nyuuzyou/wikis huggingface dataset:
Areas of Similar Polygons Inscribed in Circles are as Squares on Diameters

== Theorem ==
== Proof ==
:
Let $ABC$ and $FGH$ be circles.
Let $ABCDE$ and $FGHKL$ be similar polygons inscribed in $ABC$ and $FGH$ respectively.
Let $BM$ and $GN$ be diameters of $ABC$ and $FGH$ respectively.
It is to be demonstrated that the ratio of the square on $BM$ to the square on $GN$ equals the ratio of the area of $ABCDE$ to the area of $FGHKL$.
Let $BE, AM, GL, FN$ be joined.
We have that $ABCDE$ is similar to $FGHKL$.
Thus:
:$\angle BAE = \angle GFL$
and so from :
:$BA : AE = GF : FL$
Thus from :
:$\triangle ABE$ is equiangular with $\triangle GFL$.
Therefore:
:$\angle AEB = \angle FLG$
But from :
:$\angle AEB = \angle AMB$
and:
:$\angle FLG = \angle FNG$
Therefore:
:$\angle AMB = \angle FNG$
But from :
:the right angle $\angle BAM$ equals the right angle $\angle GFN$.
Therefore from :
:$\angle ABM = \angle FGN$
Therefore $\triangle ABM$ is equiangular with $\triangle FGN$.
Therefore from :
:$BM : GN = BA : GF$
But the ratio of the square on $BM$ to the square on $GN$ is the duplicate ratio of $BM$ to $GN$.
From :
:$ABCDE : FGHKL = BA^2 : GF^2$
Hence:
:$ABCDE : FGHKL = BM^2 : GN^2$
Hence the result.
== Sources ==
* : Book $\text{XII}$. Propositions
Category:Circles
Category:Polygons
Areas of Triangles and Parallelograms Proportional to Base

== Theorem ==
Let $ABC$ and $ACD$ be triangles.
Let $EC, CF$ be parallelograms under the same height.
Then:
:$AC : CD = \triangle ABC : \triangle ACD = \Box EC : \Box CF$
where:
:$AC : CD$ denotes the ratio of the length of $AC$ to that of $CD$
:$\triangle ABC : \triangle ACD$ denotes the ratio of the area of $\triangle ABC$ to that of $\triangle ACD$
:$\Box EC : \Box CF$ denotes the ratio of the area of parallelogram $EC$ to that of parallelogram $CF$.
== Proof ==
:
Let $BD$ be produced in both directions to the points $H, L$ and let any number of straight lines, for example, $BG, GH$ be made equal to the base $BC$, and any number of straight lines, for example, $DK, KL$ be made equal to the base $CD$.
Let $AG, AH, AK, AL$ be joined.
Since $CB = BG = GH$ it follows from Triangles with Equal Base and Same Height have Equal Area that $\triangle ABC = \triangle AGB = \triangle AHG$.
Therefore, whatever multiple the base $HC$ is of the base $BC$, that multiple also is $\triangle AHC$ of $\triangle ABC$.
For the same reason, whatever multiple the base $LC$ is of the base $CD$, that multiple also is $\triangle ALC$ of $\triangle ACD$.
If the base $HC$ is equal to the base $CL$, from Triangles with Equal Base and Same Height have Equal Area, $\triangle AHC = \triangle ACL$.
If the base $HC$ is greater than the base $CL$, $\triangle AHC > \triangle ACL$.
If the base $HC$ is less than the base $CL$, $\triangle AHC < \triangle ACL$.
So we have four magnitudes, two bases $BC, CD$ and two triangles $ABC, ACD$.
We also have that equimultiples have been taken of the base $BC$ and $\triangle ABC$, namely the base $HC$ and $\triangle AHC$.
Also we have equimultiples of the base $CD$ and $\triangle ADC$, namely the base $LC$ and $\triangle ALC$.
It has been proved that:
:$HC > CL \implies \triangle AHC > ALC$
:$HC = CL \implies \triangle AHC = ALC$
:$HC < CL \implies \triangle AHC < ALC$
Therefore from $BC : CD = \triangle ABC : \triangle ACD$.
Next, from Parallelogram on Same Base as Triangle has Twice its Area, the parallelogram $EC$ is twice the area of $\triangle ABC$.
Similarly, the parallelogram $FC$ is twice the area of $\triangle ACD$.
So from Ratio Equals its Multiples, $\triangle ABC : \triangle ACD = \Box FC : \Box RC$.
The result follows from Equality of Ratios is Transitive.
== Sources ==
* : Book $\text{VI}$. Propositions
Category:Area of Parallelogram
Category:Areas of Triangles
Arens-Fort Space is Completely Hausdorff

