avewright/proofwiki-math huggingface dataset:
Theorem: Power to Characteristic of Field is Monomorphism

Let $F$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let the characteristic of $F$ be $p$ where $p \ne 0$.
Let $\phi: F \to F$ be the mapping on $F$ defined as:
:$\forall x \in F: \map \phi x = x^p$
Then $\phi$ is a (field) monomorphism.

Proof:
Let $a, b \in F$.
First note that:
{{begin-eqn}}
{{eqn | q = \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}}
Theorem: Power to Characteristic Power of Field is Monomorphism

Let $F$ be a field whose characteristic is $p$ where $p \ne 0$.
Let $n \in \Z_{\ge 0}$ be any positive integer.
Let $\phi_n: F \to F$ be the mapping on $F$ defined as:
:$\forall x \in F: \map {\phi_n} x = x^{p^n}$
Then $\phi_n$ is a (field) monomorphism.

Proof:
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.
Theorem: Scalar Product with Identity

:$\lambda \ast e = 0_R \ast x = e$

Proof:
From {{Module-axiom|1}}, $y \to \lambda \ast y$ is an endomorphism of $\struct {G, +_G}$.
From {{Module-axiom|2}}, $\mu \to \mu \ast x$ is a homomorphism from $\struct {R, +_R}$ to $\struct {G, +_G}$.
The result follows from Homomorphism with Cancellable Codomain Preserves Identity.
{{qed}}
Theorem: Scalar Product with Inverse

:$\lambda \ast \struct {-x} = \struct {-\lambda} \ast x = -\struct {\lambda \ast x}$

Proof:
From {{Module-axiom|1}}, $y \to \lambda \ast y$ is an endomorphism of $\struct {G, +_G}$.
From {{Module-axiom|2}}, $\mu \to \mu \ast x$ is a homomorphism from $\struct {R, +_R}$ to $\struct {G, +_G}$.
The result follows from Homomorphism with Identity Preserves Inverses.
{{qed}}
Theorem: Scalar Product with Sum

:$\ds \lambda \ast \paren {\sum_{k \mathop = 1}^m x_k} = \sum_{k \mathop = 1}^m \paren {\lambda \ast x_k}$

Proof:
This follows by induction from {{Module-axiom|1}}, as follows:
For all $m \in \N_{>0}$, let $\map P m$ be the proposition:
:$\ds \lambda \ast \paren {\sum_{k \mathop = 1}^m x_k} = \sum_{k \mathop = 1}^m \paren {\lambda \ast x_k}$
Theorem: Product with Sum of Scalar

:$\ds \paren {\sum_{k \mathop = 1}^m \lambda_k} \ast x = \sum_{k \mathop = 1}^m \paren {\lambda_k \ast x}$

Proof:
This follows by induction from {{Module-axiom|2}}, as follows.
For all $m \in \N_{>0}$, let $\map P m$ be the proposition:
:$\ds \paren {\sum_{k \mathop = 1}^m \lambda_k} \ast x = \sum_{k \mathop = 1}^m \paren {\lambda_k \ast x}$
Theorem: Scalar Product with Product

:$\lambda \ast \paren {n \cdot x} = n \cdot \paren {\lambda \ast x} = \paren {n \cdot \lambda} \ast x$

Proof:
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}}
Theorem: Zero Vector is Linearly Dependent

Let $G$ be a group whose identity is $e$.
Let $R$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $\struct {G, +_G, \circ}_R$ be a unitary $R$-module.
Then the singleton set $\set e$ consisting of the zero vector is linearly dependent.

