nyuuzyou/wikis huggingface dataset:
Mapping from Finite Set to Itself is Injection iff Surjection

== Theorem ==
Let $S$ be a finite set.
Let $f: S \to S$ be a mapping.
Then $f$ is injective $f$ is surjective.
== Proof ==
Let $f$ be an injection.
From Injection to Image is Bijection, $S$ is equivalent to the image $\Img f$ of $f$.
We are given that $S$ is finite.
It follows from Infinite Set is Equivalent to Proper Subset that $\Img f = S$.
It follows by definition that $f$ is surjective.
Let $f$ be a surjection.
Then by Surjection iff Right Inverse there exists a mapping $g: S \to S$ such that:
:$f \circ g = I_S$
where $I_S$ is the identity mapping.
By Right Inverse Mapping is Injection, $g$ is an injection.
By the above, it follows that $g$ is also a surjection.
Thus $g$ is a bijection.
It follows that $f = g^{-1}$ is also a bijection and so by definition an injection.
== Sources ==
* : $\S 8$: Theorem $8.11$
Category:Injections
Category:Surjections
Mapping from Group Element to Inner Automorphism is Homomorphism

== Theorem ==
Let $G$ be a group.
Let $\kappa: G \to \Aut G$ be the mapping from $G$ to the automorphism group of $G$ defined as:
:$\forall x \in G: \map \kappa x := \kappa_x$
where $\kappa_x$ is the inner automorphism on $x$:
:$\forall g \in G: \map {\kappa_x} g = x g x^{-1}$
Then $\kappa$ is a homomorphism.
== Proof ==
Let $x, y \in G$.
By definition of automorphism group, we have that:
:$\map \kappa x \map \kappa y = \kappa_x \circ \kappa_y$
where $\circ$ denotes composition of mappings.
Then $\forall g \in G$:
And so:
:$\forall g \in G: \map {\kappa_{x y} } g = \map {\kappa_x \circ \kappa_y} g$
Thus by definition of $\kappa$:
:$\map \kappa x \map \kappa y = \map \kappa{x y}$
demonstrating that $\kappa$ is a homomorphism.
== Sources ==
* : Chapter $7$: Homomorphisms: Exercise $10$
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $25$
Category:Inner Automorphisms
Category:Group Homomorphisms
Mapping from L1 Space to Real Number Space is Continuous

== Theorem ==
Let $\struct {\R, d}$ be the real number line under the usual metric $d$.
Let $X$ be the set of continuous real functions $f: \closedint a b \to \R$.
Let $d_1$ be the $L^1$ metric on $X$.
Let $I: X \to \R$ be the real-valued function defined as:
:$\ds \forall f \in X: \map I f := \int_a^b \map f t \ \mathop d t$
Then the mapping:
:$I: \struct {X, d_1} \to \struct {\R, d}$
is continuous.
== Proof ==
The $L^1$ metric on $X$ is defined as:
:$\ds \forall f, g \in S: \map {d_1} {f, g} := \int_a^b \size {\map f t - \map g t} \rd t$
Let $\epsilon \in \R_{>0}$.
Let $f \in X$.
Let $\delta = \epsilon$.
Then:
Thus it has been demonstrated that:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall g \in X: \map {d_1} {f, g} < \delta \implies \map d {\map I f, \map I g} < \epsilon$
Hence by definition of continuity at a point, $I$ is continuous at $f$.
As $f$ is chosen arbitrarily, it follows that $I$ is continuous for all $f \in X$.
The result follows by definition of continuous mapping.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 3$: Continuity: Exercise $1$
Category:Continuous Mappings
Category:L1 Metric
Mapping from Preordering reflects Ordering

