avewright/proofwiki-math huggingface dataset:
Theorem: Restricted P-adic Metric is Metric

Let $p \in \N$ be a prime.
Let $d^\Z_p$ be the $p$-adic metric on $\Z$:
:$\forall x, y \in \Z: \map {d^\Z_p} {x, y} = \norm {x - y}_p$
where $\norm {x - y}_p$ denotes the $p$-adic norm.
Then $d^\Z_p$ is a metric.

Proof:
From $p$-adic Metric is Metric, the $p$-adic metric on $\Q$:
:$\forall x, y \in \Q: \map {d_p} {x, y} = \norm {x - y}_p$
forms a metric space $\struct {\Q, d_p}$.
The mapping:
:$\forall x, y \in \Z: \map {d^\Z_p} {x, y} = \norm {x - y}_p$
is the restriction of $d_p$ to the integers.
Hence the $p$-adic metric on $\Z$ is a metric subspace $\struct {\Z, d^\Z_p}$ of $\struct {\Q, d_p}$.
The result follows from Subspace of Metric Space is Metric Space.
{{qed}}
Theorem: Constant Function is Continuous/Metric Space/Proof 2

Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $f_c: A_1 \to A_2$ be the constant mapping from $A_1$ to $A_2$:
:$\exists c \in A_2: \forall a \in A_1: \map {f_c} a = c$
That is, every point in $A_1$ maps to the same point $c$ in $A_2$.
Then $f_c$ is continuous throughout $A_1$ with respect to $d_1$ and $d_2$.

Proof:
Let $f_c: A_1 \to A_2$ be the constant mapping between two metric spaces $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$.
Let $\epsilon \in \R_{>0}$.
Let $x \in A_1$.
Pick any $\delta \in \R_{>0}$.
Let $y \in A_1$ such that $\map {d_1} {x, y} < \delta$.
Now we have:
:$\map {f_c} x = c = \map {f_c} y$
Hence:
:$\map {d_2} {\map {f_c} x, \map {f_c} y} = \map {d_2} {c, c}$
By definition of a metric:
:$\map {d_2} {c, c} = 0 < \epsilon$
Thus we have that:
:$\forall \epsilon > 0: \forall x \in A_1: \exists \delta \in \R: \map {d_1} {x, y} < \delta \implies \map {d_2} {\map {f_c} x, \map {f_c} y} < \epsilon$
Hence by definition of continuity on a metric space, $f_c$ is continuous throughout $A_1$ with respect to $d_1$ and $d_2$.
{{qed}}
Theorem: Constant Function is Continuous/Metric Space/Proof 1

Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be metric spaces.
Let $f_c: A_1 \to A_2$ be the constant mapping from $A_1$ to $A_2$:
:$\exists c \in A_2: \forall a \in A_1: f_c \left({a}\right) = c$
That is, every point in $A_1$ maps to the same point $c$ in $A_2$.
Then $f_c$ is continuous throughout $A_1$ with respect to $d_1$ and $d_2$.

Proof:
Let $f_c: A_1 \to A_2$ be the constant mapping between two metric spaces $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$.
From Constant Function is Uniformly Continuous, $f_c$ is uniformly continuous throughout $A_1$ with respect to $d_1$ and $d_2$.
The result follows from Uniformly Continuous Function is Continuous.
{{qed}}
Theorem: Constant Function is Continuous/Metric Space

Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be metric spaces.
Let $f_c: A_1 \to A_2$ be the constant mapping from $A_1$ to $A_2$:
:$\exists c \in A_2: \forall a \in A_1: f_c \left({a}\right) = c$
That is, every point in $A_1$ maps to the same point $c$ in $A_2$.
Then $f_c$ is continuous throughout $A_1$ with respect to $d_1$ and $d_2$.

Proof:
Let $f_c: A_1 \to A_2$ be the constant mapping between two metric spaces $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$.
From Constant Function is Uniformly Continuous, $f_c$ is uniformly continuous throughout $A_1$ with respect to $d_1$ and $d_2$.
The result follows from Uniformly Continuous Function is Continuous.
{{qed}}
Theorem: Constant Function is Continuous/Metric Space

Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be metric spaces.
Let $f_c: A_1 \to A_2$ be the constant mapping from $A_1$ to $A_2$:
:$\exists c \in A_2: \forall a \in A_1: f_c \left({a}\right) = c$
That is, every point in $A_1$ maps to the same point $c$ in $A_2$.
Then $f_c$ is continuous throughout $A_1$ with respect to $d_1$ and $d_2$.

