nyuuzyou/wikis huggingface dataset:
Infinite Set is Equivalent to Proper Subset/Proof 3

== Theorem ==
== Proof ==
Let $X$ be a set which has a proper subset $Y$ such that:
:$\card X = \card Y$
where $\card X$ denotes the cardinality of $X$.
Then:
:$\exists \alpha \in \complement_X \paren Y$
and
:$Y \subsetneqq Y \cup \set \alpha \subseteq X$
The inclusion mappings:
:$i_Y: Y \to X: \forall y \in Y: i \paren y = y$
:$i_{Y \cup \set \alpha}: Y \cup \set \alpha \to X: \forall y \in Y: i \paren y = y$
give:
:$\card X = \card Y \le \card Y + \mathbf 1 \le \card X $
from which:
:$\card X = \card Y + \mathbf 1 = \card X + \mathbf 1$
So by definition $X$ is infinite.
Now suppose $X$ is infinite.
That is:
:$\card X = \card X + \mathbf 1$
Let $\alpha$ be any object such that $\alpha \notin X$.
Then there is a bijection $f: X \cup \set \alpha \to X$.
Let $f_{\restriction X}$ be the restriction of $f$ to $X$.
Then from Injection to Image is Bijection:
:$\image {f_{\restriction X} } = X \setminus \set {f \paren \alpha}$
which is a proper subset of $X$ which is equivalent to $X$.
== Sources ==
* : $\S 8$: Theorem $8.10$
Category:Infinite Set is Equivalent to Proper Subset
Infinite Set of Natural Numbers is Countably Infinite

== Theorem ==
Let $\N$ be the set of natural numbers.
Let $S$ be an infinite subset of $\N$.
Then $S$ is countably infinite.
That is, there is a bijection $f: \N \to S$.
== Proof ==
By Infinite Set has Countably Infinite Subset, we have an injection $g: \N \to S$
But by Cantor-Bernstein-Schröder Theorem/Lemma this produces a bijection $f: \N \to S$
== Sources ==
* : $1$: Set Theory and Logic: $\S 7$: Countable and Uncountable Sets: Theorem $7.1$
Category:Natural Numbers
Infinite Subset of Finite Complement Space Intersects Open Sets

== Theorem ==
Let $T = \struct {S, \tau}$ be a finite complement topology on an infinite set $S$.
Let $H \subseteq S$ be an infinite subset of $S$.
Then the intersection of $H$ with any non-empty open set of $T$ is infinite.
== Proof ==
Let $U \in \tau$ be any non-empty open set of $T$.
Then by definition of finite complement topology:
:$\relcomp S U$ is finite.
We have that:
:$H = H \cap \paren {U \cup \relcomp S U} = \paren {H \cap U} \cup \paren {H \cap \relcomp S U}$
 $H \cap U$ is finite.
From Intersection is Subset:
:$H \cap \relcomp S U \subseteq \relcomp S U$
From Subset of Finite Set is Finite:
:$H \cap \relcomp S U$ is also finite.
Thus we have:
:$H = \paren {H \cap U} \cup \paren {H \cap \relcomp S U}$
which is the union of two finite sets.
From Union of Finite Sets is Finite, $H$ is finite.
It is a contradiction that $H$ is infinite and finite at the same time.
Hence by Proof by Contradiction, $H \cap U$ must be infinite.
== Sources ==
* : Part $\text {II}$: Counterexamples: $18 \text { - } 19$. Finite Complement Topology: $1$
Category:Finite Complement Topology
Infinite Union of Closed Sets of Metric Space may not be Closed

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $V_1, V_2, V_3, \ldots$ be an infinite set of closed sets of $M$.
Then it is not necessarily the case that $\ds \bigcup_{n \mathop \in \N} V_n$ is itself a closed set of $M$.
== Proof ==
Consider the closed real interval:
:$\closedint {\dfrac 1 n} 1 \subseteq \R$
From Closed Real Interval is Closed Set, $\closedint {\dfrac 1 n} 1$ is closed in $\R$ for all $n \in \N$.
But:
:$\ds \bigcup_{n \mathop \in \N} \closedint {\dfrac 1 n} 1 = \hointl 0 1$
The result follows from Half-Open Real Interval is neither Open nor Closed.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 6$: Open Sets and Closed Sets
Category:Closed Sets
Category:Set Union
Infinite if Injection from Natural Numbers