== Theorem ==
Let $\struct {S, \RR}$ be a preordered set.
Then there exists:
:an ordered set $\struct {T, \preccurlyeq}$
:a mapping $f: S \to T$ such that:
::$\RR: = \set {\tuple {x, y}: \map f x \preccurlyeq \map f y}$
== Proof ==
Let $\sim_\RR$ denote the equivalence on $S$ induced by $\RR$:
:$x \sim_\RR y$ $x \mathrel \RR y$ and $y \mathrel \RR x$
Let $\preccurlyeq_\RR$ be the ordering on the quotient set $S / {\sim_\RR}$ by $\RR$:
:$\eqclass x {\sim_\RR} \preccurlyeq_\RR \eqclass y {\sim_\RR} \iff x \mathrel \RR y$
where $\eqclass x {\sim_\RR}$ denotes the equivalence class of $x$ under $\sim_\RR$.
Let $f: S \to S / {\sim_\RR}$ be the quotient mapping induced by $\RR$:
:$\forall x \in S: \map f x = \eqclass x \RR$
From Preordering induces Ordering we have that $\struct {T, \preccurlyeq}$ is an set with exactly the properties required.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $41 \ \text {(e)}$
Category:Orderings
Category:Preorderings
Mapping from Quotient Set when Defined is Unique

== Theorem ==
Let $S$ and $T$ be sets.
Let $\RR$ be an equivalence relation on $S$.
Let $f: S \to T$ be a mapping from $S$ to $T$.
Let $S / \RR$ be the quotient set of $S$ induced by $\RR$.
Let $q_\RR: S \to S / \RR$ be the quotient mapping induced by $\RR$.
Let the mapping $\phi: S / \RR \to T$ defined as:
:$\phi \circ q_\RR = f$
be well-defined.
Then $\phi$ is unique.
== Proof ==
From Condition for Mapping from Quotient Set to be Well-Defined, $\phi$ is well-defined :
:$\forall x, y \in S: \tuple {x, y} \in \RR \implies \map f x = \map f y$
From Quotient Mapping is Surjection, $q_\RR$ is a surjection.
Suppose $\psi: S / \RR \to T$ is another well-defined mapping defined as:
:$\psi \circ q_\RR = f$
Then we have:
Hence the result.
== Motivation ==
== Also see ==
* Definition:Well-Defined Mapping
* Condition for Mapping from Quotient Set to be Well-Defined
* Condition for Mapping from Quotient Set to be Injection
* Condition for Mapping from Quotient Set to be Surjection
== Sources ==
* : $\S 6$. Indexed families; partitions; equivalence relations: Theorem $6.5$
Category:Quotient Mappings
Category:Quotient Sets
Mapping from Set to Class of All Ordinals is Bounded Above

== Theorem ==
Let $x$ be a set.
Let $\On$ be the class of all ordinals.
Let $f: x \to \On$ be a mapping.
Then $f$ has an upper bound.
=== Corollary ===
== Proof ==
Let $I$ be the image of $f$.
By Union of Set of Ordinals is Ordinal: Corollary, $\bigcup I$ is an ordinal.
But by Union is Smallest Superset, each element of $I$ is a subset of $\bigcup I$.
Thus $\bigcup I$ is an upper bound of $f$.
Category:Class of All Ordinals
Category:Class Mappings
Mapping from Set to Class of All Ordinals is Bounded Above/Sequence Corollary

== Corollary to Mapping from Set to Class of All Ordinals is Bounded Above ==
Let $\sequence {x_n}$ be any ordinal-valued sequence.
Then $\sequence {x_n}$ is bounded above.
== Proof ==
By definition, a sequence $\sequence {x_n}$ is a mapping whose domain is a subset of the natural numbers and is thus a set.
Thus by Mapping from Set to Class of All Ordinals is Bounded Above, $\sequence {x_n}$ has an upper bound.
Category:Class of All Ordinals
Category:Class Mappings
Mapping from Set to Ordinate of Cartesian Product is Injection