Proof:
By Scalar Product with Identity we have:
:$\forall \lambda \in R: \lambda \circ e = e$
Hence the result by definition of linearly dependent.
{{qed}}
Theorem: Expression of Vector as Linear Combination from Basis is Unique

Let $V$ be a vector space of dimension $n$.
Let $\BB = \set {\mathbf x_1, \mathbf x_2, \ldots, \mathbf x_n}$ be a basis for $V$.
Let $\mathbf x \in V$ be any vector of $V$.
Then $\mathbf x$ can be expressed as a unique linear combination of elements of $\BB$.
{{explain|The definition we have of linear combination doesn't really work for this. We probably need to expand that definition.}}

Proof:
=== Proof of Existence ===
By the definition of basis, $\BB$ is a spanning set.
Hence the result, by the definition of spanning set.
{{qed|lemma}}
Theorem: Generator of Vector Space Contains Basis

:$G$ contains a basis for $E$.

Proof:
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 $\ds \bigcup_{k \mathop = 1}^n S_k$ 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}}
Theorem: Cardinality of Linearly Independent Set is No Greater than Dimension

:$H$ has at most $n$ elements.

Proof:
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}}
Theorem: Cardinality of Generator of Vector Space is not Less than Dimension

Let $V$ be a vector space over a field $F$.
Let $\BB$ be a generator for $V$ containing $m$ elements.
Then:
:$\map {\dim_F} V \le m$
where $\map {\dim_F} V$ is the dimension of $V$.

Proof:
Let $\BB = \set {x_1, x_2, \ldots, x_m}$ be a generator for $G$.
Let $\set {y_1, y_2, \ldots, y_n}$ be a subset of $G$ such that $n > m$.
As $\BB$ generates $G$, there exist $\alpha_{i j} \in F$ where $1 \le i \le m, 1 \le j \le n$ such that:
:$\ds \forall j: 1 \le j \le n: y_j = \sum_{i \mathop = 1}^m \alpha_{i j} x_i$
Now let $\beta_1, \ldots, \beta_n$ be elements of $F$.
Then:
:$\ds \sum_{j \mathop = 1}^n \beta_j y_j = \sum_{i \mathop = 1}^m \sum_{j \mathop = 1}^n \paren {\alpha_{i j} \beta_j} x_i$
As $n > m$ the result Homogeneous Simultaneous Linear Equations with More Unknowns than Equations can be applied.
That is, there exist $\beta_1, \ldots, \beta_n \in F$ which are not all zero such that:
:$\ds \forall i: 1 \le i \le m: y_j = \sum_{j \mathop = 1}^n \alpha_{i j} \beta_j = 0$
That is, such that:
:$\ds \sum_{j \mathop = 1}^n \beta_j y_j = 0$
So $\set {y_1, y_2, \ldots, y_n}$ is linearly dependent.
The result follows by definition of dimension.
{{qed}}
Theorem: Cardinality of Generator of Vector Space is not Less than Dimension

Let $V$ be a vector space over a field $F$.
Let $\BB$ be a generator for $V$ containing $m$ elements.
Then:
:$\map {\dim_F} V \le m$
where $\map {\dim_F} V$ is the dimension of $V$.

Proof:
From Generator of Vector Space Contains Basis there exists a basis $B$ of $E$ such that $B \subseteq G$.
From Cardinality of Basis of Vector Space, $\card B = n$.
The result follows.
{{qed}}
Theorem: Trichotomy Law for Real Numbers

The real numbers obey the '''trichotomy law'''.
That is, $\forall a, b \in \R$, exactly one of the following holds:
{{begin-axiom}}
{{axiom | n = 1
        | lc= $a$ is greater than $b$:
        | m = a > b
}}
{{axiom | n = 2
        | lc= $a$ is equal to $b$:
        | m = a = b
}}
{{axiom | n = 3
        | lc= $a$ is less than $b$:
        | m = a < b
}}
{{end-axiom}}

Proof:
This follows directly Real Numbers form Totally Ordered Field.
{{Qed}}
Theorem: Trichotomy Law for Real Numbers

