nyuuzyou/wikis huggingface dataset:
Method of Undetermined Coefficients/Sine and Cosine/Particular Solution/i b is not Root of Auxiliary Equation/Exponential Form

== Proof Technique ==
Consider the nonhomogeneous linear second order ODE with constant coefficients:
:$(1): \quad y' ' + p y' + q y = \map R x$
Let $\map R x$ be a linear combination of sine and cosine:
:$\map R x = \alpha \sin b x + \beta \cos b x$
such that $i b$ is not a root of the auxiliary equation to $(1)$.
The Method of Undetermined Coefficients can be used to find a particular solution to $(1)$ in the following manner.
== Method and Proof ==
Let $\map {y_g} x$ be the general solution to:
:$y' ' + p y' + q y = 0$
From General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$\map {y_g} x + \map {y_p} x$
is the general solution to $(1)$.
It remains to find $\map {y_p} x$.
Let $\map R x = \alpha \sin b x + \beta \cos b x$.
Consider the auxiliary equation to $(1)$:
:$(2): \quad m^2 + p m + q = 0$
We are given that $i b$ is not a root of $(2)$.
Assume that there is a particular solution to $(1)$ of the form:
:$y_p = A \sin b x + B \cos b x$
From Euler's Formula:
:$\cos b x + i \sin b x = e^{i b x}$
and so:
:$A \sin b x + B \cos b x$ is the real part of $\paren {A - i B} \paren {\cos b x + i \sin b x} = \paren {A - i B} e^{i b x}$
It is assumed that $A$, $B$, $p$ and $q$ are all real numbers.
Suppose we have found a solution $y$ of $(1)$ where:
:$\map f x = \map {f_1} x + i \, \map {f_2} x$
where $\map y x$ and $\map f x$ are complex-valued.
Letting $\map y x = \map {y_1} x + \map {y_2} x$, where $y_1$ and $y_2$ are the real and imaginary parts of $\map y x$, we have:
:${y_1}' ' + p {y_1}' + q y_1 + i \paren { {y_2}' ' + p {y_2}' + q y_2} = \map {f_1} x + i \, \map {f_2} x$
Equating real parts:
:${y_1}' ' + p {y_1}' + q y_1 = \map {f_1} x$
Equating imaginary parts:
:${y_2}' ' + p {y_2}' + q y_2 = \map {f_2} x$
Thus if $y$ is a particular solution to $(1)$ when the is $\map f x$:
:$\map \Re y$ is a particular solution to $(1)$ when the is $\map \Re {\map f x}$
:$\map \Im y$ is a particular solution to $(1)$ when the is $\map \Im {\map f x}$
So to find a particular solution when the is $K \cos x$ or $K \sin x$, we can first find a particular solution when the is $K e^{i b x}$ and then take its real part or imaginary part as necessary.
Hence, when we have $A \cos b x + B \sin b x$ on the :
:replace it with $\paren {A - i B} e^{i b x}$
:use the Method of Undetermined Coefficients for Exponential functions
and then take its real part.
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.5$ Particular solution: trigonometric $\map f x$
Category:Method of Undetermined Coefficients
Method of Undetermined Coefficients/Sine and Cosine/Particular Solution/i b is not Root of Auxiliary Equation/Trigonometric Form

== Proof Technique ==
Consider the nonhomogeneous linear second order ODE with constant coefficients:
:$(1): \quad y + p y' + q y = \map R x$
Let $\map R x$ be a linear combination of sine and cosine:
:$\map R x = \alpha \sin b x + \beta \cos b x$
such that $i b$ is not a root of the auxiliary equation to $(1)$.
The Method of Undetermined Coefficients can be used to find a particular solution to $(1)$ in the following manner.
== Method and Proof ==
Let $\map {y_g} x$ be the general solution to:
:$y + p y' + q y = 0$
From General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$\map {y_g} x + \map {y_p} x$
is the general solution to $(1)$.
It remains to find $\map {y_p} x$.
Let $\map R x = \alpha \sin b x + \beta \cos b x$.
Consider the auxiliary equation to $(1)$:
:$(2): \quad m^2 + p m + q = 0$
We are given that $i b$ is not a root of $(2)$.
Assume that there is a particular solution to $(1)$ of the form:
:$y_p = A \sin b x + B \cos b x$
We have:
Inserting into $(1)$:
Hence $A$ and $B$ can be expressed in terms of $\alpha$ and $\beta$:
Hence:
:$y_p = \dfrac {\alpha \paren {q - b^2} + \beta b p} {\paren {q - b^2}^2 + b^2 p^2} \sin b x + \dfrac {\beta \paren {q - b^2} - \alpha b p} {\paren {q - b^2}^2 + b^2 b^2} \cos b x$
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.5$ Particular solution: exponential $\map f x$: Example $5$
* : $\S 3.18$: The Method of Undetermined Coefficients
Category:Method of Undetermined Coefficients
Method of Undetermined Coefficients/Sum of Several Terms

