avewright/proofwiki-math huggingface dataset:
Theorem: Combination Theorem for Sequences/Complex/Product Rule

:$\ds \lim_{n \mathop \to \infty} \paren {z_n w_n} = c d$

Proof:
Let $z_n = x_n + i y_n$.
Let $w_n = u_n + i v_n$.
Let $c = a + i b$
Let $d = e + i f$.
By definition of convergent complex sequence:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} z_n
      | r = c
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \lim_{n \mathop \to \infty} x_n + i \lim_{n \mathop \to \infty} y_n
      | r = a + i b
      | c = {{Defof|Convergent Complex Sequence}}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} w_n
      | r = d
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \lim_{n \mathop \to \infty} u_n + i \lim_{n \mathop \to \infty} v_n
      | r = e + i f
      | c = {{Defof|Convergent Complex Sequence}}
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} z_n w_n
      | r = \lim_{n \mathop \to \infty} \paren {\paren {x_n u_n - y_n v_n} + i \paren {y_n u_n + x_n v_n} }
      | c = {{Defof|Complex Multiplication}}
}}
{{eqn | r = \lim_{n \mathop \to \infty} \paren {x_n u_n - y_n v_n} + i \lim_{n \mathop \to \infty} \paren {y_n u_n + x_n v_n}
      | c = {{Defof|Convergent Complex Sequence}}
}}
{{eqn | r = \paren {\lim_{n \mathop \to \infty} \paren {x_n u_n} - \lim_{n \mathop \to \infty} \paren {y_n v_n} } + i \paren {\lim_{n \mathop \to \infty} \paren {y_n u_n} + \lim_{n \mathop \to \infty} \paren {x_n v_n} }
      | c = Sum Rule for Real Sequences
}}
{{eqn | r = \paren {\lim_{n \mathop \to \infty} \paren {x_n} \lim_{n \mathop \to \infty} \paren {u_n} - \lim_{n \mathop \to \infty} \paren {y_n} \lim_{n \mathop \to \infty} \paren {v_n} } + i \paren {\lim_{n \mathop \to \infty} \paren {y_n} \lim_{n \mathop \to \infty} \paren {u_n} + \lim_{n \mathop \to \infty} \paren {x_n} \lim_{n \mathop \to \infty} \paren {v_n} }
      | c = Product Rule for Real Sequences
}}
{{eqn | r = \paren {a e - b f} + i \paren {b e + a f}
      | c = 
}}
{{eqn | r = \paren {a + i b} \paren {e + i f}
      | c = {{Defof|Complex Multiplication}}
}}
{{eqn | r = c d
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Combination Theorem for Sequences/Complex/Quotient Rule

:$\ds \lim_{n \mathop \to \infty} \frac {z_n} {w_n} = \frac c d$
provided that $d \ne 0$.

Proof:
As $z_n \to c$ as $n \to \infty$, it follows from Modulus of Limit that $\size {w_n} \to \size d$ as $n \to \infty$.
As $d \ne 0$, it follows from the definition of the modulus of $d$ that $\size d > 0$.
From Sequence Converges to Within Half Limit, we have $\exists N: \forall n > N: \size {w_n} > \dfrac {\size d} 2$.
Now, for $n > N$, consider:
{{begin-eqn}}
{{eqn | l = \size {\frac {z_n} {w_n} - \frac c d}
      | r = \size {\frac {d z_n - w_n c} {d w_n} }
      | c = 
}}
{{eqn | o = <
      | r = \frac 2 {\size m^2} \size {d z_n - w_n c}
      | c = 
}}
{{end-eqn}}
By the above, $d z_n - w_n c \to d c - d c = 0$ as $n \to \infty$.
The result follows by the Squeeze Theorem for Complex Sequences (which applies as well to real as to complex sequences).
{{qed}}
Theorem: Combination Theorem for Sequences/Normed Division Ring

Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\sequence {x_n}$,  $\sequence {y_n} $ be sequences in $R$.
Let $\sequence {x_n}$ and $\sequence {y_n}$ be convergent in the norm $\norm {\, \cdot \,}$ to the following limits:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
:$\ds \lim_{n \mathop \to \infty} y_n = m$
Let $\lambda, \mu \in R$.
Then the following results hold:
=== Sum Rule ===
{{:Combination Theorem for Sequences/Normed Division Ring/Sum Rule}}
=== Difference Rule ===
{{:Combination Theorem for Sequences/Normed Division Ring/Difference Rule}}
=== Multiple Rule ===
{{:Combination Theorem for Sequences/Normed Division Ring/Multiple Rule}}
=== Combined Sum Rule ===
{{:Combination Theorem for Sequences/Normed Division Ring/Combined Sum Rule}}
=== Product Rule ===
{{:Combination Theorem for Sequences/Normed Division Ring/Product Rule}}
=== Inverse Rule ===
{{:Combination Theorem for Sequences/Normed Division Ring/Inverse Rule}}
=== Quotient Rule ===
{{:Combination Theorem for Sequences/Normed Division Ring/Quotient Rule}}

Proof:
By Convergent Sequence in Normed Division Ring is Bounded, $\sequence {x_n}$ is bounded.
Suppose $\norm {x_n} \le K$ for $n = 1, 2, 3, \ldots$.
Let $M = \max \set {K, \norm m}$.
Then:
:$\norm m \le M$
and:
:$\forall n: \norm{x_n} \le M$
Let $\epsilon > 0$ be given.
Then $\dfrac \epsilon {2 M} > 0$.
As $\sequence {x_n}$ converges to $l$, we can find $N_1$ such that:
:$\forall n > N_1: \norm {x_n - l} < \dfrac \epsilon {2 M}$
Similarly, for $\sequence {y_n}$ we can find $N_2$ such that:
:$\forall n > N_2: \norm {y_n - m} < \dfrac \epsilon {2 M}$
Now let $N = \max \set {N_1, N_2}$.
Then if $n > N$, both the above inequalities will be true.
Thus $\forall n > N$:
{{begin-eqn}}
{{eqn | l = \norm {x_n y_n - l m}
      | r = \norm {x_n y_n - x_n m + x_n m - l m}
      | c = 
}}
{{eqn | o = \le
      | r = \norm {x_n y_n - x_n m} + \norm {x_n m - l m}
      | c = Axiom (N3) of norm (Triangle Inequality).
}}
{{eqn | r = \norm {x_n \paren {y_n - m } } + \norm {\paren {x_n - l } m}
      | c = 
}}
{{eqn | o = \le
      | r = \norm {x_n} \cdot \norm {y_n - m} + \norm {x_n - l} \cdot \norm m
      | c = Axiom (N2) of norm (Multiplicativity).
}}
{{eqn | o = \le
      | r = M \cdot \norm {y_n - m} + \norm {x_n - l} \cdot M
      | c = since $\sequence {x_n}$ is bounded by $M$ and $\norm m \le M$
}}
{{eqn | o = \le
      | r = M \cdot \dfrac \epsilon {2 M} + \dfrac \epsilon {2 M} \cdot M
      | c = 
}}
{{eqn | r = \dfrac \epsilon 2 + \dfrac \epsilon 2
      | c = 
}}
{{eqn | r = \epsilon
      | c = 
}}
{{end-eqn}}
Hence:
:$\sequence {x_n y_n}$ is convergent.
It follows that:
:$\ds \lim_{n \mathop \to \infty} \paren {x_n y_n} = l m$
{{qed}}
Theorem: Combination Theorem for Sequences/Normed Division Ring

Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\sequence {x_n}$,  $\sequence {y_n} $ be sequences in $R$.
Let $\sequence {x_n}$ and $\sequence {y_n}$ be convergent in the norm $\norm {\, \cdot \,}$ to the following limits:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
:$\ds \lim_{n \mathop \to \infty} y_n = m$
Let $\lambda, \mu \in R$.
Then the following results hold:
=== Sum Rule ===
{{:Combination Theorem for Sequences/Normed Division Ring/Sum Rule}}
=== Difference Rule ===
{{:Combination Theorem for Sequences/Normed Division Ring/Difference Rule}}
=== Multiple Rule ===
{{:Combination Theorem for Sequences/Normed Division Ring/Multiple Rule}}
=== Combined Sum Rule ===
{{:Combination Theorem for Sequences/Normed Division Ring/Combined Sum Rule}}
=== Product Rule ===
{{:Combination Theorem for Sequences/Normed Division Ring/Product Rule}}
=== Inverse Rule ===
{{:Combination Theorem for Sequences/Normed Division Ring/Inverse Rule}}
=== Quotient Rule ===
{{:Combination Theorem for Sequences/Normed Division Ring/Quotient Rule}}

Proof:
By Convergent Sequence in Normed Division Ring is Bounded, $\sequence {x_n}$ is bounded.
Suppose $\norm {x_n} \le K$ for $n = 1, 2, 3, \ldots$.
Then for $n = 1, 2, 3, \ldots$:
{{begin-eqn}}
{{eqn | l = \norm {x_n y_n - l m}
      | r = \norm {x_n y_n - x_n m + x_n m - l m}
      | c = 
}}
{{eqn | o = \le
      | r = \norm {x_n y_n - x_n m} + \norm {x_n m - l m}
      | c = {{Norm-axiom-mult|3}}
}}
{{eqn | r = \norm {x_n} \norm {y_n - m} + \norm {x_n - l} \norm m 
      | c = {{Norm-axiom-mult|2}}
}}
{{eqn | o = \le
      | r = K \norm {y_n - m} + \norm m \norm {x_n - l}
      | c = as $\sequence {x_n}$ is bounded by $K$
}}
{{eqn | o = =:
      | r = z_n
      | c = 
}}
{{end-eqn}}
We note that $\sequence {z_n}$ is a real sequence.
But $x_n \to l$ as $n \to \infty$.
So from Definition:Convergent Sequence in Normed Division Ring:
:$\norm {x_n - l} \to 0$ as $n \to \infty$
Similarly $\norm {y_n - m} \to 0$ as $n \to \infty$.
From the Combined Sum Rule for Real Sequences:
:$\ds \lim_{n \mathop \to \infty} \paren {\lambda x'_n + \mu y'_n} = \lambda l' + \mu m'$, $z_n \to 0$ as $n \to \infty$
By applying the Squeeze Theorem for Complex Sequences (which applies as well to real as to complex sequences):
:$\sequence {\norm {x_n y_n - l m}}$ converges to $0$ in $\R$.
By definition of a convergent sequence in a normed division ring:
:$\sequence{x_n y_n}$ is convergent in $R$
It follows that:
:$\ds \lim_{n \mathop \to \infty} \paren {x_n y_n} = l m$
{{qed}}
Theorem: Combination Theorem for Sequences/Normed Division Ring/Product Rule

:$\sequence {x_n y_n}$ is convergent to the limit $\ds \lim_{n \mathop \to \infty} \paren {x_n y_n} = l m$

