hcju/proofwikiqa huggingface dataset:
The additive groups of $\struct {R_1, +_1, \circ_1}$ and $\struct {R_2, +_2, \circ_2}$ are $\struct {R_1, +_1}$ and $\struct {R_2, +_2}$ respectively. Their identities are $0_1$ and $0_2$ respectively. Thus from the Constant Mapping to Group Identity is Homomorphism we have that $\zeta: \struct {R_1, +_1} \to \struct {R_2, +_2}$ is a group homomorphism: :$\map \zeta {x +_1 y} = \map \zeta x +_2 \map \zeta y$ Then we have: {{begin-eqn}} {{eqn | l = \map \zeta {x \circ_1 y}       | r = 0_2       | c = as $x \circ_1 y \in R_1$ }} {{eqn | r = \map \zeta x \circ_2 \map \zeta y       | c = as $\map \zeta x = 0_2$ and $\map \zeta y = 0_2$ }} {{end-eqn}} The results about image and kernel follow directly by definition. {{qed}}
Let $x, y \in G_1$. Then: {{begin-eqn}} {{eqn | l = \map {\phi_e} {x \circ_1 y}       | r = e_2       | c = as $x \circ_1 y \in G_1$ }} {{eqn | r = \map {\phi_e} x \circ_2 \map {\phi_e} y       | c = as $\map {\phi_e} x = e_2$ and $\map {\phi_e} y = e_2$ }} {{end-eqn}} So $\phi_e$ is a group homomorphism. The results about image and kernel follow directly by definition. {{qed}}
The main result Identity Mapping is Automorphism holds directly. As $I_G$ is a bijection, the only element that maps to $e$ is $e$ itself. Thus the kernel is $\set e$. {{qed}}
The result Identity Mapping is Automorphism holds directly, for both $+$ and $\circ$. As $I_R$ is a bijection, the only element that maps to $0$ is $0$ itself. Thus the kernel is $\set 0$. {{qed}}
First note that: :$\forall a, b \in \Z:  a \ne b \implies \dfrac a 1 = \map \phi a \ne \map \phi b = \dfrac b 1$ and so clearly $\phi$ is an injection. However, take for example $\dfrac 1 2$: :$\not \exists a \in \Z: \map \phi a = \dfrac 1 2$ as $\dfrac 1 2 \notin \Img \phi$. So $\phi$ is not a surjection. Next let $a, b \in \Z$. {{begin-eqn}} {{eqn | l = \map \phi {a + b}       | r = \frac {a + b} 1       | c =  }} {{eqn | r = \frac a 1 + \frac b 1       | c =  }} {{eqn | r = \map \phi a + \map \phi b       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \map \phi {a \times b}       | r = \frac {a \times b} 1       | c =  }} {{eqn | r = \frac {a \times b} {1 \times 1}       | c =  }} {{eqn | r = \frac a 1 \times \frac b 1       | c =  }} {{eqn | r = \map \phi a \times \map \phi b       | c =  }} {{end-eqn}} So $\phi$ can be seen to be an injective, but not surjective, ring homomorphism. Hence the result by definition of ring monomorphism and ring epimorphism. {{qed}}
By definition, if $\struct {F_1, +_1, \times_1}$ and $\struct {F_2, +_2, \times_2}$ are fields then $\struct {F_1^*, \times_1}$ and $\struct {F_2^*, \times_2}$ are groups. Again by definition: :the unity of $\struct {F_1, +_1, \times_1}$ is the identity of $\struct {F_1^*, \times_1}$ :the unity of $\struct {F_2, +_2, \times_2}$ is the identity of $\struct {F_2^*, \times_2}$. The result follows from Group Homomorphism Preserves Identity. {{Qed}}
By definition, if $\struct {F_1, +_1, \times_1}$ and $\struct {F_2, +_2, \times_2}$ are fields then $\struct {F_1^*, \times_1}$ and $\struct {F_2^*, \times_2}$ are groups. Again by definition: :the product inverse of $x$ in $\struct {F_1, +_1, \times_1}$ for $\times_1$ is the product inverse of $x$ in $\struct {F_1^*, \times_1}$ :the product inverse of $x$ in $\struct {F_2, +_2, \times_2}$ for $\times_2$ is the product inverse of $x$ in $\struct {F_2^*, \times_2}$ The result follows from Group Homomorphism Preserves Inverses. {{qed}}
