hcju/proofwikiqa huggingface dataset:
First we establish the following: {{begin-eqn}} {{eqn | l = \paren {r - \frac 1 2}^{\underline k}       | r = \paren {r - \frac 1 2} \paren {r - \frac 3 2} \paren {r - \frac 5 2} \dotsm \paren {r - \frac 1 2 - k + 1}       | c = {{Defof|Falling Factorial}} }} {{eqn | r = \dfrac {2^k \paren {r - \frac 1 2} \paren {r - \frac 3 2} \paren {r - \frac 5 2} \cdots \paren {r - \frac 1 2 - k + 1} } {2^k}       | c =  }} {{eqn | r = \dfrac {\paren {2 r - 1} \paren {2 r - 3} \paren {2 r - 5} \dotsm \paren {2 r - 2 k + 1} } {2^k}       | c =  }} {{eqn | r = \dfrac {2 r \paren {2 r - 1} \paren {2 r - 2} \paren {2 r - 3} \dotsm \paren {2 r - 2 k + 2} \paren {2 r - 2 k + 1} } {2^k \paren {2 r \paren {2 r - 2} \paren {2 r - 4} \dotsm \paren {2 r - 2 k + 2} } }       | c =  }} {{eqn | n = 1       | r = \dfrac {\paren {2 r}^{\underline{2 k} } } {2^k \times 2^k r^{\underline k} }       | c = {{Defof|Falling Factorial}} }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \dbinom r k \dbinom {r - \frac 1 2} k       | r = \dfrac {r^{\underline k} } {k!} \dfrac {\paren {r - \frac 1 2}^{\underline k} } {k!}       | c = {{Defof|Binomial Coefficient|subdef = Real Numbers}} }} {{eqn | r = \dfrac {r^{\underline k} } {k!} \dfrac {\paren {2 r}^{\underline {2 k} } } {k! \, 2^{2 k} r^{\underline k} }       | c = from $(1)$ }} {{eqn | r = \dfrac {\paren {2 r}^{\underline {2 k} } } {\paren {k!}^2 \, 4^k}       | c = simplifying }} {{eqn | r = \dfrac {\paren {2 r}^{\underline k} \paren {2 r - k}^{\underline k} } {\paren {k!}^2 \, 4^k}       | c = Falling Factorial of Sum of Integers }} {{eqn | r = \dfrac 1 {4^k} \dfrac {\paren {2 r}^{\underline k} } {k!} \dfrac {\paren {2 r - k}^{\underline k} } {k!}       | c = separating out }} {{eqn | r = \dfrac 1 {4^k} \dbinom {2 r} k \dbinom {2 r - k} k       | c = {{Defof|Binomial Coefficient|subdef = Real Numbers}} }} {{end-eqn}}
We have: {{begin-eqn}} {{eqn | l = \paren {1 - x^2}       | r = \paren {1 - x} \paren {1 + x}       | c = Difference of Two Squares }} {{eqn | ll= \leadsto       | l = \paren {1 - x}^r       | r = \dfrac {\paren {1 - x^2}^r} {\paren {1 + x}^r}       | c =  }} {{eqn | r = \paren {1 - x^2}^r \paren {1 + x}^{-r}       | c =  }} {{end-eqn}} Thus we have: {{begin-eqn}} {{eqn | l = \sum_{m \mathop \in \mathop \Z} \binom r m x^m       | r = \paren {1 - x}^r       | c = Binomial Theorem }} {{eqn | r = \paren {1 - x^2}^r \paren {1 + x}^{-r}       | c =  }} {{eqn | r = \sum_{k \mathop \in \mathop \Z} \binom r k \paren {-x^2}^k \sum_{j \mathop \in \mathop \Z} \binom {-r} j x^j       | c =  }} {{end-eqn}} Comparing the coefficients of $x^m$ on both sides yields the result. {{qed}}
$\sigma_\FF$ is a $\sigma$-algebra containing $\EE$. $\map \sigma \EE$ is a subset of ''all'' $\sigma$-algebras containing $\FF$, by definition of a generated $\sigma$-algebra. Therefore it contains $\map \sigma \EE$. {{qed}}
Apply Sigma-Algebra Contains Generated Sigma-Algebra of Subset to  $\sigma\left({\mathcal F}\right)$. {{qed}}