Proof:
By Convergent Sequence in Normed Division Ring is Bounded, $\sequence {x_n}$ is bounded.
Suppose $\norm {x_n} \le K$ for $n = 1, 2, 3, \ldots$.
Let $M = \max \set {K, \norm m}$.
Then:
:$\norm m \le M$
and:
:$\forall n: \norm{x_n} \le M$
Let $\epsilon > 0$ be given.
Then $\dfrac \epsilon {2 M} > 0$.
As $\sequence {x_n}$ converges to $l$, we can find $N_1$ such that:
:$\forall n > N_1: \norm {x_n - l} < \dfrac \epsilon {2 M}$
Similarly, for $\sequence {y_n}$ we can find $N_2$ such that:
:$\forall n > N_2: \norm {y_n - m} < \dfrac \epsilon {2 M}$
Now let $N = \max \set {N_1, N_2}$.
Then if $n > N$, both the above inequalities will be true.
Thus $\forall n > N$:
{{begin-eqn}}
{{eqn | l = \norm {x_n y_n - l m}
      | r = \norm {x_n y_n - x_n m + x_n m - l m}
      | c = 
}}
{{eqn | o = \le
      | r = \norm {x_n y_n - x_n m} + \norm {x_n m - l m}
      | c = Axiom (N3) of norm (Triangle Inequality).
}}
{{eqn | r = \norm {x_n \paren {y_n - m } } + \norm {\paren {x_n - l } m}
      | c = 
}}
{{eqn | o = \le
      | r = \norm {x_n} \cdot \norm {y_n - m} + \norm {x_n - l} \cdot \norm m
      | c = Axiom (N2) of norm (Multiplicativity).
}}
{{eqn | o = \le
      | r = M \cdot \norm {y_n - m} + \norm {x_n - l} \cdot M
      | c = since $\sequence {x_n}$ is bounded by $M$ and $\norm m \le M$
}}
{{eqn | o = \le
      | r = M \cdot \dfrac \epsilon {2 M} + \dfrac \epsilon {2 M} \cdot M
      | c = 
}}
{{eqn | r = \dfrac \epsilon 2 + \dfrac \epsilon 2
      | c = 
}}
{{eqn | r = \epsilon
      | c = 
}}
{{end-eqn}}
Hence:
:$\sequence {x_n y_n}$ is convergent.
It follows that:
:$\ds \lim_{n \mathop \to \infty} \paren {x_n y_n} = l m$
{{qed}}
Theorem: Combination Theorem for Sequences/Normed Division Ring/Product Rule

:$\sequence {x_n y_n}$ is convergent to the limit $\ds \lim_{n \mathop \to \infty} \paren {x_n y_n} = l m$

Proof:
By Convergent Sequence in Normed Division Ring is Bounded, $\sequence {x_n}$ is bounded.
Suppose $\norm {x_n} \le K$ for $n = 1, 2, 3, \ldots$.
Then for $n = 1, 2, 3, \ldots$:
{{begin-eqn}}
{{eqn | l = \norm {x_n y_n - l m}
      | r = \norm {x_n y_n - x_n m + x_n m - l m}
      | c = 
}}
{{eqn | o = \le
      | r = \norm {x_n y_n - x_n m} + \norm {x_n m - l m}
      | c = {{Norm-axiom-mult|3}}
}}
{{eqn | r = \norm {x_n} \norm {y_n - m} + \norm {x_n - l} \norm m 
      | c = {{Norm-axiom-mult|2}}
}}
{{eqn | o = \le
      | r = K \norm {y_n - m} + \norm m \norm {x_n - l}
      | c = as $\sequence {x_n}$ is bounded by $K$
}}
{{eqn | o = =:
      | r = z_n
      | c = 
}}
{{end-eqn}}
We note that $\sequence {z_n}$ is a real sequence.
But $x_n \to l$ as $n \to \infty$.
So from Definition:Convergent Sequence in Normed Division Ring:
:$\norm {x_n - l} \to 0$ as $n \to \infty$
Similarly $\norm {y_n - m} \to 0$ as $n \to \infty$.
From the Combined Sum Rule for Real Sequences:
:$\ds \lim_{n \mathop \to \infty} \paren {\lambda x'_n + \mu y'_n} = \lambda l' + \mu m'$, $z_n \to 0$ as $n \to \infty$
By applying the Squeeze Theorem for Complex Sequences (which applies as well to real as to complex sequences):
:$\sequence {\norm {x_n y_n - l m}}$ converges to $0$ in $\R$.
By definition of a convergent sequence in a normed division ring:
:$\sequence{x_n y_n}$ is convergent in $R$
It follows that:
:$\ds \lim_{n \mathop \to \infty} \paren {x_n y_n} = l m$
{{qed}}
Theorem: Combination Theorem for Sequences/Normed Division Ring/Multiple Rule

