hcju/proofwikiqa huggingface dataset:
=== Setting up a Recurrence Rule === First we consider the plane with no lines at all. This has one region, so $L_0 = 1$. Now when we have one line, we divide the plane into two regions, so $L_1 = 2$. Now consider the $n$th line. This increase the number of regions by $k$ iff it splits $k$ of the old regions. It can split $k$ of the old regions {{iff}} it hits the existing lines on the plane in $k-1$ places. Two straight lines can intersect in at most one point. So the new line can intersect the $n - 1$ old ones in at most $n - 1$ different points. Therefore $k \le n$. So we see that $L_n \le L_{n - 1} + n$. Now, it is always possible to place the $n$th line so that: :$(1): \quad$ It is not parallel to any of the others, and therefore intersects all the other $n - 1$ lines :$(2): \quad$ It does not go through any of the existing intersection points (so intersects them all in different places). Thus we see that $L_n \ge L_{n - 1} + n$. Hence the recurrence: :$L_n = L_{n - 1} + n$ === Solution of Recurrence === Using induction, we show that: :$L_n = \dfrac {n \paren {n + 1} } 2 + 1$ The base case is straightforward: :$L_0 = 1 = \dfrac {0 \paren {0 + 1} } 2 + 1$ :$L_1 = 2 = \dfrac {1 \paren {1 + 1} } 2 + 1$ Now assume the induction hypothesis: :$L_k = \dfrac {k \paren {k + 1} } 2 + 1$ and try to show: :$L_{k + 1} = \dfrac {\paren {k + 1} \paren {k + 2} } 2 + 1$ Hence the induction step: {{begin-eqn}} {{eqn | l = L_{k + 1}       | r = L_k + k + 1       | c =  }} {{eqn | r = \frac {k \paren {k + 1} } 2 + 1 + k + 1       | c =  }} {{eqn | r = \frac {\paren {k + 1} \paren {k + 2} } 2 + 1       | c = after algebra }} {{end-eqn}} Hence the result by induction. {{qed}}
From the definitions: {{begin-eqn}} {{eqn | l = \map \Delta {x^{\underline m} }       | r = \paren {x + 1}^{\underline m} - x^{\underline m}       | c = {{Defof|Forward Difference Operator}} }} {{eqn | r = \prod_{k \mathop = 0}^{m - 1} \paren {x + 1 - k} - \prod_{k \mathop = 0}^{m - 1} \paren {x - k}       | c = {{Defof|Falling Factorial|$m$th Falling Factorial}} }} {{eqn | r = \paren {x + 1} \prod_{k \mathop = 1}^{m - 1} \paren {x + 1 - k} - \paren {x - m + 1} \prod_{k \mathop = 0}^{m - 2} \paren {x - k}       | c =  }} {{eqn | r = \paren {x + 1} \prod_{k \mathop = 0}^{m - 2} \paren {x - k} - \paren {x - m + 1} \prod_{k \mathop = 0}^{m - 2} \paren {x - k}       | c = Translation of Index Variable of Product }} {{eqn | r = \paren {\paren {x + 1} - \paren {x - m + 1} } \prod_{k \mathop = 0}^{m - 2} \paren {x - k}       | c =  }} {{eqn | r = m x^{\underline{m - 1} }       | c = {{Defof|Falling Factorial|$m$th Falling Factorial}} }} {{end-eqn}} {{qed}} Category:Finite Calculus f61fprcnzkv94xizg5sqgl1273y41vr
Follows directly from Lucas' Theorem: :$\dbinom n k \equiv \dbinom {\floor {n / p} } {\floor {k / p} } \dbinom {n \bmod p} {k \bmod p} \pmod p$ where $k = p$. Then: :$k \bmod p = 0$ and so by Binomial Coefficient with Zero: :$\dbinom {n \bmod p} {k \bmod p} = 1$ Also: :$\floor {k / p} = 1$ and by Binomial Coefficient with One: :$\dbinom {\floor {n / p} } {\floor {k / p} } = \floor {\dfrac n p}$ Thus: :$\dbinom n p \equiv \floor {\dfrac n p} \times 1 \pmod p$ Hence the result. {{qed}}