We have by definition that a field is a division ring. The result can be seen to be an application of Ring Homomorphism from Division Ring is Monomorphism or Zero Homomorphism. {{qed}}
Let $\phi: F \to S$ be a ring homomorphism. Suppose $\phi$ is not a monomorphism. By definition, $\phi$ is not an injection. So there must exist $a, b \in F: \map \phi a = \map \phi b$. Let $k = a +_F \paren {-b}$. Then: {{begin-eqn}} {{eqn | l = \map \phi k       | r = \map \phi {a +_F \paren {-b} }       | c =  }} {{eqn | r = \map \phi a +_S \map \phi {-b}       | c =  }} {{eqn | r = \map \phi a +_S \paren {-\map \phi b}       | c =  }} {{eqn | r = 0_S       | c = as $\map \phi a = \map \phi b$ }} {{end-eqn}} As $a \ne b$ then $k \ne 0$ and so has a product inverse $\exists k^{-1} \in F$. So for any $x \in F$ we can write $x = k \circ \paren {k^{-1} \circ x}$ and so: {{begin-eqn}} {{eqn | l = \map \phi x       | r = \map \phi {k \circ \paren {k^{-1} \circ x} }       | c =  }} {{eqn | r = \map \phi k * \map \phi {k^{-1} \circ x}       | c =  }} {{eqn | r = 0_S * \map \phi {k^{-1} \circ x}       | c =  }} {{eqn | r = 0_S       | c =  }} {{end-eqn}} So if $\phi$ is not a monomorphism, it is the zero homomorphism. {{qed}}
Let $a, b \in F$. First note that: {{begin-eqn}} {{eqn | lo= \forall k: 0 < k < p:       | l = \binom p k       | o = \equiv       | r = 0       | rr= \pmod p       | c = Binomial Coefficient of Prime }} {{eqn | ll= \leadsto       | l = \binom p k       | r = r p       | c = for some $r \in \Z$ }} {{eqn | ll= \leadsto       | l = \binom p k a^k b^{p - k}       | r = r p \cdot a^k b^{p - k} }} {{eqn | ll= \leadsto       | l = \binom p k a^k b^{p - k}       | r = 0_F       | c = Characteristic of Field by Annihilator: Prime Characteristic }} {{eqn | ll= \leadsto       | l = \sum_{k \mathop = 1}^{p - 1} \binom p k a^k b^{p - k}       | r = 0_F       | c =  }} {{end-eqn}} So: {{begin-eqn}} {{eqn | l = \map \phi {a + b}       | r = \paren {a + b}^p       | c =  }} {{eqn | r = a^p + \sum_{k \mathop = 1}^{p - 1} \binom p k a^k b^{p - k} + b^p       | c = Binomial Theorem }} {{eqn | r = a^p + 0_F + b^p }} {{eqn | r = a^p + b^p }} {{eqn | r = \map \phi a + \map \phi b       | c =  }} {{end-eqn}} Multiplication is more straightforward: {{begin-eqn}} {{eqn | l = \map \phi {a b}       | r = \paren {a b}^p       | c =  }} {{eqn | r = a^p b^p       | c = Power of Product of Commutative Elements in Monoid }} {{eqn | r = \map \phi a \map \phi b       | c =  }} {{end-eqn}} Thus, $\phi$ is a (field) homomorphism.  It is clear that $\phi$ is not a zero homomorphism, since: {{begin-eqn}} {{eqn | l = \map \phi {1_F}       | r = 1_F^p }} {{eqn | r = 1_F       | c = Power of Identity is Identity }} {{eqn | o = \ne       | r = 0_F }} {{end-eqn}} Hence, from Ring Homomorphism from Field is Monomorphism or Zero Homomorphism, it follows that $\phi$ must be a monomorphism. {{qed}}
