) denotes relative complement.
In the language of direct image mappings this can be presented as:
:$\forall A \in \powerset S: \map {\paren {\complement_{\Img \RR} \circ \RR^\to} } A = \map {\paren {\RR^\to \circ \complement_S} } A$

Proof:
By definition of the image of $\RR$:
:$\Img \RR = \RR \sqbrk S$
So, when $B = S$ in {{Corollary|One-to-Many Image of Set Difference|1}}:
:$\relcomp {\Img \RR} {\RR \sqbrk A} = \relcomp {\RR \sqbrk S} {\RR \sqbrk A}$
Hence:
:$\relcomp {\Img \RR} {\RR \sqbrk A} = \RR \sqbrk {\relcomp S A}$
means exactly the same thing as:
:$\relcomp {\RR \sqbrk S} {\RR \sqbrk A} = \RR \sqbrk {\relcomp S A}$
that is:
:$\RR \sqbrk S \setminus \RR \sqbrk A = \RR \sqbrk {S \setminus A}$
Hence the result from One-to-Many Image of Set Difference.
{{Qed}}
Category:Set Difference
Category:Relative Complement
Category:One-to-Many Image of Set Difference
ao3auai2yiiclzvk78bw3vja0jbngy7
Theorem: Integer Power Function is Bijective iff Index is Odd

Let $n \in \Z_{\ge 0}$ be a positive integer.
Let $f_n: \R \to \R$ be the real function defined as:
:$\map {f_n} x = x^n$
Then $f_n$ is a bijection {{iff}} $n$ is odd.

Proof:
=== Even Index ===
Suppose $n$ is even.
Let $x \ne 0$.
Then $1^n = \paren {-1}^n$ by Power of Ring Negative, so $f_n$ is not injective.
Also, by Even Power is Non-Negative, $f_n$ is not surjective.
By definition, a bijection is both injective and surjective.
It follows that for even $n$, $f_n$ is not bijective.
{{qed|lemma}}
Theorem: Set of Infinite Sequences is Uncountable

Let $S$ be a set which contains more than one element.
Let $S^\infty$ denote the set of all sequences of elements of $S$.
Then $S^\infty$ is uncountable.

Proof:
As $S$ has more than one element, it must have at least two.
So, let $a, b \in S$ be those two elements.
Let $Z$ be the set of all sequences from $\set {a, b}$.
Suppose $S^\infty$ were countable.
From Subset of Countably Infinite Set is Countable, $Z$ is likewise countable.
So by definition, it would be possible to set up a bijection $\phi: Z \leftrightarrow \N$ between $Z$ and the set $\N$ of natural numbers:
{{begin-eqn}}
{{eqn | l = 0
      | o = \leftrightarrow
      | r = \sequence {z_0} = \tuple {z_{00}, z_{01}, z_{02}, \ldots, z_{0n}, \ldots}
}}
{{eqn | l = 1
      | o = \leftrightarrow
      | r = \sequence {z_1} = \tuple {z_{10}, z_{11}, z_{12}, \ldots, z_{1n}, \ldots}
}}
{{eqn | l = 
      | o = \vdots
      | r = 
}}
{{eqn | l = n
      | o = \leftrightarrow
      | r = \sequence {z_n} = \tuple {z_{n0}, z_{n1}, z_{n2}, \ldots, z_{nn}, \ldots}
}}
{{eqn | l = 
      | o = \vdots
      | r =  
}}
{{end-eqn}}
where each of $z_{rs}$ is:
: an element of $\set {a, b}$
: the $s$'th element of $\sequence {z_r}$, the particular sequence which is in correspondence with $r \in \N$.
Now we create the sequence $Q = \sequence {q_{dd} }$ where:
:$q_{dd} = \begin{cases} a & : z_{dd} = b \\ b & : z_{dd} = a \end{cases}$
Thus:
:$\forall n \in \N: q_{nn} \notin \sequence {z_n}$
and so:
:$\forall n \in \N: Q \ne \sequence {z_n}$
So $Q \notin Z$.
That is, we have constructed a sequence which has not been placed into correspondence with an element of $\N$.
Thus by definition $Z$ is uncountable.
Hence, by the Rule of Transposition, $S^\infty$ is likewise uncountable.
{{qed}}
Theorem: Union of Topologies is not necessarily Topology

Let $\tau_1$ and $\tau_2$ be topologies on a set $S$.
Then $\tau_1 \cup \tau_2$ is not necessarily also a topology on $S$.