:$\sequence {\lambda x_n}$ is convergent and $\ds \lim_{n \mathop \to \infty} \paren {\lambda x_n} = \lambda l$

Proof:
Let
:$\sequence {\tilde{x}_n} := \tuple {\lambda, \lambda, \lambda, \ldots}$
and:
:$\sequence {y_n} := \sequence {x_n}$
The claim follows from Product Rule for Sequences in Normed Division Ring, since:
:$\sequence {\lambda x_n} = \sequence {\tilde{x}_n y_n}$
{{qed}}
Theorem: Combination Theorem for Sequences/Normed Division Ring/Combined Sum Rule

:$\sequence {\lambda x_n + \mu y_n }$ is convergent and $\ds \lim_{n \mathop \to \infty} \paren {\lambda x_n + \mu y_n} = \lambda l + \mu m$

Proof:
From the Multiple Rule for Sequences in Normed Division Ring, we have:
:$\ds \lim_{n \mathop \to \infty} \paren {\lambda x_n} = \lambda l$
:$\ds \lim_{n \mathop \to \infty} \paren {\mu y_n} = \mu m$
The result now follows directly from the Sum Rule for Sequences in Normed Division Ring:
:$\ds \lim_{n \mathop \to \infty} \paren {\lambda x_n + \mu y_n} = \lambda l + \mu m$
{{qed}}
Theorem: Combination Theorem for Sequences/Normed Division Ring/Difference Rule

:$\sequence {x_n - y_n}$ is convergent and $\ds \lim_{n \mathop \to \infty} \paren {x_n - y_n} = l - m$

Proof:
From Sum Rule for Sequences in Normed Division Ring:
:$\ds \lim_{n \mathop \to \infty} \paren {x_n + y_n} = l + m$
From Multiple Rule for Sequences in Normed Division Ring:
:$\ds \lim_{n \mathop \to \infty} \paren {-y_n} = -m$
Hence:
:$\ds \lim_{n \mathop \to \infty} \paren {x_n + \paren {-y_n} } = l + \paren {-m}$
The result follows.
{{qed}}
Theorem: Absolutely Convergent Series is Convergent/Complex Numbers

Let $\ds \sum_{n \mathop = 1}^\infty z_n$ be an absolutely convergent series in $\C$.
Then $\ds \sum_{n \mathop = 1}^\infty z_n$ is convergent.

Proof:
Let $z_n = u_n + i v_n$.
We have that:
{{begin-eqn}}
{{eqn | l = \cmod {z_n}
      | r = \sqrt { {u_n}^2 + {v_n}^2}
      | c = 
}}
{{eqn | o = >
      | r = \sqrt { {u_n}^2}
      | c = 
}}
{{eqn | o = >
      | r = \size {u_n}
      | c = 
}}
{{end-eqn}}
and similarly:
:$\cmod {z_n} > \size {v_n}$
From the Comparison Test, the series $\ds \sum_{n \mathop = 1}^\infty u_n$ and $\ds \sum_{n \mathop = 1}^\infty v_n$ are absolutely convergent.
From Absolutely Convergent Real Series is Convergent, $\ds \sum_{n \mathop = 1}^\infty u_n$ and $\ds \sum_{n \mathop = 1}^\infty v_n$ are convergent.
By Convergence of Series of Complex Numbers by Real and Imaginary Part, it follows that $\ds \sum_{n \mathop = 1}^\infty z_n$ is convergent.
{{qed}}
Theorem: Cauchy's Convergence Criterion for Series

A series $\ds \sum_{i \mathop = 0}^\infty a_i$ is convergent {{iff}} for every $\epsilon > 0$ there is a number $N \in \N$ such that:
:$\size {a_{n + 1} + a_{n + 2} + \cdots + a_m} < \epsilon$
holds for all $n \ge N$ and $m > n$.
{{explain|What domain is $\sequence {a_n}$ in?}}

