nyuuzyou/wikis huggingface dataset:
Combination Theorem for Continuous Functions/Complex

== Theorem ==
Let $\C$ denote the complex numbers.
Let $f$ and $g$ be complex functions which are continuous on an open subset $S \subseteq \C$.
Let $\lambda, \mu \in \C$ be arbitrary complex numbers.
Then the following results hold:
=== Sum Rule ===
=== Multiple Rule ===
=== Combined Sum Rule ===
=== Product Rule ===
=== Quotient Rule ===
== Also see ==
* Combination Theorem for Limits of Complex Functions
== Sources ==
* : $4.9$
Category:Combination Theorems for Continuous Functions
Category:Named Theorems
Category:Complex Analysis
Combination Theorem for Continuous Functions/Complex/Combined Sum Rule

== Theorem ==
Let $\C$ denote the complex numbers.
Let $f$ and $g$ be complex functions which are continuous on an open subset $S \subseteq \C$.
Let $\lambda, \mu \in \C$ be arbitrary complex numbers.
Then:
:$\lambda f + \mu g$ is continuous on $S$.
== Proof ==
By definition of continuous, we have that
:$\forall c \in S: \ds \lim_{z \mathop \to c} \map f z = \map f c$
:$\forall c \in S: \ds \lim_{z \mathop \to c} \map g z = \map g c$
Let $f$ and $g$ tend to the following limits:
:$\ds \lim_{z \mathop \to c} \map f x = l$
:$\ds \lim_{z \mathop \to c} \map g x = m$
From the Combined Sum Rule for Limits of Complex Functions, we have that:
:$\ds \lim_{z \mathop \to c} \paren {\lambda \map f z + \mu \map g z} = \lambda l + \mu m$
So, by definition of continuous again, we have that $\lambda f + \mu g$ is continuous on $S$.
Category:Combination Theorems for Continuous Functions
Combination Theorem for Continuous Functions/Complex/Difference Rule

== Theorem ==
Let $\C$ denote the complex numbers.
Let $f$ and $g$ be complex functions which are continuous on an open subset $S \subseteq \C$.
Then:
:$f - g$ is ‎continuous on $S$.
== Proof ==
We have that:
:$\map {\paren {f - g} } x = \map {\paren {f + \paren {-g} } } x$
From Multiple Rule for Continuous Complex Functions:
:$-g$ is ‎continuous on $S$.
From Sum Rule for Continuous Complex Functions:
:$f + \paren {-g}$ is ‎continuous on $S$.
The result follows.
Category:Combination Theorems for Continuous Functions
Combination Theorem for Continuous Functions/Complex/Multiple Rule

== Theorem ==
Let $\C$ denote the complex numbers.
Let $f$ and $g$ be complex functions which are continuous on an open subset $S \subseteq \C$.
Let $\lambda \in \C$ be an arbitrary complex number.
Then:
:$\lambda f$ is continuous on $S$.
== Proof ==
By definition of continuous, we have that
:$\forall c \in S: \ds \lim_{z \mathop \to c} \map f z = \map f c$
Let $f$ tend to the following limit:
:$\ds \lim_{z \mathop \to c} \map f z = l$
From the Multiple Rule for Limits of Complex Functions, we have that:
:$\ds \lim_{z \mathop \to c} \paren {\lambda \map f z} = \lambda l$
So, by definition of continuous again, we have that $\lambda f$ is continuous on $S$.
Category:Combination Theorems for Continuous Functions
Combination Theorem for Continuous Functions/Complex/Product Rule

== Theorem ==
Let $\C$ denote the complex numbers.
Let $f$ and $g$ be complex functions which are continuous on an open subset $S \subseteq \C$.
Then:
:$f g$ is continuous on $S$
where $f g$ denotes the pointwise product of $f$ and $g$.
== Proof ==
By definition of continuous:
:$\forall c \in S: \ds \lim_{z \mathop \to c} \map f z = \map f c$
:$\forall c \in S: \ds \lim_{z \mathop \to c} \map g z = \map g c$
Let $f$ and $g$ tend to the following limits:
:$\ds \lim_{z \mathop \to c} \map f z = l$
:$\ds \lim_{z \mathop \to c} \map g z = m$
From the Product Rule for Limits of Complex Functions:
:$\ds \lim_{z \mathop \to c} \paren {\map f z \map g z} = l m$
So, by definition of continuous again, we have that $f g$ is continuous on $S$.
Category:Combination Theorems for Continuous Functions
Combination Theorem for Continuous Functions/Complex/Quotient Rule