== Theorem ==
Let $S$ and $T$ be sets such that $T \ne \O$.
Let $S \times T$ denote their cartesian product.
Let $t \in T$ be given.
Let $j_t: S \to S \times T$ be the mapping from $S$ to $S \times T$ defined as:
:$\forall s \in S: \map {j_t} s = \tuple {s, t}$
Then $j_t$ is an injection.
== Proof ==
It has been shown in Correspondence between Set and Ordinate of Cartesian Product is Mapping that $j_t$ is a mapping.
Now it is to be shown that $j_t$ is injective, that is:
:$\forall s_1, s_2 \in S: \map {j_t} {s_1} = \map {j_t} {s_2} \implies s_1 = s_2$
We have that:
Hence the result.
== Also see ==
* Definition:Canonical Injection (Abstract Algebra) for an instance of this construct in the context of algebraic structures
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 6$: Functions: Exercise $6$
Category:Cartesian Product
Category:Injections
Mapping from Singleton is Injection

== Theorem ==
Let $f: S \to T$ be a mapping.
Let $S$ be a singleton.
Then $f$ is an injection.
== Proof ==
Let $S = \set s$.
For $f$ to be an injection, all we need to do is show:
:$\forall x_1, x_2 \in S: \map f {x_1} = \map f {x_2} \implies x_1 = x_2$
But as $S$ is a singleton, it follows that:
:$x_1 = x_2 = s$
Hence the result.
Category:Injections
Category:Singletons
Mapping from Standard Discrete Metric on Real Number Line is Continuous

== Theorem ==
Let $\R$ be the real number line.
Let $\struct {\R, d_1}$ be the metric space such that $d_1$ be the Euclidean metric on $\R$.
Let $\struct {\R, d_2}$ be the metric space such that $d_2$ be the standard discrete metric on $\R$.
Let $f: \tuple {\R, d_2} \to \tuple {\R, d_1}$ be a real function.
Then $f$ is $\tuple {d_2, d_1}$-continuous on $\R$.
== Proof ==
Let $\epsilon \in \R: \epsilon > 0$.
Let $\delta = 1$.
Let $x \in \R$.
Let $y \in \R$ such that $\map {d_2} {x, y} < \delta$.
That is, $\map {d_2} {x, y} < 1$.
By the definition of the standard discrete metric on $\R$, that would mean that $\map {d_2} {x, y} = 0$ and so $x = y$.
Thus $\map f x = \map f y$.
By definition of a metric, that means:
:$\map {d_1} {\map f x, \map f y} = 0 < \epsilon$
Thus the conditions for $\tuple {d_2, d_1}$-continuity at a point are fulfilled.
This is true for all $x \in \R$.
So by definition $f$ is $\tuple {d_2, d_1}$-continuous on $\R$.
== Sources ==
* : Chapter $\text{III}$: Metric Spaces: The Definition
Category:Standard Discrete Metric
Category:Continuous Mappings
Mapping from Totally Ordered Set is Dual Order Embedding iff Strictly Decreasing

== Theorem ==
Let $\left({S, \preceq_1}\right)$ be a totally ordered set.
Let $\left({T, \preceq_2}\right)$ be an ordered set.
Let $\phi: S \to T$ be a mapping.
Then $\phi$ is a dual order embedding $\phi$ is strictly decreasing.
That is:
:$\forall x, y \in S: x \preceq_1 y \iff \phi \left({y}\right) \preceq_2 \phi \left({x}\right)$
:$\forall x, y \in S: x \prec_1 y \implies \phi \left({y}\right) \prec_2 \phi \left({x}\right)$
== Proof ==
=== Forward Implication ===
Let $\phi$ be a dual order embedding.
Then $\phi$ is an order embedding of $\left({S, \preceq_1}\right)$ into $\left({T, \succeq_2}\right)$, where $\succeq_2$ is the dual of $\preceq_2$.
Thus by Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing:
:$\phi: \left({S, \preceq_1}\right) \to \left({T, \succeq_2}\right)$ is strictly increasing.
Thus:
:$\forall x, y \in S: x \prec_1 y \implies \phi \left({x}\right) \succ_2 \phi \left({y}\right)$
so:
:$\forall x, y \in S: x \prec_1 y \implies \phi \left({y}\right) \prec_2 \phi \left({x}\right)$ 
Thus $\phi: \left({S, \preceq_1}\right) \to \left({T, \preceq_2}\right)$ is strictly decreasing.
=== Reverse Implication ===
Suppose that $\phi: \left({S, \preceq_1}\right) \to \left({T, \preceq_2}\right)$ is strictly decreasing.
Then by the same argument as above:
:$\phi: \left({S, \preceq_1}\right) \to \left({T, \preceq_2}\right)$ is strictly increasing.
Thus by Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing, $\phi$ is an order embedding of $\left({S, \preceq_1}\right)$ into $\left({T, \succeq_2}\right)$.
So $\phi$ is a dual order embedding of $\left({S, \preceq_1}\right)$ into $\left({T, \preceq_2}\right)$.
Category:Total Orderings
Category:Order Embeddings
Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing

== Theorem ==
Let $\struct {S, \preceq_1}$ be a totally ordered set.
Let $\struct {T, \preceq_2}$ be an ordered set.
Let $\phi: S \to T$ be a mapping.
Then $\phi$ is an order embedding $\phi$ is strictly increasing.
That is:
:$\forall x, y \in S: x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$
:
:$\forall x, y \in S: x \prec_1 y \implies \map \phi x \prec_2 \map \phi y$
== Proof ==
=== Forward Implication ===
Let $\phi$ be an order embedding.
Let $x, y \in S$ with $x \prec_1 y$, where $\prec_1$ is the strict predecessor relation.
Then:
 
So by definition, $\phi$ is strictly increasing.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Theorem $14.9: \ 2^\circ$
Category:Order Embeddings
Category:Total Orderings
Category:Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing
Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing/Reverse Implication

== Theorem ==
Let $\struct {S, \preceq_1}$ be a totally ordered set.
Let $\struct {T, \preceq_2}$ be an ordered set.
Let $\phi: S \to T$ be a strictly increasing mapping.
Then $\phi$ is an order embedding.
== Proof 1 ==
== Proof 2 ==
Category:Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing
Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing/Reverse Implication/Proof 1

== Theorem ==
== Proof ==
Let $x \preceq_1 y$.
Then $x = y$ or $x \prec_1 y$.
Let $x = y$.
Then
:$\map \phi x = \map \phi y$
so:
:$\map \phi x \preceq_2 \map \phi y$
Let $x \prec_1 y$.
Then by the definition of strictly increasing mapping:
:$\map \phi x \prec_2 \map \phi y$
so by the definition of $\prec_2$:
:$\map \phi x \preceq_2 \map \phi y$
Thus:
:$x \preceq_1 y \implies \map \phi x \preceq_2 \map \phi y$
It remains to be shown that:
:$\map \phi x \preceq_2 \map \phi y \implies x \preceq_1 y$
Suppose that $x \npreceq_1 y$.
Since $\preceq_1$ is a total ordering:
:$y \prec_1 x$
Thus since $\phi$ is strictly increasing:
:$\map \phi y \prec_1 \map \phi x$
Thus:
:$\map \phi x \npreceq_1 \map \phi y$
Therefore:
:$x \npreceq_1 y \implies \map \phi x \npreceq_2 \map \phi y$
By the Rule of Transposition:
:$\map \phi x \preceq_2 \map \phi y \implies x \preceq y$
Category:Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing
Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing/Reverse Implication/Proof 2

== Theorem ==
== Proof ==
Let $\phi$ be strictly increasing.
Let $\map \phi x \preceq_2 \map \phi y$.
As $\struct {S, \prec_1}$ is a strictly totally ordered set:
:Either $y \prec_1 x$, $y = x$, or $x \prec_1 y$.
 that $y \prec_1 x$.
By the definition of a strictly increasing mapping:
:$\map \phi y \prec_2 \map \phi x$
which contradicts the fact that $\map \phi x \preceq_2 \map \phi y$.
Therefore $y \nprec_1 x$.
Thus $y = x$, or $x \prec_1 y$, so $x \preceq_1 y$.
Hence:
:$\map \phi x \preceq_2 \map \phi y \iff x \preceq_1 y$
and $\phi$ has been proved to be an order embedding.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Theorem $14.9: \ 2^\circ$
Category:Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing
Mapping from Unit Circle defines Periodic Function