As $x + y = 0$, we can substitute $y = -x$, and so: :$\displaystyle \sum_k \binom n k x \paren {x - k z}^{k - 1} \paren {-x + k z}^{n - k} = 0$ is to be proved. So: {{begin-eqn}} {{eqn | l = \sum_k \binom n k x \paren {x - k z}^{k - 1} \paren {-x + k z}^{n - k}       | r = \sum_k \binom n k x \paren {x - k z}^{k - 1} \paren {-1}^{n - k} \paren {x - k z}^{n - k}       | c =  }} {{eqn | r = \sum_k \binom n k \paren {-1}^{n - k} x \paren {x - k z}^{n - 1}       | c =  }} {{end-eqn}} Then Sum over $k$ of $\dbinom r k$ by $\paren {-1}^{r - k}$ by Polynomial can be applied: :$\displaystyle \sum_k \binom r k \paren {-1}^{r - k} \map {P_r} k = r! \, b_r$ where: :$\map {P_r} k = b_0 + b_1 k + \cdots + b_r k^r$ is a polynomial in $k$ of degree $r$. As the coefficient of $k^n = 0$, we have: :$\displaystyle \sum_k \binom n k \paren {-1}^{n - k} \paren {b_0 + b_1 k + \cdots + b_{n - 1} k^n + b_n 0} = 0$ Hence the result. {{qed}}
By admitting $y = \left({x + y}\right) - x$, we have that: :$\left({x + y}\right)^n = \left({x + \left({x + y}\right) - x}\right)^n$ Expanding the {{RHS}} in powers of $\left({x + y}\right)$: {{begin-eqn}} {{eqn | o =        | r = \sum_k \binom n k x \left({x - k z}\right)^{k - 1} \left({y + k z}\right)^{n - k}       | c =  }} {{eqn | r = \sum_k \binom n k x \left({x - k z}\right)^{k - 1} \left({x + \left({x + y}\right) + k z}\right)^{n - k}       | c =  }} {{eqn | r = \sum_k \binom n k x \left({x - k z}\right)^{k - 1} \sum_j \left({x + y}\right)^j \left({-x + k z}\right)^{n - k - j} \binom {n - k} j       | c =  }} {{eqn | r = \sum_j \binom n j \left({x + y}\right)^j \sum_k \binom {n - j} {n - j - k} x \left({x - k z}\right)^{k - 1} \left({-x + k z}\right)^{n - k - j}       | c =  }} {{eqn | r = \sum_{j \mathop \le n} \binom n j \left({x + y}\right)^j 0^{n - j}       | c = Abel's Generalisation of Binomial Theorem: Special Case $x + y = 0$ }} {{eqn | r = \left({x + y}\right)^n       | c = Binomial Theorem }} {{end-eqn}} {{qed}}
By the axiom of powers, there exists the power set $\powerset \N$, where $\N$ is the set of natural numbers. By Power Set of Natural Numbers is not Countable, this set is uncountable. By the well-ordering theorem, $\powerset \N$ can be endowed with a well-ordering. Denote such an ordering with the symbol $\preccurlyeq$. Let $\powerset \N_a$ denote the initial segments of $\powerset \N$ determined by $a \in \powerset \N$   Suppose $\left({\powerset \N, \preccurlyeq}\right)$ has the property: :''$\powerset \N_a$ is countable for every $a \in \powerset \N$'' Then set $\Omega = \powerset \N$. Otherwise, suppose $\left({\powerset \N, \preccurlyeq}\right)$ does not have the above property. Consider the subset of $\powerset \N$: :$P \subseteq \set {a \in \powerset \N : \powerset \N_a \text{ is uncountable} }$ Then $P$ has a smallest element, by the definition of a well-ordered set. Call such an element $a_0$. That is, $a_0 \in \powerset \N$ is the smallest $a$ such that $\powerset \N_{a_0}$ is uncountable. Then the segment $\powerset \N_{a_0}$ is itself uncountable, by virtue of $a_0$ being in $P$. Thus every initial segment in $\powerset \N_{a_0}$ is countable, because it is not uncountable. Then set $\Omega = \powerset \N_{a_0}$ {{qed}}
