avewright/proofwiki-math huggingface dataset:
Theorem: Largest Integer Expressible by 3 Digits/Number of Digits

:$9^{9^9}$ has $369 \, 693 \, 100$ digits when expressed in decimal notation.

Proof:
Let $n$ be the number of digits in $9^{9^9}$
From Number of Digits in Number:
:$n = 1 + \floor {\map {\log_{10} } {9^{9^9} } }$
where $\floor {\ldots}$ denotes the floor function.
Then:
{{begin-eqn}}
{{eqn | l = \map {\log_{10} } {9^{9^9} }
      | o = \approx
      | r = 369 \, 693 \, 099 \cdotp 63157 \, 03685 \, 87876 \, 1
      | c = Largest Integer Expressible by 3 Digits: Logarithm Base 10
}}
{{eqn | ll= \leadsto
      | l = n
      | r = 1 + \floor {369 \, 693 \, 099 \cdotp 63157 \, 03685 \, 87876 \, 1}
      | c = 
}}
{{eqn | r = 1 + 369 \, 693 \, 099
      | c = {{Defof|Floor Function}}
}}
{{eqn | r = 369 \, 693 \, 100
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Compact Subset of Compact Space is not necessarily Closed

A compact subset of a compact space is not necessarily closed.

Proof:
Let $S$ be a set containing more than one element.
Let $\tau = \set {S, \O}$ be the indiscrete topology on $S$.
Let $x \in S$.
Let $H = S \setminus \set x$.
Then $H$ is a proper subset of $S$.
Then from Subset of Indiscrete Space is Compact and Sequentially Compact, the subspace induced by $\tau$ on $H$ is a compact subspace of $\struct {S, \tau}$.
But from Closed Sets in Indiscrete Topology, the only closed subsets of $\struct {S, \tau}$ are $S$ and $\O$.
Thus $H$ is a compact subspace of $\struct {S, \tau}$ which is not closed in $S$.
{{qed}}
Theorem: Separability is not Weakly Hereditary

The property of separability is not weakly hereditary.

Proof:
It needs to be demonstrated that there exists a separable topological space which has a subspace which is closed but not separable.
Consider an uncountable particular point space $T = \struct {S, \tau_p}$.
From Particular Point Space is Separable, $T$ is separable.
By definition, the particular point $p$ is an open point of $T$.
Thus the subset $S \setminus \set p$ is by definition closed in $T$.
But from Separability in Uncountable Particular Point Space, $S \setminus \set p$ is not separable.
Thus by Proof by Counterexample, separability is not weakly hereditary.
{{qed}}
Theorem: Filter is Finer iff Sets of Basis are Subsets

Let $S$ be a set.
Let $\powerset S$ denote the power set of $S$.
Let $\FF, \FF' \subset \powerset S$ be two filters on $S$.
Let $\FF$ have a basis $\BB$.
Let $\FF'$ have a basis $\BB'$.
$\FF$ is finer than $\FF'$ {{iff}} for every set of $\BB'$, there is a set of $\BB$ subset to it.

Proof:
=== Necessary Condition ===
Suppose for every set of $\BB'$, there is a set of $\BB$ subset to it.
Pick any $U \in \FF'$.
Then from definition of a basis:
:$\exists V' \in \BB': V' \subseteq U$
By our assumption:
:$\exists V \in \BB: V \subseteq V' \subseteq U \subseteq S$
By definition of a filter:
:$U \in \FF$.
Hence we have $\FF' \subseteq \FF$.
{{qed|lemma}}
Theorem: Existence of Topological Space which satisfies no Separation Axioms

There exists at least one example of a topological space for which none of the Tychonoff separation axioms are satisfied.

Proof:
;Proof by Counterexample
Let $T$ be the topological space consisting of the double pointed topology on the countable complement topology on an uncountable set.
From Double Pointed Countable Complement Topology fulfils no Separation Axioms, we have that $T$ satisfies none of the Tychonoff separation axioms.
{{qed}}
Theorem: T0 Space is not necessarily T1, T2, T3, T4 or T5

There exists at least one example of a topological space for which none of the Tychonoff separation axioms are satisfied except for the $T_0$ (Kolmogorov) axiom.