== Theorem ==
Let $\SS$ denote the unit circle whose center is at the origin of the Cartesian plane $\R^2$.
Let $p: \R \to \SS$ be the mapping defined as:
:$\forall x \in \R: \map p x = \tuple {\cos x, \sin x}$
Let $f': \SS \to \R$ be a real-valued function.
Then the composition $f' \circ p$ defines a periodic real function whose period is $2 \pi$.
== Proof ==
Let $f := f' \circ p$ denote the composition of $f$ with $p$.
We have:
Hence the result.
== Also see ==
* Periodic Function as Mapping from Unit Circle
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.8$: Quotient spaces
Category:Periodic Functions
Mapping is Bijection iff Composite with Direct Image Mapping with Complementation Commutes

== Theorem ==
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Then:
:$f$ is a bijection
:
:$f^\to \circ \complement_S = \complement_T \circ f^\to$
where:
:$f^\to: \powerset S \to \powerset T$ denotes the direct image mapping of $f$
:$\complement_S: \powerset S \to \powerset S$ denotes the complement relative to $S$
:$\complement_T: \powerset T \to \powerset T$ denotes the complement relative to $T$
:$\powerset S$ and $\powerset T$ denote the power sets of $S$ and $T$ respectively.
== Proof ==
=== Sufficient Condition ===
Let $f$ be a bijection.
Thus a fortiori $f$ is:
:a surjection
:a one-to-many relation
From One-to-Many Image of Set Difference: Corollary 2 we have:
:$\forall X \in \powerset S: \map {\paren {f^\to \circ \complement_S} } X = \map {\paren {\complement_{\Img f} \circ f^\to} } X$
By definition of surjection:
:$\Img f = T$
and so:
:$\forall X \in \powerset S: \map {\paren {f^\to \circ \complement_S} } X = \map {\paren {\complement_T \circ f^\to} } X$
That is:
:$f^\to \circ \complement_S = \complement_T \circ f^\to$
=== Necessary Condition ===
Let $f$ be a mapping such that:
:$f^\to \circ \complement_S = \complement_T \circ f^\to$
We have that:
Hence, by definition, $f$ is a surjection.
Now consider $A, B \subseteq S$:
It follows from Image of Intersection under Injection that $f^\to$ is an injection.
Hence from Mapping is Injection iff Direct Image Mapping is Injection:
:$f$ is an injection.
We have therefore that $f$ is both an injection and a surjection.
Hence, by definition, $f$ is a bijection.
== Sources ==
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Exercise $3$
Category:Set Complement
Category:Bijections
Category:Direct Image Mappings
Mapping is Bijection iff Direct Image Mapping is Bijection

== Theorem ==
Let $f: S \to T$ be a mapping.
Let $f^\to: \powerset S \to \powerset T$ be the direct image mapping of $f$.
Then:
:$f^\to$ is a bijection
:$f: S \to T$ is also a bijection.
== Proof ==
Follows immediately from:
* Mapping is Injection iff Direct Image Mapping is Injection
* Mapping is Surjection iff Direct Image Mapping is Surjection
Category:Bijections
Category:Direct Image Mappings
Mapping is Bounded on Union iff Bounded on Each Component/Real-Valued Function