== Theorem ==
Let $\C$ denote the complex numbers.
Let $f$ and $g$ be complex functions which are continuous on an open subset $S \subseteq \C$.
Then:
:$\dfrac f g$ is continuous on $S \setminus \set {z \in S: \map g z = 0}$
that is, on all the points $z$ of $S$ where $\map g z \ne 0$.
== Proof ==
By definition of continuous:
:$\forall c \in S: \ds \lim_{z \mathop \to c} \map f z = \map f c$
:$\forall c \in S: \ds \lim_{z \mathop \to c} \map g z = \map g c$
Let $f$ and $g$ tend to the following limits:
:$\ds \lim_{z \mathop \to c} \map f z = l$
:$\ds \lim_{z \mathop \to c} \map g z = m$
From the Quotient Rule for Limits of Complex Functions, we have that:
:$\ds \lim_{z \mathop \to c} \frac {\map f z} {\map g z} = \frac l m$
wherever $m \ne 0$.
So, by definition of continuous again, we have that $\dfrac f g$ is continuous on all points $z$ of $S$ where $\map g z \ne 0$.
Category:Combination Theorems for Continuous Functions
Combination Theorem for Continuous Functions/Complex/Sum Rule

== Theorem ==
Let $\C$ denote the complex numbers.
Let $f$ and $g$ be complex functions which are continuous on an open subset $S \subseteq \C$.
Then:
:$f + g$ is ‎continuous on $S$.
== Proof ==
We have:
Let $f$ and $g$ tend to the following limits:
From the Sum Rule for Limits of Complex Functions, we have that:
:$\ds \lim_{z \mathop \to c} \paren {\map f z + \map g z} = l + m$
So, by definition of ‎continuous again, we have that $f + g$ is continuous on $S$.
Category:Combination Theorems for Continuous Functions
Combination Theorem for Continuous Functions/Difference Rule

== Theorem ==
=== Real Functions ===
Let $f$ and $g$ be real functions which are continuous on an open subset $S \subseteq \R$.
=== Complex Functions ===
Let $f$ and $g$ be complex functions which are continuous on an open subset $S \subseteq \C$.
== Sources ==
* 
* 
Category:Combination Theorems for Continuous Functions
Combination Theorem for Continuous Functions/Product Rule

== Theorem ==
=== Real Functions ===
Let $f$ and $g$ be real functions which are continuous on an open subset $S \subseteq \R$.
=== Complex Functions ===
Let $f$ and $g$ be complex functions which are continuous on an open subset $S \subseteq \C$.
== Sources ==
* 
* 
Category:Combination Theorems for Continuous Functions
Combination Theorem for Continuous Functions/Quotient Rule

== Theorem ==
=== Real Functions ===
Let $f$ and $g$ be real functions which are continuous on an open subset $S \subseteq \R$.
=== Complex Functions ===
Let $f$ and $g$ be complex functions which are continuous on an open subset $S \subseteq \C$.
== Sources ==
* 
* 
Category:Combination Theorems for Continuous Functions
Combination Theorem for Continuous Functions/Real

== Theorem ==
Let $\R$ denote the real numbers.
Let $f$ and $g$ be real functions which are continuous on an open subset $S \subseteq \R$.
Let $\lambda, \mu \in \R$ be arbitrary real numbers.
Then the following results hold:
=== Sum Rule ===
=== Difference Rule ===
=== Multiple Rule ===
=== Combined Sum Rule ===
=== Product Rule ===
=== Quotient Rule ===
== Also see ==
* Combination Theorem for Limits of Real Functions
== Sources ==
* : $4.9$
Category:Combination Theorems for Continuous Functions
Category:Combination Theorems for Continuous Real Functions
Category:Named Theorems
Category:Continuous Real Functions
Combination Theorem for Continuous Functions/Real/Combined Sum Rule

