hcju/proofwikiqa huggingface dataset:
As $\left({S, \preceq}\right)$ is a totally ordered set, have either $a \preceq b$ or $b \preceq a$. Since both sides are seen to be invariant upon interchanging $a$ and $b$, WLOG let $a \preceq b$. Then it follows by definition of $\max$ that: : $\max \left({a, b}\right) = b$ Thus, from Intersection with Subset is Subset, it suffices to show that $b^\succ \subseteq a^\succ$. By definition of strict upper closure, this comes down to showing that: :$\forall c \in S: b \prec c \implies a \prec c$ So let $c \in S$ with $b \prec c$, and recall that $a \preceq b$. By Strictly Precedes is Strict Ordering, $a \prec c$. {{qed}}
As $\left({S, \preccurlyeq}\right)$ is a totally ordered set, either $a \preccurlyeq b$ or $b \preccurlyeq a$. Since both sides are seen to be invariant upon interchanging $a$ and $b$, WLOG let $a \preccurlyeq b$. Then it follows by definition of $\max$ that: :$\max \left({a, b}\right) = b$ Thus, from Intersection with Subset is Subset, it suffices to show that: :$b^\succcurlyeq \subseteq a^\succcurlyeq$ By definition of weak upper closure, this comes down to showing that: :$\forall c \in S: b \preccurlyeq c \implies a \preccurlyeq c$ So let $c \in S$ with $b \preccurlyeq c$. Recall that $a \preccurlyeq b$. Now as $\preccurlyeq$ is a total ordering, it is in particular transitive. Hence $a \preccurlyeq c$. {{qed}}
As $\struct {S, \preccurlyeq}$ is a totally ordered set, either $a \preccurlyeq b$ or $b \preccurlyeq a$. Both sides are seen to be invariant upon interchanging $a$ and $b$. {{WLOG}}, let $b \preccurlyeq a$. Then it follows by definition of $\min$ that $\min \set {a, b} = b$. Thus, from Intersection with Subset is Subset, it suffices to show that: :$b^\preccurlyeq \subseteq a^\preccurlyeq$ By definition of weak lower closure, this comes down to showing that: :$\forall c \in S: c \preccurlyeq b \implies c \preccurlyeq a$ So let $c \in S$ with $c \preccurlyeq b$. Recall that $b \preccurlyeq a$. Now as $\preccurlyeq$ is a total ordering, it is in particular transitive. Hence $c \preccurlyeq a$. {{qed}}
As $\left({S, \preceq}\right)$ is a totally ordered set, have either $a \preceq b$ or $b \preceq a$. Since both sides are seen to be invariant upon interchanging $a$ and $b$, let WLOG $b \preceq a$. Then it follows by definition of $\min$ that $\min \left({a, b}\right) = b$. Thus, from Intersection with Subset is Subset, it suffices to show that $b^\prec \subseteq a^\prec$. By the definition of strict lower closure, this comes down to showing that: :$\forall c \in S: c \prec b \implies c \prec a$ So let $c \in S$ with $c \prec b$, and recall that $b \preceq a$. By Strictly Precedes is Strict Ordering, $c \prec a$. {{qed}}
{{questionable}} We take as axiomatic the Zermelo-Fraenkel axioms. From these, Existence of Minimal Infinite Successor Set is demonstrated. This proves the existence of a minimal infinite successor set. Then we have that the Minimal Infinite Successor Set forms Peano Structure. It follows that the existence of a Peano structure depends upon the existence of such a minimal infinite successor set. Then we have that a Naturally Ordered Semigroup forms Peano Structure. Hence the result. {{qed}}
