hcju/proofwikiqa huggingface dataset:
We have by definition of vacuous summation that: :$\displaystyle \forall n \in \Z: n < 0: \sum_{i \mathop = 0}^n \binom n 1 = 0$ Then from Zero Choose Zero: :$\displaystyle \sum_{i \mathop = 0}^0 \binom 0 0 = 1$ Let $n > 0$. The assertion can be expressed: :$\displaystyle \sum_{i \mathop \le n} \left({-1}\right)^i \binom n i = 0$ for all $n > 0$ as $\dbinom n i = 0$ when $i < 0$ by definition of binomial coefficient. From Alternating Sum and Difference of r Choose k up to n we have: :$\displaystyle \sum_{i \mathop \le n} \left({-1}\right)^i \binom r i = \left({-1}\right)^n \binom {r - 1} n$ Putting $r = n$ we have: :$\displaystyle \sum_{i \mathop \le n} \left({-1}\right)^i \binom n i = \left({-1}\right)^n \binom {n - 1} n$ As $n - 1 < n$ it follows from the definition of binomial coefficient that: :$\dbinom {n - 1} n = 0$ {{qed}}
Follows directly from the definition of binomial coefficient, as follows. If $k < 0$ then $n - k > n$. Similarly, if $k > n$, then $n - k < 0$. In both cases: :$\dbinom n k = \dbinom n {n - k} = 0$ Let $0 \le k \le n$. {{begin-eqn}} {{eqn | l = \binom n k       | r = \frac {n!} {k! \paren {n - k}!}       | c =  }} {{eqn | r = \frac {n!} {\paren {n - k}! k!}       | c =  }} {{eqn | r = \frac {n!} {\paren {n - k}! \paren {n - \paren {n - k} } !}       | c =  }} {{eqn | r = \binom n {n - k}       | c =  }} {{end-eqn}} {{qed}}
From the definition of Cardinality of Set of Subsets, $\dbinom n k$ is the number of subsets of cardinality $k$ of a set with cardinality $n$. Let $S$ be a set with cardinality $n$. Let $\powerset S$ denote the power set of $S$. Let $\mathcal A_k \subseteq \powerset S$ be the set of elements of $\powerset S$ which have cardinality $k$. For every $T \in \mathcal A_k$ there exists its relative complement: :$\relcomp S T \in \mathcal A_{n - k}$ Define the mapping $f_k: \mathcal A_k \to \mathcal A_{n - k}$ as: :$\forall T \in \mathcal A_k: \map f T = \relcomp S T$ From Relative Complement Mapping on Powerset is Bijection it follows that $f_k$ is a bijection. By definition of set equivalence, that means: :$\card {A_k} = \card {A_{n - k} }$ That is, from the definition of Cardinality of Set of Subsets: :$\dbinom n k = \dbinom n {n - k}$ {{qed}}
If $k = 0$ then $\dbinom r k = r \dbinom {r - 1} {k - 1} = 0$ by definition. Otherwise: {{begin-eqn}} {{eqn | l = k \binom r k       | r = k \frac {r^{\underline k} } {k!}       | c =  }} {{eqn | r = k \frac {r \paren {r - 1} \paren {r - 2} \dotsm \paren {r - k + 1} } {k \paren {k - 1} \paren {k - 2} \dotsm 1}       | c =  }} {{eqn | r = \frac {r \paren {r - 1} \paren {r - 2} \dotsm \paren {r - k + 1} } {\paren {k - 1} \paren {k - 2} \dotsm 1}       | c =  }} {{eqn | r = r \frac {\paren {r - 1} \paren {r - 2} \dotsm \paren {\paren {r - 1} - \paren {k - 1} + 1} } {\paren {k - 1} \paren {k - 2} \dotsm 1}       | c =  }} {{eqn | r = r \binom {r - 1} {k - 1}       | c =  }} {{end-eqn}} {{qed|lemma}} If $k \ne 0$, we can divide both sides of: :$k \dbinom r k = r \dbinom {r - 1} {k - 1}$ by $k$ to obtain: :$\dbinom r k = \dfrac r k \dbinom {r - 1} {k - 1}$ {{qed|lemma}} If $k \ne 0$ and $r \ne 0$, we can divide both sides of: :$\dbinom r k = \dfrac r k \dbinom {r - 1} {k - 1}$ by $r$ to obtain: :$\dfrac 1 r \dbinom r k = \dfrac 1 k \dbinom {r - 1} {k - 1}$ {{qed}}
