hcju/proofwikiqa huggingface dataset:
Consider the set: :$\paren {a \circ N} \circ \paren {b \circ N} = \set {x \circ y: x \in a \circ N, y \in b \circ N}$ As $e \in N$, have: :$\paren {a \circ b} \circ N = \paren {a \circ e} \circ \paren {b \circ N} \subseteq \paren {a \circ N} \circ \paren {b \circ N}$ by {{GroupAxiom|1}} of $\circ$. Hence $\paren {a \circ b} \circ N \subseteq \paren {a \circ N} \circ \paren {b \circ N}$. Now let $x \in a \circ N$ and $y \in b \circ N$. Then by the definition of subset product: :$\exists n_1 \in N: x = a \circ n_1$ :$\exists n_2 \in N: y = b \circ n_2$ It follows that: {{begin-eqn}} {{eqn | l = x \circ y       | r = \paren {a \circ n_1} \circ \paren {b \circ n_2}       | c =  }} {{eqn | r = \paren {a \circ n_1} \circ \paren {n_3 \circ b}       | c = for some $n_3 \in N$: {{Defof|Normal Subgroup}} }} {{eqn | r = a \circ \paren {\paren {n_1 \circ n_3} \circ b}       | c = {{GroupAxiom|1}} }} {{eqn | r = a \circ \paren {n_4 \circ b}       | c = for some $n_4 \in N$: {{Defof|Subgroup}} }} {{eqn | r = a \circ \paren {b \circ n_5}       | c = for some $n_5 \in N$: {{Defof|Normal Subgroup}} }} {{eqn | r = \paren {a \circ b} \circ n_5       | c = {{GroupAxiom|1}} }} {{eqn | o = \in       | r = \paren {a \circ b} \circ N       | c = {{Defof|Subset Product}} }} {{end-eqn}} So the definition by subset product: :$\paren {a \circ N} \circ \paren {b \circ N} = \set {x \circ y: x \in a \circ N, y \in b \circ N}$ leads to the definition of coset product as: :$\paren {a \circ N} \circ \paren {b \circ N} = \paren {a \circ b} \circ N$ {{qed}} Category:Coset Product pjn937tkvtw3xejhfkt4cvrm8xhyaod
=== Necessary Condition === Let $\RR$ ibe a congruence relation for $\circ$. That is: :$\forall x_1, x_2, y_1, y_2 \in S: x_1 \mathrel \RR x_2 \land y_1 \mathrel \RR y_2 \implies \paren {x_1 \circ y_1} \mathrel \RR \paren {x_2 \circ y_2}$ As $\RR$ is an equivalence relation it is by definition reflexive. That is: :$\forall z \in S: z \mathrel \RR z$ Make the substitutions: {{begin-eqn}} {{eqn | l = x_1       | o = \to       | r = x }} {{eqn | l = x_2       | o = \to       | r = y }} {{eqn | l = y_1       | o = \to       | r = z }} {{eqn | l = y_2       | o = \to       | r = z }} {{end-eqn}} It follows that: :$\forall x, y, z \in S: x \mathrel \RR y \implies \paren {x \circ z} \mathrel \RR \paren {y \circ z}$ Similarly, make the substitutions: {{begin-eqn}} {{eqn | l = x_1       | o = \to       | r = z }} {{eqn | l = x_2       | o = \to       | r = z }} {{eqn | l = y_1       | o = \to       | r = x }} {{eqn | l = y_2       | o = \to       | r = y }} {{end-eqn}} It follows that: :$\forall x, y, z \in S: x \mathrel \RR y \implies \paren {z \circ x} \mathrel \RR \paren {z \circ y}$ {{qed|lemma}} === Sufficient Condition === Now let $\RR$ have the nature that: {{begin-eqn}} {{eqn | ll= \forall x, y, z \in S:       | l = x \mathrel \RR y       | o = \implies       | r = \paren {x \circ z} \mathrel \RR \paren {y \circ z} }} {{eqn | l = x \mathrel \RR y       | o = \implies       | r = \paren {z \circ x} \mathrel \RR \paren {z \circ y} }} {{end-eqn}} Then we have: {{begin-eqn}} {{eqn | ll= \forall x_1, x_2, y_1, y_2 \in S:       | l = x_1 \mathrel \RR y_1       | o = \implies       | r = \paren {x_1 \circ x_2} \mathrel \RR \paren {y_1 \circ x_2} }} {{eqn | l = x_2 \mathrel \RR y_2       | o = \implies       | r = \paren {y_1 \circ x_2} \mathrel \RR \paren {y_1 \circ y_2} }} {{end-eqn}} As $\RR$ is an equivalence relation it is by definition transitive. Thus it follows that: :$\paren {x_1 \circ x_2} \mathrel \RR \paren {y_1 \circ y_2}$ {{qed|lemma}} The result follows. {{qed}}
We have that an equivalence relation is a (symmetric) preordering. Thus the result Preordering of Products under Operation Compatible with Preordering can be applied directly. {{qed}}
By the monotonicity of $\mu^*$, it suffices to prove that: :$\ds \map {\mu^*} {A \cap S} \le \lim_{n \mathop \to \infty} \map {\mu^*} {A \cap S_n}$ Assume that $\map {\mu^*} {A \cap S_n}$ is finite for all $n \in \N$, otherwise the statement is trivial by the monotonicity of $\mu^*$. Let $S_0 = \O$. Then $x \in S$ {{iff}} there exists an integer $n \ge 0$ such that $x \in S_{n + 1}$.  Taking the least possible $n$, it follows that $x \notin S_n$, and so: :$x \in S_{n + 1} \setminus S_n$ Therefore: :$\ds S = \bigcup_{n \mathop = 0}^\infty \paren {S_{n + 1} \setminus S_n}$ From Intersection Distributes over Union: :$\ds A \cap S = A \cap \bigcup_{n \mathop = 0}^\infty \paren {S_{n + 1} \setminus S_n} = \bigcup_{n \mathop = 0}^\infty \paren {A \cap \paren {S_{n + 1} \setminus S_n} }$ Therefore: {{begin-eqn}} {{eqn | l = \map {\mu^*} {A \cap S}       | o = \le       | r = \sum_{n \mathop = 0}^\infty \map {\mu^*} {A \cap \paren {S_{n + 1} \setminus S_n} }       | c = {{Defof|Countably Subadditive Function}} }} {{eqn | r = \sum_{n \mathop = 0}^\infty \paren {\map {\mu^*} {A \cap S_{n + 1} } - \map {\mu^*} {A \cap S_{n + 1} \cap S_n} }       | c = {{Defof|Measurable Set of Arbitrary Outer Measure}} }} {{eqn | r = \sum_{n \mathop = 0}^\infty \paren {\map {\mu^*} {A \cap S_{n + 1} } - \map {\mu^*} {A \cap S_n} }       | c = Intersection with Subset is Subset }} {{eqn | r = \lim_{n \mathop \to \infty} \map {\mu^*} {A \cap S_n} - \map {\mu^*} {A \cap \O}       | c = Telescoping Series }} {{eqn | r = \lim_{n \mathop \to \infty} \map {\mu^*} {A \cap S_n}       | c = Intersection with Empty Set and {{Defof|Outer Measure}} }} {{end-eqn}} {{qed}}
By definition, $\O$ is considered to be a half-open $n$-rectangle. That $\JJ_{ho}^n$ is $\cap$-stable follows from Half-Open Rectangles Closed under Intersection. Thus, it remains to show condition $(3')$ for a semiring of sets: :$(3'):\quad$ If $A, B \in \JJ_{ho}^n$, then there exists a finite sequence of pairwise disjoint half-open $n$-rectangles $A_1, A_2, \ldots, A_m \in \JJ_{ho}^n$ such that $\displaystyle A \setminus B = \bigcup_{k \mathop = 1}^m A_k$. To prove this, proceed by induction on $n$: === Basis for the Induction === Suppose that $n = 1$, and let $I := \hointr a b$ and $J := \hointr c d$ be half-open intervals. It is to be demonstrated that $I \setminus J$ is a finite union of pairwise disjoint half-open intervals. Equivalently, by the above verification of the other