== Theorem ==
Let $T = \struct {S, \tau}$ be the Arens-Fort space.
Then $T$ is a $T_{2 \frac 1 2}$ (completely Hausdorff) space.
== Proof ==
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.
== Sources ==
* : Part $\text {II}$: Counterexamples: $26$. Arens-Fort Space: $1$
Category:Completely Hausdorff Spaces
Category:Arens-Fort Space
Arens-Fort Space is Completely Normal

== Theorem ==
Let $T = \struct {S, \tau}$ be the Arens-Fort space.
Then $T$ is a completely normal space.
Consequently, $T$ satisfies all weaker separation axioms.
== Proof ==
We have:
:Arens-Fort Space is $T_1$
:Arens-Fort Space is $T_5$
and so by definition $T$ is completely normal.
See Sequence of Implications of Separation Axioms for confirmation of the statement about weaker separation axioms.
== Sources ==
* : Part $\text {II}$: Counterexamples: $26$. Arens-Fort Space: $2$
Category:Completely Normal Spaces
Category:Arens-Fort Space
Arens-Fort Space is Countable

== Theorem ==
Let $T = \left({S, \tau}\right)$ be the Arens-Fort space.
Then $S$ is countably infinite.
== Proof ==
From the definition of the Arens-Fort space:
$S$ is the the Cartesian product $\Z_{\ge 0} \times \Z_{\ge 0}$ of the set of all positive integers:
: $S = \left\{{0, 1, 2, \ldots}\right\} \times \left\{{0, 1, 2, \ldots}\right\}$
We have by definition that $\Z_{\ge 0}$ is countable.
From Cartesian Product of Countable Sets is Countable it follows that $S = \Z_{\ge 0} \times \Z_{\ge 0}$ is likewise countable.
Category:Arens-Fort Space
Arens-Fort Space is Expansion of Countable Fort Space

== Theorem ==
Let $T = \struct {S, \tau}$ be the Arens-Fort space, where $S = \Z_{\ge 0} \times \Z_{\ge 0}$.
Let $T_p = \struct {S, \tau_p}$ be the Fort space on $S$ where $p = \left({0, 0}\right)$.
Then $\tau$ is an expansion of $\tau_p$.
Furthermore, $S$ is countably infinite, so $T_p$ is a countable Fort space.
== Proof ==
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.
== Sources ==
* : Part $\text {II}$: Counterexamples: $26$. Arens-Fort Space: $1$
Category:Fort Spaces
Category:Arens-Fort Space
Arens-Fort Space is Lindelöf

== Theorem ==
Let $T = \struct {S, \tau}$ be the Arens-Fort space.
Then $T$ is a Lindelöf space.
== Proof ==
We have:
:The Arens-Fort Space is Sigma-Compact.
:A $\sigma$-compact space is Lindelöf.
== Sources ==
* : Part $\text {II}$: Counterexamples: $26$. Arens-Fort Space: $4$
Category:Lindelöf Spaces
Category:Arens-Fort Space
Arens-Fort Space is Non-Meager

== Theorem ==
Let $T = \struct {S, \tau}$ be the Arens-Fort space.
Then $T$ is a non-meager space.
== Proof ==
From Meager Sets in Arens-Fort Space, we have that $A \subseteq S$ is meager in $T$ $A = \set {\tuple {0, 0} }$.
So as $\set {\tuple{0, 0} } \ne S \subseteq S$, it follows that $T$ is non-meager.
== Sources ==
* : Part $\text {II}$: Counterexamples: $26$. Arens-Fort Space: $8$
Category:Non-Meager Spaces
Category:Arens-Fort Space
Arens-Fort Space is Paracompact/Proof 1

== Theorem ==
== Proof ==
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.
== Sources ==
* : Part $\text {II}$: Counterexamples: $26$. Arens-Fort Space: $5$
Category:Arens-Fort Space is Paracompact
Arens-Fort Space is Paracompact/Proof 2

== Theorem ==
== Proof ==
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.
== Sources ==
* : Part $\text {II}$: Counterexamples: $26$. Arens-Fort Space: $5$
Category:Arens-Fort Space is Paracompact
Arens-Fort Space is Scattered

== Theorem ==
Let $T = \struct {S, \tau}$ be the Arens-Fort space.
Then $T$ is a scattered space.
== Proof ==
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.
== Sources ==
* : Part $\text {II}$: Counterexamples: $26$. Arens-Fort Space: $8$
Category:Scattered Spaces
Category:Arens-Fort Space
Arens-Fort Space is Separable