Let $t \in T$, and suppose that $t' \in t^{\succ T}$. By definition of strict upper closure, this is equivalent to: :$t \preceq \restriction_T t' \land t \ne t'$ By definition of $\preceq \restriction_T$, the first condition comes down to: :$t \preceq t' \land t' \in T$ as it is assumed that $t \in T$. In conclusion, $t' \in t^{\succ T}$ is equivalent to: :$t' \in T \land t \preceq t' \land t \ne t'$ These last two conjuncts precisely express that $t' \in t^{\succ S}$. By definition of set intersection, it also holds that: :$t' \in T \cap t^{\succ S}$ precisely when $t' \in T$ and $t' \in t^{\succ S}$. Thus, it follows that the following are equivalent: :$t' \in t^{\succ T}$ :$t' \in T \cap t^{\succ S}$ and hence the result follows, by definition of set equality. {{qed}} Category:Upper Closures 97c1pqbo4j3pty83ysgosc5ligkdtqx
Let $t \in T$. Suppose that: :$t' \in t^{\succcurlyeq T}$ By definition of weak upper closure $t^{\succcurlyeq T}$, this is equivalent to: :$t \preccurlyeq \restriction_T t'$ By definition of $\preccurlyeq \restriction_T$, this comes down to: :$t \preccurlyeq t' \land t' \in T$ as it is assumed that $t \in T$. The first conjunct precisely expresses that $t' \in t^{\succcurlyeq S}$. By definition of set intersection, it also holds that: :$t' \in T \cap t^{\succcurlyeq S}$ precisely when $t' \in T$ and $t' \in t^{\succcurlyeq S}$. Thus it follows that the following are equivalent: :$t' \in t^{\succcurlyeq T}$ :$t' \in T \cap t^{\succcurlyeq S}$ and hence the result follows, by definition of set equality. {{qed}} Category:Upper Closures heto2ge1ez8yoh0zh3d3awhm3t9hqdl
Let $t \in T$, and suppose that $t' \in t^{\prec T}$. By definition of strict lower closure, this is equivalent to: :$t' \preceq \restriction_T t \land t \ne t'$ By definition of $\preceq \restriction_T$, the first condition comes down to: :$t' \preceq t \land t' \in T$ as it is assumed that $t \in T$. In conclusion, $t' \in t^{\prec T}$ is equivalent to: :$t' \in T \land t' \preceq t \land t \ne t'$ These last two conjuncts precisely express that $t' \in t^{\prec S}$. By definition of set intersection, it also holds that: :$t' \in T \cap t^{\prec S}$ precisely when $t' \in T$ and $t' \in t^{\prec S}$. Thus, it follows that the following are equivalent: :$t' \in t^{\prec T}$ :$t' \in T \cap t^{\prec S}$ and hence the result follows, by definition of set equality. {{qed}} Category:Lower Closures 34yt591thyko8sxodcflxre3r6hpca2
Let $t \in T$, and suppose that $t' \in t^{\preccurlyeq T}$. By definition of weak lower closure $t^{\preccurlyeq T}$, this is equivalent to: :$t' \preccurlyeq \restriction_T t$ By definition of $\preccurlyeq \restriction_T$, this comes down to: :$t' \preccurlyeq t \land t' \in T$ as it is assumed that $t \in T$. The first conjunct precisely expresses that $t' \in t^{\preccurlyeq S}$. By definition of set intersection, it also holds that: :$t' \in T \cap t^{\preccurlyeq S}$ precisely when $t' \in T$ and $t' \in t^{\preccurlyeq S}$. Thus, it follows that the following are equivalent: :$t' \in t^{\preccurlyeq T}$ :$t' \in T \cap t^{\preccurlyeq S}$ and hence the result follows, by definition of set equality. {{qed}} Category:Lower Closures 19b16ktdq3w4i6qg86a4kvm7n18ezou
By Topology Discrete iff All Singletons Open, it suffices to show that for all $n \in \N$, the singleton $\left\{{n}\right\}$ is an open of $\tau$. Now observe that $\mathop{\downarrow} \left({1}\right) = \left\{{0}\right\}$, since for all $n \in \N$, $n < 1 \implies n = 0$. It follows that $\left\{{0}\right\}$ is an open set of $\tau$. Suppose now that $n \in \N$ and $n \ne 0$. Then it is known that for all $m \in \N$, $n - 1 < m < n + 1$ implies $m = n$. Thus, $\mathop{\uparrow} \left({n - 1}\right) \cap \mathop{\downarrow} \left({n + 1}\right) = \left\{{n}\right\}$. It follows that $\left\{{n}\right\}$ is an open set of $\tau$. {{explain|arrows, and why this inference holds (using the def'n of order top as gen. top)}} Hence the result, from Proof by Cases. {{qed}} Category:Natural Numbers Category:Order Topology Category:Discrete Topology n0sfjedpy5ex8x29wfrthuqby6y0v0i
