nyuuzyou/wikis huggingface dataset:
Combination Theorem for Continuous Mappings/Normed Division Ring/Inverse Rule

== 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 $g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ 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:
:$g^{-1}: \struct {U, \tau_{_U} } \to \struct {R, \tau_{_R} }$ is continuous.
== Proof ==
From Corollary to Normed Division Ring Operations are Continuous:
:$\struct {R, +, *, \tau_{_R} }$ is a topological division ring.
From Inverse Rule for Continuous Mappings to Topological Division Ring:
:$g^{-1}: \struct {U, \tau_{_U} } \to \struct {R, \tau_{_R} }$ is a continuous mapping.
== Also see ==
* Inverse Rule for Continuous Mappings to Topological Division Ring
Category:Combination Theorem for Continuous Mappings to Normed Division Ring
Combination Theorem for Continuous Mappings/Normed Division Ring/Multiple Rule

== 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 $f: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ be a continuous mappings.
Let $\lambda * f: S \to R$ be the mapping defined by:
:$\forall x \in S: \map {\paren {\lambda * f} } x = \lambda * \map f x$
Let $f * \lambda: S \to R$ be the mapping defined by:
:$\forall x \in S: \map {\paren {f * \lambda} } x = \map f x * \lambda$
Then:
:$\lambda * f: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is continuous
:$f * \lambda: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is continuous.
== Proof ==
From Corollary to Normed Division Ring Operations are Continuous:
:$\struct {R, +, *, \tau_{_R} }$ is a topological division ring.
From Multiple Rule for Continuous Mappings to Topological Division Ring:
:$\lambda * f, f * \lambda: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ are continuous mappings.
== Also see ==
* Multiple Rule for Continuous Mappings to Topological Division Ring
Category:Combination Theorem for Continuous Mappings to Normed Division Ring
Combination Theorem for Continuous Mappings/Normed Division Ring/Negation Rule

== 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 $g : \struct{S, \tau_{_S}} \to \struct{R, \tau_{_R}}$ be a continuous mapping.
Let $- g : S \to R$ be the mapping defined by:
:$\forall x \in S: \map {\paren{- g}} x = - \map g x$
Then 
:$- g : \struct{S, \tau_{_S}} \to \struct{R, \tau_{_R}}$ is continuous.
== Proof ==
From Corollary to Normed Division Ring Operations are Continuous, $\struct{R, +, *, \tau_{_R}}$ is a topological division ring.
From Negation Rule for Continuous Mappings to Topological Division Ring, $- g : \struct{S, \tau_{_S}} \to \struct{R, \tau_{_R}}$ is a continuous mapping.
== Also see ==
* Negation Rule for Continuous Mappings to Topological Division Ring
Category:Combination Theorem for Continuous Mappings to Normed Division Ring
Combination Theorem for Continuous Mappings/Normed Division Ring/Product Rule

== 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 $f, g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ be continuous mappings.
Let $f * g: S \to R$ be the mapping defined by:
:$\forall x \in S: \map {\paren {f * g} } x = \map f x * \map g x$
Then:
:$f * g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is continuous.
== Proof ==
From Corollary to Normed Division Ring Operations are Continuous:
:$\struct {R, +, *, \tau_{_R} }$ is a topological division ring.
From Product Rule for Continuous Mappings to Topological Division Ring:
:$f * g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is a continuous mapping.
== Also see ==
* Product Rule for Continuous Mappings to Topological Division Ring
Category:Combination Theorem for Continuous Mappings to Normed Division Ring
Combination Theorem for Continuous Mappings/Normed Division Ring/Sum Rule

== 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 $f, g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ be continuous mappings.
Let $f + g: S \to R$ be the mapping defined by:
:$\forall x \in S: \map {\paren {f + g} } x = \map f x + \map g x$
Then:
:$f + g: \struct {S, \tau_{_S} } \to \struct{R, \tau_{_R} }$ is continuous.
== Proof ==
From Corollary to Normed Division Ring Operations are Continuous:
:$\struct {R, +, *, \tau_{_R} }$ is a topological division ring.
From Sum Rule for Continuous Mappings to Topological Division Ring:
:$f + g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is a continuous mapping.
== Also see ==
* Sum Rule for Continuous Mappings to Topological Division Ring
Category:Combination Theorem for Continuous Mappings to Normed Division Ring
Combination Theorem for Continuous Mappings/Normed Division Ring/Translation Rule