== Theorem ==
Let $\R$ denote the real numbers.
Let $f$ and $g$ be real functions which are continuous on an open subset $S \subseteq \R$.
Let $\lambda, \mu \in \R$ be arbitrary real numbers.
Then:
:$\lambda f + \mu g$ is continuous on $S$.
== Proof ==
By definition of continuous, we have that
:$\forall c \in S: \ds \lim_{x \mathop \to c} \map f x = \map f c$
:$\forall c \in S: \ds \lim_{x \mathop \to c} \map g x = \map g c$
Let $f$ and $g$ tend to the following limits:
:$\ds \lim_{x \mathop \to c} \map f x = l$
:$\ds \lim_{x \mathop \to c} \map g x = m$
From the Combined Sum Rule for Limits of Real Functions, we have that:
:$\ds \lim_{x \mathop \to c} \paren {\lambda \map f x + \mu \map g x} = \lambda l + \mu m$
So, by definition of continuous again, we have that $\lambda f + \mu g$ is continuous on $S$.
== Sources ==
* : $\S 9.4 \ \text{(i)}$
Category:Combination Theorems for Continuous Real Functions
Combination Theorem for Continuous Functions/Real/Difference Rule

== Theorem ==
Let $\R$ denote the real numbers.
Let $f$ and $g$ be real functions which are continuous on an open subset $S \subseteq \R$.
Then:
:$f - g$ is ‎continuous on $S$.
== Proof ==
We have that:
:$\map {\paren {f - g} } x = \map {\paren {f + \paren {-g} } } x$
From Multiple Rule for Continuous Real Functions:
:$-g$ is ‎continuous on $S$.
From Sum Rule for Continuous Real Functions:
:$f + \paren {-g}$ is ‎continuous on $S$.
The result follows.
== Sources ==
* : $1$: Review of some real analysis: Exercise $1.5: 17$
Category:Combination Theorems for Continuous Real Functions
Combination Theorem for Continuous Functions/Real/Multiple Rule

== Theorem ==
Let $\R$ denote the real numbers.
Let $f$ and $g$ be real functions which are continuous on an open subset $S \subseteq \R$.
Let $\lambda \in \R$ be an arbitrary real number.
Then:
:$\lambda f$ is continuous on $S$.
== Proof ==
By definition of continuous, we have that
:$\forall c \in S: \ds \lim_{x \mathop \to c} \map f x = \map f c$
Let $f$ tend to the following limit:
:$\ds \lim_{x \mathop \to c} \map f x = l$
From the Multiple Rule for Limits of Real Functions, we have that:
:$\ds \lim_{x \mathop \to c} \paren {\lambda \map f x} = \lambda l$
So, by definition of continuous again, we have that $\lambda f$ is continuous on $S$.
Category:Combination Theorems for Continuous Real Functions
Combination Theorem for Continuous Functions/Real/Product Rule

== Theorem ==
Let $\R$ denote the real numbers.
Let $f$ and $g$ be real functions which are continuous on an open subset $S \subseteq \R$.
Then:
:$f g$ is continuous on $S$
where $f g$ denotes the pointwise product of $f$ and $g$.
== Proof ==
By definition of continuous:
:$\forall c \in S: \ds \lim_{x \mathop \to c} \map f x = \map f c$
:$\forall c \in S: \ds \lim_{x \mathop \to c} \map g x = \map g c$
Let $f$ and $g$ tend to the following limits:
:$\ds \lim_{x \mathop \to c} \map f x = l$
:$\ds \lim_{x \mathop \to c} \map g x = m$
From the Product Rule for Limits of Real Functions:
:$\ds \lim_{x \mathop \to c} \paren {\map f x \map g x} = l m$
So, by definition of continuous again, we have that $f g$ is continuous on $S$.
== Sources ==
* : $1$: Review of some real analysis: Exercise $1.5: 19$
* : $\S 9.4 \ \text{(ii)}$
* 
Category:Combination Theorems for Continuous Real Functions
Combination Theorem for Continuous Functions/Real/Quotient Rule