By Natural Numbers under Multiplication form Semigroup, $\left({\N, \times, \le}\right)$ is a semigroup. By Natural Number Multiplication is Commutative, $\times$ is commutative. By Ordering on Natural Numbers is Compatible with Multiplication, $\le$ is compatible with $\times$. The result follows. {{qed}}
Suppose $m \in \N$ is invertible for $\times$. Let $n \in \N: m \times n = 1$. Then from Natural Numbers have No Proper Zero Divisors: :$m \ne 0$ and $n \ne 0$ Thus, $1 \le m$ and $1 \le n$. If $1 \le m$ then from Ordering on Natural Numbers is Compatible with Multiplication: : $1 \le n < m \times n$ This contradicts $m \times n = 1$. The result follows. {{qed}}
Let $G = \left({V, E}\right)$ be a graph with closed walk whose length is odd. From Closed Walk of Odd Length contains Odd Circuit, such a walk contains a circuit whose length is odd. Let $C_1 = \left({v_1, \ldots, v_{2n+1} = v_1}\right)$ be such a circuit. Aiming for a contradiction, suppose $G$ has no odd cycles. Then $C_1$ is not a cycle. Hence, there exist a vertex $v_i$ where $2 \le i \le 2n-1$ and an integer $k$ such that $i+1 \le k \le 2n$ and $v_i = v_k$. If $k-i$ is odd, then we have an odd circuit $\left({v_i, \ldots, v_k = v_i}\right)$ smaller in length than $C_1$. If $k-i$ is even, then $\left({v_1, \ldots, v_i, v_{k+1}, \ldots, v_{2n+1}}\right)$ is a circuit whose length is odd smaller in length than $C_2$. This new odd length circuit is named $C_2$, and the same argument is applied as to $C_1$. Thus at each step a circuit whose length is odd is reduced. At the $n$th step for some $n \in \N$, either: :$(1): \quad C_n$ is a cycle, which contradiction the supposition that $G$ has no odd cycles or: :$(2): \quad C_n$ is a circuit whose length is $3$. But from Circuit of Length 3 is Cycle, $C_n$ is a cycle, which by definition has odd length. From this contradiction it follows that $G$ has at least one odd cycle. {{qed}} Category:Graph Theory m6r2rn15sbnqoq4fuq7vbnb3ukjz4tu
The proof proceeds by the Principle of Mathematical Induction for a Naturally Ordered Semigroup. Let $A := \set {n \in S^*: \forall a \in T_1: \map \phi {\map {\odot^n} a} = \map {\oplus^n} {\map \phi a} }$ That is, $A$ is defined as the set of all $n$ such that: :$\forall a \in T_1 \map \phi {\map {\odot^n} a} = \map {\oplus^n} {\map \phi a}$ === Basis for the Induction === We have that: {{begin-eqn}} {{eqn | l = \map \phi {\map {\odot^1} a}       | r = \map \phi a       | c = {{Defof|Power of Element of Magma}} }} {{eqn | r = \map {\oplus^1} {\map \phi a}       | c = {{Defof|Power of Element of Magma}} }} {{end-eqn}} So $1 \in A$. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $k \in A$ where $k \ge 1$, then it logically follows that $k \circ 1 \in A$. So this is our induction hypothesis: :$\forall a \in T_1: \map \phi {\map {\odot^k} a} = \map {\oplus^k} {\map \phi a}$ Then we need to show: :$\forall a \in T_1: \map \phi {\map {\odot^{k \circ 1} } a} = \map {\oplus^{k \circ 1} } {\map \phi a}$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \map \phi {\map {\odot^{k \circ 1} } a}       | r = \map \phi {\paren {\map {\odot^k} a} \odot a}       | c = {{Defof|Power of Element of Magma}} }} {{eqn | r = \paren {\map \phi {\map {\odot^k} a} } \oplus \paren {\map \phi a}       | c = {{Defof|Semigroup Homomorphism}} }} {{eqn | r = \paren {\map {\oplus^k} {\map \phi a} } \oplus \paren {\map \phi a}       | c = Induction Hypothesis }} {{eqn | r = \map {\oplus^{k \circ 1} } {\map \phi a}       | c = {{Defof|Power of Element of Magma}} }} {{end-eqn}} So $k \in A \implies k \circ 1 \in A$ and the result follows by the Principle of Mathematical Induction: :$\forall n \in \struct {S^*, \circ, \preceq}: \map \phi {\map {\odot^n} a} = \map {\oplus^n} {\map \phi a}$ {{qed}}
Consider the natural numbers $\N$ defined as a naturally ordered semigroup. Then the result follows from Homomorphism of Powers: Naturally Ordered Semigroup. {{qed}} Category:Homomorphisms Category:Natural Numbers 81mfc6y9nis59bgz4ybp8hhj79dl9m1
By Homomorphism of Powers: Natural Numbers, we need show this only for negative $n$, that is: :$\forall n \in \N^*: \map \phi {\map {\odot^{-n} } a} = \map {\oplus^{-n} } {\map \phi a}$ But by Homomorphism with Identity Preserves Inverses: :$\map \phi {a^{-1} } = \paren {\map \phi a}^{-1}$ Hence by Homomorphism of Powers: Natural Numbers: :$\map {\oplus^{-n} } {\map \phi a} = \map {\oplus^n} {\map \phi {a^{-1} } } = \map \phi {\map {\odot^n} {a^{-1} } } = \map \phi {\map {\odot^{-n} } a}$ {{qed}}
By definition of the right operation: {{begin-eqn}} {{eqn | l = a \rightarrow \paren {b \circ c}       | r = b \circ c       | c =  }} {{eqn | r = \paren {a \rightarrow b} \circ \paren {a \rightarrow c}       | c =  }} {{end-eqn}} The result follows by definition of left distributivity. {{qed}}
By definition of the left operation: {{begin-eqn}} {{eqn | l = \paren {a \circ b} \leftarrow c       | r = a \circ b       | c =  }} {{eqn | r = \paren {a \leftarrow c} \circ \paren {b \leftarrow c}       | c =  }} {{end-eqn}} The result follows by definition of right distributivity. {{qed}}
From Right Operation is Left Distributive over All Operations: :$\forall a, b, c \in S: a \rightarrow \paren {b \circ c} = \paren {a \rightarrow b} \circ \paren {a \rightarrow c}$ for all binary operations $\circ$. It remains to show that $\rightarrow$ is right distributive over $\circ$ {{iff}} $\circ$ is idempotent. === Necessary Condition === Let $\circ$ be idempotent. Then: {{begin-eqn}} {{eqn | l = \paren {a \circ b} \rightarrow c       | r = c       | c = {{Defof|Right Operation}} }} {{eqn | r = c \circ c       | c = {{Defof|Idempotent Operation}} }} {{eqn | r = \paren {a \rightarrow c} \circ \paren {b \rightarrow c}       | c = {{Defof|Right Operation}} }} {{end-eqn}} Thus $\rightarrow$ is right distributive over $\circ$. {{qed|lemma}} === Sufficient Condition === Let $\rightarrow$ be right distributive over $\circ$. Let $c \in S$ be arbitrary. Then: {{begin-eqn}} {{eqn | l = c       | r = \paren {a \circ b} \rightarrow c       | c = {{Defof|Right Operation}} }} {{eqn | r = \paren {a \rightarrow c} \circ \paren {b \rightarrow c}       | c = {{Defof|Right Distributive Operation}} }} {{eqn | r = c \circ c       | c = {{Defof|Right Operation}} }} {{end-eqn}} Hence $\circ$ is idempotent. {{qed}}