Proof by induction: For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition :$\displaystyle \sum_{k \mathop = 0}^n \binom {r + k} k = \binom {r + n + 1} n$ === Basis for the Induction === $\map P 0$ is true, as $\dbinom r 0 = 1 = \dbinom {r + 1} 0$ by definition. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P m$ is true, where $m \ge 0$, then it logically follows that $\map P {m + 1}$ is true. So this is our induction hypothesis: :$\displaystyle \sum_{k \mathop = 0}^m \binom {r + k} k = \binom {r + m + 1} m$ Then we need to show: :$\displaystyle \sum_{k \mathop = 0}^{m + 1} \binom {r + k} k = \binom {r + m + 2} {m + 1}$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 0}^{m + 1} \binom {r + k} k       | r = \sum_{k \mathop = 0}^m \binom {r + k} k + \binom {r + m + 1} {m + 1}       | c = }} {{eqn | r = \binom {r + m + 1} m + \binom {r + m + 1} {m + 1}       | c = from the induction hypothesis }} {{eqn | r = \binom {r + m + 2} {m + 1}       | c = Pascal's Rule }} {{end-eqn}} So $\map P m \implies \map P {m + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \Z_{\ge 0}: \sum_{k \mathop = 0}^n \binom {r + k} k = \binom {r + n + 1} n$ {{Qed}}
{{begin-eqn}} {{eqn | l = \binom r k       | r = \frac {r^{\underline k} } {k!}       | c = {{Defof|Binomial Coefficient}} }} {{eqn | r = \frac 1 {k!} \prod_{j \mathop = 0}^{k - 1} \paren {r - j}       | c = {{Defof|Falling Factorial}} }} {{eqn | r = \frac {\paren {-1}^k} {k!} \prod_{j \mathop = 0}^{k - 1} \paren {-\paren {r - j} }       | c =  }} {{eqn | r = \frac {\paren {-1}^k} {k!} \prod_{j \mathop = 0}^{k - 1} \paren {j - r}       | c =  }} {{eqn | r = \frac {\paren {-1}^k} {k!} \prod_{j \mathop = 0}^{k - 1} \paren {\paren {k - 1} - j - r}       | c = Permutation of Indices of Product }} {{eqn | r = \frac {\paren {-1}^k} {k!} \prod_{j \mathop = 0}^{k - 1} \paren {\paren {k - r - 1} - j}       | c =  }} {{eqn | r = \paren {-1}^k \frac {\paren {k - r - 1}^{\underline j} } {k!}       | c = {{Defof|Falling Factorial}} }} {{eqn | r = \paren {-1}^k \binom {k - r - 1} k       | c = {{Defof|Binomial Coefficient}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sum_{k \mathop \le n} \paren {-1}^k \binom r k       | r = \sum_{k \mathop \le n} \binom {k - r - 1} k       | c = Negated Upper Index of Binomial Coefficient }} {{eqn | r = \binom {-r + n} n       | c = Sum of r+k Choose k up to n }} {{eqn | r = \paren {-1}^n \binom {r - 1} n       | c = Negated Upper Index of Binomial Coefficient }} {{end-eqn}} {{qed}}