Proof:
Let $S := \set {0, 1, 2}$ be a set.
Let:
:$\tau_1 := \set {\O, \set 0, \set 1, \set {0, 1}, S}$ 
:$\tau_2 := \set {\O, \set 0, \set 2, \set {0, 2}, S}$
be topologies on $S$. 
Then:
:$\tau := \tau_1 \cup \tau_2 = \set {\O, \set 0, \set 1, \set 2, \set {0, 1} \set {0, 2}, S}$
For $\tau$ to be a topology the union of any number of elements of $\tau$ should also be in $\tau$.
But:
:$\set 1 \cup \set 2 = \set {1, 2} \not \in \tau$ 
Therefore $\tau$ is not a topology on $S$.
Hence the result. 
{{qed}}
Theorem: Niven's Theorem

Consider the angles $\theta$ in the range $0 \le \theta \le \dfrac \pi 2$.
The only values of $\theta$ such that both $\dfrac \theta \pi$ and $\sin \theta$ are rational are:
:$\theta = 0: \sin \theta = 0$
:$\theta = \dfrac \pi 6: \sin \theta = \dfrac 1 2$
:$\theta = \dfrac \pi 2: \sin \theta = 1$

Proof:
We will prove that if both $\dfrac \theta \pi$ and $\cos \theta$ are rational then:
:$\theta \in \set{ {0, \dfrac \pi 3, \dfrac \pi 2} }$
Theorem: Relative Frequency is Probability Measure

The relative frequency model is a probability measure.

Proof:
We check all the Kolmogorov axioms in turn:
Theorem: Sorgenfrey Line is Topology

The Sorgenfrey Line is a topological space.

Proof:
We have to check that $\BB = \set {\hointr a b: a, b \in \R}$ fulfills the axioms of being a basis for a topology.
By definition of synthetic basis we only have to check that:
:$(1): \quad \bigcup \BB = \R$
:$(2): \quad \forall B_1, B_2 \in \BB: \exists V \in \BB: V \subseteq B_1 \cap B_2$
We have that:
:$\forall n \in \Z: \hointr n {n + 1} \in \BB$
:$\R = \ds \bigcup_{n \mathop \in \Z} \hointr n {n + 1} \subseteq \bigcup \BB$
Hence $\R = \bigcup \BB$ and condition $(1)$ is fulfilled.
Now take $ B_1, B_2 \in \BB$ where:
:$B_1 = \hointr {a_1} {b_1}$
:$B_2 = \hointr {a_2} {b_2}$
Let $B_3$ be constructed as:
:$B_3 := \hointr {\max \set {a_1, a_2} } {\min \set {b_1, b_2} } \in \BB$
From the method of construction, it is clear that $B_3 = B_1 \cap B_2$.
Thus taking $V = B_3$, condition $(2)$ is fulfilled.
{{qed}}
Theorem: Sorgenfrey Line is Hausdorff

Let $T = \struct {\R, \tau}$ be the Sorgenfrey line.
Then $T$ is Hausdorff.

Proof:
Take $x, y \in \R$ such that $x \ne y$.
{{WLOG}}, assume that $x < y$.
From Real Numbers are Densely Ordered:
:$\exists t \in \R: x < t < y$
Then:
:$\hointr x t \cap \hointr t {y + 1} = \O$
We also have that $x \in \hointr x t$ by definition of half-open interval.
Also, as $t < y < y+1$ it is clear that $y \in \hointr t {y + 1}$.
By definition of the Sorgenfrey line, both are open in $T$.
Thus we have found two disjoint subsets of $\R$ which are open in $T$, such that one contains $x$ and the other contains $y$.
Hence the Sorgenfrey Line is Hausdorff by definition.
{{qed}}
Theorem: Convergent Sequence in Set of Integers

Let $\sequence{x_n}_{n \in \N}$ be a sequence in the set $\Z$ of integers considered as a subspace of the real number line $\R$ under the Euclidean metric.
Then $\sequence{x_n}_{n \in \N}$ converges in $\R$ to a limit {{iff}}:
:$\exists k \in \N: \forall m \in \N: m > k: x_m = x_k$
That is, {{iff}} the sequence reaches some value of $\Z$ and "stays there".