By the definition of relation, any relation $R$ is associated with a subset $R \subseteq S \times T$. Thus $\mathcal R$ is a subset of the power set $\mathcal P\left({S \times T}\right)$. The result follows from Subset Relation is Ordering. {{qed}}
By the definition of mapping, a mapping is a specific type of relation. The result then follows from Set of Relations can be Ordered by Inclusion. {{qed}} Category:Order Theory 2r44v85fi7xt28q5hb5czn37g7yssbi
From Hurwitz's Generalisation of Binomial Theorem: :$(1): \quad \left({x + y}\right)^n = \displaystyle \sum x \left({x + \epsilon_1 z_1 + \cdots + \epsilon_n z_n}\right)^{\epsilon_1 + \cdots + \epsilon_n - 1} \left({y - \epsilon_1 z_1 - \cdots - \epsilon_n z_n}\right)^{n - \epsilon_1 - \cdots - \epsilon_n}$ Setting $z = z_1 = z_2 = \cdots z_n$ we have: {{begin-eqn}} {{eqn | o =        | r = \sum x \left({x + \epsilon_1 z + \cdots + \epsilon_n z}\right)^{\epsilon_1 + \cdots + \epsilon_n - 1} \left({y - \epsilon_1 z - \cdots - \epsilon_n z}\right)^{n - \epsilon_1 - \cdots - \epsilon_n}       | c =  }} {{eqn | r = \sum \binom n k x \left({x + k z}\right)^{k - 1} \left({y - k z}\right)^{n - k}       | c = where $\epsilon_1 + \cdots + \epsilon_n = k$ }} {{eqn | r = \sum \binom n {n - k} x \left({x + k z}\right)^{k - 1} \left({y - k z}\right)^{n - k}       | c = Symmetry Rule for Binomial Coefficients }} {{eqn | r = \sum \binom n k x \left({x - k z}\right)^{k - 1} \left({y + k z}\right)^{n - k}       | c =  }} {{end-eqn}} Hence the result.
Putting $n = x = -1$ and $y = z = 1$ into the {{LHS}} :$\left({-1 + 1}\right)^{-1} = \dfrac 1 0$ which is undefined. Putting the same values into the {{RHS}} gives: {{begin-eqn}} {{eqn | o =        | r = \sum_k \dbinom {-1} k \left({-1}\right) \left({-1 - k}\right)^{k - 1} \left({1 + k}\right)^{-1 - k}       | c =  }} {{eqn | r = \sum_k \left({-1}\right)^k \dbinom {-1} k \left({1 + k}\right)^{-2}       | c =  }} {{eqn | r = \sum_{k \mathop \ge 0} \left({1 + k}\right)^{-2}       | c =  }} {{eqn | r = \sum_{k \mathop \ge 1} \dfrac 1 {k^2}       | c = Translation of Index Variable of Summation: Corollary }} {{eqn | r = \dfrac {\pi^2} 6       | c = Basel Problem }} {{end-eqn}} {{qed}}
We have from Sum over $k$ of $\dbinom r k$ by $\dbinom {s + k} n$ by $\left({-1}\right)^{r - k}$: Corollary: {{:Sum over k of r Choose k by s+k Choose n by -1^r-k/Corollary}} By definition of matrix multiplication, this is the element $a_{r n}$ of the matrix formed by multiplying the two matrices above. As can be seen, this results in the identity matrix . {{qed}}