=== Integral Index === Let $r \in \Z$. Then: {{begin-eqn}} {{eqn | l = \binom r m \binom m k       | r = \frac {r^{\underline m} } {m!} \frac {m^{\underline k} } {k!}       | c =  }} {{eqn | r = \frac {r! m!} {m! \left({r - m}\right)! k! \left({m - k}\right)!}       | c =  }} {{eqn | r = \frac {r! \left({r - k}\right)!} {k! \left({r - k}\right)! \left({m - k}\right)! \left({r - m}\right)!}       | c =  }} {{eqn | r = \binom r k \binom {r - k} {m - k}       | c =  }} {{end-eqn}} {{qed|lemma}} === Real Index === Both sides of the above equation are polynomials in $r$. Since these polynomials agree for all $r \in \Z$, they must agree for all $r \in \R$. {{qed}}
Consider the trinomial coefficient: :$\dbinom r {k, m - k, r - m}$ We use Multinomial Coefficient expressed as Product of Binomial Coefficients: :$\dbinom {k_1 + k_2 + k_3} {k_1, k_2, k_3} = \dbinom {k_1 + k_2} {k_1} \dbinom {k_1 + k_2 + k_3} {k_1 + k_2}$ and substitute as appropriate for $k_1, k_2, k_3$. We have: {{begin-eqn}} {{eqn | r = \dbinom r {k, m - k, r - m}       | o =        | c =  }} {{eqn | r = \dbinom {\left({m - k}\right) + \left({r - m}\right) + k} {m - k, r - m, k}       | c =  }} {{eqn | r = \dbinom {\left({m - k}\right) + \left({r - m}\right)} {m - k} \dbinom {\left({m - k}\right) + \left({r - m}\right) + k} {\left({m - k}\right) + \left({r - m}\right)}       | c = Multinomial Coefficient expressed as Product of Binomial Coefficients }} {{eqn | r = \dbinom {r - k} {m - k} \dbinom r m       | c =  }} {{end-eqn}} Similarly: {{begin-eqn}} {{eqn | r = \dbinom r {k, m - k, r - m}       | o =        | c =  }} {{eqn | r = \dbinom {k + \left({m - k}\right) + \left({r - m}\right)} {k, m - k, r - m}       | c =  }} {{eqn | r = \dbinom {k + \left({m - k}\right)} k \dbinom {k + \left({m - k}\right) + \left({r - m}\right)} {\left({m - k}\right) + k}       | c = Multinomial Coefficient expressed as Product of Binomial Coefficients }} {{eqn | r = \dbinom m k \dbinom r m       | c =  }} {{end-eqn}} {{qed}} The result follows on equating the two expressions.
{{begin-eqn}} {{eqn | l = \sum_n \binom {r + s} n x^n       | r = \paren {1 + x}^{r + s}       | c = Binomial Theorem }} {{eqn | r = \paren {1 + x}^r \paren {1 + x}^s       | c = Exponent Combination Laws }} {{eqn | r = \sum_k \binom r k x^k \sum_m \binom s m x^m       | c = Binomial Theorem }} {{eqn | r = \sum_k \binom r k x^k \sum_{n - k} \binom s {n - k} x^{n - k}       | c =  }} {{eqn | r = \sum_n \paren {\sum_k \binom r k \binom  s {n - k} } x^n       | c =  }} {{end-eqn}} As this has to be true for all $x$, we have that: :$\displaystyle \binom {r + s} n = \sum_k \binom r k \binom s {n - k}$ {{qed}}
This is a special case of Gauss's Hypergeometric Theorem: :${}_2F_1 \left({a, b; c; 1}\right) = \dfrac{\Gamma \left({c}\right) \Gamma \left({c - a - b}\right)} {\Gamma \left({c - a}\right) \Gamma \left({c - b}\right)}$ where: :${}_2F_1$ is the hypergeometric series :$\Gamma \left({n + 1}\right) = n!$ is the Gamma function. One regains the Chu-Vandermonde Identity by taking $a = -n$ and applying Negated Upper Index of Binomial Coefficient: :$\dbinom n k = (-1)^k \dbinom {k - n - 1} k$ throughout. {{qed}}
