avewright/proofwiki-math huggingface dataset:
Theorem: Schönemann-Eisenstein Theorem

Let $\map f x = a_d x^d + a_{d - 1} x^{d - 1} + \dotsb + a_0 \in \Z \sqbrk x$ be a polynomial over the ring of integers $\Z$.
Let $p$ be a prime such that:
:$(1): \quad p \divides a_i \iff i \ne d$
:$(2): \quad p^2 \nmid a_0$
where $p \divides a_i$ signifies that $p$ is a divisor of $a_i$.
Then $f$ is irreducible in $\Q \sqbrk x$.

Proof:
By Gauss's Lemma on Irreducible Polynomials, it suffices to show that $f$ is irreducible in $\Z \sqbrk x$.
{{AimForCont}} that $f = g h$ where $g, h \in \Z \sqbrk x$ are both non-constant.
Let:
:$\map g x = b_e x^e + b_{e - 1} x^{e - 1} + \dotsb + b_0$
:$\map h x = c_f x^f + c_{f - 1} x^{f - 1} + \dotsb + c_0$
Then we have for each $i$:
:$\ds a_i = \sum_{j + k \mathop = i} {b_j c_k}$
In particular, it follows that:
:$a_0 = b_0 c_0$
Possibly after exchanging $g$ and $h$, we may arrange that:
:$p \nmid c_0$
by condition $(2)$.
From condition $(1)$, it follows that then necessarily:
:$p \divides b_0$
We also have:
:$a_d = b_e c_f$
and by condition $(1)$:
:$p \nmid a_d$
and hence:
:$p \nmid b_e$
It follows that there exists a smallest positive $i$ such that:
:$p \nmid b_i$
Naturally, $i \le e$. 
By assumption, both $g$ and $h$ are non-constant.
Hence by Degree of Product of Polynomials over Integral Domain:
:$i < d$
{{explain|Re-evaluate the above link - may need to be Degree of Product of Polynomials over Integral Domain not Less than Degree of Factors. Clarification needed.}}
Consider:
:$a_i = b_0 c_i + b_1 c_{i - 1} + \dotsb + b_i c_0$
with the convention that $c_j = 0$ if $j > f$.
By the minimality of $i$, it follows that:
:$p \divides b_k$
for $0 \le k < i$.
Also, since neither $c_0$ nor $b_i$ is divisible by $p$, the last term $b_i c_0$ is '''not''' divisible by $p$.
Thus, we conclude that:
:$p \nmid a_i$
which contradicts condition $(1)$.
Therefore, $f$ is irreducible.
{{qed}}
Theorem: Isolated Point of Closure of Subset is Isolated Point of Subset

Let $\struct {T, \tau}$ be a topological space.
Let $H \subseteq T$ be a subset of $T$.
Let $\map \cl H$ denote the closure of $H$.
Let $x \in \map \cl H$ be an isolated point of $\map \cl H$.
Then $x$ is also an isolated point of $H$.

Proof:
Let $x \in \map \cl H$ be an isolated point of $\map \cl H$.
{{AimForCont}} that $x$ is not an isolated point of $H$.
Then by the definition of a limit point, it follows that $x$ must be a limit point of $H$.
From Set is Subset of its Topological Closure:
:$H \subseteq \map \cl H$
So by Limit Point of Subset is Limit Point of Set it follows that $x$ is a limit point of $\map \cl H$.
But then $x$ cannot be an isolated point of $\map \cl H$.
This is a contradiction.
Therefore every isolated point of $\map \cl H$ is also an isolated point of $H$.
{{qed}}
Theorem: Isolated Point of Closure of Subset is Isolated Point of Subset

Let $\struct {T, \tau}$ be a topological space.
Let $H \subseteq T$ be a subset of $T$.
Let $\map \cl H$ denote the closure of $H$.
Let $x \in \map \cl H$ be an isolated point of $\map \cl H$.
Then $x$ is also an isolated point of $H$.