{{begin-eqn}} {{eqn | l = \dbinom {\frac 1 2} k       | r = \dfrac {1/2} {1/2 - k} \dbinom {1/2 - 1} k       | c = Factors of Binomial Coefficient: Corollary 1 }} {{eqn | r = \dfrac 1 {1 - 2 k} \dbinom {-1/2} k       | c =  }} {{eqn | r = \dfrac 1 {1 - 2 k} \dfrac {\left({-1}\right)^k} {4^k} \dbinom {2 k} k       | c = Binomial Coefficient of Minus Half }} {{eqn | r = \dfrac {\left({-1}\right)^{k - 1} } {4^k \left({2 k - 1}\right)} \dbinom {2 k} k       | c = simplifying }} {{end-eqn}} {{qed}}
When $k > 0$ we have: {{begin-eqn}} {{eqn | l = \dfrac {\left({-1}\right)^{k - 1} } {4^k \left({2 k - 1}\right)} \dbinom {2 k} k       | r = \dfrac {\left({-1}\right)^{k - 1} } {4^k \left({2 k - 1}\right)} \dfrac {2 k} {2 k - k} \dbinom {2 k - 1} k       | c = Factors of Binomial Coefficient: Corollary 1 }} {{eqn | r = \dfrac {\left({-1}\right)^{k - 1} } {4^k \left({2 k - 1}\right)} \dfrac {2 k} k \dbinom {2 k - 1} k       | c =  }} {{eqn | r = \dfrac {2 \left({-1}\right)^{k - 1} } {4^k \left({2 k - 1}\right)} \dbinom {2 k - 1} k       | c =  }} {{end-eqn}} When $k = 0$ we have: {{begin-eqn}} {{eqn | l = \dfrac {\left({-1}\right)^{k - 1} } {4^k \left({2 k - 1}\right)} \dbinom {2 k} k       | r = \dfrac {\left({-1}\right)^{-1} } {\left({-1}\right)} \dbinom 0 0       | c =  }} {{eqn | r = 1       | c =  }} {{end-eqn}} while: {{begin-eqn}} {{eqn | l = \dfrac {\left({-1}\right)^{k - 1} } {2^{2 k - 1} \left({2 k - 1}\right)} \dbinom {2 k - 1} k       | r = \dfrac {\left({-1}\right)^{-1} } {2^{-1} \left({-1}\right)} \dbinom {-1} 0       | c =  }} {{eqn | r = \dfrac 1 {2^{-1} } \dbinom {-1} 0       | c =  }} {{eqn | r = 2 \dbinom {-1} 0       | c =  }} {{eqn | r = 2 \left({-1}\right)^0 \dbinom {0 - \left({-1}\right) - 1} 0       | c = Negated Upper Index of Binomial Coefficient }} {{eqn | r = 2 \dbinom 0 0       | c =  }} {{eqn | r = 2       | c =  }} {{end-eqn}} Hence: :$(1): \quad \dbinom {1/2} k = \dfrac {2 \left({-1}\right)^{k - 1} } {4^k \left({2 k - 1}\right)} \dbinom {2 k - 1} k - \delta_{k 0}$ where $\delta_{k 0}$ denotes the Kronecker delta. {{qed}}
First note that from Relation between Signed and Unsigned Stirling Numbers of the First Kind: :$\displaystyle \left[{n \atop m}\right] = \left({-1}\right)^{n + m} s \left({n, m}\right)$ From First Inversion Formula for Stirling Numbers: :$\displaystyle \sum_k \left[{n \atop k}\right] \left\{ {k \atop m}\right\} \left({-1}\right)^{n - k} = \delta_{m n}$ From Second Inversion Formula for Stirling Numbers: :$\displaystyle \sum_k \left\{ {n \atop k}\right\} \left[{k \atop m}\right] \left({-1}\right)^{n - k} = \delta_{m n}$ By definition of matrix multiplication, the element $a_{r n}$ of the matrix formed by multiplying the two matrices above. As can be seen, this results in the identity matrix . {{qed}}