== Theorem ==
Let $S$ be a set.
Let there exist an injection $\phi: \N \to S$ from the natural numbers to $S$.
Then $S$ is infinite.
== Proof ==
 that $S$ is finite.
Let $k \in \N$ be such that there exists a bijection $\psi: S \to \N_k$.
Note that $\N_k \subset \N$ since $k \in \N$, $k \notin \N_k$.
Consider the restriction $\phi \restriction \N_k$ of $\phi$ to $\N_k$.
Then $\phi \restriction \N_k: \N_k \to S$ is an injection by Restriction of Injection is Injection.
From Equivalence of Mappings between Finite Sets of Same Cardinality, it is also a bijection.
Consider $\map \phi k \in S$.
As $\phi \restriction \N_k$ is a surjection, there exists a $j \in \N_k$ such that:
:$\map \phi j = \map {\phi \restriction \N_K} j = \map \phi k$
Since $j \ne k$, it follows that $\phi$ cannot be an injection.
So there can be no such $k$.
Hence $S$ is infinite.
Category:Injections
Category:Infinite Sets
Infinitely Many Programs for URM Computable Function

== Theorem ==
Let $g: \N^k \to \N$ be a URM computable function.
Then there is an infinite number of URM programs which compute $g$.
== Proof ==
As $g$ is URM computable, there exists a URM program which computes it.
Let $Q$ be such a program.
Let $n \in \N$.
Increment the Jumps in $Q$ by $n$ linesTo increment the Jumps by $r$ for any normalized URM program is done by changing all Jumps of the form $J \left({m, n, q}\right)$ to $J \left({m, n, q+r}\right)$.. Call this amended version $Q'$.
This is because, as $Q$ was written so as to start from line 1, we need to move all the Jumps so as to point to the same lines relative to the start of $Q'$.
Then we can build the following URM program:
Line 		Command
$1$ 		$C \left({1, 1}\right)$
$2$ 		$C \left({1, 1}\right)$
$\vdots$		$\vdots$
$n$ 		$C \left({1, 1}\right)$
$n + 1$ 		The start of $Q'$
...etc.
Each of the $C \left({1, 1}\right)$ instructions codes the identity function.
So adding it to the front of $Q$ has no effect on the behaviour of $Q$
It is clear that for each $n \in \N$ there is a different URM program which computes $f$.
The result follows from Infinite if Injection from Natural Numbers.
== Notes ==
Category:URM Programs
Infinity/Examples/Limit of Reciprocal

== Example of Infinity ==
The real function $f: \R \to \R$ defined as:
:$\forall x \in \R: \map f x = \dfrac 1 x$
becomes larger in absolute value as $x$ approaches $0$.
Thus:
:$\ds \lim_{x \mathop \to 0^+} \map f x \to +\infty$
That is, for every $C \in \R_{>0}$ there exists $a \in \R_{>0}$ such that $\map f x > C$ when $0 < x < a$.
When $\map f x \to +\infty$, $\map f x$ is said to be positively infinite.
Similarly:
:$\ds \lim_{x \mathop \to 0^-} \map f x \to -\infty$
That is, for every $C \in \R_{>0}$ there exists $a \in \R_{>0}$ such that $\map f x > -C$ when $-a < x < 0$.
When $\map f x \to -\infty$, $\map f x$ is said to be negatively infinite.
== Sources ==
* 
* 
Category:Examples of Infinity
Information Contained in Letter of Alphabet

== Theorem ==
Let $\psi$ be a letter of the English alphabet.
To a first degree of approximation, the quantity of information contained in $\psi$ is $4 \cdotp 7$.
That is, there is approximately $4 \cdotp 7$ times as much information conveyed by transmission of a single letter as a single bit.
== Further Analysis ==
== Proof ==
Let it be assumed for the purposes of this exercise that each letter has an equal probability of occurring as an element of a message.
There are $26$ letters of the English alphabet.
The amount of information contained in $\psi$ is therefore:
:$\map I \psi = \dfrac {\lg 26} {\lg 2} \approx 4 \cdotp 7$
where $\lg$ denotes the binary logarithm.
== Sources ==
* 
* 
Category:Information Contained in Letter of Alphabet
Category:Information
Category:Information Theory
Information of Sample from Given Distribution