Proof:
Let $x \in \map \cl H$ be isolated in $\map \cl H$.
By definition of isolated point, $x$ is not a limit point of $\map \cl H$.
From Set is Subset of its Topological Closure:
:$H \subseteq \map \cl H$
We have that Limit Point of Subset is Limit Point of Set.
But $x$ is not a limit point of $\map \cl H$ .
So by the Rule of Transposition, $x$ cannot be a limit point of $H$.
As $x$ is not a limit point of $H$, it follows that $x$ must be an isolated point of $H$.
{{qed}}
Theorem: Divisors obey Distributive Law

{{:Euclid:Proposition/VII/5}}
In modern algebraic language:
:$a = \dfrac 1 n b, c = \dfrac 1 n d \implies a + c = \dfrac 1 n \paren {b + d}$

Proof:
Let the (natural) number $A$ be an aliquot part of the (natural) number $BC$.
Let the (natural) number $D$ be the same aliquot part of another (natural) number $EF$ that $A$ is of $BC$.
We need to show that $A + D$ is the same aliquot part of $BC + EF$.
:250px
We have that whatever aliquot part $A$ is of $BC$, then $D$ is also the same aliquot part of $EF$.
Therefore as many numbers as there are in $BC$ equal to $A$, so many numbers are there in $EF$ equal to $D$.
Let $BC$ be divided into the numbers equal to $A$, that is, $BG, GC$.
Let $EF$ be divided into the numbers equal to $D$, that is, $EH, HF$.
Then the multitude of $BG, GC$ will be equal to the multitude of $EH, HF$.
Since $BG = A$ and $EH = D$, then $BG + EH = A + D$.
For the same reason, $GC + HF = A + D$.
Therefore, as many numbers as there are in $BC$ equal to $A$, so many are there also in $BC + EF$ equal to $A + D$.
Therefore, whatever multiple $BC$ is of $A$, the same multiple also is $BC + EF$ of $A + D$.
Therefore, whatever aliquot part $A$ is of $BC$, the same aliquot part also is $A + D$ of $BC + EF$.
{{qed}}
Theorem: Divisors obey Distributive Law

{{:Euclid:Proposition/VII/5}}
In modern algebraic language:
:$a = \dfrac 1 n b, c = \dfrac 1 n d \implies a + c = \dfrac 1 n \paren {b + d}$

Proof:
A direct application of the Distributive Property:
:$\dfrac 1 n b + \dfrac 1 n d = \dfrac 1 n \paren {b + d}$
{{qed}}
Theorem: Divisors obey Distributive Law

{{:Euclid:Proposition/VII/5}}
In modern algebraic language:
:$a = \dfrac 1 n b, c = \dfrac 1 n d \implies a + c = \dfrac 1 n \paren {b + d}$

Proof:
Let the (natural) number $AB$ be an aliquant part of the (natural) number $C$.
Let the (natural) number $DE$ be the same aliquant part of another (natural) number $F$ that $AB$ is of $C$.
We need to show that $AB + DE$ is the same aliquant part of $C + F$.
:250px
We have that whatever aliquant part $AB$ is of $C$, $DE$ is also the same aliquant part as $F$.
It follows that as many aliquant parts of $C$ as there are in $AB$, so many aliquant parts of $F$ are there also in $DE$.
Let $AB$ be divided into the aliquant parts of $C$, namely $AG, GB$, and $DE$ into the aliquant parts of $F$, namely $DH, HE$.
Thus the multitude of $AG, GB$ will be equal to the multitude of $DH, HE$.
We have that whatever aliquot part $AG$ is of $C$, the same aliquot part is $DH$ of $F$ also.
Therefore, from {{EuclidPropLink|book = VII|prop = 5|title = Divisors obey Distributive Law}}, whatever aliquot part $AG$ is of $C$, the same aliquot part also is $AG + DH$ of $C + F$ also.
For the same reason, whatever aliquot part $GB$ is of $C$, the same aliquot part also is $GB + HE$ of $C + F$.
Therefore whatever aliquant part $AB$ is of $C$, the same aliquant part also is $AB + DE$ of the $C + F$
{{qed}}
Theorem: Divisors obey Distributive Law

{{:Euclid:Proposition/VII/5}}
In modern algebraic language:
:$a = \dfrac 1 n b, c = \dfrac 1 n d \implies a + c = \dfrac 1 n \paren {b + d}$

Proof:
A direct application of the Distributive Property:
:$\dfrac m n b + \dfrac m n d = \dfrac m n \paren {b + d}$
{{qed}}
Theorem: Open Set Disjoint from Set is Disjoint from Closure

Let $T = \struct {S, \tau}$ be a topological space.
Let $A \subseteq S$. 
Let $B$ be an open set of $T$ disjoint from $A$. 
Then:
: $A^- \cap B = \O$
where $A^-$ denotes the closure of $A$.