From Sum over $k$ of $\dbinom {r - t k} k \dbinom {s - t \left({n - k}\right)} {n - k} \dfrac r {r - t k}$: {{:Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk}} where $r, s, t \in \R, n \in \Z$. Setting $t = 0$: :$\displaystyle \sum_{k \mathop \ge 0} \binom r k \binom s {n - k} = \binom {r + s} n$ which is the result required. {{qed}}
First we show that: :$\dbinom n k \equiv \dbinom {\floor {n / p} } {\floor {k / p} } \dbinom {n \bmod p} {k \bmod p} \pmod p$ Consider $\dbinom n k$ as the fraction: :$\dfrac {n \paren {n - 1} \paren {n - 2} \cdots \paren {n - k + 1} } {k \paren {k - 1} \paren {k - 2} \cdots 1}$ This can be expressed as: :$(1): \quad \dbinom n k = \paren {\dfrac n k} \paren {\dfrac {n - 1} {k - 1} } \paren {\dfrac {n - 2} {k - 2} } \cdots \paren {\dfrac {n - k + 1} 1}$ Let $k = s p + t$ from the Division Theorem. Thus: : $t = k \bmod p$ The denominators of the first $t$ factors in $(1)$ do not have $p$ as a divisor. Now let $n = u p + v$, again from the Division Theorem. Thus: :$v = n \bmod p$ Now, when dealing with non-multiples of $p$, we can work modulo $p$ in both the numerator and denominator, from Common Factor Cancelling in Congruence. So we consider the first $t$ factors of $(1)$ modulo $p$. These are: :$\paren {\dfrac {u p + v} {s p + t} } \paren {\dfrac {u p + v - 1} {s p + t - 1} } \cdots \paren {\dfrac {u p + v - t + 1} {s p + 1} } \equiv \paren {\dfrac v t} \paren {\dfrac {v - 1} {t - 1} } \cdots \paren {\dfrac {v - t + 1} 1} \pmod p$ So, these first $t$ terms of $(1)$ taken together are congruent modulo $p$ to the corresponding terms of: :$\dbinom {n \bmod p} {k \bmod p}$ These differ by multiples of $p$. So we are left with $k - k \bmod p$ factors. These fall into $\floor {k / p}$ groups, each of which has $p$ consecutive values. Each of these groups contains exactly one multiple of $p$. The other $p - 1$ factors in a given group are congruent (modulo $p$) to $\paren {p - 1}!$ so they cancel out in numerator and denominator. We now need to investigate the $\left \lfloor {k / p} \right \rfloor$ multiples of $p$ in the numerator and denominator. We divide each of them by $p$ and we are left with the binomial coefficient: :$\dbinom {\floor {\paren {n - k \bmod p} / p} } {\floor {k / p} }$ Now, if $k \bmod p \le n \bmod p$, this equals: :$\dbinom {\floor {n / p} } {\floor {k / p} }$ Otherwise, if $k \bmod p > n \bmod p$, the other factor: :$\dbinom {n \bmod p} {k \bmod p}$ is zero. So the formula holds in general. {{qed}} {{Namedfor|François Édouard Anatole Lucas|cat = Lucas}}