Proof:
Let $f_c: A_1 \to A_2$ be the constant mapping between two metric spaces $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$.
Let $\epsilon \in \R_{>0}$.
Let $x \in A_1$.
Pick any $\delta \in \R_{>0}$.
Let $y \in A_1$ such that $\map {d_1} {x, y} < \delta$.
Now we have:
:$\map {f_c} x = c = \map {f_c} y$
Hence:
:$\map {d_2} {\map {f_c} x, \map {f_c} y} = \map {d_2} {c, c}$
By definition of a metric:
:$\map {d_2} {c, c} = 0 < \epsilon$
Thus we have that:
:$\forall \epsilon > 0: \forall x \in A_1: \exists \delta \in \R: \map {d_1} {x, y} < \delta \implies \map {d_2} {\map {f_c} x, \map {f_c} y} < \epsilon$
Hence by definition of continuity on a metric space, $f_c$ is continuous throughout $A_1$ with respect to $d_1$ and $d_2$.
{{qed}}
Theorem: Constant Function is Continuous/Real Function

Let $f_c: \R \to \R$ be the constant mapping:
:$\exists c \in \R: \forall a \in \R: \map {f_c} a = c$
Then $f_c$ is continuous on $\R$.

Proof:
Follows directly from:
:Constant Real Function is Uniformly Continuous
:Uniformly Continuous Real Function is Continuous.
{{qed}}
Theorem: Identity Mapping is Continuous/Metric Space

Let $M = \struct {A, d}$ be a metric space.
The identity mapping $I_A: A \to A$ defined as:
:$\forall x \in A: \map {I_A} x = x$
is a continuous mapping.

Proof:
Let $a \in A$.
Let $\epsilon \in \R_{>0}$.
Let $\delta  = \epsilon$.
Then:
{{begin-eqn}}
{{eqn | l = \map d {x, a}
      | o = <
      | r = \delta
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map d {\map {I_A} x, \map {I_A} a}
      | r = \map d {x, a}
      | c = 
}}
{{eqn | o = <
      | r = \delta
      | c = 
}}
{{eqn | r = \epsilon
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Identity Mapping on Real Vector Space from Chebyshev to Euclidean Metric is Continuous

Let $\R^n$ be an $n$-dimensional real vector space.
Let $d_2$ be the Euclidean metric on $\R^n$.
Let $d_\infty$ be the Chebyshev distance on $\R^n$.
Let $I: \R^n \to \R^n$ be the identity mapping from $\R^n$ to itself.
Then the mapping:
:$I: \struct {\R^n, d_\infty} \to \struct {\R^n, d_2}$
is $\tuple {d_\infty, d_2}$-continuous.

Proof:
Let $a = \tuple {a_1, a_2, \ldots, a_n} \in \R^n$.
Let $\epsilon \in \R_{>0}$.
Let $\delta = \dfrac {\epsilon} {\sqrt n}$.
Let $x = \tuple {x_1, x_2, \ldots, x_n}$ be such that $\map {d_\infty} {x, a} < \delta$.
That is:
:$\ds \max_{i \mathop \le i \mathop \le n} \set {\size {a_i - x_i} } < \delta$
Then:
{{begin-eqn}}
{{eqn | l = \map {d_2} {x, a}
      | r = \sqrt {\sum_{i \mathop = 1}^n \paren {a_i - x_i} }
      | c = 
}}
{{eqn | o = \le
      | r = \sqrt {n \max_{i \mathop \le i \mathop \le n} \set {\size {a_i - x_i} } }
      | c = 
}}
{{eqn | o = <
      | r = \sqrt {n \delta^2}
      | c = 
}}
{{eqn | r = \sqrt {\epsilon^2}
      | c = 
}}
{{eqn | r = \epsilon
      | c = 
}}
{{end-eqn}}
The result follows by definition of $\tuple {d_\infty, d_2}$-continuity.
{{qed}}
Theorem: Identity Mapping on Real Vector Space from Euclidean to Chebyshev Distance is Continuous

Let $\R^n$ be an $n$-dimensional real vector space.
Let $d_2$ be the Euclidean metric on $\R^n$.
Let $d_\infty$ be the Chebyshev distance on $\R^n$.
Let $I: \R^n \to \R^n$ be the identity mapping from $\R^n$ to itself.
Then the mapping:
:$I: \struct {\R^n, d_2} \to \struct {\R^n, d_\infty}$
is $\tuple {d_2, d_\infty}$-continuous.

Proof:
Let $a = \tuple {a_1, a_2, \ldots, a_n} \in \R^n$.
Let $\epsilon \in \R_{>0}$.
Let $\delta = \epsilon$.
Let $x = \tuple {x_1, x_2, \ldots, x_n}$ be such that $\map {d_2} {x, a} < \delta$.
That is:
:$\ds \sqrt {\sum_{i \mathop = i}^n \paren {a_i - x_i} } < \delta$
Then:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = i}^n \paren {a_i - x_i}
      | o = <
      | r = \delta^2
      | c = 
}}
{{eqn | ll= \leadsto
      | q = \forall i \in \set {1, 2, \ldots, n}
      | l = \paren {a_i - x_i}^2
      | o = <
      | r = \delta^2
      | c = 
}}
{{eqn | ll= \leadsto
      | q = \forall i \in \set {1, 2, \ldots, n}
      | l = \size {a_i - x_i}
      | o = <
      | r = \delta
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \max_{i \mathop \le i \mathop \le n} \set {\size {a_i - x_i} }
      | o = <
      | r = \delta
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map {d_\infty} {x, a}
      | o = <
      | r = \delta
      | c = 
}}
{{end-eqn}}
The result follows by definition of $\tuple {d_2, d_\infty}$-continuity.
{{qed}}
Theorem: Composite of Continuous Mappings at Point between Metric Spaces is Continuous at Point

Let $M_1 = \struct {X_1, d_1}, M_2 = \struct {X_2, d_2}, M_3 = \struct {X_3, d_3}$ be metric spaces.
Let $f: M_1 \to M_2$ be continuous at $a \in X_1$.
Let $g: M_2 \to M_3$ be continuous at $\map f a \in X_2$.
Then their composite $g \circ f: M_1 \to M_3$ is continuous at $a \in X_1$.

