nyuuzyou/wikis huggingface dataset:
Continuity Test for Real-Valued Functions/Sufficient Condition

== Theorem ==
Let $\struct {S, \tau}$ be a topological space.
Let $f: S \to \R$ be a real-valued function.
Let $x \in S$.
Let $f$ satisfy:
:$\forall \epsilon \in \R_{>0} : \exists U \in \tau : x \in U : \map {f^\to} U \subseteq \openint {\map f x - \epsilon} {\map f x + \epsilon}$
Then $f$ is continuous at $x$
== Proof ==
Let $f$ satisfy:
:$\forall \epsilon \in \R_{>0} : \exists U \in \tau : x \in U : \map {f^\to} U \subseteq \openint {\map f x - \epsilon} {\map f x + \epsilon}$
Let $N$ be a neighborhood of $\map f x$.
By definition of a neighborhood of $\map f x$:
:there exists $U$ such that $z \in U \subseteq N$
By definition of the topology induced by the Euclidean Metric on Real Number Line:
:there exists an open $\epsilon$-ball $\map {B_\epsilon} {\map f x}$ such that $x \in \map {B_\epsilon} {\map f x} \subseteq U$
From Open Ball in Real Number Line is Open Interval:
:$\map {B_\epsilon} {\map f x} = \openint {\map f x - \epsilon} {\map f x + \epsilon}$
Thus:
:$\exists U \in \tau : x \in U : \map {f^\to} U \subseteq \openint {\map f x - \epsilon} {\map f x + \epsilon}$
From Set is Open iff Neighborhood of all its Points, $U$ is a neighborhood of $x$.
Since $N$ was an arbitrary neighborhood of $\map f x$ it follows that $f$ is continuous at $x$.
Category:Continuity Test for Real-Valued Functions
Continuity Test using Basis

== Theorem ==
Let $\struct {X_1, \tau_1}$ and $\struct {X_2, \tau_2}$ be topological spaces.
Let $f: X_1 \to X_2$ be a mapping.
Let $\BB$ be an analytic basis for $\tau_2$.
Suppose that:
:$\forall B \in \BB: f^{-1} \sqbrk B \in \tau_1$
where $f^{-1} \sqbrk B$ denotes the preimage of $B$ under $f$.
Then $f$ is continuous.
== Proof ==
Let $U \in \tau_2$.
By the definition of an analytic basis, it follows that:
:$\ds \exists \AA \subseteq \BB: U = \bigcup \AA$
Hence:
The result follows from the definition of continuity.
== Also see ==
* Continuity Test using Sub-Basis for a general version.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.2$: Bases: Application $3.2.5$
Category:Topological Bases
Category:Continuous Mappings
Continuity Test using Sub-Basis

== Theorem ==
Let $\struct {X_1, \tau_1}$ and $\struct {X_2, \tau_2}$ be topological spaces.
Let $f: X_1 \to X_2$ be a mapping.
Let $\SS$ be an analytic sub-basis for $\tau_2$.
Suppose that:
:$\forall S \in \SS: f^{-1} \sqbrk S \in \tau_1$
where $f^{-1} \sqbrk S$ denotes the preimage of $S$ under $f$.
Then $f$ is continuous.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Analytic Basis is Analytic Sub-Basis, of which Continuity Test using Basis is seen as a special case.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.3$: Sub-bases and weak topologies
Category:Topological Bases
Category:Continuous Mappings
Category:Continuity Test using Sub-Basis
Continuity Test using Sub-Basis/Proof 1

== Theorem ==
== Proof ==
Define:
:$\ds \BB = \set {\bigcap \AA: \AA \subseteq \SS, \AA \text{ is finite} } \subseteq \powerset {X_2}$
Let $B \in \BB$.
Then there exists a finite subset $\AA \subseteq \SS$ such that:
:$\ds B = \bigcap \AA$
Hence:
Define:
:$\ds \tau = \set {\bigcup \AA: \AA \subseteq \BB} \subseteq \powerset {X_2}$
By the definition of an analytic sub-basis, we have $\tau_2 \subseteq \tau$.
Let $U \in \tau_2$.
Then $U \in \tau$; therefore:
:$\ds \exists \AA \subseteq \BB: U = \bigcup \AA$
Hence:
That is, $f$ is continuous.
Category:Continuity Test using Sub-Basis
Continuity Test using Sub-Basis/Proof 2