== Proof Technique ==
Consider the nonhomogeneous linear second order ODE with constant coefficients:
:$(1): \quad y + p y' + q y = \map R x$
Let $\map R x$ be of the form:
:$\map R x = \ds \sum_{k \mathop = 1}^n \map {f_k} x$
where each of the $f_k$ is either:
:a real polynomial function: $\map {f_k} x = \ds \sum_{j \mathop = 0}^m a_j x^j$ for some integer $m$
:a function of the form $\map {f_k} x = A e^r x$
:a function of the form $\map {f_k} x = A \cos r x + B \sin r x$
:the product of a combination of the three above.
The Method of Undetermined Coefficients can be used to solve $(1)$ in the following manner.
== Method and Proof ==
Let $\map {y_g} x$ be the general solution to:
:$(2): \quad y + p y' + q y = 0$
From Solution of Constant Coefficient Homogeneous LSOODE, $\map {y_g} x$ can be found systematically.
Let $\map {y_p} x$ be a particular solution to $(1)$.
Then from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$\map {y_g} x + \map {y_p} x$
is the general solution to $(1)$.
It remains to find $\map {y_p} x$.
For each $k$, use the Method of Undetermined Coefficients to find a particular solution to the nonhomogeneous linear second order ODE:
:$y + p y' + q y = \map {f_k} x$
Then from Combination of Solutions to Non-Homogeneous LSOODE with same Homogeneous Part, all that remains to be done is to add them all up.
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.6$ Particular solution: some further cases $\text{(iv)}$
Category:Method of Undetermined Coefficients
Method of Variation of Parameters

== Proof Technique ==
The method of variation of parameters is a technique for finding a particular solution to a nonhomogeneous linear second order ODE:
:$(1): \quad y + \map P x y' + \map Q x y = \map R x$
provided that the general solution of the corresponding homogeneous linear second order ODE:
:$(2): \quad y + \map P x y' + \map Q x y = 0$
is already known.
== Method ==
Let the general solution of $(2)$ be:
:$y = C_1 \map {y_1} x + C_2 \map {y_2} x$
Then a particular solution of $(1)$ is:
:$\ds y = y_1 \int -\frac {\map {y_2} x \map R x} {\map W {y_1, y_2} } \rd x + y_2 \int \frac {\map {y_1} x \map R x} {\map W {y_1, y_2} } \rd x$
where $\map W {y_1, y_2}$ denotes the Wronskian of $\map {y_1} x$ and $\map {y_2} x$.
== Proof ==
Let the general solution of $(2)$ be:
:$(3): \quad y = C_1 \map {y_1} x + C_2 \map {y_2} x$
Let the arbitrary constants $C_1$ and $C_2$ be replaced by functions $\map {v_1} x$ and $\map {v_2} x$.
It is required that $v_1$ and $v_2$ be determined so as to make:
:$(4): \quad y = \map {v_1} x \map {y_1} x + \map {v_2} x \map {y_2} x$
a particular solution of $(1)$.
Then:
Suppose ${v_1}' y_1 + {v_2}' y_2$ were made to vanish:
:$(6): \quad {v_1}' y_1 + {v_2}' y_2 = 0$
Then:
Hence:
Because $y_1$ and $y_2$ are both particular solutions of $(2)$:
:${y_1} + \map P x {y_1}' + \map Q x y_1 = {y_2} + \map P x {y_2}' + \map Q x y_2 = 0$
and so from $(9)$:
:$(10): \quad {v_1}' {y_1}' + {v_2}' {y_2}' = \map R x$
In summary:
We started with the assumption that:
:$(3): \quad y = C_1 \map {y_1} x + C_2 \map {y_2} x$
and so $y_1$ and $y_2$ are linearly independent.
Thus by Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE iff Linearly Dependent:
:$\map W {y_1, y_2} \ne 0$
and so $(11)$ is defined.
Thus:
and so as required:
:$\ds y = y_1 \int -\frac {\map {y_2} x \map R x} {\map W {y_1, y_2} } \rd x + y_2 \int \frac {\map {y_1} x \map R x} {\map W {y_1, y_2} } \rd x$
== Source of Name ==
The name method of variation of parameters derives from the method of operation: the parameters $C_1$ and $C_2$ are made to vary by replacing them with the functions $\map {v_1} x$ and $\map {v_2} x$.
== Sources ==
* : $\S 3.19$: The Method of Variation of Parameters
Category:Linear Second Order ODEs
Category:Proof Techniques
Method of the Auxiliary Hypothesis