Proof by induction: For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition: :$\phi_n$ is a (field) monomorphism. $\map P 0$ is trivially true: :$\map {\phi_0} x = x^{p^0} = x^1 = x$ and we see that $\phi_0$ is the identity automorphism. This is not the zero homomorphism. So from Ring Homomorphism from Field is Monomorphism or Zero Homomorphism, it follows that $\phi_0$ is a ring monomorphism. === Basis for the Induction === First we need to show that $\map P 1$ is true: :$\map {\phi_1} x = x^{p^1} = x^p$ is a (field) monomorphism. This is demonstrated to be a monomorphism in Power to Characteristic of Field is Monomorphism. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\map {\phi_k} x = x^{p^k}$ is a (field) monomorphism. Then we need to show: :$\map {\phi_{k + 1} } x = x^{p^{k + 1} }$ is a (field) monomorphism. === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \map {\phi_{k + 1} } {a + b}       | r = \paren {a + b}^{p^{k + 1} }       | c =  }} {{eqn | r = \paren {\paren {a + b}^{p^k} }^p       | c =  }} {{eqn | r = \paren {a^{p^k} + b^{p^k} }^p       | c = Induction Hypothesis }} {{eqn | r = \paren {a^{p^k} }^p + \paren {b^{p^k} }^p       | c = Basis for the Induction }} {{eqn | r = a^{p^{k + 1} } + b^{p^{k + 1} }       | c =  }} {{eqn | r = \map {\phi_{k + 1} } a + \map {\phi_{k + 1} } b       | c =  }} {{end-eqn}} Multiplication is more straightforward: {{begin-eqn}} {{eqn | l = \map {\phi_{k + 1} } {a b}       | r = \paren {a b}^{p^{k + 1} }       | c =  }} {{eqn | r = a^{p^{k + 1} } b^{p^{k + 1} }       | c =  }} {{eqn | r = \map {\phi_{k + 1} } a \map {\phi_{k + 1} } b       | c =  }} {{end-eqn}} and does not rely on the induction process. Thus, $\phi_{k + 1}$ is a homomorphism. $\phi_{k + 1}$ is not the zero homomorphism, since $\map {\phi_{k + 1} } 1 = 1^{p^{k + 1} } = 1 \ne 0$. So from Ring Homomorphism from Field is Monomorphism or Zero Homomorphism, it follows that $\phi_{k + 1}$ is a ring monomorphism. So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{\ge 0}: \phi_n$ is a (field) monomorphism. {{qed}}
From {{Module-axiom|1}}, $y \to \lambda \circ y$ is an endomorphism of $\struct {G, +_G}$. From {{Module-axiom|2}}, $\mu \to \mu \circ x$ is a homomorphism from $\struct {R, +_R}$ to $\struct {G, +_G}$. The result follows from Homomorphism with Cancellable Codomain Preserves Identity. {{qed}}
From {{Module-axiom|1}}, $y \to \lambda \circ y$ is an endomorphism of $\struct {G, +_G}$. From {{Module-axiom|2}}, $\mu \to \mu \circ x$ is a homomorphism from $\struct {R, +_R}$ to $\struct {G, +_G}$. The result follows from Homomorphism with Identity Preserves Inverses. {{qed}}
This follows by induction from {{Module-axiom|1}}, as follows: For all $m \in \N_{>0}$, let $\map P m$ be the proposition: :$\ds \lambda \circ \paren {\sum_{k \mathop = 1}^m x_k} = \sum_{k \mathop = 1}^m \paren {\lambda \circ x_k}$ === Basis for the Induction === $\map P 1$ is true, as this just says: :$\lambda \circ x_1 = \lambda \circ x_1$ This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P n$ is true, where $n \ge 1$, then it logically follows that $\map P {n + 1}$ is true. So this is our induction hypothesis: :$\ds \lambda \circ \paren {\sum_{k \mathop = 1}^n x_k} = \sum_{k \mathop = 1}^n \paren {\lambda \circ x_k}$ Then we need to show: :$\ds \lambda \circ \paren {\sum_{k \mathop = 1}^{n + 1} x_k} = \sum_{k \mathop = 1}^{n + 1} \paren {\lambda \circ x_k}$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \lambda \circ \paren {\sum_{k \mathop = 1}^{n + 1} x_k}       | r = \lambda \circ \paren {\sum_{k \mathop = 1}^n x_k + x_{n + 1} }       | c =  }} {{eqn | r = \lambda \circ \paren {\sum_{k \mathop = 1}^n x_k} + \lambda \circ x_{n + 1}       | c = {{Module-axiom|1}} }} {{eqn | r = \sum_{k \mathop = 1}^n \paren {\lambda \circ x_k} + \lambda \circ x_{n + 1}       | c = Induction hypothesis }} {{eqn | r = \sum_{k \mathop = 1}^{n + 1} \paren {\lambda \circ x_k}       | c =  }} {{end-eqn}} So $\map P n \implies \map P {n + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\ds \forall m \in \N_{>0}: \lambda \circ \paren {\sum_{k \mathop = 1}^m x_k} = \sum_{k \mathop = 1}^m \paren {\lambda \circ x_k}$ {{qed}}
This follows by induction from {{Module-axiom|2}}, as follows. For all $m \in \N_{>0}$, let $\map P m$ be the proposition: :$\ds \paren {\sum_{k \mathop = 1}^m \lambda_k} \circ x = \sum_{k \mathop = 1}^m \paren {\lambda_k \circ x}$ === Basis for the Induction === $\map P 1$ is true, as this just says: :$\lambda_1 \circ x = \lambda_1 \circ x$ This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P n$ is true, where $n \ge 1$, then it logically follows that $\map P {n + 1}$ is true. So this is our induction hypothesis: :$\ds \paren {\sum_{k \mathop = 1}^n \lambda_k} \circ x = \sum_{k \mathop = 1}^n \paren {\lambda_k \circ x}$ Then we need to show: :$\ds \paren {\sum_{k \mathop = 1}^{n + 1} \lambda_k} \circ x = \sum_{k \mathop = 1}^{n + 1} \paren {\lambda_k \circ x}$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \paren {\sum_{k \mathop = 1}^{n + 1} \lambda_k} \circ x       | r = \paren {\sum_{k \mathop = 1}^n \lambda_k + \lambda_{n + 1} } \circ x       | c =  }} {{eqn | r = \paren {\sum_{k \mathop = 1}^n \lambda_k \circ x} + \lambda_{n + 1} \circ x       | c = {{Module-axiom|2}} }} {{eqn | r = \sum_{k \mathop = 1}^n \paren {\lambda_k \circ x} + \lambda_{n + 1} \circ x       | c = Induction hypothesis }} {{eqn | r = \sum_{k \mathop = 1}^{n + 1} \paren {\lambda_k \circ x}       | c =  }} {{end-eqn}} So $\map P n \implies \map P {n + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\ds \forall m \in \N_{>0}: \paren {\sum_{k \mathop = 1}^m \lambda_k} \circ x = \sum_{k \mathop = 1}^m \paren {\lambda_k \circ x}$ {{qed}}
First let $n = 0$. The assertion follows directly from Scalar Product with Identity. Next, let $n > 0$. The assertion follows directly from Scalar Product with Sum and Product with Sum of Scalar, by letting $m = n$ and making all the $\lambda$'s and $x$'s the same. Finally, let $n < 0$. The assertion follows from Scalar Product with Product for positive $n$, Scalar Product with Inverse, and from Negative Index Law for Monoids. {{qed}}
By Scalar Product with Identity we have: :$\forall \lambda \in R: \lambda \circ e = e$ Hence the result by definition of linearly dependent. {{qed}}
=== Proof of Existence === By the definition of basis, $\mathcal B$ is a spanning set. Hence the result, by the definition of a spanning set. {{qed|lemma}} === Proof of Uniqueness === {{AimForCont}} otherwise, that: :$\displaystyle \sum_{k \mathop = 1}^n \alpha_k \mathbf x_k  = \mathbf x = \sum_{k \mathop = 1}^n \beta_k \mathbf x_k$ where $\alpha_i \ne \beta_i$ for some $1 \le i \le n$. Then: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 1}^n \alpha_k \mathbf x_k - \sum_{k \mathop = 1}^n \beta_k \mathbf x_k       | r = \mathbf x - \mathbf x }} {{eqn | ll = \leadsto       | l = \sum_{k \mathop = 1}^n \paren {\alpha_k - \beta_k} \mathbf x_k       | r = \mathbf 0 }} {{end-eqn}} However, we have that $\mathcal B = \set {\mathbf x_1, \mathbf x_2, \ldots, \mathbf x_n}$ is a basis for $V$. So, by definition, $\mathcal B$ is a linearly independent set. This means that, for $1 \le i \le n$: :$\alpha_i - \beta_i = 0$ and hence $\alpha_i = \beta_i$ for all $1 \le i \le n$. This contradicts our assumption that $\alpha_i \ne \beta_i$ for some $i$. Hence the result, from Proof by Contradiction. {{qed}}