Proof:
Suppose $\sequence{x_n}_{n \in \N}$ converges to a limit $l$.
Consider the open set in $\R$:
:$U := \openint{l - \dfrac 1 2}{l + \dfrac 1 2}$
Then $\forall x \in \Z: x \in U \implies x = l$
It follows by definition of convergence that:
:$\exists k \in \N: \forall m \in \N: m > k: x_m = x_k$
where $x_k = l$.
Now suppose that:
:$\exists k \in \N: \forall m \in \N: m > k: x_m = x_k$
Then trivially $\sequence{x_n}_{n \in \N}$ converges to the limit $x_k$.
{{qed}}
Category:Sequences
Category:Convergence
1z18fqvhzb3y9kbq8dy4nq7tr1r76zo
Theorem: Sorgenfrey Line is Perfectly Normal

Let $T = \struct {\R, \tau}$ be the Sorgenfrey line.
Then $T$ is perfectly normal.

Proof:
From the definition of perfectly normal space, it is necessary to prove that:
:$T$ is a $T_1$ space
and that:
:an arbitrary closed set is a $G_\delta$ set.
From $T_2$ Space is $T_1$ Space and Sorgenfrey Line is Hausdorff:
:the Sorgenfrey line is a $T_1$ space.
From Complement of $F_\sigma$ Set is $G_\delta$ Set it is sufficient to prove that an open set of $T$ is an $F_\sigma$ set.
Let $W$ be an arbitrary open set in $T$.
Let $O \subseteq W$ be the interior of $W$ with respect to the metric space:
:$\R = \struct {\R, d}$
where $d$ is the usual metric on $\R$.
From the definition of $T$, for each $x \in W \setminus O$, we can choose $h_x \in W$ such that $\hointr x {h_x} \subseteq W$.
Suppose $\hointr x {h_x} \cap \hointr y {h_y} \ne \O$ for some distinct points $x, y \in W \setminus O$.
Then either $x < y < h_x$ or $y < x < h_y$.
If $x < y < h_x$, then $y \in \openint x {h_x} \subseteq O$, contradicting $y \in W \setminus O$.
Similarly, if $y < x < h_y$, then $x \in \openint y {h_y}) \subseteq O$, contradicting $x \in W \setminus O$.
Thus $\family {\hointr x {h_x} }_{x \mathop \in W \setminus O}$ is a pairwise disjoint indexed family of open sets of $T$.
From Sorgenfrey Line is Separable and Separable Space satisfies Countable Chain Condition:
:$\set {\hointr x {h_x} : x \in W \setminus O}$ is countable
and thus:
:$W \setminus O$ is countable.
From Metric Space is Perfectly $T_4$ Space:
:$O$ is an $F_\sigma$ set in $\R$.
Thus from Sorgenfrey Line is Expansion of Real Line:
:$O$ is an $F_\sigma$ set in the Sorgenfrey line.
Since $W \setminus O$ is a countable union of singletons and $T$ is a $T_1$ space:
:$W \setminus O$ is an $F_\sigma$ set in $T$.
Since $W = O \cup \paren {W \setminus O}$ and $F_\sigma$ sets are closed under unions:
:$W$ is an $F_\sigma$ set in $T$.
{{qed}}
Category:Sorgenfrey Line
Category:Examples of Perfectly Normal Spaces
2hn500zfplrqmk5h3ia9vr9ogddmjxn
Theorem: Sorgenfrey Line is Expansion of Real Line

Let $\R = \struct {\R, d}$ be the metric space defined in Real Number Line is Metric Space.
Let $T = \struct {\R, \tau}$ be the Sorgenfrey line.
Then $T$ is an expansion of $\R$ as a topological space.

Proof:
It is enough to prove that any open set in $\R$ is open in $T$.
Let $a, b \in \R$.
Then:
:$\ds \openint a b = \bigcup_{\epsilon \mathop > 0} \hointr {a + \epsilon} b$
Since $\hointr {a + \epsilon} b$ are open in $T$, $\openint a b$ is also open in $T$.
{{qed}}
Category:Sorgenfrey Line
owmja931ica1kc7utbvp5gb94dcptg0
Theorem: Sorgenfrey Line satisfies all Separation Axioms

Let $T = \struct {\R, \tau}$ be the Sorgenfrey line.
Then $T$ satisfies all separation axioms.

Proof:
We have Sorgenfrey Line is Perfectly Normal.
The result follows from Sequence of Implications of Separation Axioms.
{{qed}}
Category:Sorgenfrey Line
Category:Separation Axioms
7utbg8yppc8vgkrn3ll9eeir5fu5de9
Theorem: Argument of Product equals Sum of Arguments

Let $z_1, z_2 \in \C$ be complex numbers.
Let $\arg$ be the argument operator.
Then:
:$\map \arg {z_1 z_2} = \map \arg {z_1} + \map \arg {z_2} + 2 k \pi$
where $k$ can be $0$, $1$ or $-1$.