== Theorem ==
Let $\R$ denote the real numbers.
Let $f$ and $g$ be real functions which are continuous on an open subset $S \subseteq \R$.
Then:
:$\dfrac f g$ is continuous on $S \setminus \set {x \in S: \map g x = 0}$
that is, on all the points $x$ of $S$ where $\map g x \ne 0$.
== Proof ==
By definition of continuous:
:$\forall c \in S: \ds \lim_{x \mathop \to c} \map f x = \map f c$
:$\forall c \in S: \ds \lim_{x \mathop \to c} \map g x = \map g c$
Let $f$ and $g$ tend to the following limits:
:$\ds \lim_{x \mathop \to c} \map f x = l$
:$\ds \lim_{x \mathop \to c} \map g x = m$
From the Quotient Rule for Limits of Real Functions, we have that:
:$\ds \lim_{x \mathop \to c} \frac {\map f x} {\map g x} = \frac l m$
wherever $m \ne 0$.
So, by definition of continuous again, we have that $\dfrac f g$ is continuous on all points $x$ of $S$ where $\map g x \ne 0$.
== Sources ==
* : $1$: Review of some real analysis: Exercise $1.5: 19$
* : $\S 9.4 \ \text{(iii)}$
* 
Category:Combination Theorems for Continuous Real Functions
Combination Theorem for Continuous Functions/Real/Sum Rule

== Theorem ==
Let $\R$ denote the real numbers.
Let $f$ and $g$ be real functions which are continuous on an open subset $S \subseteq \R$.
Then:
:$f + g$ is ‎continuous on $S$.
== Proof ==
By definition of ‎continuous:
:$\forall c \in S: \ds \lim_{x \mathop \to c} \map f x = \map f c$
:$\forall c \in S: \ds \lim_{x \mathop \to c} \map g x = \map g c$
Let $f$ and $g$ tend to the following limits:
:$\ds \lim_{x \mathop \to c} \map f x = l$
:$\ds \lim_{x \mathop \to c} \map g x = m$
From the Sum Rule for Limits of Real Functions, we have that:
:$\ds \lim_{x \mathop \to c} \paren {\map f x + \map g x} = l + m$
So, by definition of ‎continuous again, we have that $f + g$ is continuous on $S$.
== Sources ==
* : $1$: Review of some real analysis: Exercise $1.5: 17$
* 
Category:Combination Theorems for Continuous Real Functions
Combination Theorem for Continuous Functions/Sum Rule

== Theorem ==
=== Real Functions ===
Let $f$ and $g$ be real functions which are continuous on an open subset $S \subseteq \R$.
=== Complex Functions ===
Let $f$ and $g$ be complex functions which are continuous on an open subset $S \subseteq \C$.
== Sources ==
* 
* 
Category:Combination Theorems for Continuous Functions
Combination Theorem for Continuous Mappings/Metric Space

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $\R$ denote the real numbers.
Let $f: M \to \R$ and $g: M \to \R$ be real-valued functions from $M$ to $\R$ which are continuous on $M$.
Let $\lambda, \mu \in \R$ be arbitrary real numbers.
Then the following results hold:
=== Sum Rule ===
=== Difference Rule ===
=== Multiple Rule ===
=== Combined Sum Rule ===
=== Product Rule ===
=== Quotient Rule ===
=== Absolute Value Rule ===
=== Maximum Rule ===
=== Minimum Rule ===
== Sources ==
* : $2$: Continuity generalized: metric spaces: Exercise $2.6: 14$
Category:Combination Theorem for Continuous Mappings
Category:Named Theorems
Category:Metric Spaces
Combination Theorem for Continuous Mappings/Metric Space/Absolute Value Rule