From: :Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set :Bases of Finitely Generated Vector Space have Equal Cardinality and :Sufficient Conditions for Basis of Finite Dimensional Vector Space all we need to do is show that every infinite generator $S$ for $E$ contains a finite generator. Let $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ be an ordered basis of $E$. For each $k \in \closedint 1 n$ there is a finite subset $S_k$ of $S$ such that $a_k$ is a linear combination of $S_k$. Hence $\displaystyle \bigcup_{k \mathop = 1}^n S_k$ is a finite subset of $S$ generating $E$, for the subspace it generates contains $\set {a_1, \ldots, a_n}$ and hence is $E$. {{qed}}
Let $H$ be a linearly independent subset of $E$. By definition of dimension of vector space, $E$ has a basis with exactly $n$ elements. By Sufficient Conditions for Basis of Finite Dimensional Vector Space, $B$ is a generator for $E$. Then by Size of Linearly Independent Subset is at Most Size of Finite Generator, $H$ has at most $n$ elements. {{Qed}}
From Real Numbers form Ordered Integral Domain, $\struct {\R, +, \times, \le}$ forms an ordered integral domain. By definition of ordered integral domain, the usual ordering $\le$ is compatible with ring addition. {{Qed}}
{{begin-eqn}} {{eqn | l = a       | o = >       | r = b       | c = }} {{eqn | ll= \leadsto       | l = a + c       | o = >       | r = b + c       | c = Real Number Ordering is Compatible with Addition }} {{end-eqn}} {{begin-eqn}} {{eqn | l = c       | o = >       | r = d       | c =  }} {{eqn | ll= \leadsto       | l = b + c       | o = >       | r = b + d       | c = Real Number Ordering is Compatible with Addition }} {{end-eqn}} Finally: {{begin-eqn}} {{eqn | l = a + c       | o = >       | r = b + c }} {{eqn | l = b + c       | o = >       | r = b + d }} {{eqn | ll= \leadsto       | l = a + c       | o = >       | r = b + d       | c = Trichotomy Law for Real Numbers }} {{end-eqn}} {{qed}}
Note that as $d > 0$ it follows that $c > d > 0$ and so $c > 0$. {{begin-eqn}} {{eqn | l = a       | o = >       | r = b       | c =  }} {{eqn | ll= \leadsto       | l = a c       | o = >       | r = b c       | c = Real Number Ordering is Compatible with Multiplication, as $c > 0$ }} {{end-eqn}} {{begin-eqn}} {{eqn | l = c       | o = >       | r = d       | c =  }} {{eqn | ll= \leadsto       | l = b c       | o = >       | r = b d       | c = Real Number Ordering is Compatible with Multiplication, as $b > 0$ }} {{end-eqn}} Finally: {{begin-eqn}} {{eqn | l = a c       | o = >       | r = b c }} {{eqn | l = b c       | o = >       | r = b d }} {{eqn | ll= \leadsto       | l = a c       | o = >       | r = b d       | c = Trichotomy Law for Real Numbers }} {{end-eqn}} {{qed}} === Disproof for Negative Parameters === {{:Positive Real Number Inequalities can be Multiplied/Disproof for Negative Parameters}}