axioms, that $\JJ_{ho}^1$ is a semiring of sets. By swapping the roles of $I$ and $J$ if necessary, it may be arranged that $a \le c$. Now if $b \le c$ as well, then $I \cap J = \O$. Subsequently, this implies $I \setminus J = I$, and the statement trivially holds. Next, consider the other case, i.e., $c < b$. If also $d < b$, then we may write $I$ as the following disjoint union: :$I = \hointr a b = \hointr a c \cup \hointr c d \cup \hointr d b$ The middle term equals $J$, and we immediately obtain: :$I \setminus J = \hointr a c \cup \hointr d b$ verifying the statement in this case. Lastly, suppose that $b \le d$. Then $I \cap J = \hointr a b \cap \hointr c d = \hointr c b$. Therefore, by Set Difference with Intersection is Difference: :$I \setminus J = I \setminus \paren {I \cap J} = \hointr a b \setminus \hointr c b = \hointr a c$ and, having verified this last case, the result follows from Proof by Cases. This constitutes the induction basis. {{qed|lemma}} === Induction Hypothesis === Now assume the induction hypothesis. That is, for some fixed $n \ge 1$, assume that $\JJ_{ho}^n$ is a semiring of sets. Next, it is to be shown that $\JJ_{ho}^{n + 1}$ is also a semiring of sets. === Induction Step === The induction step goes as follows. By definition of half-open rectangle, it holds that: :$\JJ_{ho}^{n+1} = \JJ_{ho}^n \times \JJ_{ho}^1$ Further, we have that $\JJ_{ho}^n$ and $\JJ_{ho}^1$ are semirings of sets. Hence by Cartesian Product of Semirings of Sets, $\JJ_{ho}^{n + 1}$ is a semiring of sets, too. {{qed}}
We employ Characterization of Pre-Measures. It is known that $\lambda^n \left({\varnothing}\right) = 0$ by definition of Lebesgue pre-measure. The only possibility for two disjoint half-open $n$-rectangles to constitute a single, large half-open $n$-rectangle is when they are of the form: :$\left[[{\mathbf a \,.\,.\, \mathbf b}\right)) \quad \left[[{\mathbf a' \,.\,.\, \mathbf b'}\right))$ such that we have for some $i$ with $1 \le i \le n$: :$i \ne j \implies a_j = a'_j$ :$i \ne j \implies b_j = b'_j$ :$i = j \implies a'_j = b_j$ which intuitively can be visualised as two cubes that together form one large bar, namely $\left[[{\mathbf a \,.\,.\, \mathbf b'}\right))$. In this situation, we have: {{begin-eqn}} {{eqn|l = \lambda^n \left({\left[\left[{\mathbf a \,.\,.\, \mathbf b}\right)\right)}\right) + \lambda^n \left({\left[\left[{\mathbf a' \,.\,.\, \mathbf b'}\right)\right)}\right)      |r = \prod_{j \mathop = 1}^n \left({b_j - a_j}\right) + \prod_{j \mathop = 1}^n \left({b'_j - a'_j}\right)      |c = By definition of Lebesgue pre-measure }} {{eqn|r = \left({b_i - a_i + b'_i - a'_i}\right) \prod_{\substack{j \mathop = 1 \\ j \mathop \ne i} } \left({b_j - a_j}\right)      |c = By the noted properties of $a_j, b_j, a'_j, b'_j$ }} {{eqn|r = \lambda^n \left({\left[\left[{\mathbf a \,.\,.