== Theorem ==
Let $T = \struct {S, \tau}$ be the Arens-Fort space.
Then $T$ is a separable space.
== Proof ==
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.
== Sources ==
* : Part $\text {II}$: Counterexamples: $26$. Arens-Fort Space: $4$
Category:Separable Spaces
Category:Arens-Fort Space
Arens-Fort Space is Sigma-Compact

== Theorem ==
Let $T = \left({S, \tau}\right)$ be the Arens-Fort space.
Then $T$ is a $\sigma$-compact space.
== Proof 1 ==
== Proof 2 ==
Category:Sigma-Compact Spaces
Category:Arens-Fort Space
Category:Arens-Fort Space is Sigma-Compact
Arens-Fort Space is Sigma-Compact/Proof 1

== Theorem ==
== Proof ==
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 $I_S: \struct {S, \tau_d} \to \struct {S, \tau}$.
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.
Category:Arens-Fort Space is Sigma-Compact
Arens-Fort Space is Sigma-Compact/Proof 2

== Theorem ==
== Proof ==
The result follows from Arens-Fort Space is Countable and Countable Space is Sigma-Compact.
== Sources ==
* : Part $\text {II}$: Counterexamples: $26$. Arens-Fort Space: $4$
Category:Arens-Fort Space is Sigma-Compact
Arens-Fort Space is T1

== Theorem ==
Let $T = \struct {S, \tau}$ be the Arens-Fort space.
Then $T$ is a $T_1$ (Fréchet) space.
== Proof ==
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.
== Sources ==
* : Part $\text {II}$: Counterexamples: $26$. Arens-Fort Space: $1$
Category:T1 Spaces
Category:Arens-Fort Space
Arens-Fort Space is T5

== Theorem ==
Let $T = \struct {S, \tau}$ be the Arens-Fort space.
Then $T$ is a $T_5$ space.
== Proof ==
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.
, 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.
== Sources ==
* : Part $\text {II}$: Counterexamples: $26$. Arens-Fort Space: $2$
Category:T5 Spaces
Category:Arens-Fort Space
Arens-Fort Space is Totally Separated

== Theorem ==
Let $T = \struct {S, \tau}$ be the Arens-Fort space.
Then $T$ is totally separated.
== Proof ==
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.
== Sources ==
* : Part $\text {II}$: Counterexamples: $26$. Arens-Fort Space: $8$
Category:Totally Separated Spaces
Category:Arens-Fort Space
Arens-Fort Space is Zero Dimensional

== Theorem ==
Let $T = \struct {S, \tau}$ denote the Arens-Fort space.
Then $T$ is zero dimensional.
== Proof ==
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.
== Sources ==
* : Part $\text {II}$: Counterexamples: $26$. Arens-Fort Space: $8$
Category:Zero Dimensional Spaces
Category:Arens-Fort Space
Arens-Fort Space is not Compact

== Theorem ==
Let $T = \struct {S, \tau}$ be the Arens-Fort space.
Then $T$ is not a compact space.
== Proof ==
We have:
: The Arens-Fort Space is not Weakly Locally Compact.
: A Compact Space is Weakly Locally Compact.
== Sources ==
* : Part $\text {II}$: Counterexamples: $26$. Arens-Fort Space: $4$
Category:Compact Spaces
Category:Arens-Fort Space
Arens-Fort Space is not Connected

== Theorem ==
Let $T = \struct {S, \tau}$ be the Arens-Fort space.
Then $T$ is not a connected space.
== Proof ==
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.
== Sources ==
* : Part $\text {II}$: Counterexamples: $26$. Arens-Fort Space: $7$
Category:Connected Spaces
Category:Arens-Fort Space
Arens-Fort Space is not Countably Compact

== Theorem ==
Let $T = \struct {S, \tau}$ be the Arens-Fort space.
Then $T$ is not a countably compact space.
== Proof ==
 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.
== Sources ==
* : Part $\text {II}$: Counterexamples: $26$. Arens-Fort Space: $4$
Category:Countably Compact Spaces
Category:Arens-Fort Space
Arens-Fort Space is not Extremally Disconnected