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $\R$ denote the real numbers.
Let $f: M \to \R$ be a real-valued function from $M$ to $\R$
Let $f$ be continuous at $a \in M$.
Then:
:$\size f$ is continuous at $a$
where:
:$\map {\size f} x$ is defined as $\size {\map f x}$.
== Proof ==
Let $\epsilon \in \R_{>0}$ be a positive real number.
Because $f$ is continuous at $a$:
:$\exists \delta \in \R_{>0}: \map d {x, a} < \delta \implies \size {\map f x - \map f a} < \epsilon$
From Reverse Triangle Inequality:
:$\size {\size {\map f x} - \size {\map f a} } \le \size {\map f x - \map f a}$
and so:
:$\exists \delta \in \R_{>0}: \map d {x, a} < \delta \implies \size {\size {\map f x} - \size {\map f a} } < \epsilon$
$\epsilon$ is arbitrary, so:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \map d {x, a} < \delta \implies \size {\size {\map f x} - \size {\map f a} } < \epsilon$
That is, by definition:
:$\size f$ is continuous at $a$.
== Sources ==
* : $2$: Continuity generalized: metric spaces: Exercise $2.6: 14$
Category:Combination Theorem for Continuous Mappings
Combination Theorem for Continuous Mappings/Metric Space/Combined Sum Rule

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $\R$ denote the real numbers.
Let $f: M \to \R$ and $g: M \to \R$ be real-valued functions from $M$ to $\R$ which are continuous on $M$.
Let $\lambda, \mu \in \R$ be arbitrary real numbers.
Then:
:$\lambda f + \mu g$ is ‎continuous on $M$.
== Proof ==
From the Multiple Rule for Continuous Mappings on Metric Space, we have that:
:$\lambda \map f x$ and $\mu \map g x$ are ‎continuous.
From the Sum Rule for Continuous Mappings on Metric Space, we have that:
:$\lambda \map f x + \mu \map g x$ is ‎continuous.
So, by definition of ‎continuous again, we have that $\lambda f + \mu g$ is continuous on $M$.
Category:Combination Theorem for Continuous Mappings
Combination Theorem for Continuous Mappings/Metric Space/Difference Rule

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $\R$ denote the real numbers.
Let $f: M \to \R$ and $g: M \to \R$ be real-valued functions from $M$ to $\R$ which are continuous on $M$.
Then:
:$f - g$ is ‎continuous on $M$.
== Proof ==
:$\map {\paren {f - g} } x = \map {\paren {f + \paren {-g} } } x$
From Multiple Rule for Continuous Mappings on Metric Space:
:$-g$ is ‎continuous on $M$.
From Sum Rule for Continuous Mappings on Metric Space:
:$f + \paren {-g}$ is ‎continuous on $M$.
The result follows.
== Sources ==
* : $2$: Continuity generalized: metric spaces: Exercise $2.6: 14$
Category:Combination Theorem for Continuous Mappings
Combination Theorem for Continuous Mappings/Metric Space/Maximum Rule

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $\R$ denote the real numbers.
Let $f: M \to \R$ and $g: M \to \R$ be real-valued functions from $M$ to $\R$ which are continuous on $M$.
Let $\max \set {f, g}: M \to \R$ denote the pointwise maximum of $f$ and $g$.
Then:
:$\max \set {f, g}$ is ‎continuous on $M$.
== Proof ==
Let $a \in M$ be arbitrary.
From Max Operation Representation on Real Numbers
:$\max \set {x, y} = \dfrac 1 2 \paren {x + y + \size {x - y} }$
Hence:
:$\max \set {\map f x, \map g x} = \dfrac 1 2 \paren {\map f x + \map g x + \size {\map f x - \map g x} }$
From Difference Rule for Continuous Mappings on Metric Space:
:$\map f x - \map g x$ is continuous at $a$.
From Absolute Value Rule for Continuous Mappings on Metric Space:
:$\size {\map f x - \map g x}$ is continuous at $a$.
From Sum Rule for Continuous Mappings on Metric Space:
:$\map f x + \map g x$ is continuous at $a$
and hence:
:$\map f x + \map g x + \size {\map f x - \map g x}$ is continuous at $a$
From Multiple Rule for Continuous Mappings on Metric Space:
:$\dfrac 1 2 \paren {\map f x + \map g x + \size {\map f x - \map g x} }$ is continuous at $a$.
As $a$ is arbitrary:
:$\max \set {f, g}$ is ‎continuous on $M$.
== Sources ==
* : $2$: Continuity generalized: metric spaces: Exercise $2.6: 14$
Category:Combination Theorem for Continuous Mappings
Combination Theorem for Continuous Mappings/Metric Space/Minimum Rule