Proof:
Since $B \in \tau$, it follows by definition that $S \setminus B$ is closed.
Thus:
{{begin-eqn}}
{{eqn | l = A \cap B
      | r = \O
      | c = {{Defof|Disjoint Sets}}
}}
{{eqn | ll= \leadsto
      | l = A
      | o = \subseteq
      | r = S \setminus B
      | c = Empty Intersection iff Subset of Complement
}}
{{eqn | ll= \leadsto
      | l = A^-
      | o = \subseteq
      | r = \paren {S \setminus B}^-
      | c = Topological Closure of Subset is Subset of Topological Closure
}}
{{eqn | ll= \leadsto
      | l = A^-
      | o = \subseteq
      | r = S \setminus B
      | c = Closed Set Equals its Closure
}}
{{eqn | ll= \leadsto
      | l = A^- \cap B
      | r = \O
      | c = Intersection with Complement is Empty iff Subset
}}
{{end-eqn}}
{{qed}}
Category:Set Closures
Category:Open Sets
1di2et6wmn3ul4hk5r32903mod4wtkg
Theorem: Gauss's Lemma on Primitive Rational Polynomials

Let $\Q$ be the field of rational numbers.
Let $\Q \sqbrk X$ be the ring of polynomials over $\Q$ in one indeterminate $X$.
Let $\map f X, \map g X \in \Q \sqbrk X$ be primitive polynomials.
Then their product $f g$ is also a primitive polynomial.

Proof:
Let $f$ and $g$ be as follows:
{{begin-eqn}}
{{eqn | l = f
      | r = \sum_{k \mathop \in \Z} a_k X^k
}}
{{eqn | l = g
      | r = \sum_{k \mathop \in \Z} b_k X^k
}}
{{end-eqn}}
From the definition of primitive polynomial, the coefficients of $f$ and $g$ are all integers.
From the definition of polynomial product:
:$\ds f g = \sum_{k \mathop \in \Z} c_k \mathbf X^k$ 
where:
:$\ds c_k = \sum_{\substack {p \mathop + q \mathop = k \\ p, q \mathop \in \Z} } a_p b_q$
it is clear that the coefficients of $f g$ are also all integers.
{{AimForCont}} $f g$ is not primitive.
Then its coefficients must have a GCD greater than $1$.
Therefore there exists some prime $p$ which divides all the coefficients of $fg$.
Now $p$ can not divides all the coefficients of either $f$ or $g$, because they are primitive polynomials.
So:
:Let $i$ be the smallest integer such that $p$ does not divide $a_i$
:Let $j$ be the smallest integer such that $p$ does not divide $b_j$.
Consider the coefficient $c_{i+j}$ of $f g$:
:$\ds c_{i + j} = \sum_{k \mathop = 0}^{i + j} a_k b_{i + j - k} = a_0 b_{i + j} + a_1 b_{i + j - 1} + \cdots + a_i b_j + \cdots + a_{i + j - 1} b_1 + a_{i + j} b_0$
From the assumption, it follows that $p \divides c_{i + j}$, and so:
:$\ds p \divides \sum_{k \mathop = 0}^{i + j} a_k b_{i + j - k}$
where $\divides$ denotes divisibility.
Also, from the choice of $i$ and $j$, we have:
:$p \divides a_m$ whenever $m < i$
:$p \divides b_n$ whenever $n < j$
Now all the terms of $\ds \sum_{k \mathop = 0}^{i + j} a_k b_{i + j - k}$, except for $a_i b_j$, contain a factor from either $\set {a_0, a_1, \ldots, a_{i - 1} }$ or $\set {b_0, b_1, \ldots, b_{j - 1} }$.
It follows that we have:
:$p \divides \paren {\ds \sum_{k \mathop = 0}^{i - 1} a_k b_{i + j - k} + \sum_{k \mathop = i + 1}^{i + j} a_k b_{i + j - k} }$
But then, as also $p \divides c_{i + j}$, it follows that $p \divides a_i b_j$ as well.
From Euclid's Lemma for Prime Divisors, $p$ has to divide one or the other.
This contradicts the definition of $i$ and $j$.
So $i$ and $j$ cannot both exist.
It follows that $p$ divides at least one of $f$ and $g$, one of which is therefore not primitive.
From this contradiction, we conclude that $f g$ must be primitive.
Hence the result.
{{qed}}
Theorem: Gauss's Lemma on Primitive Rational Polynomials