== 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 $f: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ be a continuous mapping.
Let $\lambda + f : S \to R$ be the mapping defined by:
:$\forall x \in S: \map {\paren {\lambda + f} } x = \lambda + \map f x$
Then 
:$\lambda + f: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is continuous.
== Proof ==
From Corollary to Normed Division Ring Operations are Continuous:
:$\struct{R, +, *, \tau_{_R} }$ is a topological division ring.
From Translation Rule for Continuous Mappings to Topological Division Ring:
:$\lambda + f: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is a continuous mapping.
== Also see ==
* Translation Rule for Continuous Mappings to Topological Division Ring
Category:Combination Theorem for Continuous Mappings to Normed Division Ring
Combination Theorem for Continuous Mappings/Topological Division Ring

== Theorem ==
Let $\struct{S, \tau_{_S}}$ be a topological space.
Let $\struct{R, +, *, \tau_{_R}}$ be a topological division ring.
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 Ring
Category:Combination Theorem for Continuous Mappings
Category:Combination Theorem for Continuous Mappings to Topological Division Ring
Combination Theorem for Continuous Mappings/Topological Division Ring/Inverse Rule

== Theorem ==
Let $\struct {S, \tau_{_S} }$ be a topological space.
Let $\struct {R, +, *, \tau_{_R} }$ be a topological division ring.
Let $g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ be a continuous mapping.
Let $U = S \setminus \set {x : \map g x = 0}$
Let $g^{-1}: U \to R$ be 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:
:$g^{-1}: \struct {U, \tau_{_U} } \to \struct {R, \tau_{_R} }$ is continuous.
== Proof ==
Let $R^* = R \setminus \set 0$.
Let $\tau^*$ be the subspace topology on $R^*$.
By definition of a topological division ring:
:$\phi: \struct {R^*, \tau^*} \to \struct {R, \tau_{_R} }$ such that $\forall x \in R^*: \map \phi x = x^{-1}$ is a continuous mapping
Let $g^*: \struct {U, \tau_{_U} } \to \struct {R^*, \tau^*}$ be the restriction of $g$ to $U \subseteq R$.
From Restriction of Continuous Mapping is Continuous, $g^*$ is a continuous mapping.
 
From Composite of Continuous Mappings is Continuous, the composition $\phi \circ g^* : \struct {U, \tau_{_U} } \to \struct {R, \tau_{_R} }$ is continuous.
Now:
From Equality of Mappings:
:$g^{-1} = \phi \circ g^*$
The result follows.
Category:Combination Theorem for Continuous Mappings to Topological Division Ring
Combination Theorem for Continuous Mappings/Topological Division Ring/Multiple Rule

== Theorem ==
Let $\struct {S, \tau_{_S} }$ be a topological space.
Let $\struct {R, +, *, \tau_{_R} }$ be a topological division ring.
Let $\lambda \in R$.
Let $f: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ be a continuous mapping.
Let $\lambda * f: S \to R$ be the mapping defined by:
:$\forall x \in S: \map {\paren {\lambda * f} } x = \lambda * \map f x$
Let $f * \lambda : S \to R$ be the mapping defined by:
:$\forall x \in S: \map {\paren {f * \lambda} } x = \map f x * \lambda$
Then 
:$\lambda * f: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is continuous
:$f * \lambda: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is continuous.
== Proof ==
By definition of a topological division ring:
:$\struct {R, +, *, \tau_{_R} }$ is a topological ring.
From Multiple Rule for Continuous Mappings to Topological Ring:
:$\lambda * f, f * \lambda: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ are continuous mappings.
== Also see ==
* Multiple Rule for Continuous Mappings to Topological Ring
Category:Combination Theorem for Continuous Mappings to Topological Division Ring
Combination Theorem for Continuous Mappings/Topological Division Ring/Negation Rule

== Theorem ==
Let $\struct {S, \tau_{_S} }$ be a topological space.
Let $\struct {R, +, *, \tau_{_R} }$ be a topological division ring.
Let $g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ be a continuous mapping.
Let $-g : S \to R$ be the mapping defined by:
:$\forall x \in S: \map {\paren {-g} } x = -\map g x$
Then:
:$-g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is continuous.
== Proof ==
By definition of a topological division ring:
:$\struct {R, +, *, \tau_{_R} }$ is a topological ring.
From Negation Rule for Continuous Mappings to Topological Ring:
:$-g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is a continuous mapping.
== Also see ==
* Negation Rule for Continuous Mappings to Topological Ring
Category:Combination Theorem for Continuous Mappings to Topological Division Ring
Combination Theorem for Continuous Mappings/Topological Division Ring/Product Rule