Proof:
Let:
:$\ds s_n = \sum_{i \mathop = 0}^n a_i$
Then $\sequence {s_n}$ is a sequence in $\R$. 
From Cauchy's Convergence Criterion on Real Numbers, $\sequence {s_n}$ is convergent {{iff}} it is a Cauchy sequence.
For $m > n$ we have:
:$\size {s_m - s_n} = \size {a_{n + 1} + a_{n + 2} + \cdots + a_m}$
{{qed}}
Category:Cauchy's Convergence Criterion
Category:Real Analysis
pp79uqj7s0mgycsq1gb6gt0zp27uab1
Theorem: Combination Theorem for Sequences/Normed Division Ring/Product Rule/Proof 1

Let $\struct {R, \norm {\,\cdot\,}}$ be a normed division ring.
Let $\sequence {x_n}$,  $\sequence {y_n} $ be sequences in $R$.
Let $\sequence {x_n}$ and $\sequence {y_n}$ be convergent in the norm $\norm {\,\cdot\,}$ to the following limits:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
:$\ds \lim_{n \mathop \to \infty} y_n = m$
Then:
{{:Combination Theorem for Sequences/Normed Division Ring/Product Rule}}

Proof:
By Convergent Sequence in Normed Division Ring is Bounded, $\sequence {x_n}$ is bounded.
Suppose $\norm {x_n} \le K$ for $n = 1, 2, 3, \ldots$.
Let $M = \max \set {K, \norm m}$.
Then:
:$\norm m \le M$
and:
:$\forall n: \norm{x_n} \le M$
Let $\epsilon > 0$ be given.
Then $\dfrac \epsilon {2 M} > 0$.
As $\sequence {x_n}$ converges to $l$, we can find $N_1$ such that:
:$\forall n > N_1: \norm {x_n - l} < \dfrac \epsilon {2 M}$
Similarly, for $\sequence {y_n}$ we can find $N_2$ such that:
:$\forall n > N_2: \norm {y_n - m} < \dfrac \epsilon {2 M}$
Now let $N = \max \set {N_1, N_2}$.
Then if $n > N$, both the above inequalities will be true.
Thus $\forall n > N$:
{{begin-eqn}}
{{eqn | l = \norm {x_n y_n - l m}
      | r = \norm {x_n y_n - x_n m + x_n m - l m}
      | c = 
}}
{{eqn | o = \le
      | r = \norm {x_n y_n - x_n m} + \norm {x_n m - l m}
      | c = Axiom (N3) of norm (Triangle Inequality).
}}
{{eqn | r = \norm {x_n \paren {y_n - m } } + \norm {\paren {x_n - l } m}
      | c = 
}}
{{eqn | o = \le
      | r = \norm {x_n} \cdot \norm {y_n - m} + \norm {x_n - l} \cdot \norm m
      | c = Axiom (N2) of norm (Multiplicativity).
}}
{{eqn | o = \le
      | r = M \cdot \norm {y_n - m} + \norm {x_n - l} \cdot M
      | c = since $\sequence {x_n}$ is bounded by $M$ and $\norm m \le M$
}}
{{eqn | o = \le
      | r = M \cdot \dfrac \epsilon {2 M} + \dfrac \epsilon {2 M} \cdot M
      | c = 
}}
{{eqn | r = \dfrac \epsilon 2 + \dfrac \epsilon 2
      | c = 
}}
{{eqn | r = \epsilon
      | c = 
}}
{{end-eqn}}
Hence:
:$\sequence {x_n y_n}$ is convergent.
It follows that:
:$\ds \lim_{n \mathop \to \infty} \paren {x_n y_n} = l m$
{{qed}}
Theorem: Combination Theorem for Sequences/Normed Division Ring/Product Rule/Proof 2

Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring.
Let $\sequence {x_n}$ and $\sequence {y_n}$ be sequences in $R$.
Let $\sequence {x_n}$ and $\sequence {y_n}$ be convergent in the norm $\norm {\,\cdot\,}$ to the following limits:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
:$\ds \lim_{n \mathop \to \infty} y_n = m$
Then:
{{:Combination Theorem for Sequences/Normed Division Ring/Product Rule}}