== Theorem ==
== Proof ==
Let $\tau$ be the final topology on $X_2$ with respect to $f$.
By hypothesis, $\SS \subseteq \tau$.
By Synthetic Sub-Basis and Analytic Sub-Basis are Compatible, we have that $\tau_2$ is the topology generated by the synthetic sub-basis $\SS$.
By the definition of the generated topology, we have $\tau_2 \subseteq \tau$.
By the definition of the final topology, it follows that $f$ is continuous.
Category:Continuity Test using Sub-Basis
Continuity in Initial Topology

== Theorem ==
Let $\struct {Z, \tau}$ be a topological space. 
Let $X$ be a set.
Let $I$ be an indexing set.
Let $\family {\struct {Y_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of topological spaces indexed by $I$.
Let $\family {f_i: X \to Y_i}_{i \mathop \in I}$ be an indexed family of mappings indexed by $I$.
Let $\tau'$ be the initial topology on $Y$ with respect to $\family {f_i \restriction_Y}_{i \mathop \in I}$.
Let $g : Z \to X$ be a mapping. 
Then $g$ is $\tuple {\tau, \tau'}$-continuous : 
:$f_i \circ g : Z \to Y_i$ is $\tuple {\tau, \tau_i}$-continuous for each $i \in I$.
== Proof ==
=== Necessary Condition ===
Suppose that: 
:$g$ is $\tuple {\tau, \tau'}$-continuous.
From the definition of the initial topology:
:$f_i$ is $\tuple {\tau', \tau_i}$-continuous for each $i \in I$. 
From Composite of Continuous Mappings is Continuous:
:$f_i \circ g$ is $\tuple {\tau, \tau_i}$-continuous for each $i \in I$.
=== Sufficient Condition ===
Suppose that: 
:$f_i \circ g$ is $\tuple {\tau, \tau_i}$-continuous for each $i \in I$.
From the definition of the initial topology:
:$\SS = \set {f_i^{-1} \sqbrk U : i \in I, \, U \in \tau_i}$ 
is a synthetic sub-basis for $\tau'$. 
From Continuity Test using Sub-Basis, it suffices to show that: 
:$g^{-1} \sqbrk U \in \tau$ for $U \in \SS$.
Let $U \in \SS$.
Then there exists $i \in I$ and $V \in \tau_i$ such that $U = f_i^{-1} \sqbrk V$.
So:
:$g^{-1} \sqbrk U = g^{-1} \sqbrk {f_i^{-1} \sqbrk V}$
From Preimage of Subset under Composite Mapping:
:$g^{-1} \sqbrk {f_i^{-1} \sqbrk V} = \paren {f_i \circ g}^{-1} \sqbrk V$
From hypothesis, $f_i \circ g$ is $\struct {\tau, \tau_i}$-continuous, so since $V \in \tau_i$, we have: 
:$\paren {f_i \circ g}^{-1} \sqbrk V \in \tau$
So, for each $U \in \SS$ we have: 
:$g^{-1} \sqbrk U \in \tau$
So $g$ is $\tuple {\tau, \tau'}$-continuous from Continuity Test using Sub-Basis.
Category:Initial Topology
Category:Continuous Mappings
Continuity of Composite with Identification Mapping

== Theorem ==
Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $f_1: T_1 \to T_2$ be a mapping.
Let $S_3$ be a set.
Let $p: S_1 \to S_3$ be a mapping.
Let $\tau_3$ be the identification topology on $S_3$ with respect to $p$ and $T_1$.
Let $T_3 = \struct {S_3, \tau_3}$ be the resulting topological space.
Let $f_2: S_3 \to S_2$ be a mapping such that:
:$f_1 = f_2 \circ p$
where $f_2 \circ p$ is the composition of $f_2$ with $p$.
Then $f_1$ is a continuous mapping $f_2$ is a continuous mapping.
::$\begin{xy} \xymatrix@L+2mu@+1em{
 S_1 \ar[r]^*{p}
 \ar@{-->}[rd]_*{f_2 \circ p}
&
 S_3 \ar[d]^*{f_2}
\\ &
 S_2
}\end{xy}$
== Proof ==
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.8$: Quotient spaces: Proposition $3.8.2$
Category:Identification Topology
Continuity of Composite with Inclusion