== Theorem ==
Let $S$ be a sample of $n$ observations from a probability distribution with frequency function $\map f {x, \theta}$.
The information $I$ is given by:
:$I = n \map E {\paren {\dfrac {\partial \ln f} {\partial \theta} }^2}$
== Proof ==
== Sources ==
* 
* 
Category:Information (Estimation Theory)
Information of Sample from Given Distribution under Regularity Conditions

== Theorem ==
Let $S$ be a sample of $n$ observations from a probability distribution with frequency function $\map f {x, \theta}$.
Let it be assumed that certain regularity conditions apply.
Let it also be assumed that the extremes do not depend on $\theta$.
The information $I$ is given by:
:$I = -n \map E {\dfrac {\partial^2 \ln f} {\partial \theta^2} }$
== Proof ==
== Sources ==
* 
* 
Category:Information (Estimation Theory)
Ingham's Theorem on Convergent Dirichlet Series

== Theorem ==
Let $\sequence {a_n} \le 1$
For a complex number $z \in \C$, let $\map \Re z$ denote the real part of $z$. 
Form the series $\ds \sum_{n \mathop = 1}^\infty a_n n^{-z}$ which converges to an analytic function $\map F z$ for $\map \Re z > 1$.
Let $\map F z$ be analytic throughout $\map \Re z \ge 1$.
Then $\ds \sum_{n \mathop = 1}^\infty a_n n^{-z}$ converges throughout $\map \Re z \ge 1$.
== Proof ==
Fix a $w$ in $\map \Re w \ge 1$.
Then $\map F {z + w}$ is analytic in $\map \Re z \ge 0$.
We note that since $\map F {z + w}$ is analytic on $\map \Re z = 0$, it must be analytic on an open set containing $\map \Re z = 0$. 
Choose some $R \ge 1$.
We have that $\map F {z + w}$ is analytic on such an open set.
Thus we can determine $\delta = \map \delta R > 0, \delta \le \dfrac 1 2$ such that $\map F {z + w}$ is analytic in $\map \Re z \ge -\delta, \size {\map \Im z} \le R$.
We also choose an $M = \map M R$ so that $\map F {z + w}$ is bounded by $M$ in $-\delta \le \map \Re z, \cmod z \le R$.
Now form the counterclockwise contour $\Gamma$ as the arc $\cmod z = R, \map \Re z > - \delta$ and the segment $\map \Re z = -\delta, \cmod z \le R$.
We denote by $A, B$ respectively, the parts of $\Gamma$ in the right and left half-planes.
By Cauchy's Residue Theorem:
:$\ds 2 \pi i \map F w = \oint_\Gamma \map F {z + w} N^z \paren {\frac 1 z + \frac z {R^2} } \rd z$
Since $\map F {z + w}$ converges to its series on $A$, we may split it into the partial sum and remainder after $N$ terms:
:$\map {s_N} {z + w}, \map {r_N} {z + w}$
respectively.
Again, by Cauchy's Residue Theorem:
:$\ds \int_A \map {s_N} {z + w} N^z \paren {\frac 1 z + \frac z {R^2} } \rd z = 2 \pi i \map {s_N} w - \int_{-A} \map {s_N} {z + w} N^z \paren {\frac 1 z + \frac z {R^2} } \rd z$
where $-A$ is the reflection of $A$ through the origin.
Changing $z \to -z$, we have:
:$\ds \int_A \map {s_N} {z + w} N^z \paren {\frac 1 z + \frac z {R^2} } \rd z = 2 \pi i \map {s_N} w - \int_A \map {s_N} {w - z} N^{-z} \paren {\frac 1 z + \frac z {R^2} } \rd z$
Combining these results gives:
For what follows, allow $z = x + i y$ and observe that on $A, \cmod z = R$.
So:
and on $B$:
Already we can place an upper bound on one of these integrals:
:$\ds \size {\int_B \map F {z + w} N^z \paren {\frac 1 z + \frac z {R^2} } \rd z} \le \int_{-R}^R M N^x \frac 2 \delta \rd y + 2 M \int_{-\delta}^0 N^x \frac {2 x} {R^2} \rd x$
Category:Dirichlet Series
Initial Object is Unique