Proof:
Let $\theta_1 = \map \arg {z_1}, \theta_2 = \map \arg {z_2}$.
Then the polar forms of $z_1, z_2$ are:
{{begin-eqn}}
{{eqn | l = z_1
      | r = \cmod {z_1} \paren {\cos \theta_1 + i \sin \theta_1}
}}
{{eqn | l = z_2
      | r = \cmod {z_2} \paren {\cos \theta_2 + i \sin \theta_2}
}}
{{end-eqn}}
By the definition of complex multiplication, factoring $\cmod {z_1} \cmod {z_2}$ from all terms, we have:
:$z_1 z_2 = \cmod {z_1} \cmod {z_2} \paren {\paren {\cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2} + i \paren {\cos \theta_1 \sin \theta_2 + \sin \theta_1 \cos \theta_2} }$
Using Sine of Sum and Cosine of Sum, we have:
:$z_1 z_2 = \cmod {z_1} \cmod {z_2} \paren {\map \cos {\theta_1 + \theta_2} + i \, \map \sin {\theta_1 + \theta_2} }$
The theorem follows from the definition of $\map \arg z$, which says that $\map \arg {z_1 z_2}$ satisfies the equations:
{{begin-eqn}}
{{eqn | n = 1
      | l = \dfrac {\cmod {z_1} \cmod {z_2} \map \cos {\theta_1 + \theta_2} } {\cmod {z_1} \cmod {z_2} }
      | r = \map \cos {\map \arg {z_1 z_2} }
}}
{{eqn | n = 2
      | l = \dfrac {\cmod {z_1} \cmod {z_2} \map \sin {\theta_1 + \theta_2} } {\cmod {z_1} \cmod {z_2} }
      | r = \map \sin {\map \arg {z_1 z_2} }
}}
{{end-eqn}}
which in turn means that:
{{begin-eqn}}
{{eqn | l = \map \cos {\theta_1 + \theta_2}
      | r = \map \cos {\map \arg {z_1 z_2} }
}}
{{eqn | l = \map \sin {\theta_1 + \theta_2}
      | r = \map \sin {\map \arg {z_1 z_2} }
}}
{{end-eqn}}
There are $3$ possibilities for the size of $\theta_1 + \theta_2$:
:$(1): \quad \theta_1 + \theta_2 > \pi$
Then:
:$-\pi < \theta_1 + \theta_2 - 2 \pi \le \pi$
and we have:
{{begin-eqn}}
{{eqn | l = \map \cos {\theta_1 + \theta_2}
      | r = \map \cos {\theta_1 + \theta_2 - 2 \pi}
      | c = Cosine of Angle plus Full Angle
}}
{{eqn | l = \map \sin {\theta_1 + \theta_2}
      | r = \map \sin {\theta_1 + \theta_2 - 2 \pi}
      | c = Sine of Angle plus Full Angle
}}
{{end-eqn}}
and so $\theta_1 + \theta_2 - 2 \pi$ is the argument of $z_1 z_2$ within its principal range.
:$(2): \quad \theta_1 + \theta_2 \le -\pi$
Then:
:$-\pi < \theta_1 + \theta_2 + 2 \pi \le \pi$
and we have:
{{begin-eqn}}
{{eqn | l = \map \cos {\theta_1 + \theta_2}
      | r = \map \cos {\theta_1 + \theta_2 + 2 \pi}
      | c = Cosine of Angle plus Full Angle
}}
{{eqn | l = \map \sin {\theta_1 + \theta_2}
      | r = \map \sin {\theta_1 + \theta_2 + 2 \pi}
      | c = Sine of Angle plus Full Angle
}}
{{end-eqn}}
and so $\theta_1 + \theta_2 + 2 \pi$ is within the principal range of $z_1 z_2$.
:$(3): \quad -\pi < \theta_1 + \theta_2 \le \pi$
Then $\theta_1 + \theta_2$ is already within the principal range of $z_1 z_2$.
Therefore:
:$\map \arg {z_1 z_2} = \theta_1 + \theta_2 = \map \arg {z_1} + \map \arg {z_2} + 2 k \pi$
where $k$ can be $0$, $1$ or $-1$.
{{qed}}
Theorem: Convergence of Sequence in Discrete Space

Let $T = \struct {S, \tau}$ be a discrete topological space.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $S$.
Then $\sequence {x_n}_{n \mathop \in \N}$ converges in $T$ to a limit {{iff}}:
:$\exists N \in \N: \forall n \in \N: n > N \implies x_n = x_N$
That is, {{iff}} the sequence reaches some value of $S$ and "stays there".