The real numbers obey the '''trichotomy law'''.
That is, $\forall a, b \in \R$, exactly one of the following holds:
{{begin-axiom}}
{{axiom | n = 1
        | lc= $a$ is greater than $b$:
        | m = a > b
}}
{{axiom | n = 2
        | lc= $a$ is equal to $b$:
        | m = a = b
}}
{{axiom | n = 3
        | lc= $a$ is less than $b$:
        | m = a < b
}}
{{end-axiom}}

Proof:
$\le$ is a total ordering on $\R$.
The trichotomy follows directly from Trichotomy Law.
{{Qed}}
Theorem: Real Number Ordering is Compatible with Addition

:$\forall a, b, c \in \R: a < b \implies a + c < b + c$

Proof:
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}}
Theorem: Real Number Inequalities can be Added

Let $a, b, c, d \in \R$ such that $a > b$ and $c > d$.
Then:
:$a + c > b + d$

Proof:
{{begin-eqn}}
{{eqn | l = a
      | o = >
      | r = b
      | c = {{hypothesis}}
}}
{{eqn | ll= \leadsto
      | l = a + c
      | o = >
      | r = b + c
      | c = Real Number Ordering is Compatible with Addition
}}
{{eqn | l = c
      | o = >
      | r = d
      | c = {{hypothesis}}
}}
{{eqn | ll= \leadsto
      | l = b + c
      | o = >
      | r = b + d
      | c = Real Number Ordering is Compatible with Addition
}}
{{eqn | ll= \leadsto
      | l = a + c
      | o = >
      | r = b + d
      | c = Transitive Law
}}
{{end-eqn}}
{{qed}}
Theorem: Positive Real Number Inequalities can be Multiplied

Let $a, b, c, d \in \R$ such that $a > b$ and $c > d$.
Let $b > 0$ and $d > 0$.
Then $a c > b d$.
If $b < 0$ or $d < 0$ the inequality does not hold.

Proof:
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}}
Theorem: Positive Real Number Inequalities can be Multiplied

Let $a, b, c, d \in \R$ such that $a > b$ and $c > d$.
Let $b > 0$ and $d > 0$.
Then $a c > b d$.
If $b < 0$ or $d < 0$ the inequality does not hold.

Proof:
Proof by Counterexample:
Let $a = c = -1, b = d = -2$.
Then $a c = 1$ but $b d = 2$.
{{qed}}
Theorem: Real Ordering is not Compatible with Subtraction

Let $a, b, c, d \in R$ be real numbers such that $a > b$ and $c > d$.
Then it does not necessarily hold that:
:$a - c > b - d$
That is, the usual ordering is not compatible with subtraction.

Proof:
Proof by Counterexample:
For example, set $a = 5, b = 3, c = 4, d = 1$
Then $a - c = 1$ while $b - d = 2$.
{{qed}}
Theorem: Real Ordering is not Compatible with Division

Let $a, b, c, d \in \R$ be real numbers such that $a > b$ and $c > d$.
Then it does not necessarily hold that:
:$\dfrac a c > \dfrac b d$
That is, the usual ordering is not compatible with division.

Proof:
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}}
Theorem: Reverse Triangle Inequality/Real and Complex Fields/Corollary 1

Let $x$ and $y$ be elements of either the real numbers $\R$ or the complex numbers $\C$.
Then:
:$\size {x - y} \ge \size x - \size y$
where $\size x$ denotes either the absolute value of a real number or the complex modulus of a complex number.

Proof:
From the Reverse Triangle Inequality:
:$\cmod {x - y} \ge \cmod {\cmod x - \cmod y}$
By the definition of both absolute value and complex modulus:
:$\cmod {\cmod x - \cmod y} \ge 0$
As:
:$\cmod x - \cmod y = \pm \cmod {\cmod x - \cmod y}$
it follows that:
:$\cmod {\cmod x - \cmod y} \ge \cmod x - \cmod y$
Hence the result.
{{qed}}
Theorem: Reverse Triangle Inequality/Real and Complex Fields/Corollary 1

Let $x$ and $y$ be elements of either the real numbers $\R$ or the complex numbers $\C$.
Then:
:$\size {x - y} \ge \size x - \size y$
where $\size x$ denotes either the absolute value of a real number or the complex modulus of a complex number.