From Left Operation is Right Distributive over All Operations: :$\forall a, b, c \in S: \paren {a \circ b} \leftarrow c = \paren {a \leftarrow c} \circ \paren {b \leftarrow c}$ for all binary operations $\circ$. It remains to show that $\leftarrow$ is left distributive over $\circ$ {{iff}} $\circ$ is idempotent. === Necessary Condition === Let $\circ$ be idempotent. Then: {{begin-eqn}} {{eqn | l = a \leftarrow \paren {b \circ c}       | r = a       | c = {{Defof|Left Operation}} }} {{eqn | r = a \circ a       | c = {{Defof|Idempotent Operation}} }} {{eqn | r = \paren {a \leftarrow b} \circ \paren {a \leftarrow c}       | c = {{Defof|Left Operation}} }} {{end-eqn}} Thus $\leftarrow$ is left distributive over $\circ$. {{qed|lemma}} === Sufficient Condition === Let $\leftarrow$ be left distributive over $\circ$. Let $a \in S$ be arbitrary. Then: {{begin-eqn}} {{eqn | l = a       | r = a \leftarrow \paren {b \circ c}       | c = {{Defof|Left Operation}} }} {{eqn | r = \paren {a \leftarrow b} \circ \paren {a \leftarrow c}       | c = {{Defof|Left Distributive Operation}} }} {{eqn | r = a \circ a       | c = {{Defof|Left Operation}} }} {{end-eqn}} Hence $\circ$ is idempotent. {{qed}}
By Integral of Characteristic Function, have: :$I_\mu \left({\chi_E}\right) = \mu \left({E}\right)$ where $I_\mu \left({\chi_E}\right)$ is the $\mu$-integral of $\chi_E$. From Integral of Positive Measurable Function Extends Integral of Positive Simple Function, it also holds that: :$\displaystyle \int \chi_E \, \mathrm d\mu = I_\mu \left({\chi_E}\right)$ Combining these equalities gives the result. {{qed}}
Let $\mathbf x = \begin {bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end {bmatrix}$. Let $\mathbf I_n$ be the unit matrix of order $n$. Then: {{begin-eqn}} {{eqn | l = \mathbf x_{n \times 1}       | r = \mathbf I_n \mathbf x_{n \times 1}       | c = {{Defof|Left Identity}} }} {{eqn | r = \begin {bmatrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \\ \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end {bmatrix}       | c = Unit Matrix is Identity:Lemma }} {{eqn | r = \begin {bmatrix} \mathbf e_1 & \mathbf e_2 & \cdots & \mathbf e_n \end {bmatrix} \begin {bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end {bmatrix}       | c = {{Defof|Standard Ordered Basis on Vector Space}} }} {{eqn | r = \sum_{i \mathop = 1}^n \mathbf e_i x_i       | c = {{Defof|Matrix Product (Conventional)}} }} {{eqn | l = \map T {\mathbf x}       | r = \map T {\sum_{i \mathop =1}^n \mathbf e_i x_i} }} {{eqn | r = \sum_{i \mathop = 1}^n \map T {\mathbf e_i} x_i       | c = {{Defof|Linear Transformation on Vector Space}} }} {{eqn | r = \begin {bmatrix} \map T {\mathbf e_1} & \map T {\mathbf e_2} & \cdots & \map T {\mathbf e_n} \end {bmatrix} \begin {bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end {bmatrix}       | c = {{Defof|Matrix Product (Conventional)}} }} {{end-eqn}} That $\mathbf A_T$ is $m \times n$ follows from each $\map T {\mathbf e_i}$ being an element of $\R^m$ and thus having $m$ rows. {{qed}}
Let $\mathbf A = \sqbrk a_{m n}, \mathbf B = \sqbrk b_{n p}$ {{begin-eqn}} {{eqn | lo= \forall i \in \closedint 1 m, j \in \closedint 1 p:       | l = \mathbf A \paren {\lambda \mathbf B}       | r = \lambda \sum_{k \mathop = 1}^n a_{i k} b_{k j}       | c = {{Defof|Matrix Product (Conventional)}} and {{Defof|Matrix Scalar Product}} }} {{eqn | r = \sum_{k \mathop = 1}^n a_{i k} \paren {\lambda b_{k j} } }} {{eqn | r = \mathbf A \paren {\lambda \mathbf B}       | c = {{Defof|Matrix Product (Conventional)}} and {{Defof|Matrix Scalar Product}} }} {{end-eqn}} {{qed}} {{proofread}} {{expand|proof literally carries over for any commutative ring in place of $\Bbb F$}} Category:Conventional Matrix Multiplication b2tvzdjy8an71tgdj8k0a3t4rceaq58