== Theorem ==
Let $\mathbf C$ be a metacategory.
Let $0$ and $0'$ be two initial objects of $\mathbf C$.
Then there is a unique isomorphism $u: 0 \to 0'$.
Hence, initial objects are unique up to unique isomorphism.
== Proof ==
Consider the following commutative diagram:
$\quad\quad \begin{xy}
<-4em,0em>*+{0} = "M",
<0em,0em> *+{0'}= "N",
<0em,-4em>*+{0} = "M2",
<4em,-4em>*+{0'}= "N2",
"M";"N" **@{-} ?>*@{>} ?*!/_.6em/{u},
"M";"M2" **@{-} ?>*@{>} ?*!/^.6em/{\operatorname{id}_0},
"N";"M2" **@{-} ?>*@{>} ?*!/_.6em/{v},
"N";"N2" **@{-} ?>*@{>} ?*!/_1em/{\operatorname{id}_{0'}},
"M2";"N2"**@{-} ?>*@{>} ?*!/^.6em/{u},
\end{xy}$
It commutes as each of the morphisms in it originates from an initial object, and hence is unique.
Thus, $v$ is an inverse to $u$, and so $u$ is an isomorphism.
== Also see ==
* Terminal Object is Unique
== Sources ==
* : $\S 2.2$: Proposition $2.10$
Category:Category Theory
Initial Segment Determined by Smallest Element is Empty

== Theorem ==
Let $\struct {S, \preceq}$ be a well-ordered set, where $S$ is non-empty.
Let $s_0 = \min S$, the smallest element of $S$.
Then the initial segment determined by $s_0$, $S_{s_0}$, is empty.
== Proof ==
By the definition of initial segment:
:$S_{s_0} := \set {b \in S: b \prec s_0}$
By the definition of smallest element:
:$\forall b \in S: s_0 \preceq b$
Thus $S_{s_0}$ is empty.
Category:Initial Segments
Category:Well-Orderings
Initial Segment of Canonical Order is Set

== Theorem ==
Let $R_0$ denote the canonical ordering of $\paren {\On \times \On}$.
Then, for all $\tuple {x, y} \in \paren {\On \times \On}$, the $R_0$-initial segment is a set.
== Proof ==
Let $z = \map \max {x, y}^+$.
Let $\tuple {v, w} \mathrel R_0 \tuple {x, y}$.
Then:
Thus, the initial segment:
: $\paren {\On \times \On}_{\tuple {x, y} } \subseteq \paren {z \times z}$
By Axiom of Subsets Equivalents, the initial segment of $\tuple {x, y}$ is a set.
== Sources ==
* : $\S 7.56 \ (2)$
Category:Ordinals
Category:Order Theory
Initial Segment of Natural Numbers determined by Zero is Empty

== Theorem ==
Let $\N_k$ denote the initial segment of the natural numbers determined by $k$:
:$\N_k = \set {0, 1, 2, 3, \ldots, k - 1}$
Then $\N_0 = \O$.
== Proof ==
From the definition of $\N_0$:
:$\N_0 = \set {n \in \N: n < 0}$ 
From the definition of zero, $0$ is the minimal element of $\N$.
So there is no element $n$ of $\N$ such that $n < 0$.
Thus $\N_0 = \O$.
Category:Natural Numbers
Initial Segment of Natural Numbers forms Complete Residue System

== Theorem ==
Let $m \in \Z_{\ne 0}$ be a non-zero integer.
Let $\N_m = \set {0, 1, 2, \ldots, m - 1}$ denote the initial segment of $\N$
Then $\N_m$ is a complete residue system modulo $m$.
== Proof ==
Let $n \in \Z$.
From the Division Theorem there exist unique integers $q$ and $u$ such that:
:$n = m q + u$
such that $0 \le u < m$.
That is:
:$n \equiv u \pmod m$
and $u$ is one of $0, 1, \ldots, m - 1$.
Also, since $\forall i, j \in \N_m: \size {i - j} < m$, no two elements of $\N_m$ are congruent.
Thus $\N_m = \set {0, 1, 2, \ldots, m - 1}$ is a complete residue system modulo $m$.
== Sources ==
* : $\text {4-2}$ Residue Systems: Corollary $\text {4-1}$
Category:Residue Classes
Initial Segment of One-Based Natural Numbers determined by Zero is Empty