Proof:
Let $T$ be the overlapping interval space.
From Overlapping Interval Space fulfils no Separation Axioms but $T_0$, we have that $T$ satisfies none of the Tychonoff separation axioms except for the $T_0$ axiom.
{{qed}}
Theorem: T1 Space is not necessarily T2, T3, T4 or T5

There exists at least one example of a topological space for which none of the Tychonoff separation axioms are satisfied except for the $T_0$ axiom and $T_1$ axiom.

Proof:
;Proof by Counterexample
Let $T$ be the finite complement topology on a countable space.
From Finite Complement Space is $T_1$, $T$ is a $T_1$ (Fréchet) space.
Hence from $T_1$ Space is $T_0$, also a $T_0$ (Kolmogorov) space.
The rest of the result follows from:
:Finite Complement Space is not $T_2$
and:
:Finite Complement Space is not $T_3$
:Finite Complement Space is not $T_4$
:Finite Complement Space is not $T_5$
{{qed}}
Theorem: T2 Space is not necessarily T3, T4 or T5

A topological space which is a $T_2$ (Hausdorff) space (and hence also a $T_0$ space and $T_1$ space) is not necessarily a $T_3$ space, a $T_4$ space or a $T_5$ space.

Proof:
Let $T$ be an irrational slope topological space.
We have:
:Irrational Slope Space is $T_2$
:Irrational Slope Space is $T_1$
:Irrational Slope Space is $T_0$
Thus $T$ is a $T_2$ (Hausdorff) space, a $T_1$ space and a $T_0$ space.
The rest of the result follows from:
:Irrational Slope Space is not $T_3$
:Irrational Slope Space is not $T_4$
:Irrational Slope Space is not $T_5$.
{{qed}}
Theorem: T3 Space is not necessarily T0, T1, T2, T4 or T5

There exists at least one example of a topological space for which none of the Tychonoff separation axioms are satisfied except for the $T_3$ axiom.

Proof:
Proof by Counterexample:
Let $T$ be the topological space consisting of the double pointed topology on the Tychonoff corkscrew topology.
From Double Pointed Tychonoff Corkscrew fulfils no Separation Axioms but $T_3$, we have that $T$ satisfies none of the Tychonoff separation axioms except for the $T_3$ axiom.
{{qed}}
Theorem: T4 Space is not necessarily T0, T1, T2, T3 or T5

There exists at least one example of a topological space for which none of the Tychonoff separation axioms are satisfied except for the $T_4$ axiom.

Proof:
Let $T_S = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Let $D = \struct {\set {0, 1}, \vartheta}$ be the indiscrete topology on two points.
Let $T_S \times D$ be the double pointed topology on $T$.
Let $\paren {T_S \times D}^*_{\bar p}$ be the open extension topology on $S \times \set {0, 1} \cup \set p$ where $p \notin S \times \set {0, 1}$.
From Open Extension of Double Pointed Countable Complement Topology is $T_4$, we have that $\paren {T_S \times D}^*_{\bar p}$ satisfies none of the Tychonoff separation axioms except for the $T_4$ axiom.
{{qed}}
Theorem: Compact Hausdorff Space is not necessarily T5

A compact $T_2$ (Hausdorff) space is not necessarily a $T_5$ space.

Proof:
Let $T$ be the Tychonoff plank.
From Tychonoff Plank is Compact, $T$ is a compact topological space.
From Tychonoff Plank is $T_2$, $T$ is a $T_2$ (Hausdorff) space.
But from Tychonoff Plank is not $T_5$, $T$ is not a $T_5$ space.
{{qed}}
Theorem: Regular Space is not necessarily Completely Regular

A topological space which is a regular space is not necessarily also a completely regular space.

Proof:
Let $T$ be a Tychonoff corkscrew.
From Tychonoff Corkscrew is Regular, $T$ is a regular space.
From Tychonoff Corkscrew is not Completely Regular, $T$ is not a completely regular space.
Hence the result.
{{qed}}
Theorem: Completely Regular Space is not necessarily Normal

A topological space which is a completely regular space is not necessarily also a normal space.

Proof:
Let $T$ be the Niemytzki plane.
From Niemytzki Plane is Completely Regular, $T$ is a completely regular space.
From Niemytzki Plane is not Normal, $T$ is not a normal space.
Hence the result.
{{qed}}
Theorem: T4 Space is not necessarily T3.5

A topological space which is a $T_4$ space is not necessarily also a $T_{3 \frac 1 2}$ space.