\, \mathbf b'}\right)\right)}\right)      |c = By definition of Lebesgue pre-measure }} {{end-eqn}} Thus it is verified that $\lambda^n$ is finitely additive. Finally, suppose that $\leftdecreasing sequence of sets, with limit $\varnothing$. Then there exists at least one $j$ with $1 \le j \le n$ such that: :$\displaystyle \lim_{m \to \infty} a_{m,j} = \lim_{m \to \infty} b_{m,j}$ which by Combination Theorem for Sequences is equivalent to: :$\displaystyle \lim_{m \to \infty} b_{m,j} - a_{m,j} = 0$ The fact that the sequence is decreasing means that, from Cartesian Product of Subsets, for all $m \in \N$, for all $1 \le i \le n$: :$\left[{a_{m,i} \,.,\,.\, b_{m,i}}\right) \subseteq \left[{a_{1,i} \,.\,.\, b_{1,i} }\right)$ and whence $b_{m,i} - a_{m,i} \le b_{1,i} - a_{m,1}$. Hence we have: {{begin-eqn}} {{eqn|l = \lim_{m \to \infty} \lambda^n \left({\left[\left[{\mathbf a_m \,.\,.\, \mathbf b_m}\right)\right)}\right)      |r = \lim_{m \to \infty} \prod_{i \mathop = 1}^n \left({b_{m,i} - a_{m,i} }\right)      |c = By definition of Lebesgue pre-measure }} {{eqn|o = \le      |r = \lim_{m \to \infty} \left({b_{m,j} - a_{m,j} }\right) \prod_{\substack{i \mathop = 1 \\ i \mathop \ne j} } \left({b_{1,i} - a_{1,i} }\right)      |c = By the above discussion }} {{eqn|r = 0      |c = Combination Theorem for Sequences }} {{end-eqn}} This verifies the last condition for Characterization of Pre-Measures, since $\lambda^n$ only takes finite values. Hence $\lambda^n$ is a pre-measure. {{qed}}
From Lebesgue Pre-Measure is Pre-Measure, $\lambda^n$ is a pre-measure on $\mathcal{J}_{ho}^n$. By Half-Open Rectangles form Semiring of Sets, $\mathcal{J}_{ho}^n$ is a semiring of sets. Also, from Characterization of Euclidean Borel Sigma-Algebra, $\mathcal B \left({\R^n}\right) = \sigma \left({\mathcal{J}_{ho}^n}\right)$. Now observe that the half-open $n$-rectangles $\leftincreasing sequence of sets with limit $\R^n$. Also, by definition of $\lambda^n$, have: :$\lambda^n \left({\left[[{-n, n}\right))}\right) = \displaystyle \prod_{i \mathop = 1}^n n - \left({-n}\right) = \left({2n}\right)^n < +\infty$ Hence, Carathéodory's Theorem and its corollary apply. These yield existence and uniqueness of Lebesgue measure, the extension of $\lambda^n$ to $\mathcal B \left({\R^n}\right)$. {{qed}}
Any singleton $\left\{{\mathbf x}\right\} \subseteq \R^n$ is closed by combining: * Euclidean Space is Complete Metric Space * Metric Space is Hausdorff * Corollary to Compact Subspace of Hausdorff Space is Closed Whence by Closed Set Measurable in Borel Sigma-Algebra, $\left\{{\mathbf x}\right\} \in \mathcal B \left({\R^n}\right)$. Here, $\mathcal B \left({\R^n}\right)$ is the Borel $\sigma$-algebra on $\R^n$. Write $\mathbf x + \epsilon = \left({x_1 + \epsilon, \ldots, x_n + \epsilon}\right)$ for $\epsilon > 0$. Then: :$\displaystyle \left\{{\mathbf x}\right\} = \bigcap_{m \mathop \in \N} \left[\left[{\mathbf x \,.\,.\, \mathbf x + \frac 1 m}\right)\right)$ where $\left[\left[{\mathbf x \,.\,.\, \mathbf x + \dfrac 1 m}\right)\right)$ is a half-open $n$-rectangle. {{handwaving|justify equality}} By definition of Lebesgue measure, we have (for all $m \in \N$): :$\displaystyle \lambda^n \left({\left[\left[{\mathbf x \,.\,.\, \mathbf x + \frac 1 m}\right)\right)}\right) = \prod_{i \mathop = 1}^n \frac 1 m = m^{-n}$ From Characterization of Measures, it follows that: :$\displaystyle \lambda^n \left({\left\{{\mathbf x}\right\}}\right) = \lim_{m \to \infty} m^{-n}$ which equals $0$ from Sequence of Powers of Reciprocals is Null Sequence. Therefore, for each $\mathbf x \in \R^n$: :$\lambda^n \left({\left\{{\mathbf x}\right\}}\right) = 0$ that is, $\lambda^n$ is a diffuse measure. {{qed}}