Proof by Counterexample: Let $a = c = -1, b = d = -2$. Then $a c = 1$ but $b d = 2$. {{qed}}
Proof by Counterexample: For example, set $a = 5, b = 3, c = 4, d = 1$ Then $a - c = 1$ while $b - d = 2$. {{qed}}
Proof by Counterexample: For example, set $a = 5, b = 3, c = 4, d = 1$ Then $\dfrac a c = 1 \frac 1 4$ while $\dfrac b d = 3$. {{qed}}
Let $\epsilon > 0$ be given. Then $\dfrac \epsilon 2 > 0$. We are given that: : $\displaystyle \lim_{n \mathop \to \infty} x_n = l$ : $\displaystyle \lim_{n \mathop \to \infty} y_n = m$ By definition of the limit of a real sequence, we can find $N_1$ such that: :$\forall n > N_1: \size {x_n - l} < \dfrac \epsilon 2$ where $\size {x_n - l}$ denotes the absolute value of $x_n - l$ Similarly we can find $N_2$ such that: :$\forall n > N_2: \size {y_n - m} < \dfrac \epsilon 2$ Let $N = \max \set {N_1, N_2}$. Then if $n > N$, both the above inequalities will be true: :$n > N_1$ :$n > N_2$ Thus $\forall n > N$: {{begin-eqn}} {{eqn | l = \size {\paren {x_n + y_n} - \paren {l + m} }       | r = \size {\paren {x_n - l} + \paren {y_n - m} }       | c =  }} {{eqn | o = \le       | r = \size {x_n - l} + \size {y_n - m}       | c = Triangle Inequality for Real Numbers }} {{eqn | o = <       | r = \frac \epsilon 2 + \frac \epsilon 2       | c =  }} {{eqn | r = \epsilon       | c =  }} {{end-eqn}} Hence the result: : $\displaystyle \lim_{n \mathop \to \infty} \paren {x_n + y_n} = l + m$ {{qed}}
Let $\epsilon > 0$. We need to find $N$ such that: :$\forall n > N: \size {\lambda x_n - \lambda l} < \epsilon$ If $\lambda = 0$ the result is trivial. So, assume $\lambda \ne 0$. Then $\size \lambda > 0$ from the definition of the absolute value of $\lambda$. Hence $\dfrac \epsilon {\size \lambda} > 0$. We have that $x_n \to l$ as $n \to \infty$. Thus it follows that: :$\exists N: \forall n > N: \size {x_n - l} < \dfrac \epsilon {\size \lambda}$ That is: :$\forall n > N: \size \lambda \size {x_n - l} < \epsilon$ But we have: {{begin-eqn}} {{eqn | l = \size \lambda \size {x_n - l}       | r = \size {\lambda \paren {x_n - l} }       | c = Absolute Value of Product }} {{eqn | r = \size {\lambda x_n - \lambda l}       | c =  }} {{end-eqn}} Hence: : $\displaystyle \lim_{n \mathop \to \infty} \paren {\lambda x_n} = \lambda l$ {{qed}}
From the Multiple Rule for Real Sequences, we have: :$\displaystyle \lim_{n \mathop \to \infty} \paren {\lambda x_n} = \lambda l$ :$\displaystyle \lim_{n \mathop \to \infty} \paren {\mu y_n} = \mu m$ The result now follows directly from the Sum Rule for Real Sequences: :$\displaystyle \lim_{n \mathop \to \infty} \paren {\lambda x_n + \mu y_n} = \lambda l + \mu m$ {{qed}}
Because $\sequence {x_n}$ converges, it is bounded by Convergent Sequence is Bounded. Suppose $\size {x_n} \le K$ for $n = 1, 2, 3, \ldots$. Then: {{begin-eqn}} {{eqn | l = \size {x_n y_n - l m}       | r = \size {x_n y_n - x_n m + x_n m - l m}       | c =  }} {{eqn | o = \le       | r = \size {x_n y_n - x_n m} + \size {x_n m - l m}       | c = Triangle Inequality for Real Numbers }} {{eqn | r = \size {x_n} \size {y_n - m} + \size m \size {x_n - l}       | c = Absolute Value of Product }} {{eqn | o = \le       | r = K \size {y_n - m} + \size m \size {x_n - l}       | c =  }} {{eqn | o = =:       | r = z_n       | c =  }} {{end-eqn}} But $x_n \to l$ as $n \to \infty$. So $\size {x_n - l} \to 0$ as $n \to \infty$ from Convergent Sequence Minus Limit. Similarly $\size {y_n - m} \to 0$ as $n \to \infty$. From the Combined Sum Rule for Real Sequences: :$\displaystyle \lim_{n \mathop \to \infty} \paren {\lambda x_n + \mu y_n} = \lambda l + \mu m$, $z_n \to 0$ as $n \to \infty$ The result follows by the Squeeze Theorem for Real Sequences. {{qed}}