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $\R$ denote the real numbers.
Let $f: M \to \R$ and $g: M \to \R$ be real-valued functions from $M$ to $\R$ which are continuous on $M$.
Let $\min \set {f, g}: M \to \R$ denote the pointwise maximum of $f$ and $g$.
Then:
:$\min \set {f, g}$ is ‎continuous on $M$.
== Proof ==
Let $a \in M$ be arbitrary.
From Min Operation Representation on Real Numbers
:$\min \set {x, y} = \dfrac 1 2 \paren {x + y - \size {x - y} }$
Hence:
:$\min \set {\map f x, \map g x} = \dfrac 1 2 \paren {\map f x + \map g x - \size {\map f x - \map g x} }$
From Difference Rule for Continuous Mappings on Metric Space:
:$\map f x - \map g x$ is continuous at $a$.
From Absolute Value Rule for Continuous Mappings on Metric Space:
:$\size {\map f x - \map g x}$ is continuous at $a$.
From Sum Rule for Continuous Mappings on Metric Space:
:$\map f x + \map g x$ is continuous at $a$
From Difference Rule for Continuous Mappings on Metric Space:
:$\map f x + \map g x - \size {\map f x - \map g x}$ is continuous at $a$
From Multiple Rule for Continuous Mappings on Metric Space:
:$\dfrac 1 2 \paren {\map f x + \map g x - \size {\map f x - \map g x} }$ is continuous at $a$.
As $a$ is arbitrary:
:$\min \set {f, g}$ is ‎continuous on $M$.
== Sources ==
* : $2$: Continuity generalized: metric spaces: Exercise $2.6: 14$
Category:Combination Theorem for Continuous Mappings
Combination Theorem for Continuous Mappings/Metric Space/Multiple Rule

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $\R$ denote the real numbers.
Let $f: M \to \R$ be a real-valued function from $M$ to $\R$ which is continuous on $M$.
Let $\lambda \in \R$ be an arbitrary real number.
Then:
:$\lambda f$ is ‎continuous on $M$.
== Proof ==
By definition of ‎continuous:
:$\forall a \in M: \ds \lim_{x \mathop \to a} \map f x = \map f a$
Let $f$ tend to the following limit:
:$\ds \lim_{x \mathop \to a} \map f x = l$
From the Multiple Rule for Limits of Real Functions, we have that:
:$\ds \lim_{x \mathop \to a} \paren {\lambda \map f x} = \lambda l$
So, by definition of ‎continuous again, we have that $\lambda \map f x$ is continuous on $M$.
== Sources ==
* : $2$: Continuity generalized: metric spaces: Exercise $2.6: 14$
Category:Combination Theorem for Continuous Mappings
Combination Theorem for Continuous Mappings/Metric Space/Product Rule

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $\R$ denote the real numbers.
Let $f: M \to \R$ and $g: M \to \R$ be real-valued functions from $M$ to $\R$ which are continuous on $M$.
Then:
:$f g$ is ‎continuous on $M$.
== Proof ==
By definition of ‎continuous:
:$\forall a \in M: \ds \lim_{x \mathop \to a} \map f x = \map f a$
:$\forall a \in M: \ds \lim_{x \mathop \to a} \map g x = \map g a$
Let $f$ and $g$ tend to the following limits:
:$\ds \lim_{x \mathop \to a} \map f x = l$
:$\ds \lim_{x \mathop \to a} \map g x = m$
From the Product Rule for Limits of Real Functions, we have that:
:$\ds \lim_{x \mathop \to a} \paren {\map f x \map g x} = l m$
So, by definition of ‎continuous again, we have that $f g$ is continuous on $M$.
== Sources ==
* : $2$: Continuity generalized: metric spaces: Exercise $2.6: 14$
Category:Combination Theorem for Continuous Mappings
Combination Theorem for Continuous Mappings/Metric Space/Quotient Rule