== Theorem ==
Let $T = \struct {A, \tau}$ and $T' = \struct {A', \tau'}$ be topological spaces.
Let $H \subseteq A$.
Let $T_H = \struct {H, \tau_H}$ be a topological subspace of $T$.
Let $i: H \to A$ be the inclusion mapping.
Let $f: A \to A'$ and $g: A' \to H$ be mappings.
Then the following apply:
=== Mapping on Inclusion ===
=== Inclusion on Mapping ===
=== Uniqueness of Induced Topology ===
Category:Continuity of Composite with Inclusion
Category:Topological Subspaces
Category:Continuous Mappings
Category:Inclusion Mappings
Category:Composite Mappings
Continuity of Composite with Inclusion/Inclusion on Mapping

== Theorem ==
Let $T = \struct {A, \tau}$ and $T' = \struct {A', \tau'}$ be topological spaces.
Let $H \subseteq A$.
Let $T_H = \struct {H, \tau_H}$ be a topological subspace of $T$.
Let $i: H \to A$ be the inclusion mapping.
Let $g: A' \to H$ be a mapping.
$g$ is $\tuple {\tau', \tau_H}$-continuous $i \circ g$ is $\tuple {\tau', \tau}$-continuous.
== Proof ==
=== Necessary Condition ===
Suppose $g$ is $\tuple {\tau', \tau_H}$-continuous.
From Inclusion Mapping is Continuous, $i$ is $\tuple {\tau_H, \tau}$-continuous.
It follows from Composite of Continuous Mappings is Continuous that $i \circ g$ is $\tuple {\tau', \tau}$-continuous.
=== Sufficient Condition ===
Suppose $i \circ g$ is $\tuple {\tau', \tau}$-continuous.
Let $V \in \tau_H$.
Then from the definition of topological subspace:
:$V = U \cap H$ for some $U \in \tau$
and by Preimage of Subset under Inclusion Mapping:
:$U \cap H = i^{-1} \sqbrk U$
So by Preimage of Subset under Composite Mapping:
:$g^{-1} \sqbrk V = g^{-1} \sqbrk {i^{-1} \sqbrk U} = \paren {i \circ g}^{-1} \sqbrk U$
Thus:
:$g^{-1} \sqbrk V \in \tau'$
by continuity of $i \circ g$.
Hence $g$ is $\tuple {\tau', \tau_H}$-continuous.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.4$: Subspaces: Proposition $3.4.2 \ \text{(b)}$
* : $\S 3$: New Spaces From Old: Subspaces
Category:Continuity of Composite with Inclusion
Continuity of Composite with Inclusion/Mapping on Inclusion

== Theorem ==
Let $T = \struct {A, \tau}$ and $T' = \struct {A', \tau'}$ be topological spaces.
Let $H \subseteq A$.
Let $T_H = \struct {H, \tau_H}$ be a topological subspace of $T$.
Let $i: H \to A$ be the inclusion mapping.
Let $f: A \to A'$ be a mapping.
If $f$ is $\tuple {\tau, \tau'}$-continuous, then $f \circ i$ is $\tuple {\tau_H, \tau'}$-continuous
== Proof ==
From Inclusion Mapping is Continuous, $i$ is $\tuple {\tau_H, \tau}$-continuous.
It follows from Composite of Continuous Mappings is Continuous that $f \circ i$ is $\tuple {\tau_H, \tau'}$-continuous.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.4$: Subspaces: Proposition $3.4.2 \ \text{(a)}$
Category:Continuity of Composite with Inclusion
Continuity of Composite with Inclusion/Uniqueness of Induced Topology