== Theorem ==
Let $\struct {S, \tau_{_S} }$ be a topological space.
Let $\struct {R, +, *, \tau_{_R} }$ be a topological division ring.
Let $f, g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ be continuous mappings.
Let $f * g: S \to R$ be the mapping defined by:
:$\forall x \in S: \map {\paren {f * g} } x = \map f x * \map g x$
Then:
:$f * g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is continuous.
== Proof ==
By definition of a topological division ring:
:$\struct {R, +, *, \tau_{_R} }$ is a topological ring.
From Product Rule for Continuous Mappings to Topological Ring:
:$f * g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is a continuous mapping.
== Also see ==
* Product Rule for Continuous Mappings to Topological Ring
Category:Combination Theorem for Continuous Mappings to Topological Division Ring
Combination Theorem for Continuous Mappings/Topological Division Ring/Sum Rule

== Theorem ==
Let $\struct {S, \tau_{_S} }$ be a topological space.
Let $\struct {R, +, *, \tau_{_R} }$ be a topological division ring.
Let $f, g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ be continuous mappings.
Let $f + g : S \to R$ be the mapping defined by:
:$\forall x \in S: \map {\paren {f + g} } x = \map f x + \map g x$
Then 
:$f + g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is continuous.
== Proof ==
By definition of a topological division ring, $\struct {R, +, *, \tau_{_R} }$ is a topological ring.
From Sum Rule for Continuous Mappings into Topological Ring:
:$f + g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is a continuous mapping.
== Also see ==
* Sum Rule for Continuous Mappings into Topological Ring
Category:Combination Theorem for Continuous Mappings to Topological Division Ring
Combination Theorem for Continuous Mappings/Topological Division Ring/Translation Rule

== Theorem ==
Let $\struct {S, \tau_{_S} }$ be a topological space.
Let $\struct {R, +, *, \tau_{_R} }$ be a topological division ring.
Let $\lambda \in R$.
Let $f: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ be a continuous mapping.
Let $\lambda + f : S \to R$ be the mapping defined by:
:$\forall x \in S: \map {\paren {\lambda + f} } x = \lambda + \map f x$
Then 
:$\lambda + f: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is continuous.
== Proof ==
By definition of a topological division ring:
:$\struct {R, +, *, \tau_{_R} }$ is a topological ring.
From Translation Rule for Continuous Mappings to Topological Ring:
:$\lambda + f: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is a continuous mapping.
== Also see ==
* Translation Rule for Continuous Mappings to Topological Ring
Category:Combination Theorem for Continuous Mappings to Topological Division Ring
Combination Theorem for Continuous Mappings/Topological Group

== Theorem ==
Let $\struct {S, \tau_{_S} }$ be a topological space.
Let $\struct {G, *, \tau_{_G} }$ be a topological group.
Let $\lambda \in G$.
Let $f, g : \struct {S, \tau_{_S} } \to \struct {G, \tau_{_G} }$ be continuous mappings.
Then the following results hold:
=== Product Rule ===
=== Multiple Rule ===
=== Inverse Rule ===
== Also see ==
* Combination Theorem for Continuous Mappings to Topological Semigroup
Category:Combination Theorem for Continuous Mappings
Category:Combination Theorem for Continuous Mappings to Topological Group
Combination Theorem for Continuous Mappings/Topological Group/Inverse Rule

== Theorem ==
Let $\struct {S, \tau_S}$ be a topological space.
Let $\struct {G, *, \tau_G}$ be a topological group.
Let $g: \struct {S, \tau_S} \to \struct {G, \tau_G}$ be a continuous mapping.
Let $g^{-1}: S \to G$ be the mapping defined by:
:$\forall x \in S: \map {\paren {g^{-1} } } x = \paren {\map g x}^{-1}$
Then:
:$g^{-1}: \struct {S, \tau_S} \to \struct {G, \tau_G}$ is a continuous mapping.
== Proof ==
By definition of a topological group:
:$\phi: \struct {G, \tau_G} \to \struct {G, \tau_G}$ such that $\forall x \in G: \map \phi x = x^{-1}$ is a continuous mapping
From Composite of Continuous Mappings is Continuous:
:the composition $\phi \circ g: \struct {S, \tau_S} \to \struct {G, \tau_G}$ is continuous.
Now:
From Equality of Mappings:
:$g^{-1} = \phi \circ g$
The result follows.
Category:Combination Theorem for Continuous Mappings to Topological Group
Combination Theorem for Continuous Mappings/Topological Group/Multiple Rule