Proof:
Let $T$ be a Hjalmar Ekdal space.
From Hjalmar Ekdal Space is $T_4$, $T$ is a $T_4$ space.
From Hjalmar Ekdal Space is not $T_{3 \frac 1 2}$, $T$ is not a $T_{3 \frac 1 2}$ space.
Hence the result.
{{qed}}
Theorem: Normal Space is not necessarily Completely Normal

A topological space which is a normal space is not necessarily also a completely normal space.

Proof:
Let $T$ be a Tychonoff plank.
From Tychonoff Plank is Normal, $T$ is a normal space.
From Tychonoff Plank is not Completely Normal, $T$ is not a completely normal space.
Hence the result.
{{qed}}
Theorem: T2.5 Space is not necessarily Regular

There exists at least one example of a topological space which is a $T_{2 \frac 1 2}$ space, but is not also a regular space.

Proof:
Let $T$ be a half-disc space.
From Half-Disc Space is $T_{2 \frac 1 2}$, $T$ is a $T_{2 \frac 1 2}$ space.
From Half-Disc Space is not Regular, $T$ is not a regular space.
Hence the result.
{{qed}}
Theorem: T2 Space is not necessarily T2.5

A topological space which is a $T_2$ (Hausdorff) space is not necessarily also a $T_{2 \frac 1 2}$ space.

Proof:
Let $T$ be an irrational slope space.
From Irrational Slope Space is $T_2$, $T$ is a $T_2$ (Hausdorff) space.
From Irrational Slope Space is not $T_{2 \frac 1 2}$, $T$ is not a $T_{2 \frac 1 2}$ space.
{{qed}}
Theorem: Separation Properties Not Preserved under Expansion

Let $S$ be a set.
Let $T_A = \struct {S, \tau_1}$ and $T_B = \struct {S, \tau_2}$ be topological spaces such that $\tau_2$ is an expansion of $\tau_1$.
These separation properties are not generally preserved under expansion:
{{help|Can we determine whether the semiregular property is preserved under expansion? We know that $T_2$ is but $T_3$ isn't, although the demonstration that $T_3$ isn't may revolve around its ability of an open set to be written as the union of regular open sets, which may give us an angle to attack it by.}}
=== $T_3$ Property is Not Preserved under Expansion ===
{{:T3 Property is Not Preserved under Expansion}}
=== Regular Property is Not Preserved under Expansion ===
{{:Regular Property is Not Preserved under Expansion}}
=== $T_{3 \frac 1 2}$ Property is Not Preserved under Expansion ===
{{:T3.5 Property is Not Preserved under Expansion}}
=== Completely Regular Property is Not Preserved under Expansion ===
{{:Completely Regular Property is Not Preserved under Expansion}}
=== $T_4$ Property is Not Preserved under Expansion ===
{{:T4 Property is Not Preserved under Expansion}}
=== Normal Property is Not Preserved under Expansion ===
{{:Normal Property is Not Preserved under Expansion}}
=== $T_5$ Property is Not Preserved under Expansion ===
{{:T5 Property is Not Preserved under Expansion}}
=== Completely Normal Property is Not Preserved under Expansion ===
{{:Completely Normal Property is Not Preserved under Expansion}}

Proof:
Let $\struct {\R, \tau_1}$ be the set of real numbers under the usual (Euclidean) topology.
Let $\struct {\R, \tau_2}$ be the indiscrete rational extension of $\struct {\R, \tau_1}$.
By definition, $\struct {\R, \tau_2}$ is an expansion of $\struct {\R, \tau_1}$.
From Metric Space fulfils all Separation Axioms, $\struct {\R, \tau_1}$ is:
:a $T_3$ space
:a regular space
:a $T_{3 \frac 1 2}$ space
:a completely regular space
:a $T_4$ space
:a normal space
:a $T_5$ space
:a completely normal space
But we have:
:Indiscrete Extension of Reals is not $T_3$
:Indiscrete Extension of Reals is not $T_4$
:Indiscrete Extension of Reals is not $T_5$
so:
:$\struct {\R, \tau_2}$ is not a $T_3$ space
:$\struct {\R, \tau_2}$ is not a $T_4$ space
:$\struct {\R, \tau_2}$ is not a $T_5$ space
By definition:
:a regular space is a $T_3$ space
:a normal space is a $T_4$ space
:a completely normal space is a $T_5$ space
and we also have:
:$T_{3 \frac 1 2}$ is $T_3$ Space
:Completely Regular Space is Regular
Hence by the Rule of Transposition:
:$\struct {\R, \tau_2}$ is not a $T_{3 \frac 1 2}$ space
:$\struct {\R, \tau_2}$ is not a regular space
:$\struct {\R, \tau_2}$ is not a normal space
:$\struct {\R, \tau_2}$ is not a completely normal space
:$\struct {\R, \tau_2}$ is not a completely regular space.
Hence the result.
{{qed}}
Theorem: Existence of Completely Normal Space whose Product Space is Not Normal

There exists at least one example of a completely normal topological space $T$ such that the product space $T \times T$ is not a normal topological space.