Proof:
By Convergent Sequence in Normed Division Ring is Bounded, $\sequence {x_n}$ is bounded.
Suppose $\norm {x_n} \le K$ for $n = 1, 2, 3, \ldots$.
Then for $n = 1, 2, 3, \ldots$:
{{begin-eqn}}
{{eqn | l = \norm {x_n y_n - l m}
      | r = \norm {x_n y_n - x_n m + x_n m - l m}
      | c = 
}}
{{eqn | o = \le
      | r = \norm {x_n y_n - x_n m} + \norm {x_n m - l m}
      | c = {{Norm-axiom-mult|3}}
}}
{{eqn | r = \norm {x_n} \norm {y_n - m} + \norm {x_n - l} \norm m 
      | c = {{Norm-axiom-mult|2}}
}}
{{eqn | o = \le
      | r = K \norm {y_n - m} + \norm m \norm {x_n - l}
      | c = as $\sequence {x_n}$ is bounded by $K$
}}
{{eqn | o = =:
      | r = z_n
      | c = 
}}
{{end-eqn}}
We note that $\sequence {z_n}$ is a real sequence.
But $x_n \to l$ as $n \to \infty$.
So from Definition:Convergent Sequence in Normed Division Ring:
:$\norm {x_n - l} \to 0$ as $n \to \infty$
Similarly $\norm {y_n - m} \to 0$ as $n \to \infty$.
From the Combined Sum Rule for Real Sequences:
:$\ds \lim_{n \mathop \to \infty} \paren {\lambda x'_n + \mu y'_n} = \lambda l' + \mu m'$, $z_n \to 0$ as $n \to \infty$
By applying the Squeeze Theorem for Complex Sequences (which applies as well to real as to complex sequences):
:$\sequence {\norm {x_n y_n - l m}}$ converges to $0$ in $\R$.
By definition of a convergent sequence in a normed division ring:
:$\sequence{x_n y_n}$ is convergent in $R$
It follows that:
:$\ds \lim_{n \mathop \to \infty} \paren {x_n y_n} = l m$
{{qed}}
Theorem: Convergent Sequence in Normed Division Ring is Bounded

Let $\struct {R, \norm {\,\cdot\,}}$ be a normed division ring.
Let $\sequence {x_n}$ be a sequence in $R$.
Let $\sequence {x_n}$ be convergent in the norm $\norm {\,\cdot\,}$ to the following limit:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
Then $\sequence {x_n}$ is bounded.

Proof:
Let $\sequence {x_n}$ be convergent in the norm $\norm {\,\cdot\,}$ to the limit $l$, then:
:$\forall \epsilon \in \R_{>0}: \exists N \in \N : \forall n \ge N: \norm {x_n - l} < \epsilon$
Let $n_1$ satisfy:
:$\forall n \ge n_1: \norm {x_n - l} < 1$
Then $\forall n \ge n_1$:
{{begin-eqn}}
{{eqn | l = \norm {x_n}
      | r = \norm {x_n - l + l}
}}
{{eqn | o = \le
      | r = \norm {x_n - l} + \norm l
      | c = {{Norm-axiom-mult|3}}
}}
{{eqn | o = \le
      | r = 1 + \norm l
      | c = as $n \ge n_1$
}}
{{end-eqn}}
Let $K = \max \set {\norm {x_1}, \norm {x_2}, \dots, \norm {x_{n_1 - 1} }, 1 + \norm l}$.
Then:
:$\forall n < n_1: \norm {x_n} \le K$
:$\forall n \ge n_1: \norm {x_n} \le 1 + \norm l \le K$
Thus, by definition, $\sequence {x_n}$ is bounded.
{{qed}}
Theorem: Convergent Sequence in Normed Division Ring is Bounded

Let $\struct {R, \norm {\,\cdot\,}}$ be a normed division ring.
Let $\sequence {x_n}$ be a sequence in $R$.
Let $\sequence {x_n}$ be convergent in the norm $\norm {\,\cdot\,}$ to the following limit:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
Then $\sequence {x_n}$ is bounded.