== Theorem ==
Let $\N^*_k$ denote the initial segment of the one-based natural numbers determined by $k$:
:$\N^*_k = \set {1, 2, 3, \ldots, k - 1, k}$
Then $\N^*_0 = \O$.
== Proof ==
From the definition of $\N^*_0$:
:$\N^*_0 = \set {n \in \N_{>0}: n \le 0}$ 
From the definition of one, the minimal element of $\N_{>0}$ is $1$.
From Zero Strictly Precedes One we know that $0 < 1$.
So there is no element $n$ of $\N_{>0}$ such that $n \le 0$.
Thus $\N^*_0 = \O$.
== Also see ==
* Initial Segment of Natural Numbers determined by Zero is Empty
Category:Natural Numbers
Initial Segment of Ordinal is Ordinal

== Theorem ==
Let $S$ be an ordinal.
Let $a \in S$.
Then the initial segment $S_a = a$ of $S$ determined by $a$ is also an ordinal.
In other words, every element of a (non-empty) ordinal is also an ordinal.
== Proof ==
By Subset of Well-Ordered Set is Well-Ordered, $S_a$ is well-ordered.
Suppose that $b \in S_a$.
From Ordering on Ordinal is Subset Relation, and the definition of an initial segment, it follows that $b \subset a$.
Then:
The result follows from the definition of an ordinal.
== Sources ==
* : $\S 7.6$
* : $\S 1$: Naive Set Theory: $\S 1.7$: Well-Orderings and Ordinals: Theorem $1.7.6$
Category:Ordinals
Initial Segment of Ordinals under Lexicographic Order

== Theorem ==
Let $\preccurlyeq_l$ denote the lexicographic order for the set $\paren {\On \times \On}$.
Let the ordinal number $1$ denote the successor of $\O$.
Then the initial segment of $\tuple {1, \O}$ with respect to the lexicographic order $\preccurlyeq_l$ is a proper class.
This initial segment shall be denoted $\paren {\On \times \On}_{\tuple {1, \O} }$.
== Proof ==
Define the mapping $F: \On \to \On \times \On$ as:
:$\forall x \in \On: \map F x = \tuple {\O, x}$
Then, $F: \On \to \paren {\On \times \On}_{\tuple {1, \O} }$, since $\O < 1$.
That is, $F$ is a mapping from the class of all ordinals to the initial segment of $\tuple {1, \O}$ with respect to lexicographic order.
By Equality of Ordered Pairs, $F$ is injective.
But since $\On$ is a proper class by the Burali-Forti Paradox, the initial segment of $\tuple {1, \O}$ is a proper class as well.
== Sources ==
* : $\S 7.54 \ (2)$
Category:Ordinals
Category:Lexicographic Order
Initial Topology on Hausdorff Locally Convex Space is Weak Topology

== Theorem ==
Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \PP}$ be a Hausdorff locally convex space over $\GF$. 
Let $X^\ast$ be the topological dual space of $X$.
Let $w$ be the initial topology on $X$ with respect to $X^\ast$.
Then $w$ is the weak topology on $X$.
== Proof ==
From Topological Dual Space of Hausdorff Locally Convex Space Separates Points, if $x \ne y$ then there exists $f \in X^\ast$ such that $\map f x \ne \map f y$.
That is, if $x \ne \mathbf 0_X$, there exists $f \in X^\ast$ such that $\map f x \ne 0$.
Category:Weak Topologies on Topological Vector Spaces
Category:Locally Convex Spaces
Initial Topology on Normed Vector Space is Weak Topology

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space over $\GF$. 
Let $X^\ast$ be the normed dual space of $X$. 
Let $w$ be the initial topology on $X$ with respect to $X^\ast$.
Then $w$ is the weak topology on $X$. 
== Proof ==
From Normed Dual Space Separates Points, if $x \ne y$ then there exists $f \in X^\ast$ such that $\map f x \ne \map f y$.
That is, if $x \ne \mathbf 0_X$, there exists $f \in X^\ast$ such that $\map f x \ne 0$.
Category:Weak Topologies on Topological Vector Spaces
Category:Normed Vector Spaces
Initial Topology on Vector Space Generated by Linear Functionals is Locally Convex