Proof:
By the Triangle Inequality:
:$\cmod {x + y} - \cmod y \le \cmod x$
Let $z = x + y$.
Then $x = z - y$ and so:
:$\cmod z - \cmod y \le \cmod {z - y}$
Renaming variables as appropriate gives:
:$\cmod {x - y} \ge \cmod x - \cmod y$
{{qed}}
Theorem: Reverse Triangle Inequality/Real and Complex Fields/Corollary 1

Let $x$ and $y$ be elements of either the real numbers $\R$ or the complex numbers $\C$.
Then:
:$\size {x - y} \ge \size x - \size y$
where $\size x$ denotes either the absolute value of a real number or the complex modulus of a complex number.

Proof:
Let $z_1$ and $z_2$ be represented by the points $A$ and $B$ respectively in the complex plane.
From Geometrical Interpretation of Complex Subtraction, we can construct the parallelogram $OACB$ where:
:$OA$ and $OB$ represent $z_1$ and $z_2$ respectively
:$BA$ represents $z_1 - z_2$.
:400px
But $OA$, $OB$ and $BA$ form the sides of a triangle.
The result then follows directly from Sum of Two Sides of Triangle Greater than Third Side.
{{qed}}
Theorem: Combination Theorem for Sequences/Real/Sum Rule

:$\ds \lim_{n \mathop \to \infty} \paren {x_n + y_n} = l + m$

Proof:
Let $\epsilon > 0$ be given.
Then $\dfrac \epsilon 2 > 0$.
We are given that:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
:$\ds \lim_{n \mathop \to \infty} y_n = m$
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:
:$\ds \lim_{n \mathop \to \infty} \paren {x_n + y_n} = l + m$
{{qed}}
Theorem: Combination Theorem for Sequences/Real/Multiple Rule

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

Proof:
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 Function is Completely Multiplicative
}}
{{eqn | r = \size {\lambda x_n - \lambda l}
      | c = 
}}
{{end-eqn}}
Hence:
:$\ds \lim_{n \mathop \to \infty} \paren {\lambda x_n} = \lambda l$
{{qed}}
Theorem: Combination Theorem for Sequences/Real/Combined Sum Rule

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

Proof:
From the Multiple Rule for Real Sequences, 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 Real Sequences:
:$\ds \lim_{n \mathop \to \infty} \paren {\lambda x_n + \mu y_n} = \lambda l + \mu m$
{{qed}}
Theorem: Combination Theorem for Sequences/Real/Product Rule

:$\ds \lim_{n \mathop \to \infty} \paren {x_n y_n} = l m$

Proof:
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 Function is Completely Multiplicative
}}
{{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:
:$\ds \lim_{n \mathop \to \infty} \paren {\lambda x_n + \mu y_n} = \lambda l + \mu m$, $z_n \to 0$ as $n \to \infty$
The result follows by the Squeeze Theorem for Real Sequences.
{{qed}}
Theorem: Combination Theorem for Sequences/Real/Quotient Rule

:$\ds \lim_{n \mathop \to \infty} \frac {x_n} {y_n} = \frac l m$
provided that $m \ne 0$.

Proof:
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}}
Theorem: Squeeze Theorem/Sequences/Real Numbers

Let $\sequence {x_n}$, $\sequence {y_n}$ and $\sequence {z_n}$ be sequences in $\R$.
Let $\sequence {y_n}$ and $\sequence {z_n}$ both be convergent to the following limit:
:$\ds \lim_{n \mathop \to \infty} y_n = l, \lim_{n \mathop \to \infty} z_n = l$
Suppose that:
:$\forall n \in \N: y_n \le x_n \le z_n$
Then:
:$x_n \to l$ as $n \to \infty$
that is:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
Thus, if $\sequence {x_n}$ is always between two other sequences that both converge to the same limit, $\sequence {x_n} $ is said to be '''sandwiched''' or '''squeezed''' between those two sequences and itself must therefore converge to that same limit.