== Definition ==
The method rests on the Criterion of Deduction:
Let $A$ be a relation in $I$, and $I*$ be the the theory obtained by adjoining $A$ to the axioms of $I$. If $B$ is a theorem in $I*$, then $A \implies B$ is a theorem in $I$.
The proof is on the page 30 in the book Theory of Sets by Bourbaki.
=== Example ===
Suppose a field, where the inverse of the zero-element and its existence are not defined in axioms.
The auxiliary hypothesis:
'''
adjoin the inconsistent assumption (not yet proved) to the axioms that the zero-element has an inverse:
:${0}^{-1}*0 = 1$
According to the axioms of the field:
:$0^{-1}*0 = 0*0^{-1}= 0$
Contradiction, since an axiom of the field states the uniqueness of the product. So the zero-element has no inverse by the Method of the auxiliary hypothesis. Otherwise, the set of axioms would be inconsistent.
A common mistake is to assume the non-existence of the inverse for the zero-element from the axioms of the field. It is a consequence, not a definition per se. 
== Similarity to a proof method from Propositional Logic ==
If $a \land \neg b$ leads to contradiction, then the proposition $(a \implies b)$ is true.
:$(a \implies b) \iff (\neg a \lor b) \iff \neg (a \land \neg b)$
== References ==
Theory of Sets by Bourbaki
Metric Closure and Topological Closure of Subset are Equivalent

== Theorem ==
Let $M = \struct{A, d}$ be a metric space.
Let $T = \struct{A, \tau}$ be the topological space with the topology induced by $d$.
Let $H \subseteq A$.
Then:
:the metric closure of $H$ in $M$ equals the topological closure of $H$ in $T$
== Proof ==
Let $H^i$ be the set of isolated points of $H$ in $M$.
From Isolated Point in Metric Space iff Isolated Point in Topological Space:
:$H^i$ equals the set of isolated points of $H$ in the topological space $T$.
Let $H'$ be the set of limit points of $H$ in $M$.
From Limit Point in Metric Space iff Limit Point in Topological Space:
:$H'$ equals the set of limit points of $H$ in the topological space $T$.
Let $H^-$ denote the closure of $H$ in the metric space $M$.
By definition of the closure of $H$ in the metric space $M$
:$H^- = H' \cup H^i$
Let $\map \cl H$ denote the closure of $H$ in the topological space $T$.
By definition of the closure of $H$ in the topological space $T$
:$\map \cl H = H' \cup H^i$
Thus:
:$H^- = \map \cl H$
Category:Set Closures
Metric Defines Norm iff it Preserves Linear Structure

== Theorem ==
Let $\struct {k, \norm {\,\cdot\,}_k}$ be a valued field.
Let $V$ be a vector space over the valued field $\struct {k, \norm {\,\cdot\,}_k}$.
Let $d: V \times V \to k$ be a metric on $V$.
Then the function $\norm v := \map d {v, 0}$ is a norm on $V$ for all $x, y, z \in V$, $\lambda \in k$:
:$(1): \quad \map d {x + z, y + z} = \map d {x, y}$ (homogeneity or translation invariance)
:$(2): \quad \map d {\lambda x, \lambda y} = \norm \lambda_k \map d {x, y}$ (the enlargement property)
== Proof ==
Suppose first that $d$ satisfies the hypotheses $(1)$ and $(2)$.
From :
:$\forall u, v \in V: \map d {u, v} \ge 0$
Hence:
:$\forall u \in V: \norm u = \map d {u, 0} \ge 0$
Moreover, from :
:$\norm u = 0 \implies \map d {u, 0} = 0$
and hence:
:$u = 0$
Now let $\lambda \in K$, $u \in V$.
Then, using the enlargement property of $d$:
Finally if $u,v \in V$, then we have
Category:Metric Spaces
Metric Induced by Norm is Invariant Metric