== Theorem ==
Let $T = \struct {A, \tau}$ and $T' = \struct {A', \tau'}$ be topological spaces.
Let $H \subseteq A$.
Let $T_H = \struct {H, \tau_H}$ be a topological subspace of $T$.
Let $i: H \to A$ be the inclusion mapping.
Let $g: A' \to H$ be a mapping.
The induced topology $\tau_H$ is the only topology on $H$ satisfying Continuity of Composite with Inclusion: Inclusion on Mapping for all possible $g$.
== Proof ==
Suppose $\tau' '$ is a topology on $H$ such that:
:$(1) \quad$ For any topological space $T' = \struct {A', \tau'}$, and
:$(2) \quad$ For any mapping $g: A' \to H$:
$g$ is $\tuple {\tau', \tau' '}$-continuous $i \circ g$ is $\tuple {\tau', \tau}$-continuous.
It needs to be shown that $\tau' '$ must be the same as $\tau_H$.
Let $A' = H$ and $\tau' = \tau' '$.
Let $g$ be the identity mapping on $H$.
From Identity Mapping is Continuous, $g$ is $\tuple {\tau' ', \tau' '}$-continuous
Thus from Composite of Continuous Mappings is Continuous, $i \circ g$ is $\tuple {\tau' ', \tau}$-continuous.
Hence for any $U \in \tau$:
:$\map {\paren {i \circ g}^{-1} } U \in \tau' '$
But:
:$\map {\paren {i \circ g}^{-1} } U = \map {i^{-1} } U = U \cap H$
Hence $\tau_H \subseteq \tau' '$.
Next, take take $A' = H$ and $\tau' = \tau_H$.
Let $g$ be the identity mapping on $H$.
We have that $i \circ g = i$ is $\tuple {\tau_H, \tau}$-continuous,
From Continuity of Composite with Inclusion: Inclusion on Mapping, it follows that $g$ is $\tuple {\tau_H, \tau' '}$-continuous.
But by definition of continuity, this is the same as saying $\tau' ' \subseteq \tau_H$.
So $\tau' ' = \tau_H$, as required.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.4$: Subspaces: Proposition $3.4.3$
Category:Continuity of Composite with Inclusion
Continuity of Heaviside Step Function

== Theorem ==
Let $\mu_c: \R \to \R$ be the Heaviside step function:
:$\map {\mu_c} x = \begin {cases} 0 & : x < c \\ 1 & : x > c \\ \text {arbitrary} & : x = c \end {cases}$
Then $\mu_c$ is continuous at every point of $\R$ except at $c$.
== Proof ==
Let $x \in \R: x \ne c$.
Let $\epsilon \in \R_{>0}$.
Let $\delta < \size {x - c}$.
Then by definition of the Heaviside step function:
:$\forall y \in \closedint {x - \delta} {x + \delta}: \map {\mu_c} x = \begin {cases} 0 & : x < c \\ 1 & : x > c \end {cases}$
Thus:
:$\forall y \in \closedint {x - \delta} {x + \delta}: \map {\mu_c} y = \map {\mu_c} x$
Thus:
:$\size {y - x} \implies: \map {\mu_c} y - \map {\mu_c} x < \epsilon$
and so $\mu_c$ is continuous at $\R$ except at $x$.
Now suppose $x = c$.
Let $\epsilon = \dfrac 1 3$.
Let $\delta \in \R_{>0}$.
Let $\alpha \in \R_{>0}: \alpha < \delta$.
Let $y = x + \alpha$.
Let $y \in \R: \size {y - x} < \delta$ and $y \ne x$.
Suppose $\map {\mu_c} y - \map {\mu_c} x < \epsilon$.
Then:
while:
So whatever the value of $\map f x$, either:
:$\map f {x + \alpha} - \map f x > \epsilon$
or:
:$\map f {x - \alpha} - \map f x > \epsilon$
It follows that $\mu_c$ is not continuous at $c$.
== Also see ==
* Heaviside Step Function is Piecewise Continuous
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 4$: Open Balls and Neighborhoods: Exercise $2$
Category:Continuity
Category:Heaviside Step Function
Continuity of Linear Functionals in Initial Topology on Vector Space Generated by Linear Functionals