Proof:
Let $T$ be the Sorgenfrey line.
Let $T' = T \times T$ be Sorgenfrey's half-open square.
From Sorgenfrey Line satisfies all Separation Axioms, $T$ is a completely normal space.
From Sorgenfrey's Half-Open Square is not Normal, $T'$ is not a normal space.
Hence the result.
{{qed}}
Theorem: Completely Normal Space is not necessarily Perfectly Normal

A completely normal topological space is not necessarily perfectly normal.

Proof:
Let $T$ be an uncountable Fort space.
From Fort Space is Completely Normal, $T$ is a completely normal space.
From Uncountable Fort Space is not Perfectly Normal, $T$ is not a perfectly normal space.
Hence the result.
{{qed}}
Theorem: Semiregular Topological Space is not necessarily T3

A semiregular topological space is not necessarily a $T_3$ space.

Proof:
Let $T$ be a simplified Arens square.
From Simplified Arens Square is Semiregular:
:$T$ is a semiregular space.
From Simplified Arens Square is not $T_3$:
:$T$ is not a $T_3$ space.
Hence the result.
{{qed}}
Theorem: Semiregular Topological Space is not necessarily T2.5

A semiregular topological space is not necessarily a $T_{2 \frac 1 2}$ space.

Proof:
Let $T$ be a simplified Arens square.
From Simplified Arens Square is Semiregular:
:$T$ is a semiregular space.
From Simplified Arens Square is not $T_{2 \frac 1 2}$:
:$T$ is not a $T_{2 \frac 1 2}$ space.
Hence the result.
{{qed}}
Theorem: Semiregular Topological Space is not necessarily Urysohn

A semiregular topological space is not necessarily a Urysohn space.

Proof:
Let $T$ be an Arens square.
From Arens Square is Semiregular, $T$ is a semiregular space.
From Arens Square is not Urysohn, $T$ is not a Urysohn space.
Hence the result.
{{qed}}
Theorem: Topological Space is Compact iff Every Filter has Adherent Point

A topological space $T = \struct {S, \tau}$ is '''compact''' {{iff}} every filter on $S$ has an adherent point in $S$.

Proof:
{{Recall|Compact Topological Space|compact topological space|index = 2}}
{{:Definition:Compact Topological Space/Definition 2}}
Theorem: Sigma-Compact Space is not necessarily Compact

A $\sigma$-compact topological space is not necessarily also a compact space.

Proof:
Let $T = \struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
From Real Number Line is $\sigma$-Compact, $T$ is a $\sigma$-compact space.
From Real Number Line is not Countably Compact, $T$ is not a countably compact space.
The result follows from Compact Space is Countably Compact.
{{qed}}
Theorem: Lindelöf Space is not necessarily Sigma-Compact

A Lindelöf space is not necessarily also a compact space.

Proof:
Let $T$ be the Sorgenfrey line.
From Sorgenfrey Line is Lindelöf, $T$ is a Lindelöf space.
From Sorgenfrey Line is not $\sigma$-Compact, $T$ is not a $\sigma$-compact space.
Hence the result.
{{qed}}
Theorem: Compact Space is not necessarily Sequentially Compact

There exists at least one example of a compact topological space which is not also a sequentially compact space.

Proof:
Let $T = \Bbb I^\Bbb I$ be the uncountable Cartesian product of the closed unit interval under the usual (Euclidean) topology.
From Uncountable Cartesian Product of Closed Unit Interval is Compact Space, $T$ is a compact space.
From Uncountable Cartesian Product of Closed Unit Interval is not Sequentially Compact Space, $T$ is not a sequentially compact space.
Hence the result.
{{qed}}
Theorem: Compact Space is not necessarily Sequentially Compact

There exists at least one example of a compact topological space which is not also a sequentially compact space.