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Let $d$ be the metric induced by $\norm {\, \cdot \,}$.
Then $d$ is invariant.
== Proof ==
Let $x, y, z \in X$. 
Then, we have: 
Category:Metric Spaces
Category:Normed Vector Spaces
Metric Induced by Norm is Metric

== Theorem ==
Let $V$ be a normed vector space.
Let $\norm{\,\cdot\,}$ denote its norm.
Let $d$ be the metric induced by $\norm {\,\cdot\,}$.
Then $d$ is a metric.
== Proof ==
=== Proof of and ===
Let $x, y \in V$. 
Then:
:$\map d {x, y} = \norm {x - y} \ge 0$
and furthermore:
=== Proof of ===
Let $x, y, z \in V$.
Then:
=== Proof of ===
Let $x, y \in V$.
Then: 
As $d$ satisfies the metric space axioms, it is a metric.
== Sources ==
* $3.1$: Norms
Category:Norm Theory
Category:Metric Spaces
Metric Induced by Norm on Normed Division Ring is Metric

== Theorem ==
Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring with zero and unity denoted by $0_R$ and $1_R$ respectively.
Let $d$ be the metric induced by $\norm{\,\cdot\,}$.
Then $d$ is a metric.
== Proof ==
=== Proof of and ===
Let $x, y \in R$. 
Then $\map d {x, y} = \norm {x - y} \ge 0$, and furthermore:
=== Proof of ===
Let $x, y, z \in R$.
Then:
=== Proof of ===
Let $x, y \in R$.
Then: 
As $d$ satisfies the metric space axioms, it is a metric.
== Sources ==
* : $\S 2.3$: Topology: Problem $42$
Category:Normed Division Rings
Category:Metric Spaces
Metric Induced by a Pseudometric

== Theorem ==
Let $X$ be a set on which there is a pseudometric $d: X \times X \to \R$.
For any $x, y \in X$, let $x \sim y$ denote that $\map d {x, y} = 0$.
Let $\eqclass x \sim$ denote the equivalence class of $x$ under $\sim$.
Let $X^*$ be the quotient of $X$ by $\sim$.
Then the mapping $d^*: X^* \times X^* \to \R$ defined by:
:$\map {d^*} {\eqclass x \sim, \eqclass y \sim} = \map d {x, y}$
is a metric.
Hence $\struct {X^*, d^*}$ is a metric space.
== Proof ==
From Pseudometric Defines an Equivalence Relation we have that $\sim$ is indeed an equivalence relation.
First we verify that $d^*$ is well-defined.
Let $z \in \eqclass x \sim$ and $w \in \eqclass y \sim$.
From Equivalence Class Equivalent Statements:
:$\eqclass z \sim = \eqclass x \sim$ and $\eqclass w \sim = \eqclass y \sim$
So $d^*$ is independent of the elements chosen from the equivalence classes.
Now we need to show that $d^*$ is a metric.
To do that, all we need to do is show that $\map {d^*} {a, b} > 0$ where $a \ne b$.
So let $\map {d^*} {a, b} = 0$ where $a = \eqclass x \sim, b = \eqclass y \sim$.
Then $\map d {x, y} = 0$ so $y \in \eqclass x \sim$ and so $a = b$.
Hence the result.
Category:Pseudometric Spaces
Category:Metric Spaces
Metric Induces Topology

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Then the topology $\tau$ induced by the metric $d$ is a topology on $M$.
== Proof ==
We examine each of the criteria for being a topology separately.
:$(1): \quad$ By Union of Open Sets of Metric Space is Open, the union of any collection of open sets of a metric space is open.
:$(2): \quad$ By Finite Intersection of Open Sets of Metric Space is Open, a finite intersection of open sets of a metric space is open.
:$(3): \quad$ By Open Sets in Metric Space, $\O \in \tau$ and $A \in \tau$.
Hence the result.
== Motivation ==
Thus it can be seen that the concept of an open set as applied to a metric space is directly equivalent to that of an open set as applied to a topological space.
This is the motivation behind the definition of open sets in topology.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 6$: Open Sets and Closed Sets: Theorem $6.4$
* : $3$: Continuity generalized: topological spaces: $3.1$: Topological Spaces: Example $3.1.5$
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces
Category:Topology
Category:Metric Spaces
Metric Space Compact iff Complete in All Equivalent Metrics

== Theorem ==
Let $M_1 = \struct {A, d_1}$ be a metric space.
Then $M_1$ is compact $M_2 = \struct {A, d_2}$ is a complete metric space whenever $d_2$ is equivalent to $d_1$.
== Proof ==
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces: Complete Metric Spaces
Category:Complete Metric Spaces
Category:Compact Spaces
Metric Space Completeness is Preserved by Isometry