Let $\Q$ be the field of rational numbers.
Let $\Q \sqbrk X$ be the ring of polynomials over $\Q$ in one indeterminate $X$.
Let $\map f X, \map g X \in \Q \sqbrk X$ be primitive polynomials.
Then their product $f g$ is also a primitive polynomial.

Proof:
Recall Polynomial has Integer Coefficients iff Content is Integer.
By hypothesis $f$ and $g$ have content $1 \in \Z$.
Therefore:
:$f, g \in \Z \sqbrk X$
{{AimForCont}} that $f g$ is not primitive, say:
:$\cont {f g} = d \ne 1$
By the Fundamental Theorem of Arithmetic we can choose a prime $p$ dividing $d$.
From Quotient Epimorphism from Integers by Principal Ideal, let $\pi : \Z \to \Z / p \Z$ be the quotient epimorphism to the ring of integers modulo $p$.
From Ring of Integers Modulo Prime is Field, $\Z / p \Z$ is a field.
Let $\Pi : \Z \sqbrk X \to \paren {\Z / p \Z} \sqbrk X$ be the induced homomorphism of the polynomial rings.
By construction, $p$ divides each coefficient of $f g$, so:
:$\map \Pi {f g} = \map \Pi f \, \map \Pi g = 0$
From Polynomial Forms over Field form Integral Domain, $\paren {\Z / p \Z} \sqbrk X$ is an integral domain.
Thus:
:$\map \Pi f = 0$ or $\map \Pi g = 0$
After possibly exchanging $f$ and $g$, we may assume that $\map \Pi f = 0$.
Now by Kernel of Induced Homomorphism of Polynomial Forms, if:
:$f = a_0 + a_1 X + \cdots + a_n X^n$
we must have:
:$\map \pi {a_i} = 0$
for $i = 0, \ldots, n$.
That is:
:$a_i \in p \Z$
for $i = 0, \ldots, n$.
But this says precisely that $p$ divides each $a_i$, $i = 0, \ldots, n$.
Therefore $p$ divides the content of $f$, a contradiction.
Hence our assumption that $f g$ is not primitive was invalid.
The result follows by Proof by Contradiction.
{{Qed}}
Theorem: Real Numbers under Addition Modulo 1 form Group

Let $S = \set {x \in \R: 0 \le x < 1}$.
Let $\circ: S \times S \to S$ be the operation defined as:
:$x \circ y = x + y - \floor {x + y}$
That is, $\circ$ is defined as addition modulo $1$.
Then $\struct {S, \circ}$ is a group.

Proof:
First note that Modulo Addition is Well-Defined.
Taking the group axioms in turn:
Theorem: Preimage of Cover is Cover

Let $\phi: S \to T$ be a mapping between the sets $S$ and $T$.
Let $\UU$ be a cover of $T$.
Then the set:
:$\set {\map {\phi^{-1} } U: U \in \UU}$
is a cover of $S$.

Proof:
Let $x \in S$.
Then $\map \phi x \in T$.
Since $\UU$ is a cover of $T$:
:$\exists U \in \UU: \map \phi x \in U$
By definition of preimage, $x \in \map {\phi^{-1} } U$.
So:
:$\forall x \in S: \exists \map {\phi^{-1} } U \in S: x \in \map {\phi^{-1} } U$
That is, $\set {\map {\phi^{-1} } U: U \in \UU}$ is a cover of $S$.
{{qed}}
Category:Preimages under Mappings
Category:Covers
pw5x0upg6k9tkp443ls8051kxeezi7y
Theorem: Compactness is Preserved under Continuous Surjection

Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a continuous surjection.
If $T_A$ is compact, then $T_B$ is also compact.