From Binomial Coefficient of Prime, we have: :$\dbinom p k \equiv 0 \pmod p$ when $1 \le k \le p - 1$. From Pascal's Rule: :$\dbinom {p-1} k + \dbinom {p - 1} {k - 1} = \dbinom p k \equiv 0 \pmod p$ This certainly holds for $k = 1$, and so we have: :$\dbinom {p - 1} 1 + \dbinom {p - 1} 0 = \dbinom p 1 \equiv 0 \pmod p$ But from Binomial Coefficient with Zero: :$\dbinom {p - 1} 0 = 1 \equiv 1 \pmod p$ So: :$\dbinom {p - 1} 1 \equiv -1 \pmod p$ Then: :$\dbinom {p - 1} 2 + \dbinom {p - 1} 1 = \dbinom p 2 \equiv 0 \pmod p$ and so: :$\dbinom {p - 1} 2 \equiv 1 \pmod p$ The result follows. {{qed}}
From Binomial Coefficient of Prime, we have: :$\dbinom p k \equiv 0 \pmod p$ when $1 \le k \le p - 1$. From Pascal's Rule: :$\dbinom {p + 1} k = \dbinom p {k - 1} + \dbinom p k$ The result follows immediately. {{qed}}
Define a topology on the integers $\Z$ by declaring a subset $U \subseteq \Z$ to be an open set {{iff}} it is either: :the empty set $\O$ or: :a union of sequences $\map S {a, b}$, where: ::$\map S {a, b} = \set {a n + b: n \in \Z} = a \Z + b$ In other words, $U$ is open {{iff}} every $x \in U$ admits some non-zero integer $a$ such that $\map S {a, x} \subseteq U$. The open set axioms are verified as follows: $(O1): \quad$ All unions of open sets are open: For any set of open sets $U_i$ and $x$ in their union $U$, any of the numbers $a_i$ for which $\map S {a_i, x} \subseteq U_i$ also shows that $\map S {a_i, x} \subseteq U$. $(O2): \quad$ The intersection of two (and hence finitely many) open sets is open: Let $U_1$ and $U_2$ be open sets. Let $x \in U_1 \cap U_2$ (with integers $a_1$ and $a_2$ establishing membership). Set $a$ to be the lowest common multiple of $a_1$ and $a_2$. Then: :$\map S {a, x} \subseteq \map S {a_i, x} \subseteq U_1 \cap U_2$ $(O3): \quad$ By definition, $\O$ is open: $\Z$ is just the sequence $\map S {1, 0}$, and so is open as well. The topology is quite different from the usual Euclidean one, and has two notable properties: :$(1): \quad$ Since any non-empty open set contains an infinite sequence, no finite set can be open. Put another way, the complement of a finite set cannot be a closed set. :$(2): \quad$ The basis sets $\map S {a, b}$ are both open and closed: they are open by definition, and we can write $\map S {a, b}$ as the complement of an open set as follows: :$\displaystyle \map S {a, b} = \Z \setminus \bigcup_{j \mathop = 1}^{a - 1} \map S {a, b + j}$ The only integers that are not integer multiples of prime numbers are $-1$ and $+1$, that is: :$\displaystyle \Z \setminus \set {-1, + 1} = \bigcup_{\text {$p$ prime} } \map S {p, 0}$ By the first property, the set on the {{LHS}} cannot be closed. On the other hand, by the second property, the sets $\map S {p, 0}$ are closed. So, if there were only finitely many prime numbers, then the set on the {{RHS}} would be a finite union of closed sets, and hence closed. Therefore by Proof by Contradiction, there must be infinitely many prime numbers. {{qed}}
Euclid's Theorem states that: : For any finite set of prime numbers, there exists a prime number not in that set. The result follows by corollary. {{qed}}