Notice that the constant function $g$ defined as :$X \ni x' \mapsto g \left({x'}\right) := f \left({x}\right) \in \overline{\R}$  is $\delta_x$-almost everywhere equal to $f$. This follows from the fact that the set of elements of $X$ where $f$ and $g$ take different values, namely: :$\left\{ {x' \in X : f \left({x}\right) = g \left({x'}\right) \ne f \left({x'}\right)}\right\}$ does not contain $x$. So, by the very definition of $\delta_x$, its $\delta_x$-measure is $0$. Therefore: {{begin-eqn}} {{eqn | l = \int f \, \mathrm d \delta_x       | r = \int g \, \mathrm d \delta_x       | c =  }} {{eqn | r = \delta_x \left({X}\right) f \left({x}\right)       | c =  }} {{eqn | r = f \left({x}\right)       | c = }} {{end-eqn}} {{qed}}
From Matrix Multiplication is Homogeneous of Degree $1$: :$\forall \lambda \in \mathbb F \in \set {\R, \C}: \mathbf A \paren {\lambda \mathbf x} = \lambda \paren {\mathbf A \mathbf x}$ From Matrix Multiplication Distributes over Matrix Addition: :$\forall \mathbf x, \mathbf y \in \R^m: \mathbf A \paren {\mathbf x + \mathbf y} = \mathbf A \mathbf x + \mathbf A \mathbf y$ Hence the result, from the definition of linear transformation. {{qed}}
From the vector space axioms we have that $\exists \mathbf 0 \in \mathbf V$. It remains to be proved that $\map T {\mathbf 0} = \mathbf 0'$: {{begin-eqn}} {{eqn | l = \map T {\mathbf 0}       | r = \map T {\mathbf 0 + \mathbf 0} }} {{eqn | r = \map T {\mathbf 0} + \map T {\mathbf 0}       | c = {{Defof|Linear Transformation on Vector Space}} }} {{eqn | ll= \leadsto       | l = \mathbf 0'       | r = \map T {\mathbf 0}       | c = subtracting $\map T {\mathbf 0}$ from both sides }} {{end-eqn}} {{qed}}
From the vector space axioms we have that $\exists \mathbf 0 \in \mathbf V$. What remains is to prove that $\map T {\mathbf 0} = \mathbf 0'$: {{begin-eqn}} {{eqn | l = \map T {\mathbf 0}       | r = \map T {0 \, \mathbf 0}       | c = Zero Vector Scaled is Zero Vector }} {{eqn | r = 0 \, \map T {\mathbf 0}       | c = {{Defof|Linear Transformation on Vector Space}} }} {{eqn | r = \mathbf 0'       | c = Vector Scaled by Zero is Zero Vector }} {{end-eqn}} {{qed}}
Let $a \in A$, $b \in B$. Then: {{begin-eqn}} {{eqn | l = a \circ b       | o = \preceq       | r = \sup A \circ b       | c = {{Defof|Supremum of Set}} }} {{eqn | o = \preceq       | r = \sup A \circ \sup B       | c = {{Defof|Supremum of Set}} }} {{end-eqn}} Hence $\sup A \circ \sup B$ is an upper bound for $A \circ_{\PP} B$. Suppose that $u$ is an upper bound for $A \circ_{\PP} B$. Then: {{begin-eqn}} {{eqn | lo= \forall b \in B: \forall a \in A:       | l = a \circ b       | o = \preceq       | r = u }} {{eqn | ll= \leadsto       | lo= \forall b \in B: \forall a \in A:       | l = a       | o = \preceq       | r = u \circ b^{-1} }} {{eqn | ll= \leadsto       | lo= \forall b \in B:       | l = \sup A       | o = \preceq       | r = u \circ b^{-1}       | c = {{Defof|Supremum of Set}} }} {{eqn | ll= \leadsto       | lo= \forall b \in B:       | l = b       | o = \preceq       | r = \paren {\sup A}^{-1} \circ u }} {{eqn | ll= \leadsto       | l = \sup B       | o = \preceq       | r = \paren {\sup A}^{-1} \circ u       | c = {{Defof|Supremum of Set}} }} {{eqn | ll= \leadsto       | l = \sup A \circ \sup B       | o = \preceq       | r = u }} {{end-eqn}} Therefore: :$\sup \paren {A \circ_{\PP} B} = \sup A \circ \sup B$ {{qed}}