Proof:
Let $T_A$ be compact.
Take an open cover $\UU$ of $T_B$.
From Preimage of Cover is Cover, $\set {\phi^{-1} \sqbrk U: U \in \UU}$ is a cover of $S_A$.
But $\phi$ is continuous, and for all $U \in \UU$, $U$ is open in $T_B$.
It follows that $\forall U \in \UU: \phi^{-1} \sqbrk U$ is open in $T_A$.
So $\set {\phi^{-1} \sqbrk U:\ U \in \UU}$ is an open cover of $T_A$.
$T_A$ is compact, so we take a finite subcover:
:$\set {\phi^{-1} \sqbrk {U_1}, \ldots, \phi^{-1} \sqbrk {U_n} }$
We have that $\phi$ is surjective.
So from Surjection iff Right Inverse, $\phi \sqbrk {\phi^{-1} \sqbrk A} = A$
So:
:$\set {\phi \sqbrk {\phi^{-1} \sqbrk {U_1} }, \ldots, \phi \sqbrk {\phi^{-1} \sqbrk {U_n} } } = \set {U_1, \ldots, U_n} \subseteq \UU$
is a finite subcover of $\UU$ on $T_B$.
{{qed}}
Theorem: Sigma-Compactness is Preserved under Continuous Surjection

Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a continuous surjection.
If $T_A$ is $\sigma$-compact, then $T_B$ is also $\sigma$-compact.

Proof:
Let $T_A$ be $\sigma$-compact.
Then:
:$\ds S_A = \bigcup_{i \mathop = 1}^\infty S_i$
where $S_i \subseteq S_A$ are compact.
Since $\phi$ is surjective, we have from Image of Union under Relation:
:$\ds \phi \sqbrk {S_A} = S_B = \phi \sqbrk {\bigcup_{i \mathop = 1}^\infty S_i} = \bigcup_{i \mathop = 1}^\infty \phi \sqbrk {S_i}$
From Compactness is Preserved under Continuous Surjection, we have that $\phi \sqbrk {S_i}$ is compact for all $i \in \N$.
So $S_B$ is the union of a countable number of compact subsets.
Thus, by definition, $T_B$ is also $\sigma$-compact.
{{qed}}
Theorem: Countable Compactness is Preserved under Continuous Surjection

Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a continuous surjection.
If $T_A$ is countably compact, then $T_B$ is also countably compact.

Proof:
Let $T_A$ be countably compact.
Take a countable open cover $\UU$ of $T_B$.
From Preimage of Cover is Cover, $\VV := \set {\phi^{-1} \sqbrk U: U \in \UU}$ is a cover of $S_A$.
$\VV$ is a countable cover because there it is bijective with $\UU$.
{{explain|It needs to be demonstrated that "it" is bijective, and in this context, exactly what "it" is.}}
By hypothesis, $\phi$ is continuous.
For all $U \in \UU$, $U$ is open in $T_B$.
It follows that $\forall U \in \UU: \phi^{-1} \sqbrk U$ is open in $T_A$.
So $\VV$ is a countable open cover of $T_A$.
$T_A$ is countably compact, so we take a finite subcover:
:$\set {\phi^{-1} \sqbrk {U_1}, \ldots, \phi^{-1} \sqbrk {U_n} }$
We have that $\phi$ is surjective.
So from Surjection iff Right Inverse:
:$\phi \sqbrk {\phi^{-1} \sqbrk A} = A$
So:
:$\set {\phi \sqbrk {\phi^{-1} \sqbrk {U_1} }, \ldots, \phi \sqbrk {\phi^{-1} \sqbrk {U_n} } } = \set {U_1, \ldots, U_n} \subseteq \UU$
is a finite subcover of $\UU$ on $T_B$.
{{explain|It is not clear that $\set {U_1, \ldots, U_n}$ covers $T_B$.}}
{{qed}}
Theorem: Sequential Compactness is Preserved under Continuous Surjection

Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a continuous surjection.
If $T_A$ is sequentially compact, then $T_B$ is also sequentially compact.