{{AimForCont}} that there are only $N$ prime numbers. Let the set of all primes be: :$\Bbb P = \set {p_1, p_2, \ldots, p_N}$ By the Fundamental Theorem of Arithmetic, every integer $k > 1$ can be expressed in the form: :$k = {p_1}^{a_1} {p_2}^{a_2} \dotsm {p_N}^{a_N}$ Let $n > 1$ be fixed. Let $a$ be the largest exponent occurring in the prime decomposition of all positive integers $k \le n$. Then: :$\displaystyle \sum_{k \mathop = 1}^n \frac 1 k \le \prod_{j \mathop = 1}^N \paren {\sum_{k \mathop = 0}^a \frac 1 { {p_j}^k} }$ is expressible in the form: {{begin-eqn}} {{eqn | n = 1       | l = 1 + \frac 1 2 + \frac 1 3 + \dotsb + \frac 1 n       | o = \le       | r = \paren {1 + \frac 1 {p_1} + \frac 1 { {p_1}^2} + \dotsb + \frac 1 { {p_1}^a} }       | c =  }} {{eqn | o =       | ro= \times       | r = \paren {1 + \frac 1 {p_2} + \frac 1 { {p_2}^2} + \dotsb + \frac 1 { {p_2}^a} }       | c =  }} {{eqn | o =       | ro= \times       | r = \ \dotsb       | c =  }} {{eqn | o =       | ro= \times       | r = \paren {1 + \frac 1 {p_N} + \frac 1 { {p_N}^2} + \dotsb + \frac 1 { {p_N}^a} }       | c =  }} {{end-eqn}} which can be seen by multiplying out the factors on the {{RHS}}. But from Sum of Infinite Geometric Sequence: :$1 + x + x^2 + \dotsb = \dfrac 1 {1 - x}$ for all $x$ such that $\size x < 1$. Thus the factors in $(1)$ are less than the numbers: :$\dfrac 1 {1 - 1 / p_1}, \dfrac 1 {1 - 1 / p_2}, \dotsb, \dfrac 1 {1 - 1 / p_N}$ and so: :$1 + \dfrac 1 2 + \dfrac 1 3 + \dotsb + \dfrac 1 n < \dfrac {p_1} {p_1 - 1} \dfrac {p_2} {p_2 - 1} \dotsm \dfrac {p_N} {p_N - 1}$ We have chosen $n > 1$ arbitrarily, so this holds for every $n > 1$. This contradicts the result Harmonic Series is Divergent. Hence the result, by Proof by Contradiction. {{qed}}
{{AimForCont}} there exist only a finite number of primes. From Sum of Reciprocals of Powers as Euler Product: :$\displaystyle \sum_{n \mathop \ge 1} \dfrac 1 {n^z} = \prod_p \frac 1 {1 - p^{-z} }$ When $z = 1$ this gives: :$\displaystyle \sum_{n \mathop \ge 1} \dfrac 1 n = \prod_p \frac 1 {1 - 1/p}$ As by hypothesis there exist only a finite number of primes, the {{RHS}} is also finite. But from Harmonic Series is Divergent, the {{LHS}} diverges to infinity. The result follows by Proof by Contradiction. {{qed}}
A direct application of Factorial Divisible by Prime Power. {{qed}} Category:Factorials Category:Binary Notation 8595atu12hqzf4sfr9m0dvw223wlnb3
From the Quotient-Remainder Theorem, we have: :$\exists q, r \in \Z: n = q b + r$ where $0 \le b < r$. We have that: {{begin-eqn}} {{eqn | l = n       | r = \sum_{j \mathop = 0}^m {r_j b^j}       | c =  }} {{eqn | r = \sum_{j \mathop = 1}^m {r_j b^j} + r_0       | c =  }} {{eqn | r = b \sum_{j \mathop = 1}^m {r_j b^{j-1} } + r_0       | c =  }} {{end-eqn}} Hence we can express $n = q b + r$ where: : $\displaystyle q = \sum_{j \mathop = 1}^m {r_{j} b^{j-1}}$ : $r = r_0$ where: : $\displaystyle \sum_{j \mathop = 1}^m {r_j b^{j-1}} = \left[{r_m r_{m-1} \ldots r_2 r_1}\right]_b$ The result follows from the definition of the modulo operation. {{qed}}
If $p > n$, then $\map {s_p} n = n$ and we have that: :$\dfrac {n - \map {s_p} n} {p - 1} = 0$ which ties in with the fact that $\floor {\dfrac n p} = 0$. Hence the result holds for $p > n$. So, let $p \le n$. From De Polignac's Formula, we have that: {{begin-eqn}} {{eqn | l = \mu       | r = \sum_{k \mathop > 0} \floor {\frac n {p^k} }       | c =  }} {{eqn | r = \floor {\frac n p} + \floor {\frac n {p^2} } + \floor {\frac n {p^3} } + \cdots + \floor {\frac n {p^s} }       | c =  }} {{end-eqn}} where $p^s < n < p^{s + 1}$. From Quotient and Remainder to Number Base: General Result: :$\displaystyle \floor {\frac n {p^k} } = \sqbrk {r_m r_{m - 1} \ldots r_{k + 1} r_k}_p = \sum_{j \mathop = 0}^{m - k} r_j p^j$ where $0 \le r_j < p$. Hence: {{begin-eqn}} {{eqn | l = \mu       | r = \sum_{k \mathop > 0} \paren {\sum_{j \mathop = 0}^{m - k} r_j p^j}       | c =  }} {{eqn | r = r_m p^{m - 1} + r_{m - 1} p^{m - 2} + \cdots + r_2 p + r_1       | c =  }} {{eqn | o = +       | r = r_m p^{m - 2} + r_{m - 1} p^{m - 3} + \cdots + r_2       | c =  }} {{eqn | o = +       | r = \cdots       | c =  }} {{eqn | o = +       | r = r_m p + r_{m - 1}       | c =  }} {{eqn | o = +       | r = r_m       | c =  }} {{eqn | r = r_m \sum_{j \mathop = 0}^{m - 1} p^j + r_{m - 1} \sum_{j \mathop = 0}^{m - 2} p^j + \cdots + r_2 \sum_{j \mathop = 0}^1 p^j + r_1 \sum_{j \mathop = 0}^0 p^j       | c =  }} {{eqn | r = \sum_{k \mathop = 1}^m \paren {r_k \sum_{j \mathop = 0}^{k - 1} p^j}       | c =  }} {{eqn | r = \sum_{k \mathop = 1}^m \paren {r_k \frac {p^k - 1} {p - 1} }       | c = from Sum of Geometric Sequence }} {{eqn | r = \frac 1 {p - 1} \paren {\sum_{k \mathop = 1}^m r_k p^k - \sum_{k \mathop = 1}^m r_k}       | c =  }} {{end-eqn}} Hence the result. {{qed}}
Let $-m = r + k n$. Then $-m + n = r + \paren {k + 1} n$ and the result follows directly by definition.
From Complex Numbers form Field, $\C$ forms a field. By definition, a field is a ring. Thus it is possible to use the Subring Test. We note that $\Z \sqbrk i$ is not empty, as (for example) $0 + 0 i \in \Z \sqbrk i$. Let $a + b i, c + d i \in \Z \sqbrk i$. Then we have $-\paren {c + d i} = -c - d i$, and so: {{begin-eqn}} {{eqn | l = \paren {a + b i} + \paren {-\paren {c + d i} }       | r = \paren {a + b i} + \paren {- c - d i}       | c =  }} {{eqn | r = \paren {a + \paren {-c} } + \paren {b + \paren {-d} } i       | c =  }} {{eqn | r = \paren {a - c} + \paren {b - d} i       | c =  }} {{end-eqn}} We have that $a, b, c, d \in \Z$ and $\Z$ is an integral domain, therefore by definition a ring. So it follows that $a - c \in \Z$ and $b - d \in \Z$, and hence: :$\paren {a - c} + \paren {b - d} i \in \Z \sqbrk i$ Now consider $\paren {a + b i} \paren {c + d i}$. By the definition of complex multiplication, we have: :$\paren {a + b i} \paren {c + d i} = \paren {a c - b d} + \paren {a d + b c} i$ As $a, b, c, d \in \Z$ and $\Z$ is a ring, it follows that: :$a c - b d \in \Z$ and $ad + bc \in \Z$ So: :$\paren {a + b i} \paren {c + d i} \in \Z \sqbrk i$ So by the Subring Test, $\Z \sqbrk i$ is a subring of $\C$. {{qed}}
From Complex Numbers form Field, $\C$ forms a field. Thus it is possible to use the Subfield Test. $\Q \left[{i}\right]$ is not empty, as (for example) $0 + 0 i \in \Q \left[{i}\right]$. Let $a + b i, c + d i \in \Q \left[{i}\right]$. Then we have $- \left({c + d i}\right) = -c - d i$, and so: {{begin-eqn}} {{eqn | l = \left({a + b i}\right) + \left({- \left({c + d i}\right)}\right)       | r = \left({a + b i}\right) + \left({- c - d i}\right)       | c =  }} {{eqn | r = \left({a + \left({-c}\right)}\right) + \left({b + \left({-d}\right)}\right) i       | c =  }} {{eqn | r = \left({a - c}\right) + \left({b - d}\right) i       | c =  }} {{end-eqn}} As $a, b, c, d \in \Q$ and $\Q$ is a field, it follows that $a - c \in \Q$ and $b - d \in \Q$, and hence $\left({a - c}\right) + \left({b - d}\right)i \in \Q \left[{i}\right]$. Now consider $\left({a + b i}\right) \left({c + d i}\right)$. By the definition of complex multiplication, we have: :$\left({a + b i}\right) \left({c + d i}\right) = \left({a c - b d}\right) + \left({ad + bc}\right) i$ As $a, b, c, d \in \Q$ and $\Q$ is a field, it follows that $a c - b d \in \Q$ and $ad + bc \in \Q$ and so $\left({a + b i}\right) \left({c + d i}\right) \in \Q \left[{i}\right]$. Finally, let $z = x + y i \in \Q \left[{i}\right]_{\ne 0} = \Q \left[{i}\right] - \left\{{0 + 0 i}\right\}$, i.e. such that $x + y i \ne 0 + 0 i$. Then, by Inverse for Complex Multiplication: : $\dfrac 1 {x + y i} = \dfrac {x - y i} {x^2 + y^2}$ As $x$ and $y$ are not both zero, it follows that: : $x^2 + y^2 \ne 0$ and so $x^2 + y^2 \in \Q_{\ne 0}$ Thus it follows that either $\dfrac x {x^2 + y^2} \in \Q_{\ne 0}$ or $\dfrac y {x^2 + y^2} \in \Q_{\ne 0}$ (or both, or course). Thus $\dfrac 1 {x + y i} \in \Q \left[{i}\right]_{\ne 0}$. So by the Subfield Test, $\Q \left[{i}\right]$ is a subfield of $\C$. {{qed}} Category:Subfields Category:Complex Numbers pcz00meqp8cgi9714kosfve246lzw2e
Let $z_1 = x_1 + i y_1, z_2 = x_2 + i y_2$. Let us define the mapping $\phi: \C \to \C$ defined as: :$\forall z \in \C: \map \phi z = \overline z$ We check that $\phi$ has the morphism property: By Sum of Complex Conjugates: :$\map \phi {z_1 + z_2} = \map \phi {z_1} + \map \phi {z_2}$ By Product of Complex Conjugates: :$\map \phi {z_1 z_2} = \map \phi {z_1} \map \phi {z_2}$ So the morphism property holds for both complex addition and complex multiplication. Hence we can say that complex conjugation is a field homomorphism. We note that $\overline z_1 = \overline z_2 \implies z_1 = z_2$ and so complex conjugation is injective. Also, complex conjugation is trivially surjective, and hence bijective. The result then follows by definition of field automorphism. {{qed}}
Substituting $y = x + \dfrac {a_{n-1}} {n a_n}$ gives us $x = y - \dfrac {a_{n-1}} {n a_n}$. By the Binomial Theorem: :$a_n x^n = a_n \left({y^n - \dfrac {a_{n-1}} {a_n} y^{n-1} + P'_{n-2} \left({y}\right)}\right)$ where $P'_{n-2} \left({y}\right)$ is a polynomial in $y$ of order $n-2$. Now we note that: :$a_{n-1} x^{n-1} = a_{n-1} y^{n-1} - P''_{n-2} \left({y}\right)$ where $P''_{n-2} \left({y}\right)$ is another polynomial in $y$ of order $n-2$. The terms in $y^{n-1}$ cancel out. Hence the result. {{qed}} {{namedfor|Ehrenfried Walther von Tschirnhaus|cat=Tschirnhaus}} Category:Polynomial Theory p50hjazki1ntfmx77c73g8r0gooqpor
Let $A \in \mathcal R$. Suppose $A_1 \subseteq A$. Let $A_2 = A \setminus A_1$, where $A \setminus A_1$ denotes set difference. By definition, $A_2$ is then the relative complement of $A_1$ with respect to $A$. From Union with Relative Complement it then follows that $A_1 \cup A_2 = A$. By Ring of Sets Closed under Various Operations: :$A \setminus A_1 = A_2 \in \mathcal R$ But by Set Difference Intersection with Second Set is Empty Set: :$\left({A \setminus A_1}\right) \cap A_1 = \varnothing$ Hence for any given $A$ for which we have $A_1 \subseteq A$, we can represent $A$ as the finite expansion $A_1 \cup A_2$ such that $A_1 \cap A_2 = \varnothing$ and $A_1, A_2 \in \mathcal R$ Thus, by definition, $\mathcal R$ is a semiring of sets. {{qed}} Category:Rings of Sets Category:Semirings of Sets j2ghf6dq2wvl6428llxgjrv3p0plqpk