It holds trivially that: :$\displaystyle E = \bigcup_{e \mathop \in E} \left\{{e}\right\}$ and in particular, this union is countable. Also, $\mu \left({\left\{{e}\right\}}\right) = 0$ for all $e \in E$ as $\mu$ is diffuse. Hence Null Sets Closed under Countable Union applies to yield: :$\mu \left({E}\right) = 0$ {{qed}}
As $\varnothing$ is trivially a half-open $n$-rectangle, let us assume that: :$\left[[{\mathbf a \,.\,.\, \mathbf b}\right)) \cap \left[[{\mathbf c \,.\,.\, \mathbf d}\right)) \ne \varnothing$ By Cartesian Product of Intersections: General Case, it follows that: :$\displaystyle \left[[{\mathbf a \,.\,.\, \mathbf b}\right)) \cap \left[[{\mathbf c \,.\,.\, \mathbf d}\right)) = \prod_{i \mathop = 1}^n \left[{a_i \,.\,.\, b_i}\right) \cap \left[{c_i \,.\,.\, d_i}\right)$ which leaves only to prove that the $\left[{a_i \,.\,.\, b_i}\right) \cap \left[{c_i \,.\,.\, d_i}\right)$ are half-open intervals. Now let $x \in \left[{a_i \,.\,.\, b_i}\right) \cap \left[{c_i \,.\,.\, d_i}\right)$. Then $x$ is subject to: * $x \ge a_i$ and $x \ge c_i$, i.e., $x \ge \max \left\{{a_i, c_i}\right\}$ * $x < b_i$ and $x < d_i$, i.e., $x < \min \left\{{b_i, d_i}\right\}$ and we see that these conditions are satisfied precisely when: :$x \in \left[{\max \left\{{a_i, c_i}\right\} \,.\,.\, \min \left\{{b_i, d_i}\right\}}\right)$ Thus, we conclude: :$\left[{a_i \,.\,.\, b_i}\right) \cap \left[{c_i \,.\,.\, d_i}\right) = \left[{\max \left\{{a_i, c_i}\right\} \,.\,.\, \min \left\{{b_i, d_i}\right\}}\right)$ showing that indeed the intersection is a half-open interval. Combining this with the above reasoning, it follows that indeed: :$\left[[{\mathbf a \,.\,.\, \mathbf b}\right)) \cap \left[[{\mathbf c \,.\,.\, \mathbf d}\right))$ is again a half-open $n$-rectangle. {{qed}}
By hypothesis, $\struct {T, *}$ is a group. By {{GroupAxiom|2}}, $\struct {T, *}$ has an identity. Thus Homomorphism with Identity Preserves Inverses can be applied. {{Qed}}
Let $x, x', y, y' \in S$ such that: :$x \mathop {\mathcal R_\phi} x' \land y \mathop {\mathcal R_\phi} y'$ By definition of induced equivalence: {{begin-eqn}} {{eqn | l = x \mathop {\mathcal R_\phi} x'       | o = \implies       | r = \map \phi x = \map \phi {x'}       | c =  }} {{eqn | l = y \mathop {\mathcal R_\phi} y'       | o = \implies       | r = \map \phi y = \map \phi {y'}       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \map \phi {x \circ y}       | r = \map \phi x * \map \phi y       | c = {{Defof|Epimorphism (Abstract Algebra)}} }} {{eqn | r = \map \phi {x'} * \map \phi {y'}       | c = equality shown above }} {{eqn | r = \map \phi {x' \circ y'}       | c = {{Defof|Epimorphism (Abstract Algebra)}} }} {{end-eqn}} Thus $\paren {x \circ y} \mathop {\mathcal R_\phi} \paren {x' \circ y'}$ by definition of induced equivalence. So $\mathcal R_\phi$ is a congruence relation for $\circ$.