== Theorem ==
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $\phi: M_1 \to M_2$ be an isometry.
If $M_1$ is complete then so is $M_2$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces: Complete Metric Spaces
Category:Metric Space Completeness is Preserved by Isometry
Category:Complete Metric Spaces
Category:Isometries (Metric Spaces)
Metric Space Completeness is Preserved by Isometry/Proof 1

== Theorem ==
== Proof ==
Let $\tau_1$ be the topology on $A_1$ induced by $d_1$.
Let $\tau_2$ be the topology on $A_2$ induced by $d_2$.
Let $\sequence {x_n}$ be a Cauchy sequence in $A_2$.
From Inverse of Isometry of Metric Spaces is Isometry, $\phi^{-1}$ is an isometry.
Since Isometric Image of Cauchy Sequence is Cauchy Sequence, $\sequence {\map {\phi^{-1} } {x_n} }$ is a Cauchy sequence.
Since $M_1$ is a complete metric space, $\sequence {\map {\phi^{-1} } {x_n} }$ converges.
Since Isometry Preserves Sequence Convergence, $\sequence {\map \phi {\map {\phi^{-1} } {x_n} } }$ converges.
But:
:$\sequence {\map \phi {\map {\phi^{-1} } {x_n} } } = \sequence {x_n}$
so $\sequence {x_n}$ converges.
Thus each Cauchy sequence in $M_2$ converges.
It follows by definition that so $M_2$ is a complete metric space.
Category:Metric Space Completeness is Preserved by Isometry
Metric Space Completeness is Preserved by Isometry/Proof 2

== Theorem ==
== Proof ==
Let $\epsilon \in \R_{>0}$.
Let $\sequence {b_n}$ be a Cauchy sequence in $A_2$.
Thus:
:$\exists N_1 \in \N: \map {d_2} {b_n, b_m} < \epsilon$
whenever $n, m \ge N_1$ and $b_n, b_m \in A_2$.
We have that $M_1$ is isometric to $M_2$.
Isometry is Equivalence Relation and so in particular symmetric.
Hence $M_2$ is isometric to $M_1$, via $\phi^{-1}$.
Thus:
:$\map {d_1} {\map {\phi^{-1} } {b_n}, \map {\phi^{-1} } {b_m} } = \map {d_2} {b_n, b_m} < \epsilon$
whenever $n, m \ge N_1$ and $\map {\phi^{-1} } {b_n}, \map {\phi^{-1} } {b_m} \in A$.
So $\sequence {\map {\phi^{-1} } {b_n} }$ is Cauchy in $A_1$.
Since $A_1$ is complete, $\sequence {\map {\phi^{-1} } {b_n} }$ converges in $A_1$ to, say, $a$.
By definition of isometry, $\phi^{-1}$ is a bijection, and in particular surjective.
Thus there exists some $b \in A_2$ such that $\map {\phi^{-1} } b = a$.
Since $\sequence {\map {\phi^{-1} } {b_n} }$ converges to $\map {\phi^{-1} } b$, there exists some $N_2 \in \N$ such that:
:$\map {d_1} {\map {\phi^{-1} } {b_n}, \map {\phi^{-1} } b} < \epsilon$
whenever $n \ge N_2$ and $\map {\phi^{-1} } {b_n}, \map {\phi^{-1} } b \in A_1$.
Since $M_2$ is isometric to $M_1$, we have:
:$\map {d_1} {\map {\phi^{-1} } {b_n}, \map {\phi^{-1} } b} = \map {d_2} {b_n, b}$
and so:
:$\map {d_2} {b_n, b} < \epsilon$
whenever $n \ge N_2$ and $b_n, b \in A_2$.
Thus $\sequence {b_n}$ converges in $A_2$.
Since $\sequence {b_n}$ was an arbitrary Cauchy sequence, we have that $M_2$ is complete, as required.
Category:Metric Space Completeness is Preserved by Isometry
Metric Space Completeness is not Preserved by Homeomorphism