First we need to show that $\mathcal L$ is a ring of sets. Let $A, B \in \mathcal L$. Then by definition of $\mathcal L$, they have expansions: {{begin-eqn}} {{eqn | l = A       | r = \bigcup_{i \mathop = 1}^m A_i       | c = where $A_i \in \mathcal S$ }} {{eqn | l = B       | r = \bigcup_{j \mathop = 1}^n B_j       | c = where $B_j \in \mathcal S$ }} {{end-eqn}} Since $\mathcal S$ is a semiring of sets, we have: :$C_{ij} = A_i \cap B_j \in \mathcal S$ By Pairwise Disjoint Subsets in Semiring Part of Partition, there exist finite expansions: {{begin-eqn}} {{eqn | l = A_i       | r = \left({\bigcup_{j \mathop = 1}^n C_{ij} }\right) \cup \left({\bigcup_{k \mathop = 1}^{r_i} D_{ik} }\right)       | c = where $D_{ik} \in \mathcal S$ }} {{eqn | l = B_j       | r = \left({\bigcup_{i \mathop = 1}^m C_{ij} }\right) \cup \left({\bigcup_{l \mathop = 1}^{s_j} E_{jl} }\right)       | c = where $E_{jl} \in \mathcal S$ }} {{end-eqn}} From these, it follows that $A \cap B$ and $A \ast B$ have the finite expansions: {{begin-eqn}} {{eqn | l=A \cap B       | r=\bigcup_{i, \ j} C_{ij}       | c= }} {{eqn | l=A \ast B       | r=\left({\bigcup_{i, \ k} D_{ik} }\right) \cup \left({\bigcup_{j, \ l} E_{jl} }\right)       | c= }} {{end-eqn}} Hence both $A \cap B \in \mathcal L$ and $A \ast B \in \mathcal L$. So by definition, $\mathcal L$ is a ring of sets. From the details of the above construction, the fact that $\mathcal L$ is the '''minimal''' ring generated by $\mathcal S$ follows immediately. {{qed}} Category:Rings of Sets Category:Semirings of Sets 3ch44wug6e5midtrxr1pnypng9304vw
Proof by induction: For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the proposition: :$\displaystyle \bigcap_{j \mathop = 1}^n A_j \in \mathcal R$ $P(1)$ is true, as this just says $A_1 \in \mathcal R$. === Basis for the Induction === $P(2)$ is the case: : $A_1 \cap A_2 \in \mathcal R$ which is immediate, from definition 1 of ring of sets. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true. So this is our induction hypothesis: :$\displaystyle \bigcap_{j \mathop = 1}^k A_j \in \mathcal R$ Then we need to show: :$\displaystyle \bigcap_{j \mathop = 1}^{k+1} A_j \in \mathcal R$ === Induction Step === This is our induction step: We have that: :$\displaystyle \bigcap_{j \mathop = 1}^{k+1} A_j = \bigcap_{j \mathop = 1}^k A_j \cap A_{k+1}$ But from the induction hypothesis we have that: :$\displaystyle \bigcap_{j \mathop = 1}^k A_j \in \mathcal R$ Hence from the basis for the induction, it follows that: :$\displaystyle \bigcap_{j \mathop = 1}^{k+1} A_j \in \mathcal R$ So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \bigcap_{j \mathop = 1}^n A_j \in \mathcal R$
Let $\mathcal S$ be a $\sigma$-algebra whose unit is $\mathbb U$. Let $A_1, A_2, \ldots$ be a countably infinite collection of elements of $\mathcal S$. Then: {{begin-eqn}} {{eqn | l = \forall i: \mathbb U \setminus A_i       | o = \in       | r = \mathcal S       | c = $\mathcal S$ is closed under relative complement with $\mathbb U$ }} {{eqn | ll= \implies       | l = \bigcup_{i \mathop = 1}^\infty \left({\mathbb U \setminus A_i}\right)       | o = \in       | r = \mathcal S       | c = $\mathcal S$ is closed under countable unions }} {{eqn | ll= \implies       | l = \mathbb U \setminus \bigcap_{i \mathop = 1}^\infty A_i       | o = \in       | r = \mathcal S       | c = De Morgan's Laws }} {{eqn | ll= \implies       | l = \bigcap_{i \mathop = 1}^\infty A_i       | o = \in       | r = \mathcal S       | c = $\mathcal S$ is closed under relative complement with $\mathbb U$ }} {{end-eqn}} Thus $\mathcal S$ is a $\delta$-algebra. {{qed}}