Proof:
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 $\ds \lim_{n \mathop \to \infty} y_n = l$ we know that:
:$\exists N_1: \forall n > N_1: \size {y_n - l} < \epsilon$
As $\ds \lim_{n \mathop \to \infty} z_n = l$ 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}}
Theorem: Squeeze Theorem/Sequences/Complex Numbers

Let $\sequence {a_n}$ be a null sequence in $\R$, that is:
:$a_n \to 0$ as $n \to \infty$
Let $\sequence {z_n}$ be a sequence in $\C$.
Suppose $\sequence {a_n}$ dominates $\sequence {z_n}$.
That is:
: $\forall n \in \N: \cmod {z_n} \le a_n$
Then $\sequence {z_n}$ is a null sequence.

Proof:
{{begin-eqn}}
{{eqn | q = \forall n \in \N
      | l = \cmod {z_n}
      | o = \le
      | r = a_n
      | c = {{Defof|Dominate (Analysis)}}
}}
{{eqn | q = \forall n \in \N
      | l = a_n
      | o = \le
      | r = \size {a_n}
      | c = Negative of Absolute Value
}}
{{eqn | q = \forall \epsilon \in \R_{>0}: \exists N \in \R_{>0}: \forall n \in \N
      | lo= n > N \implies
      | l = \size {a_n}
      | o = <
      | r = \epsilon
      | c = {{Defof|Null Sequence|subdef = Real Numbers}}
}}
{{eqn | ll= \leadsto
      | q = \forall \epsilon \in \R_{>0}: \exists N \in \R_{>0}: \forall n \in \N
      | lo= n > N \implies
      | l = \cmod {z_n}
      | o = <
      | r = \epsilon
      | c = Extended Transitivity
}}
{{end-eqn}}
Thus $\sequence {z_n}$ is a null sequence.
{{qed}}
Theorem: Squeeze Theorem/Functions

Let $a$ be a point on an open real interval $I$.
Let $f$, $g$ and $h$ be real functions defined at all points of $I$ except for possibly at point $a$.
Suppose that:
:$\forall x \ne a \in I: \map g x \le \map f x \le \map h x$
:$\ds \lim_{x \mathop \to a} \map g x = \lim_{x \mathop \to a} \map h x = L$
Then:
:$\ds \lim_{x \mathop \to a} \ \map f x = L$

Proof:
We start by proving the special case where $\forall x: \map g x = 0$ and $L = 0$, in which case:
:$\ds \lim_{x \mathop \to a} \map h x = 0$
Let $\epsilon > 0$ be a positive real number.
Then by the definition of the limit of a function:
:$\exists \delta > 0: 0 < \size {x - a} < \delta \implies \size {\map h x} < \epsilon$
Now:
:$\forall x \ne a: 0 = \map g x \le \map f x \le \map h x$
so that:
:$\size {\map f x} \le \size {\map h x}$
Thus:
:$0 < \size {x - a} < \delta \implies \size {\map f x} \le \size {\map h x} < \epsilon$
By the transitive property of $\le$, this proves that:
:$\ds \lim_{x \mathop \to a} \map f x = 0 = L$
{{qed|lemma}}
We now move on to the general case, with $\map g x$ and $L$ arbitrary.
For $x \ne a$, we have:
:$\map g x \le \map f x \le \map h x$
By subtracting $\map g x$ from all expressions, we have:
:$0 \le \map f x - \map g x \le \map h x - \map g x$
Since as $x \to a, \map h x \to L$ and $\map g x \to L$, we have:
:$\map h x - \map g x \to L - L = 0$
From the special case, we now have:
:$\map f x - \map g x \to 0$
We conclude that:
:$\map f x = \paren {\map f x - \map g x} + \map g x \to 0 + L = L$
{{qed}}