== Theorem ==
Let $f$ be a real-valued function.
Then:
:$f$ is bounded on the union of a finite number of sets within the domain of $f$
:
:$f$ is bounded on each of the sets.
== Proof ==
Let $S_i$, $i \in \set {1, \ldots, n}$, $n \in \N_{>0}$, denote sets within the domain of $f$.
Define $S = \ds \bigcup_{i \mathop = 1}^n S_i$.
=== Necessary Condition ===
Let $f$ be bounded on each of the sets $S_i$.
We need to prove that $f$ is bounded on $S$.
Let $b_i$ be a bound for $f$ on $S_i$, $i \in \set {1, \ldots, n}$.
Let $b = \map \max {b_1, \ldots, b_n}$ where $\max$ is the |maximum operation.
We note that $b \ge b_i$ for every $i$ in $\set {1, \ldots, n}$.
Now, by definition of bound, a number greater than or equal to a bound is also a bound.
Therefore, since $b_i$ is a bound for $f$ on $S_i$ and $b \ge b_i$, $b$ is a bound for $f$ on $S_i$ for every $i$.
Since $b$ is a bound for $f$ on every set $S_i$, $b$ is a bound for $f$ on the union of the sets $S_i$.
In other words, $f$ is bounded on $S$.
=== Sufficient Condition ===
Let $f$ be bounded on $S$.
We need to prove that $f$ is bounded on each of the sets $S_i$, $i \in \set {1, \ldots, n}$.
Since $f$ is bounded on $S$, there is a bound $K$ that satisfies $K \ge \size {\map f s}$ for every element $s$ in $S$.
Pick a set $S_i$.
Because $K \ge \size {\map f s}$ is true of every element $s$ of $S$, this is also true of every element of $S_i$ as $S_i$ is a subset of $S$.
Consequently, $K$ is a bound for $f$ on $S_i$.
Since $i$ is arbitrary, we conclude that $f$ is bounded on $S_i$ for every $i$ in $\set {1, \ldots, n}$.
== Sources ==
* : $5$: Compact spaces: $5.1$: Motivation: Step $2$
Category:Bounded Real-Valued Functions
Category:Real-Valued Functions
Mapping is Constant iff Image is Singleton

== Theorem ==
A mapping is a constant mapping its image is a singleton.
== Proof ==
=== Necessary Condition ===
Let $f_c: S \to T$ be a constant mapping.
Then from the definition of the image of an element:
:$\forall x \in S: \map {f_c} x = c \implies \Img S = \set c$
Thus the image of $f_c: S \to T$ is a singleton.
=== Sufficient Condition ===
Let $f: S \to T$ such that its image is a singleton $\set c \subseteq T$.
Thus $f: S \to T$ is a constant mapping (and we can write it $f_c: S \to T$).
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Functions
Category:Constant Mappings
Category:Singletons
Mapping is Constant iff Increasing and Decreasing

== Theorem ==
Let $\left({S, \preceq_1}\right)$ and $\left({T, \preceq_2}\right)$ be ordered sets.
Let $\phi: \left({S, \preceq_1}\right) \to \left({T, \preceq_2}\right)$ be a mapping.
Then $\phi$ is a constant mapping $\phi$ is both increasing and decreasing.
== Proof ==
=== Necessary Condition ===
Suppose $\phi$ is a constant mapping.
Then:
: $\forall x, y \in S: \phi \left({x}\right) = \phi \left({y}\right)$
So:
: $\forall x, y \in S: \phi \left({x}\right) \mathop{\preceq_2} \phi \left({y}\right)$
: $\forall x, y \in S: \phi \left({y}\right) \mathop{\preceq_2} \phi \left({x}\right)$
and so $\phi$ is both increasing and decreasing.
=== Sufficient Condition ===
Suppose $\phi$ is both increasing and decreasing.
Let $x, y \in S$.
Then:
: $\phi \left({x}\right) \mathop{\preceq_2} \phi \left({y}\right)$
: $\phi \left({y}\right) \mathop{\preceq_2} \phi \left({x}\right)$
As $\preceq$ is an ordering, by definition $\preceq$ is antisymmetric.
This means thtat:
: $\phi \left({y}\right) = \phi \left({x}\right)$
As this holds for any $x, y \in S$ it follows for all $x, y \in S$ by Universal Generalisation.
Thus:
: $\forall x, y \in S: \phi \left({x}\right) = \phi \left({y}\right)$
and so $\phi$ is a constant mapping.
== Sources ==
* : $\S 12.1$
Category:Increasing Mappings
Category:Decreasing Mappings
Category:Constant Mappings
Mapping is Continuous iff Inverse Images of Open Sets are Open