Proof:
=== Sufficient Condition ===
Suppose $\sequence {x_n}_{n \mathop \in \N}$ converges to a limit $L$.
Then:
{{begin-eqn}}
{{eqn | l = \set L
      | o = \in
      | r = \tau
      | c = {{Defof|Discrete Space}}
}}
{{eqn | ll= \leadsto
      | q = \exists N \in \N: \forall n \in \N
      | lo= n > N \implies
      | l = x_n
      | o = \in
      | r = \set L
      | c = {{Defof|Convergent Sequence (Topology)}}
}}
{{eqn | ll= \leadsto
      | q = \exists N \in \N: \forall n \in \N
      | lo= n > N \implies
      | l = x_n
      | r = L
      | c = {{Defof|Singleton}}
}}
{{eqn | ll= \leadsto
      | q = \exists N \in \N
      | l = x_N
      | r = L
      | c = Universal Instantiation
}}
{{eqn | ll= \leadsto
      | q = \exists N \in \N: \forall n \in \N
      | lo= n > N \implies
      | l = x_n
      | r = x_N
}}
{{end-eqn}}
{{qed|lemma}}
Theorem: Discrete Subspace of Fortissimo Space

Let $T = \struct {S, \tau_p}$ be a Fortissimo space.
Let $T' = \struct {S \setminus \set p, \tau_p}$ be the topological subspace induced on $T$ by the subset $S \setminus \set p$.
Then $T'$ is a discrete topological space.

Proof:
By the definition of Fortissimo space, every $A \subset S \setminus \set p$ is open in $T$, because $p \notin A$.
Thus by definition of topological subspace, $A \subset S \setminus \set p$ is open in $T'$.
The result follows by the definition of discrete space.
{{qed}}
Category:Fortissimo Spaces
Category:Discrete Topologies
42vrt3urxob1tvrvs196nl8gw0xg6xy
Theorem: Countable Space is Sigma-Compact

Let $T = \struct {S, \tau}$ be a countable topological space.
Then $T$ is $\sigma$-compact.

Proof:
From Finite Space Satisfies All Compactness Properties, for every $p \in S$, $\set p$ is compact.
Then $\ds S = \bigcup_{p \mathop \in S} \set p$ is a countable union of compact sets.
Thus by definition $T$ is $\sigma$-compact.
{{qed}}
Theorem: Closed Sets of Fortissimo Space

Let $T = \struct {S, \tau_p}$ be a Fortissimo space.
Then $H \subseteq S$ is closed in $T$ {{iff}}:
:$p \in H$
or
:$H$ is countable
or both.

Proof:
By definition of a Fortissimo space, $U \subseteq S$ is open in $T$ {{iff}}:
:$p \in \relcomp S U$
or
:$\relcomp S U$ is countable
or both.
The result follows from the definition of closed set.
{{qed}}
Category:Fortissimo Spaces
Category:Examples of Closed Sets
0xomz0ygotpxyb42i8wirdwnyjypfql
Theorem: Union of Topologies on Singleton or Doubleton is Topology

Let $S$ be a set which is either a singleton or a doubleton.
Let $\family {\tau_i}_{i \mathop \in I}$ be an arbitrary non-empty indexed set of topologies on $S$.
Then $\tau := \ds \bigcup_{i \mathop \in I} {\tau_i}$ is also a topology on $S$.