Proof:
Let $T := \hointr 0 \Omega \times \Bbb I^\Bbb I$ be the Cartesian product of $\hointr 0 \Omega$ under the interval topology, with the uncountable Cartesian product of the closed unit interval under the usual (Euclidean) topology.
From Cartesian Product of Open Ordinal Space with Uncountable Cartesian Product of Closed Unit Interval is Countably Compact Space, $T$ is a countably compact space.
From Cartesian Product of Open Ordinal Space with Uncountable Cartesian Product of Closed Unit Interval is not Sequentially Compact Space, $T$ is not a sequentially compact space.
Hence the result.
{{qed}}
Theorem: Countably Compact Space is not necessarily Sequentially Compact

A countably compact topological space is not necessarily also a sequentially compact space.

Proof:
Let $T := \hointr 0 \Omega \times \Bbb I^\Bbb I$ be the Cartesian product of $\hointr 0 \Omega$ under the interval topology, with the uncountable Cartesian product of the closed unit interval under the usual (Euclidean) topology.
From Cartesian Product of Open Ordinal Space with Uncountable Cartesian Product of Closed Unit Interval is Countably Compact Space, $T$ is a countably compact space.
From Cartesian Product of Open Ordinal Space with Uncountable Cartesian Product of Closed Unit Interval is not Sequentially Compact Space, $T$ is not a sequentially compact space.
Hence the result.
{{qed}}
Theorem: Weakly Countably Compact Space is not necessarily Countably Compact

A weakly countably compact topological space is not also necessarily a countably compact space.

Proof:
Let $T$ be the deleted integer topological space.
From Deleted Integer Topology is Weakly Countably Compact, $T$ is a weakly countably compact space.
From Deleted Integer Topology is not Countably Compact, $T$ is not a countably compact space.
Hence the result.
{{qed}}
Theorem: Weakly Countably Compact Space is not necessarily Countably Compact

A weakly countably compact topological space is not also necessarily a countably compact space.

Proof:
Let $T$ be the odd-even topology on the strictly positive integers $\Z_{>0}$.
From Odd-Even Topology is Weakly Countably Compact, $T$ is a weakly countably compact space.
From Odd-Even Topology is not Countably Compact, $T$ is not a countably compact space.
Hence the result.
{{qed}}
Theorem: Pseudocompact Space is not necessarily Countably Compact

A pseudocompact topological space is not necessarily also a countably compact space.

Proof:
Let $T$ be an infinite particular point space.
From Particular Point Space is Pseudocompact, $T$ is a pseudocompact space.
From Infinite Particular Point Space is not Weakly Countably Compact, $T$ is not a weakly countably compact space.
From Countably Compact Space is Weakly Countably Compact, $T$ cannot be a countably compact space.
Hence the result.
{{qed}}
Theorem: Weakly Locally Compact Space is not necessarily Strongly Locally Compact

A weakly locally compact topological space is not necessarily also a strongly locally compact space.

Proof:
Let $T$ be the nested interval topological space.
From Nested Interval Space is Weakly Locally Compact, $T$ is a weakly locally compact space.
From Nested Interval Space is not Strongly Locally Compact, $T$ is not a strongly locally compact space.
Hence the result.
{{qed}}
Theorem: Weakly Locally Compact Space is not necessarily Strongly Locally Compact

A weakly locally compact topological space is not necessarily also a strongly locally compact space.

Proof:
Let $T$ be an infinite particular point space.
From Particular Point Space is Weakly Locally Compact, $T$ is a weakly locally compact space.
From Infinite Particular Point Space is not Strongly Locally Compact, $T$ is not a strongly locally compact space.
Hence the result.
{{qed}}
Theorem: Strongly Locally Compact Space is not necessarily Sigma-Compact

A strongly locally compact topological space is not necessarily also a $\sigma$-compact space.

Proof:
Let $T$ be an uncountable discrete space.
From Discrete Space is Strongly Locally Compact, $T$ is a strongly locally compact space.
From Uncountable Discrete Space is not $\sigma$-Compact, $T$ is not a $\sigma$-compact space.
Hence the result.
{{qed}}
Category:Strongly Locally Compact Spaces
Category:Sigma-Compact Spaces
Category:Sequence of Implications of Local Compactness Properties
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Theorem: Weakly Sigma-Locally Compact Space is not necessarily Strongly Locally Compact

A weakly $\sigma$-locally compact topological space is not necessarily also a strongly locally compact space.