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $X$ be a vector space over $\GF$. 
Let $F$ be a set of linear functionals on $X$ that form a vector space over $\GF$.
That is, for each $\lambda, \mu \in \GF$ and $f, g \in F$, we have: 
:$\lambda f + \mu g \in F$
Let $\tau$ be the initial topology on $X$ generated by $F$.
For each $f \in F$, define $p_f : X \to \R_{\ge 0}$ by: 
:$\map {p_f} x = \cmod {\map f x}$ 
and let: 
:$\PP = \set {p_f : f \in F}$
Then the standard topology on the locally convex space $\struct {X, \PP}$ is precisely $\tau$. 
== Proof ==
Let $\tau'$ be the standard topology on the locally convex space.
From definition, a sub-basis for $\tau'$ is given by: 
:$\SS' = \set {\map {B_{p_f} } {\epsilon, x} : f \in F, \, \epsilon > 0, \, x \in X}$
where: 
:$\map {B_{p_f} } {\epsilon, x} = \set {y \in X : \map {p_f} {y - x} < \epsilon}$
For each $u \in \GF$, let: 
:$\map {B_\epsilon} {u, \GF} = \set {v \in \GF : \cmod {u - v} < \epsilon}$
From Open Balls form Basis for Open Sets of Metric Space:
:$\BB_\GF = \set {\map {B_\epsilon} {u, \GF} : u \in \GF, \, \epsilon > 0}$ 
forms a basis for $\GF$.
From Sub-Basis for Initial Topology in terms of Sub-Bases of Target Spaces:
:$\SS = \set {f^{-1} \sqbrk {\map {B_\epsilon} {u, \GF} } : u \in \GF, \, \epsilon > 0}$
is a sub-basis for $\tau$.
We show that $\SS = \SS' \cup \set \O$. 
First, take $f = 0$. 
Then: 
:$\ds f^{-1} \sqbrk {\map {B_\epsilon} {u, \GF} } = \begin{cases}X & 0 \in \map {B_\epsilon} {u, \GF} \\ \O & \text{otherwise}\end{cases} \in \SS' \cup \set \O$ 
On the other hand, if $f = 0$ and $x \in X$ then: 
Now suppose that $f \ne 0$. 
Then from Linear Functional on Vector Space is Zero or Surjective, $f$ is surjective.
So, for each $u \in \GF$ there exists $x \in X$ such that $u = \map f x$.
Then we have: 
So we have: 
:$\set {f^{-1} \sqbrk {\map {B_\epsilon} {u, \GF} } : u \in \GF, \, \epsilon > 0} = \set {\map {B_{p_f} } {\epsilon, x} : f \in F \setminus \set 0, \, \epsilon > 0}$ 
Combining with our result for $f = 0$, we obtain the desired: 
:$\SS = \SS' \cup \set \O$
Since $\O$ is contained in any topology on $X$ we have: 
:$\map \tau \SS = \map \tau {\SS'}$
and hence:
:$\tau = \tau'$
Category:Locally Convex Spaces
Category:Linear Functionals
Category:Initial Topology
Initial Topology with respect to Mapping equals Set of Preimages