== Theorem ==
Let $\struct {S, \tau_{_S} }$ be a topological space.
Let $\struct {G, *, \tau_{_G} }$ be a topological group.
Let $\lambda \in G$.
Let $f: \struct {S, \tau_{_S} } \to \struct {G, \tau_{_G} }$ be a continuous mapping.
Let $\lambda * f: S \to G$ be the mapping defined by:
:$\forall x \in S: \map {\paren {\lambda * f} } x = \lambda * \map f x$
Let $f * \lambda : S \to G$ be the mapping defined by:
:$\forall x \in S: \map {\paren {f * \lambda} } x = \map f x * \lambda$
Then:
:$\lambda * f: \struct {S, \tau_{_S} } \to \struct {G, \tau_{_G} }$ is a continuous mapping
:$f * \lambda: \struct {S, \tau_{_S} } \to \struct {G, \tau_{_G} }$ is a continuous mapping.
== Proof ==
By definition, a topological group is a topological semigroup.
Hence $\struct {G, *, \tau_{_G}}$ is a topological semigroup.
From Multiple Rule for Continuous Mappings to Topological Semigroup:
:$\lambda * f, f * \lambda: \struct {S, \tau_{_S} } \to \struct {G, \tau_{_G} }$ are continuous mappings.
== Also see ==
* Multiple Rule for Continuous Mappings to Topological Semigroup
Category:Combination Theorem for Continuous Mappings to Topological Group
Combination Theorem for Continuous Mappings/Topological Group/Product Rule

== Theorem ==
Let $\struct {S, \tau_{_S} }$ be a topological space.
Let $\struct {G, *, \tau_{_G} }$ be a topological group.
Let $f, g: \struct {S, \tau_{_S} } \to \struct {G, \tau_{_G} }$ be continuous mappings.
Let $f * g: S \to G$ be the mapping defined by:
:$\forall x \in S: \map {\paren {f * g} } x = \map f x * \map g x$
Then:
:$f * g : \struct {S, \tau_{_S} } \to \struct {G, \tau_{_G} }$ is a continuous mapping.
== Proof ==
By definition, a topological group is a topological semigroup.
Hence $\struct {G, *, \tau_{_G} }$ is a topological semigroup.
From Product Rule for Continuous Mappings to Topological Semigroup:
:$f * g: \struct {S, \tau_{_S} } \to \struct {G, \tau_{_G} }$ is a continuous mapping.
== Also see ==
* Product Rule for Continuous Mappings to Topological Semigroup
Category:Combination Theorem for Continuous Mappings to Topological Group
Combination Theorem for Continuous Mappings/Topological Ring

== Theorem ==
Let $\struct {S, \tau_{_S} }$ be a topological space.
Let $\struct {R, +, *, \tau_{_R} }$ be a topological ring.
Let $\lambda \in R$.
Let $f, g : \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ be continuous mappings.
Then the following results hold:
=== Sum Rule ===
=== Translation Rule ===
=== Negation Rule ===
=== Product Rule ===
=== Multiple Rule ===
== Also see ==
* Combination Theorem for Continuous Mappings to Topological Group
* Combination Theorem for Continuous Mappings to Topological Semigroup
Category:Combination Theorem for Continuous Mappings
Category:Combination Theorem for Continuous Mappings to Topological Ring
Combination Theorem for Continuous Mappings/Topological Ring/Combined Rule

== Theorem ==
Let $\struct {S, \tau_S}$ be a topological space.
Let $\struct {R, +, *, \tau_R}$ be a topological ring.
Let $\lambda, \mu \in R$ be arbitrary elements in $R$.
Let $f, g : \struct {S, \tau_S} \to \struct {R, \tau_R}$ be continuous mappings.
Let $\lambda * f + \mu * g: S \to R$ be the mapping defined by:
:$\forall x \in S: \map {\paren {\lambda * f + \mu * g}} x = \lambda * \map f x + \mu * \map g x$
Let $f * \lambda + g * \mu : S \to R$ be the mapping defined by:
:$\forall x \in S: \map {\paren {f *\lambda + g * \mu}} x = \map f x * \lambda + \map g x * \mu$
Then:
:$\lambda * f + \mu * g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is a continuous mapping
:$f * \lambda + g * \mu: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is a continuous mapping.
== Proof ==
Category:Combination Theorem for Continuous Mappings
Combination Theorem for Continuous Mappings/Topological Ring/Multiple Rule