We assume $S \ne \O \ne T$; otherwise the theorem holds vacuously.  Define: :$S' = S \cup \text{ initial segments in } S$ :$T' = T \cup \text{ initial segments in } T$ :$\FF = \set {f: S' \to T' \mid f \text{ is an order isomorphism} }$ We note that $\FF$ is not empty, because it at least contains a trivial order isomorphism between singletons: :$f_0: \set {\text{smallest element in } S} \to \set {\text{smallest element in } T}$ Such smallest elements are guaranteed to exist by virtue of $S$ and $T$ being well-ordered. By the definition of initial segment, the initial segments of $S$ are subsets of $S$. For any initial segment $I_\alpha$ of $S$, such a segment has an upper bound by definition, namely, $\alpha$. Also, no initial segment of $S'$ is the entirety of $S$. Thus every initial segment of $S'$ has an upper bound, $S$ itself. Every chain in $S'$ has an upper bound, because it defines an initial segment.   The previous reasoning also applies to $T'$. By Ordered Product of Tosets is Totally Ordered Set, $S' \times T'$ is itself a totally ordered set, with upper bound $S \times T$. Thus the hypotheses of Zorn's Lemma are satisfied for $\FF$. Let $f_1$ be a maximal element of $\FF$. Call its domain $A$ and its codomain $B$. Suppose $A$ is an initial segment $I_a$ in $S$ and $B$ is an initial segment $I_b$ in $T$. Then $f_1$ can be extended by defining $\map {f_1} a = b$. This would contradict $f_1$ being maximal, so it cannot be the case that both $A$ and $B$ are initial segments. Then precisely one of the following hold: :$A = S$, with $S$ order isomorphic to an initial segment in $T$ or: :$B = T$, with $T$ order isomorphic to an initial segment in $S$ or: :both $A = S$ and $B = T$, with $S$ order isomorphic to $T$. The cases are distinct by No Isomorphism from Woset to Initial Segment. {{qed}}
From Wosets are Isomorphic to Each Other or Initial Segments, precisely one of the following holds: * $\Omega$ is order isomorphic to $\Omega'$ or: * $\Omega$ is order isomorphic to an initial segment in $\Omega'$ or: * $\Omega'$ is order isomorphic to an initial segment in $\Omega$. By the definition of minimal uncountable well-ordered sets, $\Omega$ and $\Omega'$ are uncountable. The initial segments of $\Omega$ and $\Omega'$ are countable. An uncountable set can't be isomorphic to a countable set. {{MissingLinks|the above claim needs proof. The theorem might or mightnot exist already on the wiki}} Thus $\Omega$ must be order isomorphic to $\Omega'$. The result follows from the definition of uniqueness. {{qed}}
The proof proceeds by induction. For all $n \in \Z_{\ge 1}$, let $\map P n$ be the proposition: :$\displaystyle \prod_{k \mathop = 1}^n \paren {1 + q^{k - 1} x} = \sum_{k \mathop \in \Z} \dbinom n k_q q^{k \paren {k - 1} / 2} x^k$ === Basis for the Induction === $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = \prod_{k \mathop = 1}^1 \paren {1 + q^{k - 1} x}       | r = 1 + x       | c =  }} {{eqn | l = \sum_{k \mathop \in \Z} \dbinom 1 k_q q^{k \paren {k - 1} / 2} x^k       | r = \dbinom 1 0_q q^{0 \paren {0 - 1} / 2} x^0 + \dbinom 1 1_q q^{1 \paren {1 - 1} / 2} x^1       | c = Gaussian Binomial Coefficient of 1: for all other $k$, $\dbinom 1 k_q = 0$ }} {{eqn | r = q^{0 \paren {-1} / 2} + q^{1 \times 0 / 2} x^1       | c = Gaussian Binomial Coefficient of 1: $\dbinom 1 0_q = \dbinom 1 1_q = 0$ }} {{eqn | r = q^0 + q^0 x       | c =  }} {{eqn | r = 1 + x       | c =  }} {{end-eqn}} Thus $\map P 1$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true. So this is the induction hypothesis: :$\displaystyle \prod_{k \mathop = 1}^r \paren {1 + q^{k - 1} x} = \sum_{k \mathop \in \Z} \dbinom r k_q q^{k \paren {k - 1} / 2} x^k$ from which it is to be shown that: :$\displaystyle \prod_{k \mathop = 1}^{r + 1} \paren {1 + q^{k - 1} x} = \sum_{k \mathop \in \Z} \dbinom {r + 1} k_q q^{k \paren {k - 1} / 2} x^k$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \prod_{k \mathop = 1}^{r + 1} \paren {1 + q^{k - 1} x}       | r = \prod_{k \mathop = 1}^r \paren {1 + q^{k - 1} x} \times \paren {1 + q^r x}       | c =  }} {{eqn | r = \sum_{k \mathop \in \Z} \dbinom r k_q q^{k \paren {k - 1} / 2} x^k \times \paren {1 + q^r x}       | c = Induction Hypothesis }} {{eqn | r = \sum_{k \mathop \in \Z} \dbinom r k_q q^{k \paren {k - 1} / 2} x^k + q^r x \sum_{k \mathop \in \Z} \dbinom r k_q q^{k \paren {k - 1} / 2} x^k       | c =  }} {{eqn | r = \sum_{k \mathop \in \Z} \dbinom r k_q q^{k \paren {k - 1} / 2} x^k + \sum_{k \mathop \in \Z} \dbinom r k_q q^r q^{k \paren {k - 1} / 2} x^{k + 1}       | c =  }} {{eqn | r = \sum_{k \mathop \in \Z} \dbinom r k_q q^{k \paren {k - 1} / 2} x^k + \sum_{k \mathop \in \Z} \dbinom r {k - 1}_q q^r q^{\paren {k - 1} \paren {k - 2} / 2} x^k       | c = Translation of Index Variable of Summation }} {{eqn | r = \sum_{k \mathop \in \Z} \paren {\dbinom r k_q q^{k \paren {k - 1} / 2} + q^r q^{-\paren {k - 1} } \dbinom r {k - 1}_q q^{k \paren {k - 1} / 2} } x^k       | c =  }} {{eqn | r = \sum_{k \mathop \in \Z} \paren {\dbinom r k_q + q^{r + 1 - k} \dbinom r {k - 1}_q} q^{k \paren {k - 1} / 2} x^k       | c =  }} {{eqn | r = \sum_{k \mathop \in \Z} \dbinom {r + 1} k_q q^{k \paren {k - 1} / 2} x^k       | c = Addition Rule for Gaussian Binomial Coefficients }} {{end-eqn}} So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{\ge 1}: \displaystyle \prod_{k \mathop = 1}^n \paren {1 + q^{k - 1} x} = \sum_{k \mathop \in \Z} \dbinom n k_q q^{k \paren {k - 1} / 2} x^k$ {{qed}}
By definition of Gaussian binomial coefficient: {{begin-eqn}} {{eqn | l = \binom n m_q       | o = :=       | r = \prod_{j \mathop = 0}^{m - 1} \dfrac {1 - q^{n - j} } {1 - q^{j + 1} }       | c =  }} {{eqn | r = \dfrac {\paren {1 - q^n} \paren {1 - q^{n - 1} } \dotsm \paren {1 - q^{n - m + 1} } } {\paren {1 - q^m} \paren {1 - q^{m - 1} } \dotsm \paren {1 - q^1} }       | c =  }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \dbinom {n - 1} m_q + \dbinom {n - 1} {m - 1}_q q^{n - m}       | r = \prod_{j \mathop = 0}^{m - 1} \dfrac {1 - q^{n - 1 - j} } {1 - q^{j + 1} } + q^{n - m} \prod_{j \mathop = 0}^{m - 2} \dfrac {1 - q^{n - 1 - j} } {1 - q^{j + 1} }       | c = {{Defof|Gaussian Binomial Coefficient}} }} {{eqn | r = \prod_{j \mathop = 0}^{m - 2} \dfrac {1 - q^{n - 1 - j} } {1 - q^{j + 1} } \paren {\dfrac {1 - q^{n - 1 - \paren {m - 1} } } {1 - q^{\paren {m - 1} + 1} } + q^{n - m} }       | c = extracting $\displaystyle \prod_{j \mathop = 0}^{m - 2} \dfrac {1 - q^{n - 1 - j} } {1 - q^{j + 1} }$ as a factor }} {{eqn | r = \prod_{j \mathop = 0}^{m - 2} \dfrac {1 - q^{n - 1 - j} } {1 - q^{j + 1} } \paren {\dfrac {1 - q^{n - m} } {1 - q^m} + q^{n - m} }       | c = simplifying }} {{eqn | r = \prod_{j \mathop = 0}^{m - 2} \dfrac {1 - q^{n - 1 - j} } {1 - q^{j + 1} } \paren {\dfrac {1 - q^{n - m} + q^{n - m} \paren {1 - q^m} } {1 - q^m} }       | c = placing over a common denominator }} {{eqn | r = \prod_{j \mathop = 0}^{m - 2} \dfrac {1 - q^{n - 1 - j} } {1 - q^{j + 1} } \paren {\dfrac {1 - q^{n - m} + q^{n - m} - q^{n - m} q^m} {1 - q^m} }       | c =  }} {{eqn | r = \prod_{j \mathop = 0}^{m - 2} \dfrac {1 - q^{n - 1 - j} } {1 - q^{j + 1} } \paren {\dfrac {1 - q^n} {1 - q^m} }       | c =  }} {{eqn | r = \prod_{j \mathop = 0}^{m - 1} \dfrac {1 - q^{n - 1 - j} } {1 - q^{j + 1} }       | c =  }} {{eqn | r = \binom n m_q       | c = {{Defof|Gaussian Binomial Coefficient}} }} {{end-eqn}} {{qed}}
By definition of Gaussian binomial coefficient: {{begin-eqn}} {{eqn | l = \binom n m_q       | o = :=       | r = \prod_{j \mathop = 0}^{m - 1} \dfrac {1 - q^{n - j} } {1 - q^{j + 1} }       | c =  }} {{eqn | r = \dfrac {\paren {1 - q^n} \paren {1 - q^{n - 1} } \dotsm \paren {1 - q^{n - m + 1} } } {\paren {1 - q^m} \paren {1 - q^{m - 1} } \dotsm \paren {1 - q^1} }       | c =  }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \dbinom {n - 1} m_q q^m + \dbinom {n - 1} {m - 1}_q       | r = q^m \prod_{j \mathop = 0}^{m - 1} \dfrac {1 - q^{n - 1 - j} } {1 - q^{j + 1} } + \prod_{j \mathop = 0}^{m - 2} \dfrac {1 - q^{n - 1 - j} } {1 - q^{j + 1} }       | c = {{Defof|Gaussian Binomial Coefficient}} }} {{eqn | r = \prod_{j \mathop = 0}^{m - 2} \dfrac {1 - q^{n - 1 - j} } {1 - q^{j + 1} } \paren {q^m \dfrac {1 - q^{n - 1 - \paren {m - 1} } } {1 - q^{\paren {m - 1} + 1} } + 1}       | c = extracting $\displaystyle \prod_{j \mathop = 0}^{m - 2} \dfrac {1 - q^{n - 1 - j} } {1 - q^{j + 1} }$ as a factor }} {{eqn | r = \prod_{j \mathop = 0}^{m - 2} \dfrac {1 - q^{n - 1 - j} } {1 - q^{j + 1} } \paren {q^m \dfrac {1 - q^{n - m} } {1 - q^m} + 1}       | c = simplifying }} {{eqn | r = \prod_{j \mathop = 0}^{m - 2} \dfrac {1 - q^{n - 1 - j} } {1 - q^{j + 1} } \paren {\dfrac {q^m \paren {1 - q^{n - m} } + \paren {1 - q^m} } {1 - q^m} }       | c = placing over a common denominator }} {{eqn | r = \prod_{j \mathop = 0}^{m - 2} \dfrac {1 - q^{n - 1 - j} } {1 - q^{j + 1} } \paren {\dfrac {q^m - q^n + 1 - q^m} {1 - q^m} }       | c =  }} {{eqn | r = \prod_{j \mathop = 0}^{m - 2} \dfrac {1 - q^{n - 1 - j} } {1 - q^{j + 1} } \paren {\dfrac {1 - q^n} {1 - q^m} }       | c =  }} {{eqn | r = \prod_{j \mathop = 0}^{m - 1} \dfrac {1 - q^{n - 1 - j} } {1 - q^{j + 1} }       | c =  }} {{eqn | r = \binom n m_q       | c = {{Defof|Gaussian Binomial Coefficient}} }} {{end-eqn}} {{qed}}