== Theorem ==
Let $T = \struct {S, \tau}$ be the Arens-Fort space.
Then $T$ is not extremally disconnected.
== Proof ==
Let $\map {S_m} V$ denote the set:
:$\map {S_m} V := \set {n: \tuple {m, n} \notin V}$
where $V \subseteq \Z_{\ge 0} \times \Z_{\ge 0}$ (the same set $S_m$ used in the definition of the Arens-Fort space).
Let $U = \set {\tuple {n, m}: \exists k: m = 2 k} \setminus \set {\tuple {0, 0} }$.
From the definition of the Arens-Fort space, $U$ is open in $T$ since $\left({0, 0}\right) \notin U$.
We have that:
:$\tuple {0, 0} \in \relcomp S U$
and:
:$\map {S_m} U$ is infinite for any $m \in \N$.
Thus by definition of the Arens-Fort space, $\relcomp S U$ is not open in $T$.
So $U$ is not closed in $T$.
Let $U^-$ denote the closure of $U$.
We have that:
:$\tuple {0, 0} \notin \relcomp S {U \cup \set {\tuple {0, 0} } }$
Therefore by definition of the Arens-Fort space, $\relcomp S {U \cup \set {\tuple {0, 0} } }$ is open in $T$.
Therefore $U \cup \set {\tuple {0, 0} }$ is closed in $T$.
From Set Closure is Smallest Closed Set in Topological Space, it follows that $U^- = U \cup \set {\tuple {0, 0} }$.
We have that:
:$\tuple {0, 0} \in U \cup \set {\tuple {0, 0} }$
and:
:$\map {S_m} {U \cup \set {\tuple {0, 0} } }$ is infinite for any $m \in \N$.
Therefore by definition of the Arens-Fort space, $U \cup \set {\tuple {0, 0} }$ is not open.
Thus we have created an open set $U$ whose closure $U^-$ is not itself open.
Thus by definition the Arens-Fort space is not extremally disconnected.
== Sources ==
* : Part $\text {II}$: Counterexamples: $26$. Arens-Fort Space: $9$
Category:Extremally Disconnected Spaces
Category:Arens-Fort Space
Arens-Fort Space is not First-Countable

== Theorem ==
Let $T = \struct {S, \tau}$ be the Arens-Fort space.
Then $T$ is not a first-countable space.
== Proof ==
 $T$ 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:
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.
== Sources ==
* : Part $\text {II}$: Counterexamples: $26$. Arens-Fort Space: $3$
Category:First-Countable Spaces
Category:Arens-Fort Space
Arens-Fort Space is not Locally Connected

== Theorem ==
Let $T = \struct {S, \tau}$ be the Arens-Fort space.
Then $T$ is not a locally connected space.
== Proof ==
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.
It is deduced that any local basis is formed with disconnected sets.
Thus, by definition, $T$ is not locally connected.
== Sources ==
* : Part $\text {II}$: Counterexamples: $26$. Arens-Fort Space: $7$
Category:Locally Connected Spaces
Category:Arens-Fort Space
Arens-Fort Space is not Weakly Locally Compact

== Theorem ==
Let $T = \struct {S, \tau}$ be the Arens-Fort space.
Then $T$ is not a weakly locally compact space.
== Proof ==
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.
== Also see ==
* Arens-Fort Space is Completely Hausdorff
== Sources ==
* : Part $\text {II}$: Counterexamples: $26$. Arens-Fort Space: $4$
Category:Weakly Locally Compact Spaces
Category:Arens-Fort Space
Arens-Fort Topology is Topology

== Theorem ==
Let $T = \struct {S, \tau}$ be the Arens-Fort space.
Then $\tau$ is a topology on $T$.
== Proof ==
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:
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: 
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:
:$\ds \relcomp S {\bigcup \UU} = \bigcap_{U \mathop \in \UU} \relcomp S U$
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:
:$\ds \tuple {0, 0} \notin \bigcap \UU$
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 $\ds \bigcup \UU \in \tau$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $26$. Arens-Fort Space
Category:Arens-Fort Space
Argument Form/Examples/Socrates is Mortal

== Example of Argument Form ==
The Socrates is Mortal argument is as follows:
The argument form is:
:$\forall x: \paren {\map M x \implies \map F x}$
:$\map M a$
:$\therefore \map F a$
In this context:
:$\map M x$ means: $x$ is a human
:$\map F x$ means: $x$ is mortal
:$a$ is a specific instance of $x$.
The following arguments have the same form:
:$(1): \quad$ All men are mortal.
:$(2): \quad$ Alfred is a man.
:$(3): \quad$ Therefore Alfred is mortal.
:$(1): \quad$ All dogs are four-legged.
:$(2): \quad$ Rover is a dog.
:$(3): \quad$ Therefore Rover is four-legged.
== Sources ==
* 
* 
Category:Examples of Argument Forms