== Theorem ==
Let $X$ be a set.
Let $\struct {Y, \tau_Y}$ be a topological space.
Let $f: X \to Y$ be a mapping.
Let $\tau_X$ be the initial topology on $X$ $f$.
Then:
:$\tau_X = \set {f^{-1} \sqbrk U: U \in \tau_Y}$
== Proof ==
Define:
:$\tau = \set {f^{-1} \sqbrk U: U \in \tau_Y}$
By definition, $\tau_X$ is the topology generated by $\tau$.
Therefore:
:$\tau \subseteq \tau_X$
If $\tau$ is a topology on $X$, then it follows from the definition of the generated topology that:
:$\tau_X \subseteq \tau$
By definition of set equality:
:$\tau_X = \tau$
Hence, it suffices to prove that $\tau$ is a topology on $X$.
We now verify the open set axioms for $\tau$ to be a topology on $X$.
=== ===
Let $\AA \subseteq \tau$.
It is to be shown that:
:$\ds \bigcup \AA \in \tau$
Define:
:$\ds \AA' = \set {V \in \tau_Y: f^{-1} \sqbrk V \subseteq \bigcup \AA} \subseteq \tau_Y$
Let:
:$\ds U = \bigcup \AA'$
By the definition of a topology, we have $U \in \tau_Y$.
By Preimage of Union under Mapping: General Result and Union is Smallest Superset: Family of Sets:
:$\ds f^{-1} \sqbrk U = \bigcup_{V \mathop \in \AA'} f^{-1} \sqbrk V \subseteq \bigcup \AA$
By the definition of $\tau$ and by Set is Subset of Union: General Result, we have:
:$\ds \forall S \in \AA: \exists V \in \tau_Y: S = f^{-1} \sqbrk V \subseteq \bigcup \AA$
That is:
:$\forall S \in \AA: \exists V \in \AA': S = f^{-1} \sqbrk V$
By Set is Subset of Union: General Result, we have:
:$\forall V \in \AA': V \subseteq U$
By Preimage of Subset is Subset of Preimage, it follows that:
:$\forall S \in \AA: S \subseteq f^{-1} \sqbrk U$
By Union is Smallest Superset: General Result, we conclude that:
:$\ds \bigcup \AA \subseteq f^{-1} \sqbrk U$
Hence, by definition of set equality:
:$\ds \bigcup \AA = f^{-1} \sqbrk U \in \tau$
=== ===
Let $A, B \in \tau$.
Let $U, V \in \tau_Y$ be such that $A = f^{-1} \sqbrk U$ and $B = f^{-1} \sqbrk V$.
By the definition of a topology, we have $U \cap V \in \tau_Y$.
Then, by Preimage of Intersection under Mapping, $A \cap B = f^{-1} \sqbrk {U \cap V} \in \tau$.
=== ===
By the definition of a topology, we have $Y \in \tau_Y$.
Hence, by Preimage of Mapping equals Domain, it follows that $X = f^{-1} \sqbrk Y \in \tau$.
All the open set axioms are fulfilled, and the result follows.
Category:Topology
Initial Topology with respect to Point-Separating Family of Mappings onto Hausdorff Spaces is Hausdorff

== Theorem ==
Let $X$ be a set.
Let $I$ be an indexing set.
Let $\family {\struct {Y_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of Hausdorff topological spaces indexed by $I$.
Let $\family {f_i: X \to Y_i}_{i \mathop \in I}$ be an indexed family of mappings indexed by $I$ that separate the points of $X$ in the sense that: 
:for each $x, y \in X$ with $x \ne y$, there exists $i \in I$ such that $\map {f_i} x \ne \map {f_i} y$. 
Let $\tau$ be the initial topology on $X$ with respect to $\family {f_i}_{i \mathop \in I}$.
Then $\struct {X, \tau}$ is Hausdorff.
== Proof ==
Let $x, y \in X$. 
Then there exists $i \in I$ such that $\map {f_i} x \ne \map {f_i} y$. 
Since $Y_i$ is Hausdorff, there exists $U, V \in \tau_i$ such that $U \cap V = \O$ and $\map {f_i} x \in U$, $\map {f_i} y \in V$. 
Then we have $x \in f_i^{-1} \sqbrk U$ and $y \in f_i^{-1} \sqbrk V$.
From the definition of the initial topology, $f_i : X \to Y_i$ is $\tuple {\tau, \tau_i}$-continuous.
So $f_i^{-1} \sqbrk U \in \tau$ and $f_i^{-1} \sqbrk V \in \tau$. 
We just need to show that $f_i^{-1} \sqbrk U \cap f_i^{-1} \sqbrk V = \O$.
This is immediate since by Preimage of Intersection under Mapping, since we have: 
:$f_i^{-1} \sqbrk U \cap f_i^{-1} \sqbrk V = f_i^{-1} \sqbrk {U \cap V} = f_i^{-1} \sqbrk \O = \O$
So $f_i^{-1} \sqbrk U$ and $f_i^{-1} \sqbrk V$ are disjoint open sets containing $x$ and $y$ respectively.
So $\struct {X, \tau}$ is Hausdorff.
Category:Initial Topology
Category:Hausdorff Spaces