== Theorem ==
Let $X$ and $Y$ be normed vector spaces.
Let $f : X \to Y$ be a mapping.
Then:
:$f$ is continuous on $X$
:
:for every $V$ open in $Y$, $\map {f^{-1}} V$ is open in $X$.
=== Corollary ===
== Proof ==
=== Sufficient Condition ===
Let $c \in X$.
Let $\epsilon \in \R_{\mathop > 0}$.
Let $V := \map {B_\epsilon} {\map f c}$ be an open ball in $Y$.
By Open Ball is Open Set in Normed Vector Space, $V$ is an open set in $Y$.
Let $\map {f^{-1}} V = \map {f^{-1}} {\map {B_\epsilon} {\map f c}}$ be an open set in $X$.
Note that $\map f c \in \map {B_\epsilon} {\map f c}$.
Hence, $c \in \map {f^{-1}} {\map {B_\epsilon} {\map f c}}$.
By definition of open set:
:$\exists \delta > 0 : \map {B_\delta} c \subseteq \map {f^{-1}} {\map {B_\epsilon} {\map f c}}$
In other words:
:$\forall x \in X : \norm {x - c} < \delta \implies x \in \map {f^{-1}} {\map {B_\epsilon} {\map f c}}$
or
:$\map f x \in \map {B_\epsilon} {\map f c}$
By definition of open ball:
:$\norm {\map f x - \map f c} < \epsilon$
But $\epsilon$ was arbitrary.
Hence:
:$\forall \epsilon \in \R_{\mathop > 0} : \exists \delta > 0 : \norm {x - c} < \delta \implies \norm {\map f x - \map f c} < \epsilon$
Therefore, $f$ is continuous at $c$.
However, $c \in X$ was arbitrary.
Hence, $f$ is continuous on $X$.
=== Necessary Condition ===
Let $f$ be continuous on $X$.
Let $V$ be an open subset of $Y$.
Let $c \in \map {f^{-1}} V$.
Hence $\map f c \in V$.
By definition of an open set:
:$\exists \epsilon \in \R_{\mathop > 0} : \map {B_\epsilon} {\map f c} \subseteq V$
By continuity of $f$ at $c$:
:$\exists \delta \in \R_{\mathop > 0} : \forall x \in X : \norm {x - c} < \delta \implies \norm {\map f x - \map f c} < \epsilon$
In other words:
:$\exists \delta \in \R_{\mathop > 0} : \forall x \in X : \norm {x - c} < \delta \implies \map f x \in \map {B_\epsilon} {\map f c} \subseteq V$
or
:$\exists \delta \in \R_{\mathop > 0} : \forall x \in X : \norm {x - c} < \delta \implies \map f x \in V$
$\map f x \in V$ is equivalent to $x \in \map {f^{-1}} V$.
Suppose $\norm {x - c} < \delta$.
By definition of open ball:
:$x \in \map {B_\delta} c$
But then:
:$x \in \map {B_\delta} c \implies x \in \map {f^{-1}} V$
Hence:
:$\map {B_\delta} c \subseteq \map {f^{-1}} V$
But $\delta$ exists for any $c \in \map {f^{-1}} V$.
By definition, $\map {f^{-1}} V$ is open in $X$.
== Also see ==
* Real-Valued Mapping is Continuous if Inverse Images of Unbounded Open Intervals are Open
== Sources ==
* : Chapter $\S 2.2$: Continuous and linear maps. Continuous maps
Category:Continuous Mappings on Normed Vector Spaces