Proof:
Let $T$ be the nested interval topological space.
From Nested Interval Space is Weakly $\sigma$-Locally Compact, $T$ is a weakly $\sigma$-locally compact space.
From Nested Interval Space is not Strongly Locally Compact, $T$ is not a strongly locally compact space.
Hence the result.
{{qed}}
Theorem: First-Countable Space is not necessarily Second-Countable

A first-countable topological space is not necessarily also a second-countable space.

Proof:
Let $T$ be the Sorgenfrey line.
From Sorgenfrey Line is First-Countable, $T$ is a first-countable space.
From Sorgenfrey Line is not Second-Countable, $T$ is not a second-countable space.
Hence the result.
{{qed}}
Theorem: Lindelöf Space is not necessarily Second-Countable

A Lindelöf space is not necessarily also a second-countable space.

Proof:
Let $T$ be the Sorgenfrey line.
From Sorgenfrey Line is Lindelöf, $T$ is a Lindelöf space.
From Sorgenfrey Line is not Second-Countable, $T$ is not a second-countable space.
Hence the result.
{{qed}}
Theorem: Separable Space is not necessarily Second-Countable

A separable topological space is not necessarily also a second-countable space.

Proof:
Let $T$ be the Sorgenfrey line.
From Sorgenfrey Line is Separable, $T$ is a separable space.
From Sorgenfrey Line is not Second-Countable, $T$ is not a second-countable space.
Hence the result.
{{qed}}
Theorem: Topological Space satisfying Countable Chain Condition is not necessarily Separable

A topological space which satisfies the countable chain condition is not necessarily also a separable space.

Proof:
Let $T$ be a countable complement topology on an uncountable set.
From Countable Complement Space Satisfies Countable Chain Condition, $T$ satisfies the countable chain condition.
From Countable Complement Space is not Separable, $T$ is not a separable space.
Hence the result.
{{qed}}
Theorem: Equivalence of Definitions of Strongly Locally Compact Space

Let $T = \struct {S, \tau}$ be a topological space.
{{TFAE|def = Strongly Locally Compact Space}}

Proof:
=== 1 implies 2 ===
Follows immediately from Topological Closure is Closed.
{{qed|lemma}}
Theorem: Existence and Uniqueness of Reduced Form of Group Word

Let $X$ be a set.
Let $w$ be a group word on $X$.
Then $w$ has a unique reduced form.

Proof:
=== Existence ===
By induction on the length of $w$.
Theorem: Abelianization of Free Group is Free Abelian Group

Let $X$ be a set.
Let $\struct {F_X, \iota}$ be a free group on $X$.
Let $F_X^{\mathrm {ab} }$ be its abelianization.
Let $\pi : F_X \to F_X^{\mathrm {ab} }$ be the quotient group epimorphism.
Then $\struct {F_X^{\mathrm {ab} }, \pi \circ \iota}$ is a free abelian group on $X$.

Proof:
:$\xymatrix {
X \ar[d]_\iota \ar[rd]^{\forall f} & \\
F_X \ar[d]_\pi \ar[r]^{\exists ! g} & H \\
F_X^{\mathrm {ab} } \ar[ru]_{\exists ! h} }$
Lef $H$ be any abelian group and $f:X \to H$ be any mapping.
By Universal Property of Free Group on Set, there exists a unique group homomorphism $g:F_X \to H$ such that $g \circ \iota = f$.
By Universal Property of Abelianization of Group, there exists a unique group homomorphism $h:F_X^{\mathrm {ab} } \to H$ such that $h \circ \pi = g$.
Therefore, $h \circ (\pi \circ \iota) = f$.
{{Handwaving|Actually we have to show that $\pi \circ \iota$ is the natural inclusion from $X$ to $F_X^{\mathrm {ab} }$...}}
Hence, $F_X^{\mathrm {ab} }$ satisfies the Universal Property of Free Abelian Group on Set.
Hence $F_X^{\mathrm {ab} }$ is a Free Abelian Group on $X$.
{{qed}}
Category:Group Theory
hq65lur9bypni916vpjsxgl3gmdoo43
Theorem: Paracompact Space is not necessarily Compact

A paracompact topological space is not necessarily also a compact topological space.

Proof:
Let $T = \struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
From Real Number Line is Paracompact, $T$ is a paracompact space.
From Real Number Line is not Compact, $T$ is not a compact space.
Hence the result.
{{qed}}