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $X$ be a vector space over $\GF$. 
Let $F$ be a set of linear functionals on $X$ that form a vector space over $\GF$.
That is, for each $\lambda, \mu \in \GF$ and $f, g \in F$, we have: 
:$\lambda f + \mu g \in F$
Let $\tau$ be the initial topology on $X$ generated by $F$.
Let $g : X \to \GF$ be a linear functional.
Then $g$ is $\tau$-continuous $g \in F$. 
== Proof ==
=== Sufficient Condition ===
From the definition of the initial topology, $\tau$ is the coarsest topology making all $g \in F$ continuous.
So if $g \in F$ then $g$ is $\tau$-continuous.
=== Necessary Condition ===
Suppose that $g$ is $\tau$-continuous.
For each $x \in \GF$ and $\epsilon > 0$, let: 
:$\map {B_\epsilon} {x, \GF} = \set {y \in \GF : \cmod {y - x} < \epsilon}$
Then since $g$ is $\tau$-continuous, we have: 
:$g^{-1} \sqbrk {\map {B_1} {0, \GF} } \in \tau$
and so $g^{-1} \sqbrk {\map {B_1} {0, \GF} }$ is an open neighborhood of ${\mathbf 0}_X$. 
From Initial Topology on Vector Space Generated by Linear Functionals is Locally Convex, $\tau$ is generated by the seminorms: 
:$\set {p_f : f \in F}$
where we define $p_f : X \to \R_{\ge 0}$ by:
:$\map {p_f} x = \cmod {\map f x}$ 
for each $x \in X$.
So, from Open Sets in Standard Topology of Locally Convex Space, there exists $g_1, \ldots, g_n \in F$ and $\epsilon > 0$ such that: 
:$\set {x \in X : \cmod {\map {g_k} x} < \epsilon \text { for each } 1 \le k \le n} \subseteq g^{-1} \sqbrk {\map {B_1} {0, \GF} }$
We will deduce that: 
:$g \in \span \set {g_1, \ldots, g_n}$ 
so that $g \in F$.
Let:
:$\ds x \in \bigcap_{k \mathop = 1}^n \ker g_k$ 
Then $\map {g_k} {\lambda x} = 0$ for each $\lambda \in \GF$ and $1 \le k \le n$. 
In particular, $\cmod {\map {g_k} {\lambda x} } < \epsilon$ for each $\lambda \in \GF$ and $1 \le k \le n$.
So $\lambda x \in g^{-1} \sqbrk {\map {B_1} {0, \GF} }$ for each $\lambda \in \GF$.
Then: 
:$\cmod {\map g {\lambda x} } < 1$ 
for each $\lambda \in \GF$. 
So, by the linearity of $g$ we have: 
:$\cmod \lambda \cmod {\map g x} < 1$ 
for each $\lambda \in \GF$.
If $\cmod {\map g x} > 0$, we could take: 
:$\ds \cmod \lambda > \frac 1 {\cmod {\map g x} }$
to get: 
:$\cmod \lambda \cmod {\map g x} > 1$
Hence we must have $\map g x = 0$, so that $g \in \ker f$.
So, we have: 
:$\ds \bigcap_{k \mathop = 1}^n \ker g_k \subseteq \ker g$ 
So from Condition for Linear Dependence of Linear Functionals in terms of Kernel, we have: 
:$g \in \span \set {g_1, \ldots, g_n}$
and so $g \in F$ as desired.
Category:Initial Topology
Continuity of Linear Transformation/Normed Vector Space