Let $x_1, x_2, z \in \R$. Let $x_1 \equiv x_2 \pmod z$ as defined by an equivalence relation. That is, let $\mathcal R_z$ be the relation on the set of all $x, y \in \R$: :$\mathcal R_z = \left\{{\left({x, y}\right) \in \R \times \R: \exists k \in \Z: x = y + k z}\right\}$ Let $\left({x_1, x_2}\right) \in \mathcal R_z$. Then by definition, $\exists k \in \Z: x_1 = x_2 + k z$. So, by definition of the modulo operation, we have: {{begin-eqn}} {{eqn | l = x_1 \mod z       | r = \left({x_2 + kz}\right) - z \left \lfloor {\frac {x_2 + kz} z}\right \rfloor       | c =  }} {{eqn | r = \left({x_2 + kz}\right) - z \left \lfloor {\frac {x_2} z + k}\right \rfloor       | c =  }} {{eqn | r = \left({x_2 + kz}\right) - z \left \lfloor {\frac {x_2} z}\right \rfloor + k z       | c =  }} {{eqn | r = x_2 - z \left \lfloor {\frac {x_2} z}\right \rfloor       | c =  }} {{eqn | r = x_2 \mod z       | c =  }} {{end-eqn}} So: :$x_1 \equiv x_2 \pmod z$ in the sense of definition by modulo operation. {{qed|lemma}} Now let $x_1 \equiv x_2 \pmod z$ in the sense of definition by modulo operation. That is, :$x_1 \equiv x_2 (\bmod\, z) \iff x_1 \mod z = x_2 \mod z$. Let $z = 0$. Then by definition, $x_1 \mod 0 = x_1$ and $x_2 \mod 0 = x_2$. So as $x_1 \mod 0 = x_2 \mod 0$ we have that $x_1 = x_2$. So $x_1 - x_2 = 0 = 0.z$ and so $x_1 \equiv x_2 \pmod z$ in the sense of definition by integer multiple. Now suppose $z \ne 0$. Then from definition of the modulo operation: :$x_1 \mod z = x_1 - z \left \lfloor {\dfrac {x_1} z}\right \rfloor$ :$x_2 \mod z = x_2 - z \left \lfloor {\dfrac {x_2} z}\right \rfloor$ Thus: :$x_1 - z \left \lfloor {\dfrac {x_1} z}\right \rfloor = x_2 - z \left \lfloor {\dfrac {x_2} z}\right \rfloor$ and so: : $x_1 - x_2 = z \left({\left \lfloor {\dfrac {x_1} z}\right \rfloor - \left \lfloor {\dfrac {x_2} z}\right \rfloor}\right)$ From the definition of the floor function, we see that both $\left \lfloor {\dfrac {x_1} z}\right \rfloor$ and $\left \lfloor {\dfrac {x_2} z}\right \rfloor$ are integers. Therefore, so is $\left \lfloor {\dfrac {x_1} z}\right \rfloor - \left \lfloor {\dfrac {x_2} z}\right \rfloor$ an integer. So $\exists k \in \Z: x_1 - x_2 = k z$. Thus $x_1 - x_2 = k z$ and: :$x_1 \equiv x_2 \pmod z$ in the sense of definition by integer multiple. {{qed|lemma}} Now let $x_1 \equiv x_2 \pmod z$ in the sense of definition by integer multiple. That is, $\exists k \in \Z: x_1 - x_2 = k z$. Then $x_1 = x_2 + k z$ and so $\left({x_1, x_2}\right) \in \mathcal R_z$ where: :$\mathcal R_z = \left\{{\left({x, y}\right) \in \R \times \R: \exists k \in \Z: x = y + k z}\right\}$ and so :$x_1 \equiv x_2 \pmod z$ in the sense of definition by equivalence relation. {{qed|lemma}} So all three definitions are equivalent: $(1) \implies (2) \implies (3) \implies (1)$. {{qed}}
Checking in turn each of the critera for equivalence: === Reflexive === We have that Equal Numbers are Congruent: :$\forall x, y, z \in \R: x = y \implies x \equiv y \pmod z$ so it follows that: :$\forall x \in \R: x \equiv x \pmod z$ and so congruence modulo $z$ is reflexive. {{qed|lemma}} === Symmetric === {{begin-eqn}} {{eqn | l = x       | o = \equiv       | r = y \pmod z       | c =  }} {{eqn | ll= \leadsto       | l = x - y       | r = k z       | c =  }} {{eqn | ll= \leadsto       | l = y - x       | r = \paren {-k} z       | c =  }} {{eqn | l = y       | o = \equiv       | r = x \pmod z       | c =  }} {{end-eqn}} So congruence modulo $z$ is symmetric. {{qed|lemma}} === Transitive === {{begin-eqn}} {{eqn | l = x_1       | o = \equiv       | r = x_2 \pmod z       | c =  }} {{eqn | ll= \land       | l = x_2       | o = \equiv       | r = x_3 \pmod z       | c =  }} {{eqn | ll= \leadsto       | l = \paren {x_1 - x_2}       | r = k_1 z       | c =  }} {{eqn | ll= \land       | l = \paren {x_2 - x_3}       | r = k_2 z       | c =  }} {{eqn | ll= \leadsto       | l = \paren {x_1 - x_2} + \paren {x_2 - x_3}       | r = \paren {k_1 + k_2} z       | c =  }} {{eqn | ll= \leadsto       | l = \paren {x_1 - x_3}       | r = \paren {k_1 + k_2} z       | c =  }} {{eqn | l = x_1       | o = \equiv       | r = x_3 \pmod z       | c =  }} {{end-eqn}} So congruence modulo $z$ is transitive. {{qed|lemma}} So we are justified in supposing that congruence, as we have defined it, is an equivalence. {{qed}}
Follows directly from the definition of congruence: :$x \equiv y \pmod z \iff x \bmod z = y \bmod z$  When $z = 0$ we have by definition: :$x \bmod 0 := x$ And so $x \bmod 0 = y \bmod 0 \iff x = y$. Hence the result. {{qed}} Category:Modulo Arithmetic cekrtgnb3larl48i6x3esxn6kisvbjy
Assume first that $\FF$ is a filter on $S$. Then $S \in \FF$ and thus $\BB \ne \O$. Because $\O \notin \FF$ we have that $\O \notin \BB$, since $\BB \subseteq \FF$. Let $V_1, V_2 \in \BB$. Then: :$V_1, V_2 \in \FF$ Because $\FF$ is a filter: :$V := V_1 \cap V_2 \in \FF$ The definition of $\FF$ implies therefore that there is $U \in \BB$ such that $U \subseteq V = V_1 \cap V_2$. {{qed|lemma}} Assume now that $\BB$ satisfies conditions $(1)$ and $(2)$. To show that $\FF$ is a filter, note that because $\BB \ne \O$ there is a $B \in \BB$. Because $\BB \subset \powerset S$: :$B \subseteq S$ and thus by the definition of filter: :$S \in \FF$ Because: :the only subset of $\O$ is $\O$ and: :since $\O \notin \BB$ it follows that: :$\O \notin \FF$ Let $V_1, V_2 \in \FF$. Then there exist $U_1, U_2 \in \BB$ such that $U_1 \subseteq V_1$ and $U_2 \subseteq V_2$. Because $\BB$ satisfies condition $(1)$ there exists a set $U \in \BB$ for which $U \subseteq U_1 \cap U_2$ holds. Because $U_1 \cap U_2 \subseteq V_1 \cap V_2$ this implies that: :$U \subseteq V_1 \cap V_2$ and therefore: :$V_1 \cap V_2 \in \FF$ {{qed}}
Let $x$ be a limit point of $\mathcal F$. Define: :$\mathcal B := \left\{{F \cap U : F \in \mathcal F \text{ and } U \text{ is a neighborhood of } x}\right\}$ Then $\mathcal B$ is filter basis by definition. Let $\mathcal F'$ be the corresponding generated filter. By construction we have $\mathcal F \subseteq \mathcal F'$ and $U \in \mathcal F'$ for every neighborhood $U$ of $x$. Thus $\mathcal F'$ converges to $x$. {{qed|lemma}} Let there be a filter $\mathcal F'$ on $X$ satisfying $\mathcal F \subseteq \mathcal F'$ which converges to $x$. Let $U \subseteq X$ be a neighborhood of $x$. Let $F \in \mathcal F$. Then: :$U, F \in \mathcal F'$ and therefore: : $U \cap F \in \mathcal F'$ Because $\varnothing \notin \mathcal F'$ it follows that: : $U \cap F \ne \varnothing$ This holds for any neighborhood $U$ of $x$. Hence $x$ is a limit point of $F$. Therefore: :$x \in \overline F$ Because this holds for all $F \in \mathcal F$: :$\displaystyle x \in \bigcap \left\{{\overline F : F \in \mathcal F}\right\}$ and so $x$ is a limit point of $\mathcal F$. {{qed}} Category:Limit Points Category:Filter Theory p1pd74w2bzjj4d2cj7l840vdq6pch9w
=== $(1) \iff (2)$: Compact Space satisfies Finite Intersection Axiom === {{:Compact Space satisfies Finite Intersection Axiom}} === $(1) \iff (3)$: Alexander's Compactness Theorem === {{:Alexander's Compactness Theorem}} === $(4) \implies (5)$: Every Filter has Limit Point implies Every Ultrafilter Converges === {{:Every Filter has Limit Point implies Every Ultrafilter Converges}} === $(5) \implies (4)$: Every Ultrafilter Converges implies Every Filter has Limit Point === {{:Every Ultrafilter Converges implies Every Filter has Limit Point}} === $(2) \implies (4)$ === Let $T = \left({S, \tau}\right)$ satisfy the Finite Intersection Axiom. Let $\mathcal F$ be a filter on $X$. {{AimForCont}} that $\mathcal F$ has no limit point. Thus: : $\bigcap \left\{{\overline F : F \in \mathcal F}\right\} = \varnothing$ By hypothesis there are therefore sets $F_1, \ldots, F_n \in \mathcal F$ such that: :$\overline F_1 \cap \ldots \cap \overline F_n = \varnothing$ Because for any set $M$ we have $M \subseteq \overline M$: :$\overline F_1, \ldots, \overline F_n \in \mathcal F$ But by definition of a filter, $\mathcal F$ must not contain the empty set. Thus $\mathcal F$ has a limit point. {{qed|lemma}} === $(4) \implies (2)$ === Let $\mathcal A \subset \mathcal P \left({S}\right)$ be a set of closed subsets of $S$. Let: : $\bigcap \tilde{\mathcal A} \ne \varnothing$ for all finite subsets $\tilde{\mathcal A}$ of $\mathcal A$. We show that this implies $\bigcap \mathcal A \ne \varnothing$. From our assumption, $\mathcal B := \left\{{\bigcap \tilde{\mathcal A} : \tilde{\mathcal A} \subseteq \mathcal A \text{ finite}}\right\}$ is a filter basis. Let $\mathcal F$ be the corresponding generated filter. By hypothesis $\mathcal F$ has a limit point. Thus: : $\varnothing \ne \bigcap \left\{{\overline F : F \in \mathcal F}\right\} \subseteq \bigcap \mathcal B \subseteq \bigcap \mathcal A$. Thus $\bigcap \mathcal A \ne \varnothing$. Hence $T = \left({S, \tau}\right)$ satisfies the Finite Intersection Axiom. {{qed}} Category:Compact Spaces 3r0mx41m3ccsr8qgnwdz3du7ngh73ni