Proof:
Let $S$ be a singleton.
Let $S = \set a$. 
From Topology on Singleton is Indiscrete Topology, the only possible topology on $S$ is the indiscrete topology. 
From Set Union is Idempotent, the union of any number of indiscrete topologies on $S$ is the indiscrete topology.
Thus the union of any number of topologies on a singleton is a topology on that singleton.
{{qed|lemma}}
Let $S$ be a doubleton.
Let $S = \set {a, b}$.
By Topologies on Doubleton, there are $4$ possible different topologies on $S$:
{{begin-eqn}}
{{eqn | l = \tau_1 
      | r = \set {\O, \set {a, b} }
      | c = Indiscrete Topology
}}
{{eqn | l = \tau_2 
      | r = \set {\O, \set a, \set {a, b} }
      | c = Sierpiński Topology
}}
{{eqn | l = \tau_3 
      | r = \set {\O, \set b, \set {a, b} }
      | c = Sierpiński Topology
}}
{{eqn | l = \tau_4 
      | r = \set {\O, \set a, \set b, \set {a, b} }
      | c = Discrete Topology
}}
{{end-eqn}}
It remains to show that every union of elements in $\set {\tau_1, \tau_2, \tau_3, \tau_4}$ is a topology on $S$.
By inspection, the following statements are seen to hold:
:$\tau_1 \cup \tau_2 = \tau_2$
:$\tau_1 \cup \tau_3 = \tau_3$
:$\tau_1 \cup \tau_4 = \tau_4$
:$\tau_2 \cup \tau_3 = \tau_4$
:$\tau_2 \cup \tau_4 = \tau_4$
:$\tau_3 \cup \tau_4 = \tau_4$
Now let $\family {\tau_i}_{i \mathop \in I}$ be an arbitrary non-empty indexed set of topologies for a set $S$.
By the above, if $\tau_4$ is one of the $\tau_i$:
:$\ds \tau = \bigcup_{i \mathop \in I} \tau_i = \tau_4$
So assume that $\tau_4$ is not one of the $\tau_i$.
If both $\tau_2$ and $\tau_3$ are in $\family {\tau_i}_{i \mathop \in I}$, then by the above:
:$\ds \tau = \bigcup_{i \mathop \in I} \tau_i = \tau_4$
Now suppose that $\tau_3$ is in $\family {\tau_i}_{i \mathop \in I}$ and $\tau_2$ is not.
Then by the above it follows that:
:$\tau = \tau_3$
Now suppose that $\tau_2$ is in $\family {\tau_i}_{i \mathop \in I}$ and $\tau_3$ is not.
Then by the above it follows that:
:$\tau = \tau_2$
Finally, assume that $\tau_2$ and $\tau_3$ are not in $\family {\tau_i}_{i \mathop \in I}$.
Then the only topology in $\family {\tau_i}_{i \mathop \in I}$ is $\tau_1$. 
In this, we find:
:$\tau = \tau_1$
Hence the result.
{{qed}}
Theorem: Meager Sets in Arens-Fort Space

Let $T = \struct {S, \tau}$ be the Arens-Fort space.
Let $A \subseteq S$.
Then $A$ is meager {{iff}} $A = \set {\tuple {0, 0} }$.

Proof:
First let $A = \set {\tuple {0, 0} }$.
From the definition of Arens-Fort space, $\set {\tuple {0, 0} }$ is closed because $S \setminus \set {\tuple {0, 0} }$ is open.
From Closed Set Equals its Closure:
:$\set {\tuple {0, 0} }^- = \set {\tuple {0, 0} }$
where $\set {\tuple {0, 0} }^-$ denotes the closure of $\set {\tuple {0, 0} }$.
From the definition of Arens-Fort space, $\set {\tuple {0, 0} }$ is not open.
Therefore the smallest open set contained in $\set {\tuple {0, 0} }$ is $\O$.
Hence:
:$\set {\tuple {0, 0} }^\circ = \O$ 
where $\set {\tuple {0, 0} }^\circ$ denotes the interior of $\set {\tuple {0, 0} }$.
Thus we have that:
:$\paren {\set {\tuple {0, 0} }^-}^\circ = \O$
and so by definition $\set {\tuple {0, 0} }$ is nowhere dense in $T$.
From Union of Singleton it trivially follows that $\set {\tuple {0, 0} }$ is the union of a countable set of subsets of $S$ (that is: just the one) which are nowhere dense in $S$.
Hence by definition $\set {\tuple {0, 0} }$ is meager.
{{qed|lemma}}
Now assume $A \ne \set {\tuple {0, 0} }$.
Then $\exists x \in A: x \ne \tuple {0, 0}$. 
By definition of the Arens-Fort space, $\set x$ is open in $T$. 
Thus, from Interior of Open Set $\set x^\circ = \set x$.
Hence $\set x \subseteq \paren {\set x^-}^\circ$ and by definition is not nowhere dense in $T$.
So $A$ is not the union of a countable set of subsets of $S$ which are nowhere dense in $S$.
So, by definition, $A$ is not meager.
{{qed}}
Category:Arens-Fort Space
Category:Examples of Meager Spaces
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