Proof:
Let $T_A$ be a sequentially compact space.
Take an infinite sequence $\sequence {x_n} \subseteq S_B$.
From the surjectivity of $\phi$, there exists another infinite sequence $\sequence {y_n} \subseteq S_A$ such that $\map \phi {y_n} = x_n$.
By the definition of sequential compactness, there exists a subsequence of $\sequence {y_n}$
Let this subsequence be named $\sequence {y_{n_k} }$.
Let $\sequence {y_{n_k} }$ converge to $y \in S_A$ with respect to $T_A$.
From the continuity of $\phi$, it is concluded that $\map \phi {y_{n_k} } = x_{n_k}$ converges to $\map \phi y \in S_B$.
Thus, $\sequence {x_n}$ has a subsequence that converges.
By definition, $T_B$ is sequentially compact.
{{qed}}
Theorem: Lindelöf Property is Preserved under Continuous Surjection

Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a continuous surjection.
If $T_A$ is a Lindelöf space, then $T_B$ is also a Lindelöf space.

Proof:
Let $T_A$ be a Lindelöf space.
Take an open cover $\UU$ of $T_B$.
From Preimage of Cover is Cover, $\set {\phi^{-1} \sqbrk U: U \in \UU}$ is a cover of $S_A$.
But $\phi$ is continuous, and for all $U \in \UU$, $U$ is open in $T_B$.
It follows that $\forall U \in \UU: \phi^{-1} \sqbrk U$ is open in $T_A$.
So $\set {\phi^{-1} \sqbrk U: U \in \UU}$ is an open cover of $T_A$.
$T_A$ is Lindelöf , so we take a countable subcover:
:$\set {\phi^{-1} \sqbrk {U_1}, \ldots, \phi^{-1} \sqbrk {U_n}, \ldots}$
We have that $\phi$ is surjective.
So from Surjection iff Right Inverse:
:$\phi \sqbrk {\phi^{-1} \sqbrk A} = A$
So:
:$\set {\phi \sqbrk {\phi^{-1} \sqbrk {U_1} }, \ldots, \phi \sqbrk {\phi^{-1} \sqbrk {U_n} }, \ldots} = \set {U_1, \ldots, U_n, \ldots} \subseteq \UU$
is a countable subcover of $\UU$ on $T_B$.
{{qed}}
Theorem: Multiples of Divisors obey Distributive Law

{{:Euclid:Proposition/VII/6}}
In modern algebraic language:
:$a = \dfrac m n b, c = \dfrac m n d \implies a + c = \dfrac m n \paren {b + d}$

Proof:
Let the (natural) number $AB$ be an aliquant part of the (natural) number $C$.
Let the (natural) number $DE$ be the same aliquant part of another (natural) number $F$ that $AB$ is of $C$.
We need to show that $AB + DE$ is the same aliquant part of $C + F$.
:250px
We have that whatever aliquant part $AB$ is of $C$, $DE$ is also the same aliquant part as $F$.
It follows that as many aliquant parts of $C$ as there are in $AB$, so many aliquant parts of $F$ are there also in $DE$.
Let $AB$ be divided into the aliquant parts of $C$, namely $AG, GB$, and $DE$ into the aliquant parts of $F$, namely $DH, HE$.
Thus the multitude of $AG, GB$ will be equal to the multitude of $DH, HE$.
We have that whatever aliquot part $AG$ is of $C$, the same aliquot part is $DH$ of $F$ also.
Therefore, from {{EuclidPropLink|book = VII|prop = 5|title = Divisors obey Distributive Law}}, whatever aliquot part $AG$ is of $C$, the same aliquot part also is $AG + DH$ of $C + F$ also.
For the same reason, whatever aliquot part $GB$ is of $C$, the same aliquot part also is $GB + HE$ of $C + F$.
Therefore whatever aliquant part $AB$ is of $C$, the same aliquant part also is $AB + DE$ of the $C + F$
{{qed}}
Theorem: Multiples of Divisors obey Distributive Law

{{:Euclid:Proposition/VII/6}}
In modern algebraic language:
:$a = \dfrac m n b, c = \dfrac m n d \implies a + c = \dfrac m n \paren {b + d}$

Proof:
A direct application of the Distributive Property:
:$\dfrac m n b + \dfrac m n d = \dfrac m n \paren {b + d}$
{{qed}}
Theorem: First-Countability is Preserved under Open Continuous Surjection

Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a surjective open mapping which is also continuous.
If $T_A$ is first-countable, then $T_B$ is also first-countable.