By definition, a probability measure is a measure. Hence, again by definition, it is a countably additive function. By Measure is Finitely Additive Function, we have that $\Pr$ is an additive function. So Additive Function is Strongly Additive can be applied directly. {{qed}}
From Set Difference and Intersection form Partition: :$A$ is the union of the two disjoint sets $A \setminus B$ and $A \cap B$ :$B$ is the union of the two disjoint sets $B \setminus A$ and $A \cap B$. So, by the definition of probability measure: :$\map \Pr A = \map \Pr {A \setminus B} + \map \Pr {A \cap B}$ :$\map \Pr B = \map \Pr {B \setminus A} + \map \Pr {A \cap B}$ From Set Difference is Disjoint with Reverse: :$\paren {A \setminus B} \cap \paren {B \setminus A} = \O$ Hence: {{begin-eqn}} {{eqn | l = \map \Pr A + \map \Pr B       | r = \map \Pr {A \setminus B} + 2 \map \Pr {A \cap B} + \map \Pr {B \setminus A}       | c =  }} {{eqn | r = \map \Pr {\paren {A \setminus B} \cup \paren {A \cap B} \cup \paren {B \setminus A} } + \map \Pr {A \cap B}       | c = Set Difference and Intersection form Partition: Corollary 1 }} {{eqn | r = \map \Pr {A \cup B} + \map \Pr {A \cap B}       | c =  }} {{end-eqn}} Hence the result. {{qed}}
We have that $\mathcal P \left({S}\right)$ is an algebra of sets. {{ProofWanted}} Category:Set Systems qv4yjug4795py7qpwx2pln7iu85zdbq
Proof by induction: In the below, we assume that $A_1, A_2, \ldots$ are all pairwise disjoint elements of $\mathcal A$. For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the proposition: :$\displaystyle f \left({\bigcup_{i \mathop = 1}^n A_i}\right) = \sum_{i \mathop = 1}^n f \left({A_i}\right)$ $P(1)$ is true, as this just says: : $f \left({A_1}\right) = f \left({A_1}\right)$ === Basis for the Induction === $P(2)$ is the case: : $f \left({A_1 \cup A_2}\right) = f \left({A_1}\right) + f \left({A_2}\right)$ which comes from the definition of an additive function. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true. So this is our induction hypothesis: :$\displaystyle f \left({\bigcup_{i \mathop = 1}^k A_i}\right) = \sum_{i \mathop = 1}^k f \left({A_i}\right)$ Then we need to show: :$\displaystyle f \left({\bigcup_{i \mathop = 1}^{k+1} A_i}\right) = \sum_{i \mathop = 1}^{k+1} f \left({A_i}\right)$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = f \left({\bigcup_{i \mathop = 1}^{k+1} A_i}\right)       | r = f \left({\bigcup_{i \mathop = 1}^k A_i \cup A_{k+1} }\right)       | c =  }} {{eqn | r = f \left({\bigcup_{i \mathop = 1}^k A_i}\right) + f \left({A_{k+1} }\right)       | c = as $A_{k+1}$ and $\displaystyle \bigcup_{i \mathop = 1}^k A_i$ are disjoint }} {{eqn | r = \sum_{i \mathop = 1}^k f \left({A_i}\right) + f \left({A_{k+1} }\right)       | c = from the induction hypothesis }} {{eqn | r = \sum_{i \mathop = 1}^{k+1} f \left({A_i}\right)       | c =  }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \N_{>0}: \left({\bigcup_{i \mathop = 1}^n A_i}\right) = \sum_{i \mathop = 1}^n f \left({A_i}\right)$ {{qed}} Category:Set Systems bb6ocatojajbbcmcsu0zkeao1ldmdvq
Follows as a corollary of Countably Additive Function also Finitely Additive. {{qed}}