From the Quotient Theorem for Surjections, there is a unique bijection from $S / \mathcal R_\phi$ onto $T$ satisfying $\psi \bullet q_{\mathcal R_\phi} = \phi$. Also: {{begin-eqn}} {{eqn | l = \forall x, y \in S: \map \psi {\eqclass x {\mathcal R_\phi} \circ_{\mathcal R_\phi} \eqclass y {\mathcal R_\phi} }       | r = \map \psi {\eqclass {x \circ y} {\mathcal R_\phi} }       | c = {{Defof|Quotient Structure}} }} {{eqn | r = \map \phi {x \circ y}       | c = {{Defof|Epimorphism (Abstract Algebra)}} }} {{eqn | r = \map \phi x * \map \phi y       | c = {{Defof|Epimorphism (Abstract Algebra)}} }} {{eqn | r = \map \psi {\eqclass x {\mathcal R_\phi} } * \map \psi {\eqclass y {\mathcal R_\phi} }       | c = {{Defof|Quotient Mapping}} }} {{end-eqn}} Therefore $\psi$ is an isomorphism. {{Qed}}
From the definition of kernel: :$\map \ker \phi = \set {x \in G: \map \phi x = e_H}$ From Group Homomorphism Preserves Identity we have that: :$\map \phi {e_G} = e_H$ Hence the result. {{Qed}}
For all $\displaystyle x \in \bigcup_{i \mathop \in I} S_i$, there exists a $i \left({x}\right) \in I$ such that $x \in S_{i \left({x}\right)}$. Thus the mapping $\displaystyle \iota : \bigcup_{i \mathop \in I} S_i \to \bigsqcup_{i \mathop \in I} S_i$ defined by: :$\iota \left({x}\right) = \left({x, i \left({x}\right)}\right)$ is an injection. {{handwaving}} {{qed}} Category:Set Theory bhdszl9s1i2q3y7mouiapo397zhhime
By definition of domination, for all $i \in I$, there exists an injection $\iota_i: A_i \to B_i$. Thus the mapping $\displaystyle \iota : \bigsqcup_{i \mathop \in I} A_i \to \bigsqcup_{i \mathop \in I} B_i$ defined by: :$\iota \left({x, i}\right) = \left({\iota_i \left({x}\right), i}\right)$ is an injection. {{handwaving|That statement needs proof. Probably best in a separate page.}} {{qed}} Category:Set Theory qraiskgupdvig3zs5ehs5bw7n3mpbbr
$f$ is a surjection from ... {{link wanted|Needs a link to a result specifying that $f$ is surjective (may already exist).}} {{qed|lemma}} Then: {{begin-eqn}} {{eqn | l = \map f x \times \map f y       | r = \paren {\cos x + i \sin x} \paren {\cos y + i \sin y}       | c = }} {{eqn | r = \cos x \cos y + i \sin x \cos y + \cos x i \sin y + i \sin x i \sin y       | c = }} {{eqn | r = \paren {\cos x \cos y - \sin x \sin y} + i \paren {\sin x \cos y + \cos x \sin y}       | c = as $i^2 = -1$ }} {{eqn | r = \map \cos {x + y} + i \map \sin {x + y}       | c = Cosine of Sum and Sine of Sum }} {{eqn | r = \map f {x + y}       | c = }} {{end-eqn}} So $f$ is a (group) homomorphism. {{qed|lemma}} Thus $f$ is seen to be a surjective homomorphism. Hence, by definition, it is a (group) epimorphism. {{qed|lemma}} From Cosine of Multiple of Pi: :$\forall n \in \Z: \cos n \pi = \paren {-1}^n$ and from Sine of Multiple of Pi: :$\forall n \in \Z: \sin n \pi = 0$ From Sine and Cosine are Periodic on Reals, it follows that these are the only values of $\Z$ for which this holds. For $\cos x + i \sin x = 1 + 0 i$ it is necessary that: :$\cos x = 1$ :$\sin x = 0$ and it can be seen that the only values of $x$ for this to happen is: :$x \in \set {2 \pi n: n \in \Z}$ Hence, by definition of kernel: :$\map \ker f = \set {2 \pi n: n \in \Z}$ {{qed}}