== Theorem ==
Let $X, Y$ be a normed vector spaces over $\R$.
Let $T : X \to Y$ be a linear mapping.
:
:$(1): \quad T$ is continuous over $X$
:$(2): \quad T$ is continuous at $\mathbf 0$
:$(3): \quad \exists M > 0 : \forall x \in X : \norm {\map T x}_Y \le M \norm x_X$
== Proof ==
=== $\paren 1 \implies \paren 2$ ===
Let $T$ be continuous on $X$.
$X$ is a vector space.
By definition, $\exists \mathbf 0 \in X$.
Hence, $T$ is continuous at $\mathbf 0$.
=== $\paren 2 \implies \paren 3$ ===
Let $T$ be continuous at $\mathbf 0$.
Let $\epsilon := 1 > 0$.
By definition of continuity:
:$\exists \delta > 0: \forall x \in X : \norm {x - \mathbf 0} = \norm x < \delta \implies \norm {\map T x - \map T {\mathbf 0} } = \norm {\map T x} < 1$
Let $y := \dfrac \delta {2 \norm x} x$.
By definition of norm:
:$\norm x \in \R$
Hence:
:$\dfrac \delta {2 \norm x} \in \R$
Because $X$ is a vector space:
:$y \in X$
Furthermore:
Then:
Suppose $x = \mathbf 0$.
Then:
All together:
:$\norm {\map T x} \le M \norm x$
where $M = \dfrac 2 \delta$.
=== $\paren 3 \implies \paren 1$ ===
Let $M > 0$ be such that:
:$\forall x \in X: \norm {\map T x} \le M \norm x$
Let $x_0 \in X$.
Let $\epsilon > 0$.
Let $\delta := \dfrac \epsilon M > 0$.
Because $X$ is a vector space:
:$x - x_0 \in X$
Suppose:
:$\forall x \in X : \norm {x - x_0} < \delta$
Then:
By definition, $T$ is continuous at $x_0$.
But $x_0$ was arbitrary.
Hence, $T$ is continuous on $X$.
== Sources ==
* : Chapter $\S 2.3$: The normed space $\map {CL} {X,Y}$
Category:Linear Transformations
Category:Normed Vector Spaces
Category:Continuous Mappings
Category:Continuous Mappings on Normed Vector Spaces
Continuity of Linear Transformations

== Theorem ==
Let $H, K$ be Hilbert spaces, and let $A: H \to K$ be a linear transformation.
Then the following four statements are equivalent:
:$(1): \quad A$ is continuous
:$(2): \quad A$ is continuous at $\mathbf 0_H$
:$(3): \quad A$ is continuous at some point
:$(4): \quad \exists c > 0: \forall h \in H: \norm {\map A h}_K \le c \norm h_H$
== Proof ==
It is clear that $(1) \implies (2) \implies (3)$ and for $(4) \implies (2)$:
For any $\epsilon > 0$, there exists $\delta = \dfrac \epsilon c$, such that when $\norm {\mathbf 0_H - h}_H < \delta$:
:$\norm {\map A h - \map A {\mathbf 0_H} }_K \le c \norm h_H < c\delta = \epsilon$
Now we prove $(3) \implies (1)$:
Let $A$ be continuous at some point $h_0$.
For any sequence $h_n \to h$ in $H$:
:$h_n - h + h_0 \to h_0$
Hence:
:$\ds \lim_{n \mathop \to \infty} \map A {h_n - h + h_0} = \lim_{n \mathop \to \infty} \map A {h_n} - \map A h + \map A {h_0} = \map A {h_0}$
We see that:
:$\ds \lim_{n \mathop \to \infty} \map A {h_n} = \map A h$
Thus $A$ is continuous.
Now for the proof $(2) \implies (4)$.
We have that $A$ is continuous at $\mathbf 0_H$.
Hence there exists an open ball of positive real radius $a$, centered at $\mathbf 0_H$, such that its image under $A$ is included in the open ball of radius $1$, centered at $\mathbf 0_K$. 
This is, if $\norm h_H < a$, then:
:$\norm {\map A h}_K < 1$
Let $h$ be an arbitrary element in $H$.
Let $\epsilon > 0$.
We have:
:$\norm {\dfrac a {\norm h_H + \epsilon} h}_H < a$
Hence:
:$\norm {\map A {\dfrac a {\norm h_H + \epsilon} h} }_K = \dfrac a {\norm h_H + \epsilon} \norm {\map A h}_K < 1$
Therefore:
:$\norm {\map A h}_K < \dfrac 1 a \norm h_H + \dfrac \epsilon a$
Let $\epsilon \to 0$.
Therefore, we can take:
:$c := \dfrac 1 a$
== Sources ==
* $\text {II}.1.1$
Category:Linear Transformations on Hilbert Spaces
Continuity of Mapping between Metric Spaces by Closed Sets