== Theorem ==
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $\phi: M_1 \to M_2$ be a homeomorphism.
If $M_1$ is complete then it is not necessarily the case that so is $M_2$.
== Proof ==
Let $\Z_{>0}$ be the set of (strictly) positive integers.
Let $d: \Z_{>0} \times \Z_{>0} \to \R$ be the usual (Euclidean) metric on $\Z_{>0}$.
Let $\delta: \Z_{>0} \times \Z_{>0} \to \R$ be the metric on $\Z_{>0}$ defined as:
:$\forall x, y \in \Z_{>0}: \map \delta {x, y} = \dfrac {\size {x - y} } {x y}$
Let $\tau_d$ denote the metric topology for $d$.
Let $\tau_\delta$ denote the metric topology for $\delta$.
From Topologies induced by Usual Metric and Scaled Euclidean Metric on Positive Integers are Homeomorphic:
:$\tau_d$ and $\tau_\delta$ are homeomorphic.
From Positive Integers under Usual Metric is Complete Metric Space:
:$\struct {\Z_{>0}, d}$ is a complete metric space.
However, from Positive Integers under Scaled Euclidean Metric is not Complete Metric Space:
:$\struct {\Z_{>0}, \delta}$ is not a complete metric space.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces: Complete Metric Spaces
Category:Complete Metric Spaces
Category:Homeomorphisms (Metric Spaces)
Metric Space Continuity by Epsilon-Delta

== 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 from $A_1$ to $A_2$.
Let $a \in A_1$ be a point in $A_1$.
Then the following definitions of continuity of $f$ at $a$ with respect to $d_1$ and $d_2$ are equivalent:
=== $\epsilon$-$\delta$ Definition ===
=== Definition by Limits ===
== Proof ==
=== Definition by Limits implies $\epsilon$-$\delta$ Definition ===
Suppose that $f$ is $\tuple {d_1, d_2}$-continuous at $a$ in the sense that:
:$(1): \quad$ The limit of $\map f x$ as $x \to a$ exists
:$(2): \quad \ds \lim_{x \mathop \to a} \map f x = \map f a$
Let $\ds \lim_{x \mathop \to a} \map f x$.
Then by the $\epsilon$-$\delta$ definition of limit:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: 0 < \map {d_1} {x, a} < \delta \implies \map {d_2} {\map f x, L} < \epsilon$
By $(2)$: 
:$\ds \lim_{x \mathop \to a} \map f x = \map f a$
that is:
:$\map f a = L$
Then:
:$0 = \map {d_1} {a, a}$
and so:
:$\map {d_2} {\map f a, L} = 0 < \epsilon$
Thus the definition by limits implies the $\epsilon$-$\delta$ definition.
=== $\epsilon$-$\delta$ Definition implies Definition by Limits ===
Suppose that $f$ is $\tuple {d_1, d_2}$-continuous at $a$ in the sense that:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in A_1: \map {d_1} {x, a} < \delta \implies \map {d_2} {\map f x, \map f a} < \epsilon$
where $\R_{>0}$ denotes the set of all strictly positive real numbers.
There are two possibilities for $\map {d_1} {x, a}$:
$(\text a): \quad 0 < \map {d_1} {x, a} < \delta$
Then by the $\epsilon$-$\delta$ definition of limit:
:$\ds \lim_{x \mathop \to a} \map f x = \map f a$
$(\text b): \quad \map {d_1} {x, a} = 0$
That is, $x = a$.
Hence:
:$\map f x = \map f a$
and so again:
:$\ds \lim_{x \mathop \to a} \map f x = \map f a$
Thus the $\epsilon$-$\delta$ definition implies the definition by limits.
== Also see ==
* Metric Space Continuity by Open Ball
Category:Continuous Mappings on Metric Spaces
Category:Metric Spaces
Metric Space Continuity by Inverse of Mapping between Neighborhoods

== 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 from $A_1$ to $A_2$.
Let $a \in A_1$ be a point in $A_1$.
$f$ is continuous at $a$ with respect to the metrics $d_1$ and $d_2$ :
:for each neighborhood $N$ of $\map f a$ in $M_2$, $f^{-1} \sqbrk N$ is a neighborhood of $a$.
== Proof ==
By definition, $f$ is continuous at $a$ with respect to the metrics $d_1$ and $d_2$ :
:for each neighborhood $N$ of $\map f a$ in $M_2$ there exists a corresponding neighborhood $N'$ of $a$ in $M_1$ such that $f \sqbrk {N'} \subseteq N$.
For a mapping $f: X \to Y$ we have:
:$f \sqbrk U \subseteq V \iff U \subseteq f^{-1} \sqbrk V$
where $U \subseteq X$ and $V \subseteq Y$.
Hence the result.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 4$: Open Balls and Neighborhoods: Theorem $4.7$
Category:Neighborhoods
Category:Continuous Mappings on Metric Spaces