As $y_n \to m$ as $n \to \infty$, it follows from Modulus of Limit that $\size {y_n} \to \size m$ as $n \to \infty$. As $m \ne 0$, it follows from the definition of the modulus of $m$ that $\size m > 0$. As the statement is given, it is possible that $y_n = 0$ for some $n$. At such $n$, the terms $\dfrac {x_n} {y_n}$ are not defined. However, from Sequence Converges to Within Half Limit, we have: : $\exists N: \forall n > N: \size {y_n} > \dfrac {\size m} 2$ Hence for all $n > N$ we have that $y_n \ne 0$. Thus we may restrict our attention to the domain of $\sequence {y_n}$ such that $n > N$, knowing that $\dfrac {x_n} {y_n}$ will be defined in that domain. So, for $n > N$, consider: {{begin-eqn}} {{eqn | l = \size {\frac {x_n} {y_n} - \frac l m}       | r = \size {\frac {m x_n - y_n l} {m y_n} }       | c =  }} {{eqn | o = <       | r = \frac 2 {\size m^2} \size {m x_n - y_n l}       | c =  }} {{end-eqn}} By the above, $m x_n - y_n l \to m l - m l = 0$ as $n \to \infty$. The result follows by the Squeeze Theorem for Real Sequences. {{qed}}
From Negative of Absolute Value: Corollary 1: :$\size {x - l} < \epsilon \iff l - \epsilon < x < l + \epsilon$ Let $\epsilon > 0$. We need to prove that: :$\exists N: \forall n > N: \size {x_n - l} < \epsilon$ As $\displaystyle \lim_{n \mathop \to \infty} y_n = l$ we know that: :$\exists N_1: \forall n > N_1: \size {y_n - l} < \epsilon$ As $\displaystyle \lim_{n \mathop \to \infty} z_n = l$ we know that: :$\exists N_2: \forall n > N_2: \size {z_n - l} < \epsilon$ Let $N = \max \set {N_1, N_2}$. Then if $n > N$, it follows that $n > N_1$ and $n > N_2$. So: :$\forall n > N: l - \epsilon < y_n < l + \epsilon$ :$\forall n > N: l - \epsilon < z_n < l + \epsilon$ But: :$\forall n \in \N: y_n \le x_n \le z_n$ So: :$\forall n > N: l - \epsilon < y_n \le x_n \le z_n < l + \epsilon$ and so: :$\forall n > N: l - \epsilon < x_n < l + \epsilon$ So: :$\forall n > N: \size {x_n - l} < \epsilon$ Hence the result. {{qed}}
In order to show that $\left \langle {z_n} \right \rangle$ is a null sequence, we want to show that: :$\forall \epsilon > 0: \exists N: \forall n > N: \left|{z_n}\right| < \epsilon$ But since $\left \langle {a_n} \right \rangle$ is a null sequence: :$\exists N: \forall n > N: a_n < \epsilon$ So, using this value of $n$, we have: $\left \vert {z_n} \right \vert \le a_n < \epsilon$ Thus $\left\langle{ z_n }\right\rangle$ is a null sequence. {{qed}}
Let $\sequence {x_n}$ be any sequence of points of $S$ such that: :$\forall n \in \N^*: x_n \ne c$ :$\displaystyle \lim_{n \mathop \to \infty} \ x_n = c$ By Limit of Function by Convergent Sequences, we have: :$\displaystyle \lim_{n \mathop \to \infty} \map f {x_n} = l$ :$\displaystyle \lim_{n \mathop \to \infty} \map g {x_n} = m$ By the Sum Rule for Sequences: :$\displaystyle \lim_{n \mathop \to \infty} \paren {\map f {x_n} + \map g {x_n} } = l + m$ Applying Limit of Function by Convergent Sequences again, we get: :$\displaystyle \lim_{x \mathop \to c} \paren {\map f x + \map g x} = l + m$ {{qed}}