We show that $\mathcal S^* = \left\{{\bigcap S : S \subseteq \mathcal S \text{ finite}}\right\}$ (cf. the definition of the generated topology) is a basis. To see that $\mathcal S^*$ is a basis, we need to prove two things: :$(1): \quad X = \bigcup \mathcal S^*$ :$(2): \quad$ For any $U_1, U_2 \in \mathcal S^*$ and $x \in U_1 \cap U_2$ there is a $U \in \mathcal S^*$ such that $x \in U \subseteq U_1 \cap U_2$ First note that $X = \bigcap \varnothing \in \mathcal S^*$ and therefore $\bigcup \mathcal S^* = X$. Additionally, if $U_1, U_2 \in \mathcal S^*$, then there exist finite sets $S_1, S_2 \subseteq \mathcal S$ such that $U_1 = \bigcap S_1$ and $U_2 = \bigcap S_2$ by the definition of $\mathcal S^*$. Thus we have: :$U_1 \cap U_2 = \bigcap (S_1 \cup S_2)$ Because $S_1 \cup S_2$ is again a finite set it follows that $U_1 \cap U_2 \in \mathcal S^*$. This implies $(2)$. {{qed}} Category:Topology kteu3tmth5alfasdd1kri8c4sqsh8j9
=== $(1)$ implies $(2)$ === Suppose that $A, B \in \BB$. By hypothesis: :$\displaystyle \exists \AA \subseteq \BB: A \cap B = \bigcup \AA$ By Set is Subset of Union: General Result, we have: :$\displaystyle \forall W \in \AA: W \subseteq \bigcup \AA = A \cap B$ Therefore: :$\displaystyle \forall x \in A \cap B: \exists W \in \AA \subseteq \BB: x \in W \subseteq A \cap B$ {{qed|lemma}} === $(2)$ implies $(1)$ === Suppose that $A, B \in \BB$. Define the set: :$\displaystyle \AA = \set {W \in \BB: W \subseteq A \cap B} \subseteq \BB$ By Union is Smallest Superset: General Result, we have: :$\displaystyle \bigcup \AA \subseteq A \cap B$ By hypothesis, we have: :$\displaystyle \forall x \in A \cap B: \exists W \in \AA: x \in W$ Hence, by the definition of a subset and the definition of set union: :$\displaystyle A \cap B \subseteq \bigcup \AA$ Therefore, by definition of set equality: :$\displaystyle A \cap B = \bigcup \AA$ {{qed}}
Since $a$ has period $b$, we have: :$\map a {n + b} = \map a n$ So if we define: :$\displaystyle \map F z = \sum_{n \mathop \ge 0} \map a n z^n$ we have: {{begin-eqn}} {{eqn | l = \map F z)       | r = \paren {\sum_{k \mathop = 0}^{b - 1} \map a k z^k} + z^b \paren {\sum_{k \mathop = 0}^{b - 1} \map a k z^k} + z^{2 b} \paren {\sum_{k \mathop = 0}^{b - 1} \map a k z^k} + \cdots       | c =  }} {{eqn | r = \frac 1 {1 - z^b} \paren {\sum_{k \mathop = 0}^{b - 1} \map a k z^k}       | c =  }} {{eqn | r = \frac {\map P z} {1 - z^b}       | c =  }} {{end-eqn}} where the last step defines the polynomial $P$. If we expand $F$ now using partial fractions, we get {{finish}} {{Namedfor|Joseph Fourier|cat = Fourier}} Category:Fourier Analysis f86y86utrio7ji8yxcgnwoj38zhl9yp
By definition of modulo operation: :$x \bmod y := \begin{cases} x - y \left \lfloor {\dfrac x y}\right \rfloor & : y \ne 0 \\ x & : y = 0 \end{cases}$ If $y = 0$ we have that: :$z \left({x \bmod 0}\right) = z x = \left({z x}\right) \bmod \left({z 0}\right)$ If $y \ne 0$ we have that: {{begin-eqn}} {{eqn | l = z \left({x \bmod y}\right)       | r = z x - z y \left \lfloor {\frac x y}\right \rfloor       | c =  }} {{eqn | r = z x - z y \left \lfloor {\frac {z x} {z y} }\right \rfloor       | c =  }} {{eqn | r = \left({z x}\right) \bmod \left({z y}\right)       | c = {{Defof|Modulo Operation}} }} {{end-eqn}} {{qed}}