Proof:
Let $\phi$ be surjective, continuous and open.
Let $T_A$ be first countable.
Let $b \in S_B$.
Since $\phi$ is surjective there is a point $a \in S_A$ such that:
:$\map \phi a = b$
From the first-countability of $T_A$, there is a local base $\BB$, say, of $a$ which is countable.
Let $\BB = \set {V_n: n \in \N}$.
We need to show that $\set {\phi \sqbrk {V_n}: n \in \N}$ is a local base for $b$.
Let $U$ be an open set of $T_B$ that contains $b$.
As $b = \map \phi a$ we have that:
:$a \in \phi^{-1} \sqbrk U$
From the continuity of $\phi$, we have that $\phi^{-1} \sqbrk U$ is open.
As $\BB$ is a local base, there is an open set $V_n \subseteq \phi^{-1} \sqbrk U$ such that $a \in V_n$.
$\phi$ is surjective, so from Surjection iff Right Inverse we have that:
:$\phi \sqbrk {\phi^{-1} \sqbrk U} = U$
So, applying $\phi$ to $V_n$, from Image of Subset under Mapping is Subset of Image we obtain:
:$\phi \sqbrk {V_n} \subseteq U$
such that $b \in \phi \sqbrk {V_n}$.
This means that $\set {\phi \sqbrk {V_n}: n \in \N}$ is a local base for $b$.
Thus, $T_B$ is first countable.
{{qed}}
Theorem: Second-Countability is Preserved under Open Continuous Surjection

Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a surjective open mapping which is also continuous.
If $T_A$ is second-countable, then $T_B$ is also second-countable.

Proof:
Let $\phi$ be surjective, continuous and open.
Let $T_A$ be second-countable.
By definition of second-countability $T_A$ has a countable basis, $\BB$, say.
Let $\BB = \set {V_n: n \in \N}$.
We need to show that $\set {\phi \sqbrk {V_n}: n \in \N}$ is a base for $T_B$.
Let $U$ be an open set of $T_B$.
$\phi$ is continuous, so $\phi^{-1} \sqbrk U$ is open in $T_A$.
As $\BB$ is a base for $T_A$, there exists an open set $V_n \subseteq \phi^{-1} \sqbrk U$.
$\phi$ is surjective, so from Surjection iff Right Inverse we have that:
:$\phi \sqbrk {\phi^{-1} \sqbrk U} = U$
So, applying $\phi$ to $V_n$, from Image of Subset under Mapping is Subset of Image we obtain:
:$\phi \sqbrk {V_n} \subseteq U$.
This means that $\set {\phi \sqbrk {V_n}: n \in \N}$ is a base for $T_B$.
Thus, $T_B$ is second-countable.
{{qed}}
Theorem: Arens-Fort Space is Countable

Let $T = \struct {S, \tau}$ 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:
:$S = \set {0, 1, 2, \ldots} \times \set {0, 1, 2, \ldots}$
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.
{{qed}}
Category:Arens-Fort Space
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Theorem: Connected iff no Proper Clopen Sets

Let $T = \struct {S, \tau}$ be a topological space.
Then:
:$T$ is connected
{{iff}}:
:there exists no non-empty proper subset of $S$ which is clopen in $T$.

Proof:
=== Sufficient Condition ===
Let $T$ be connected.
{{AimForCont}} there exists $H \subset S$ such that:
:$H$ is clopen in $T$
:$H$ is a non-empty proper subset of $S$, that is:
::$\O \ne H \ne S$
Then $H$ and $\relcomp S H$ are open sets whose union is $S$.
Thus $\set {H \mid \relcomp S H}$ form a partition of $S$.
Hence by definition, $T$ is not connected.
This contradicts our assumption.
Hence, by Proof by Contradiction, such $H$ does not exist.
{{qed|lemma}}