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $\R$ denote the real numbers.
Let $f: M \to \R$ and $g: M \to \R$ be real-valued functions from $M$ to $\R$ which are continuous on $M$.
Then:
:$\dfrac f g$ is ‎continuous on $M \setminus \set {x \in A: \map g x = 0}$.
that is, on all the points $x$ of $A$ where $\map g x \ne 0$.
== Proof ==
By definition of ‎continuous:
:$\forall a \in M: \ds \lim_{x \mathop \to a} \map f x = \map f a$
:$\forall a \in M: \ds \lim_{x \mathop \to a} \map g x = \map g a$
Let $f$ and $g$ tend to the following limits:
:$\ds \lim_{x \mathop \to a} \map f x = l$
:$\ds \lim_{x \mathop \to a} \map g x = m$
From the Product Rule for Limits of Real Functions, we have that:
:$\ds \lim_{x \mathop \to a} \paren {\dfrac {\map f x} {\map g x} = \dfrac l m$
whenever $m \ne 0$.
So, by definition of ‎continuous again, we have that $\dfrac f g$ is continuous on $M \setminus \set {x \in A: \map g x = 0}$.
== Sources ==
* : $2$: Continuity generalized: metric spaces: Exercise $2.6: 14$
Category:Combination Theorem for Continuous Mappings
Combination Theorem for Continuous Mappings/Metric Space/Sum Rule

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $\R$ denote the real numbers.
Let $f: M \to \R$ and $g: M \to \R$ be real-valued functions from $M$ to $\R$ which are continuous on $M$.
Then:
:$f + g$ is ‎continuous on $M$.
== Proof ==
By definition of ‎continuous:
:$\forall a \in M: \ds \lim_{x \mathop \to a} \map f x = \map f a$
:$\forall a \in M: \ds \lim_{x \mathop \to a} \map g x = \map g a$
Let $f$ and $g$ tend to the following limits:
:$\ds \lim_{x \mathop \to a} \map f x = l$
:$\ds \lim_{x \mathop \to a} \map g x = m$
From the Sum Rule for Limits of Real Functions, we have that:
:$\ds \lim_{x \mathop \to a} \paren {\map f x + \map g x} = l + m$
So, by definition of ‎continuous again, we have that $f + g$ is continuous on $M$.
== Sources ==
* : $2$: Continuity generalized: metric spaces: Exercise $2.6: 14$
Category:Combination Theorem for Continuous Mappings
Combination Theorem for Continuous Mappings/Normed Division Ring

== Theorem ==
Let $\struct{S, \tau_{_S}}$ be a topological space.
Let $\struct{R, +, *, \norm{\,\cdot\,}}$ be a normed division ring.
Let $\tau_{_R}$ be the topology induced by the norm $\norm{\,\cdot\,}$.
Let $\lambda \in R$.
Let $f,g : \struct{S, \tau_{_S}} \to \struct{R, \tau_{_R}}$ be continuous mappings.
Let $U = S \setminus \set{x : \map g x = 0}$
Let $g^{-1} : U \to R$ denote the mapping defined by:
:$\forall x \in U : \map {g^{-1}} x = \map g x^{-1}$
Let $\tau_{_U}$ be the subspace topology on $U$.
Then the following results hold:
=== Sum Rule ===
=== Translation Rule ===
=== Negation Rule ===
=== Product Rule ===
=== Multiple Rule ===
=== Inverse Rule ===
== Also see ==
* Combination Theorem for Continuous Mappings to Topological Division Ring
Category:Combination Theorem for Continuous Mappings
Category:Combination Theorem for Continuous Mappings to Normed Division Ring