== Theorem ==
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $f: A_1 \to A_2$ be a mapping.
Then $f$ is continuous :
:for every $V \subseteq A_2$ which is closed in $M_2$, $f^{-1} \sqbrk V$ is closed in $M_1$.
== Proof ==
First the following is established.
Let $W \in A_2$.
We note that:
:$f^{-1} \sqbrk {A_2} = A_1$
Hence, from Preimage of Set Difference under Mapping:
:$f^{-1} \sqbrk {A_2 \setminus W} = A_1 \setminus f^{-1} \sqbrk W$
=== Necessary Condition ===
Suppose that:
:for every $V \subseteq A_2$ which is closed in $M_2$, $f^{-1} \sqbrk V$ is closed in $M_1$.
Let $U$ be open in $M_2$.
Then by definition $A_2 \setminus U$ is closed in $M_2$.
By hypothesis, $f^{-1} \sqbrk {A_2 \setminus U} = A_1 \setminus f^{-1} \sqbrk U$ is closed in $M_1$.
So $f^{-1} \sqbrk U$ is open in $M_1$.
This is true for any $U \in A_2$ as $U$ is arbitrary.
It follows by definition that $f$ is continuous.
=== Sufficient Condition ===
Now let $f$ be continuous.
Let $V$ be closed in $M_2$.
Then $A_2 \setminus V$ is open in $M_2$.
As $f$ is continuous, $f^{-1} \sqbrk {A_2 \setminus V} = A_1 \setminus f^{-1} \sqbrk V$ is open in $M_1$.
So $f^{-1} \sqbrk V$ is closed in $M_1$.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 6$: Open Sets and Closed Sets: Theorem $6.10$
Category:Metric Spaces
Category:Closed Sets
Category:Continuous Mappings
Continuity of Mapping between Metric Spaces by Convergent Sequence

== Theorem ==
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $f: A_1 \to A_2$ be a mapping.
Then $f$ is continuous at $a \in X$ :
:whenever $\ds \lim_{n \mathop \to \infty} x_n = a$ for a sequence $\sequence {x_n}$ of points of $A_1$ 
it is true that:
:$\ds \lim_{n \mathop \to \infty} \map f {x_n} = \map f a$
== Proof ==
=== Necessary Condition ===
Let $f$ be continuous at $a \in A_1$.
Let $\ds \lim_{n \mathop \to \infty} x_n = a$.
Let $V$ be a neighborhood of $\map f a$.
Then by Metric Space Continuity by Inverse of Mapping between Neighborhoods $f^{-1} \sqbrk V$ is a neighborhood of $a$.
By Limit of Sequence in Metric Space in Neighborhood:
:$\exists N \in \N: n > N \implies x_n \in f^{-1} \sqbrk V$
Thus for each neighborhood $V$ of $\map f a$
:$\exists N \in \N: n > N \implies \map f {x_n} \in V$
By Limit of Sequence in Metric Space in Neighborhood:
:$\ds \lim_{n \mathop \to \infty} \map f {x_n} = \map f a$
=== Sufficient Condition ===
 $f$ is not continuous at $a \in A_1$.
Then there is a neighborhood $V$ of $\map f a$ such that for each neighborhood $U$ of $a$:
:$f \sqbrk U \nsubseteq V$
In particular, for each open $\epsilon$-ball $\map {B_\epsilon} a$ such that $\forall n \in \N_{>0}: \epsilon = \dfrac 1 n$:
:$f \sqbrk {\map {B_\epsilon} a} \nsubseteq V$
Thus for each $n \in \N_{>0}$ there exists a point $x_n$ such that:
:$x_n \subseteq \map {B_\epsilon} a$
and:
:$\map f {x_n} \notin V$
Now:
:$\map {d_1} {a, x_n} < \dfrac 1 n$
and therefore:
:$\ds \lim_{n \mathop \to \infty} x_n = a$
whereas $\ds \lim_{n \mathop \to \infty} \map f {x_n} = \map f a$ is impossible, as:
:$\forall n \in \N_{>0}: \map f {x_n} \notin V$
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 5$: Limits: Theorem $5.4$
Category:Continuous Mappings on Metric Spaces
Category:Limits of Sequences