That $\sequence {T_n}_{n \mathop \in \N}$ partitions $S$, means precisely that: :$(1):\quad$ The $T_n$ are pairwise disjoint :$(2):\quad \displaystyle \bigcup_{n \mathop \in \N} T_n = S$ It is more convenient to prove $(1)$ and $(2)$ separately: === Proof of $(1)$ === Let $l, m \in \N$ be such that $l < m$. Then by Set Difference is Subset, $T_l \subseteq S_l$. As the $S_n$ form an increasing sequence of sets, it follows that also $T_l \subseteq S_{m - 1}$ because $m - 1 \ge l$. Now compute as follows: {{begin-eqn}} {{eqn | l = T_m \cap T_l       | o = \subseteq       | r = T_m \cap S_{m-1}       | c = Set Intersection Preserves Subsets }} {{eqn | o = \subseteq       | r = \paren {S_m \setminus S_{m - 1} } \cap S_{m-1}       | c = Definition of $T_m$ }} {{eqn | r = \O       | c = Set Difference Intersection with Second Set is Empty Set }} {{end-eqn}} Hence $T_m \cap T_l = \O$. Reversing the roles of $m$ and $l$ leads to the same conclusion if $l > m$. Hence, by definition, the $T_n$ are pairwise disjoint. {{qed|lemma}} === Proof of $(2)$ === By Set Union Preserves Subsets and Set Difference is Subset, have that: :$\displaystyle \bigcup_{n \mathop \in \N} T_n \subseteq \bigcup_{n \mathop \in \N} S_n = S$ To establish $(2)$, by definition of set equality, it is now only required to show that $S \subseteq \displaystyle \bigcup_{n \mathop \in \N} T_n$. So let $s \in S$. Then by definition of union, the set: :$N_s := \set {n \in \N: s \in S_n}$ is nonempty. By Well-Ordering Principle, $N_s$ contains a minimal element, $n$, say. If $n = 1$, then $s \in S_1 = T_1$. If $n > 1$, then, by minimality of $n$, $s \notin S_{n - 1}$. Hence, by definition of set difference, $s \in T_n = S_n \setminus S_{n - 1}$. By definition of set union, it follows that: :$s \in \displaystyle \bigcup_{n \mathop \in \N} T_n$ That is, by definition of subset: :$S \subseteq \displaystyle \bigcup_{n \mathop \in \N} T_n$ {{qed}} Category:Set Theory n1ezkqegevtsxnuz00948pd2infibcm
From Non-Zero Complex Numbers under Multiplication form Abelian Group, $\struct {\C_{\ne 0}, \times}$ is a group. Therefore $\struct {\C_{\ne 0}, \times}$ is closed by group axiom $G0$. Let $w, z \in \C_{\ne 0}$. {{begin-eqn}} {{eqn | l = \map {f_n} {w \times z}       | r = \paren {w \times z}^n       | c =  }} {{eqn | r = w^n \times z^n       | c = Power of Product of Commutative Elements in Group }} {{eqn | r = \map {f_n} w \times \map {f_n} z       | c =  }} {{end-eqn}} Thus $f_n$ is a group homomorphism. Now suppose $w = r \paren {\cos \alpha + i \sin \alpha}$, expressing $w$ in polar form. Then $w = \map {f_n} z$ where: :$z = r^{1/n} \paren {\cos \dfrac \alpha n + i \sin \dfrac \alpha n}$ and so: :$\forall w: w \in \map {f_n} {\C_{\ne 0} }$ That is, $f_n$ is a surjection. Being a group homomorphism which is also a surjection, by definition $f_n$ is then a group epimorphism. The kernel of $f_n$ is the set: :$U_n = \set {e^{2 i k \pi / n}: k \in \N_n}$ which follows from Complex Roots of Unity in Exponential Form. {{qed}}
From Real Part as Mapping is Surjection, $f$ is a surjection. Let $z_1, z_2 \in \C$. Let $z_1 = x_1 + i y_1, z_2 = x_2 + i y_2$. Then: {{begin-eqn}} {{eqn | l = \map f {z_1 + z_2}       | r = \map \Re {z_1 + z_2}       | c = Definition of $f$ }} {{eqn | r = \map \Re {x_1 + i y_1 + x_2 + i y_2}       | c = Definition of $z_1$ and $z_2$ }} {{eqn | r = x_1 + x_2       | c = {{Defof|Real Part}} }} {{eqn | r = \map \Re {z_1} + \map \Re {z_2}       | c = {{Defof|Real Part}} }} {{eqn | r = \map f {z_1} + \map f {z_2}       | c = Definition of $f$ }} {{end-eqn}} So $f$ is a group homomorphism. Thus $f$ is a surjective group homomorphism and therefore by definition a group epimorphism. Finally: :$\forall y \in \R: \map \Re {0 + i y} = 0$ It follows from Complex Addition Identity is Zero that:  :$\map \ker f = \set {i x: x \in \R}$ {{qed}}