Suppose the hypotheses of the theorem hold. If $J = X$ then the theorem is satisfied. Assume $J \ne X$. Then $X \setminus J$ is non-empty. By Subset of Well-Ordered Set is Well-Ordered, $X \setminus J$ is itself well-ordered. Thus $X \setminus J$ has a smallest element; call it $b$. We claim that $b \preccurlyeq y$ for every $y \in J$. {{AimForCont}} there exists a $y \in J$ with $b \preccurlyeq y$. Then there exists some $x_0 \in A$ with $y$ in the initial segment $S_{x_0}$. Thus $b \in S_{x_0}$ and so $b \in J$. This contradicts the fact that $b \in X \setminus J$. Thus there cannot exist a $y \in J$ with $b \preccurlyeq y$. Any element strictly preceding $b$ is in $J$, by choice of $b$. Also, $b \notin J$, because $b \in X \setminus J$. From above, any element strictly succeeding $b$ is not in $J$. Thus $J = S_b$ by definition of initial segment. {{qed}}
By the minimal uncountable well-ordered set, $\Omega$ is uncountable Then $\omega \ne \Omega$, for the former is countable and the latter is not. Consider the union: :$\displaystyle \bigcup_{x \mathop \in \omega} S_x$ of initial segments $S_x$ in $\omega$. By the definition of $\Omega$, any $S_x$ is countable. Thus $\displaystyle \bigcup_{x \mathop \in \omega} S_x$ is a countable union of countable sets. This union is also countable, and so also cannot be all of $\Omega$, as $\Omega$ is uncountable. From Union of Initial Segments is Initial Segment or All of Woset, there exists some $y \in \Omega$ such that: :$\displaystyle \bigcup_{x \mathop \in \omega} S_x = S_y$ Such a $y$ is an upper bound for $\omega$ by the definition of an initial segment. {{qed}} {{AoC|Countable Union of Countable Sets is Countable}}
By definition of Gaussian binomial coefficient: :$\dbinom 1 m_q = \displaystyle \prod_{k \mathop = 0}^{m - 1} \dfrac {1 - q^{1 - k} } {1 - q^{m + 1} }$ When $m = 0$ the product on the {{RHS}} is vacuous, and so: :$\dbinom 1 0_q = 0$ Let $m > 0$. Then: {{begin-eqn}} {{eqn | l = \dbinom 1 m_q       | r = \prod_{k \mathop = 0}^{m - 1} \dfrac {1 - q^{1 - k} } {1 - q^{m + 1} }       | c = {{Defof|Gaussian Binomial Coefficient}} }} {{eqn | r = \paren {\dfrac {1 - q^{1 - 0} } {1 - q^{0 + 1} } } \paren {\dfrac {1 - q^{1 - 1} } {1 - q^{1 + 1} } } \paren {\dfrac {1 - q^{1 - 2} } {1 - q^{2 + 1} } } \cdots \paren {\dfrac {1 - q^{1 - \paren {m - 1} } } {1 - q^{\paren {m - 1} + 1} } }       | c =  }} {{eqn | r = \paren {\dfrac {1 - q} {1 - q} } \paren {\dfrac {1 - q^0} {1 - q^2} } \paren {\dfrac {1 - q^{-1} } {1 - q^3} } \cdots \paren {\dfrac {1 - q^{2 - m} } {1 - q^m} }       | c =  }} {{end-eqn}} When $m > 0$ there exists a factor $1 - q^0 = 0$ in the numerator of the {{RHS}}. Hence when $m > 0$ we have that $\dbinom 1 m_q = 0$. We are left with: {{begin-eqn}} {{eqn | l = \dbinom 1 1_q       | r = \prod_{k \mathop = 0}^0 \dfrac {1 - q^{1 - k} } {1 - q^{m + 1} }       | c = {{Defof|Gaussian Binomial Coefficient}} }} {{eqn | r = \dfrac {1 - q^{1 - 0} } {1 - q^{0 + 1} } }} {{eqn | r = \dfrac {1 - q} {1 - q}       | c =  }} {{eqn | r = 1       | c =  }} {{end-eqn}} {{qed}} Category:Gaussian Binomial Coefficients tta8hdqi87379heldry58y396mng75k