== Theorem ==
Let $\struct {S, \tau_{_S} }$ be a topological space.
Let $\struct {R, +, *, \tau_{_R} }$ be a topological ring.
Let $\lambda \in R$.
Let $f: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ be a continuous mapping.
Let $\lambda * f : S \to R$ be the mapping defined by:
:$\forall x \in S: \map {\paren {\lambda * f} } x = \lambda * \map f x$
Let $f * \lambda : S \to R$ be the mapping defined by:
:$\forall x \in S: \map {\paren {f * \lambda} } x = \map f x * \lambda$
Then 
:$\lambda * f: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is continuous
:$f * \lambda: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is continuous.
== Proof ==
By definition of a topological ring:
:$\struct {R, *, \tau_{_R} }$ is a topological semigroup.
From Multiple Rule for Continuous Mappings to Topological Semigroup:
:$\lambda * f, f * \lambda: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ are continuous mappings.
== Also see ==
* Multiple Rule for Continuous Mappings to Topological Semigroup
Category:Combination Theorem for Continuous Mappings to Topological Ring
Combination Theorem for Continuous Mappings/Topological Ring/Negation Rule

== Theorem ==
Let $\struct {S, \tau_{_S} }$ be a topological space.
Let $\struct {R, +, *, \tau_{_R} }$ be a topological ring.
Let $g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ be a continuous mappings.
Let $-g : S \to R$ be the mapping defined by:
:$\forall x \in S: \map {\paren {-g} } x = -\map g x$
Then 
:$-g : \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is continuous.
== Proof ==
By definition of a topological ring:
:$\struct {R, +, \tau_{_R} }$ is a topological group.
From Inverse Rule for Continuous Mappings to Topological Group:
:$-g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is a continuous mapping.
== Also see ==
* Inverse Rule for Continuous Mappings to Topological Group
Category:Combination Theorem for Continuous Mappings to Topological Ring
Combination Theorem for Continuous Mappings/Topological Ring/Product Rule

== Theorem ==
Let $\struct {S, \tau_{_S} }$ be a topological space.
Let $\struct {R, +, *, \tau_{_R} }$ be a topological ring.
Let $f, g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ be continuous mappings.
Let $f * g : S \to R$ be the mapping defined by:
:$\forall x \in S: \map {\paren {f * g} } x = \map f x * \map g x$
Then 
:$f * g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is continuous.
== Proof ==
By definition of a topological ring:
:$\struct {R, *, \tau_{_R} }$ is a topological semigroup.
From Product Rule for Continuous Mappings to Topological Semigroup:
:$f * g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is a continuous mapping.
== Also see ==
* Product Rule for Continuous Mappings to Topological Semigroup
Category:Combination Theorem for Continuous Mappings to Topological Ring
Combination Theorem for Continuous Mappings/Topological Ring/Sum Rule

== Theorem ==
Let $\struct{S, \tau_{_S}}$ be a topological space.
Let $\struct{R, +, *, \tau_{_R}}$ be a topological ring.
Let $f, g: \struct{S, \tau_{_S}} \to \struct{R, \tau_{_R}}$ be continuous mappings.
Let $f + g : S \to R$ be the mapping defined by:
:$\forall x \in S: \map {\paren{f + g}} x = \map f x + \map g x$
Then:
:$f + g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is continuous.
== Proof ==
By definition of a topological ring, $\struct {R, +, \tau_{_R} }$ is a topological group.
From Product Rule for Continuous Mappings to Topological Group:
:$f + g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is a continuous mapping.
== Also see ==
* Product Rule for Continuous Mappings to Topological Group
Category:Combination Theorem for Continuous Mappings to Topological Ring
Combination Theorem for Continuous Mappings/Topological Ring/Translation Rule

== Theorem ==
Let $\struct {S, \tau_{_S} }$ be a topological space.
Let $\struct {R, +, *, \tau_{_R} }$ be a topological ring.
Let $\lambda \in R$.
Let $f: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ be a continuous mapping.
Let $\lambda + f: S \to R$ be the mapping defined by:
:$\forall x \in S: \map {\paren {\lambda + f} } x = \lambda + \map f x$
Then 
:$\lambda + f: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is continuous.
== Proof ==
By definition of a topological ring:
:$\struct {R, +, \tau_{_R} }$ is a topological group.
From Multiple Rule for Continuous Mappings to Topological Group:
:$\lambda + f: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is a continuous mapping.
== Also see ==
* Multiple Rule for Continuous Mappings to Topological Group
Category:Combination Theorem for Continuous Mappings to Topological Ring
Combination Theorem for Continuous Mappings/Topological Semigroup

== Theorem ==
Let $\struct{S, \tau_{_S}}$ be a topological space.
Let $\struct{G, *, \tau_{_G}}$ be a topological semigroup.
Let $\lambda \in G$.
Let $f,g : \struct{S, \tau_{_S}} \to \struct{G, \tau_{_G}}$ be continuous mappings.
Then the following results hold:
=== Product Rule ===
=== Multiple Rule ===
Category:Combination Theorem for Continuous Mappings
Category:Combination Theorem for Continuous Mappings to Topological Semigroup