Let $x, x' \in S_1$. Suppose that: :$\map {\inj_1} x = \map {\inj_1} {x'}$ Then by definition of canonical injection: :$\tuple {x, e_2} = \tuple {x', e_2}$ By Equality of Ordered Pairs: :$x = x'$ That is, $\inj_1$ is an injection. {{qed|lemma}} Similarly, let $x, x' \in S_2$. Suppose that: :$\map {\inj_2} x = \map {\inj_2} {x'}$ Then by definition of canonical injection: :$\tuple {e_1, x} = \tuple {e_1, x'}$ Again by Equality of Ordered Pairs: :$x = x'$ That is, $\inj_2$ is an injection. {{qed|lemma}} So $\inj_1$ and $\inj_2$ are injections. {{qed}}
Let: :$x, y \in S_j: \map {\inj_j} x = \map {\inj_j} y$ Then: :$\tuple {e_1, e_2, \dotsc, e_{j - 1}, x, e_{j + 1}, \dotsc, e_n} = \tuple {e_1, e_2, \dotsc, e_{j - 1}, y, e_{j + 1}, \dotsc, e_n}$ By Equality of Ordered Tuples, it follows directly that: :$x = y$ Thus the canonical injections are injective. {{Qed}}
Let $a, b \in X$, and let $r \in Y$ lie between $\map f a$ and $\map f b$. Define the sets: :$A = f \sqbrk X \cap r^\prec$ and $B = f \sqbrk X \cap r^\succ$ where $r^\prec$ and $r^\succ$ denote the strict lower closure and strict upper closure respectively of $r$ in $Y$. $A$ and $B$ are disjoint by construction. $A$ and $B$ are also non-empty since one contains $\map f a$ and the other contains $\map f b$. $A$ and $B$ are also both open by definition as the intersection of open sets. Suppose there is no point $c$ such that $\map f c = r$. Then: :$f \sqbrk X = A \cup B$ so $A$ and $B$ constitute a separation of $X$. But this contradicts the fact that Continuous Image of Connected Space is Connected. Hence by Proof by Contradiction: :$\exists c \in X: \map f c = r$ which is what was to be proved. {{qed}}
Let $S_0, \ldots, S_{n-1}$ be countable sets. For $i \in \left\{{0, \ldots, n-1}\right\}$, let $f_i: \N \to S_i$ be a surjection. These exist by Surjection from Natural Numbers iff Countable. Now define $f: \N \to \displaystyle \bigcup_{i \mathop = 0}^{n-1} S_i$ by: :$f \left({m}\right) := f_i \left({\left\lfloor{\dfrac m n}\right\rfloor}\right)$ where $i$ is the unique element of $\left\{{0, \ldots, n-1}\right\}$ such that $m \equiv i \pmod n$, and $\left\lfloor{x}\right\rfloor$ denotes the floor of $x$. Now let $x \in \displaystyle \bigcup_{i \mathop = 0}^{n-1} S_i$ be arbitrary. Let $k$ be the smallest natural number such that $x \in S_k$. Let $l$ be the smallest natural number such that $f_k \left({l}\right) = x$. These are guaranteed to exist by definition of set union and surjectivity of $f_k$, respectively. Then we have: :$f \left({l n + k}\right) = f_k \left({\left\lfloor{\dfrac {l n + k} n}\right\rfloor}\right) = f_k \left({l}\right) = x$ since $k < n$. Since $x$ was arbitrary, $f$ is a surjection. Hence by Surjection from Natural Numbers iff Countable, $\displaystyle \bigcup_{i \mathop = 0}^{n-1} S_i$ is countable {{qed}}
From the definition of binomial coefficients: :$\dbinom r k = \dfrac {r^{\underline k}} {k!}$ for $k \ge 0$ where $r^{\underline k}$ is the falling factorial. In turn: :$\displaystyle x^{\underline k} := \prod_{j \mathop = 0}^{k-1} \left({x - j}\right)$ But when $k = 0$, we have: :$\displaystyle \prod_{j \mathop = 0}^{-1} \left({x - j}\right) = 1$ as $\displaystyle \prod_{j \mathop = 0}^{-1} \left({x - j}\right)$ is a vacuous product. From the definition of the factorial we have that $0! = 1$. Thus: :$\forall r \in \R: \dbinom r 0 = 1$ {{qed}} === Integer Coefficients === This is completely compatible with the result for natural numbers: :$\forall n \in \N: \dbinom n 0 = 1$ {{:Binomial Coefficient with Zero/Integer Coefficients}}