Proof:
Let $d$ be the metric induced on $R$ be the norm $\norm {\,\cdot\,}$.
Let $\sequence {x_n}$ be convergent to the limit $l$ in $\struct {R, \norm {\,\cdot\,}}$.
By the definition of a convergent sequence in a normed division ring, $\sequence {x_n} $ is convergent to the limit $l$ in $\struct {R, d}$.
By Convergent Sequence in Metric Space is Bounded, $\sequence {x_n} $ is a bounded sequence in $\struct {R, d}$.
By Sequence is Bounded in Norm iff Bounded in Metric, $\sequence {x_n} $ is a bounded sequence in $\struct {R, \norm {\,\cdot\,} }$.
{{qed}}
Theorem: Convergent Sequence in Normed Division Ring is Bounded

Let $\struct {R, \norm {\,\cdot\,}}$ be a normed division ring.
Let $\sequence {x_n}$ be a sequence in $R$.
Let $\sequence {x_n}$ be convergent in the norm $\norm {\,\cdot\,}$ to the following limit:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
Then $\sequence {x_n}$ is bounded.

Proof:
Let $\sequence {x_n}$ be convergent to the limit $l$ in $\struct {R, \norm {\,\cdot\,} }$.
By modulus of limit in normed division ring, $\sequence {\norm {x_n} }$ is a convergent sequence in $\R$.
By Convergent Real Sequence is Bounded, $\sequence {\norm {x_n} }$ is bounded.
That is:
:$\exists M \in \R_{> 0}: \forall n, \norm {x_n} = \size {\norm {x_n} } \le M$
Thus, by definition, $\sequence {x_n}$ is bounded.
{{qed}}
Theorem: Convergent Sequence in Normed Division Ring is Bounded

Let $\struct {R, \norm {\,\cdot\,}}$ be a normed division ring.
Let $\sequence {x_n}$ be a sequence in $R$.
Let $\sequence {x_n}$ be convergent in the norm $\norm {\,\cdot\,}$ to the following limit:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
Then $\sequence {x_n}$ is bounded.

Proof:
Let $\sequence {x_n}$ be convergent to the limit $l$ in $\struct {R, \norm {\,\cdot\,}}$.
By Convergent Sequence is Cauchy Sequence in Normed Division Ring, $\sequence {x_n}$ is a Cauchy sequence in $\struct {R, \norm {\,\cdot\,}}$.
By Cauchy Sequence in Normed Division Ring is Bounded, $\sequence {x_n}$ is a bounded sequence in $\struct {R, \norm {\,\cdot\,}}$.
{{qed}}
Theorem: Metric Induced by Norm on Normed Division Ring is Metric

Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring with zero and unity denoted by $0_R$ and $1_R$ respectively.
Let $d$ be the metric induced by $\norm{\,\cdot\,}$.
Then $d$ is a metric.

Proof:
=== Proof of {{Metric-space-axiom|1|nolink}} and {{Metric-space-axiom|4|nolink}} ===
Let $x, y \in R$. 
Then $\map d {x, y} = \norm {x - y} \ge 0$, and furthermore:
{{begin-eqn}}
{{eqn | l = \map d {x, y}
      | r = 0
}}
{{eqn | ll= \leadstoandfrom
      | l = \norm {x - y}
      | r = 0
      | c = {{Defof|Metric Induced by Norm on Division Ring}}
}}
{{eqn | ll= \leadstoandfrom
      | l = x - y
      | r = \mathbf 0_R
      | c = {{Norm-axiom-mult|1}}
}}
{{eqn | ll= \leadstoandfrom
      | l = x
      | r = y
}}
{{end-eqn}}
{{qed|lemma}}