Proof:
Let $\epsilon \in \R_{>0}$.
The strategy is to find a $\delta \in \R_{>0}$ such that:
:$\map {d_1} {x, a} < \delta \implies \map {d_3} {\map g {\map f x}, \map g {\map f a} } < \epsilon$
As $g$ is continuous at $\map f a$:
:$\exists \eta \in \R_{>0}: \forall y \in X_2: \map {d_2} {y, \map f a} < \eta \implies \map {d_3} {\map g y, \map g {\map f a} } < \epsilon$
As $f$ is continuous at $a$:
:$\forall \eta \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in X_1: \map {d_1} {x, a} < \delta \implies \map {d_2} {\map f x, \map f a} < \eta$
Hence:
:$\map {d_3} {\map g {\map f x}, \map g {\map f a} } < \epsilon$
{{qed}}
Theorem: Composite of Continuous Mappings between Metric Spaces is Continuous

Let $M_1 = \struct {X_1, d_1}$, $M_2 = \struct {X_2, d_2}$ and $M_3 = \struct {X_3, d_3}$ be metric spaces.
Let $f: M_1 \to M_2$ and $g: M_2 \to M_3$ be continuous mappings.
Then their composite $g \circ f: M_1 \to M_3$ is continuous.

Proof:
From Metric Induces Topology, the metric spaces described are topological spaces.
The result follows from Composite of Continuous Mappings is Continuous.
{{qed}}
Theorem: Composite of Continuous Mappings between Metric Spaces is Continuous

Let $M_1 = \struct {X_1, d_1}$, $M_2 = \struct {X_2, d_2}$ and $M_3 = \struct {X_3, d_3}$ be metric spaces.
Let $f: M_1 \to M_2$ and $g: M_2 \to M_3$ be continuous mappings.
Then their composite $g \circ f: M_1 \to M_3$ is continuous.

Proof:
Let $f$ and $g$ be continuous mappings.
By definition:
:$f$ is continuous at $a \in X_1$ for all $a \in X_1$
:$g$ is continuous at $\map f a \in X_2$ for all $\map f a \in X_2$.
The result follows from Composite of Continuous Mappings at Point between Metric Spaces is Continuous at Point
{{qed}}
Theorem: Metric Space Completeness is not Preserved by Homeomorphism

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.
{{explain|review link above: the topologies are equivalent but they are not themselves homeomorphic metric spaces}}
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.
{{qed}}
Theorem: Mapping from L1 Space to Real Number Space is Continuous

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:
{{begin-eqn}}
{{eqn | q = \forall g \in X
      | l = \map {d_1} {f, g}
      | o = <
      | r = \delta
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \int_a^b \size {\map f t - \map g t} \rd t
      | o = <
      | r = \delta
      | c = {{Defof|L1 Metric on Closed Real Interval|$L^1$ Metric}}
}}
{{eqn | ll= \leadsto
      | l = \size {\int_a^b \paren {\map f t - \map g t} \rd t}
      | o = <
      | r = \delta
      | c = Absolute Value of Definite Integral
}}
{{eqn | ll= \leadsto
      | l = \size {\int_a^b \map f t \rd t - \int_a^b \map g t \rd t}
      | o = <
      | r = \delta
      | c = Linear Combination of Definite Integrals
}}
{{eqn | ll= \leadsto
      | l = \size {\map I f - \map I g}
      | o = <
      | r = \delta
      | c = Definition of $I$
}}
{{eqn | ll= \leadsto
      | l = \map d {\map I f, \map I g}
      | o = <
      | r = \delta
      | c = Definition of $d$
}}
{{eqn | ll= \leadsto
      | l = \map d {\map I f, \map I g}
      | o = <
      | r = \epsilon
      | c = Definition of $\delta$
}}
{{end-eqn}}
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.
{{qed}}
Theorem: Cartesian Product under Chebyshev Distance of Continuous Mappings between Metric Spaces is Continuous

Let $n \in \N_{>0}$.
Let $M_1 = \struct {A_1, d_1}, M_2 = \struct {A_2, d_2}, \ldots, M_n = \struct {A_n, d_n}$ be metric spaces.
Let $N_1 = \struct {B_1, d'_1}, N_2 = \struct {B_2, d'_2}, \ldots, N_n = \struct {B_n, d'_n}$ be metric spaces.
Let $f_i: M_i \to N_i$ be continuous mappings for all $i \in \set {1, 2, \ldots, n}$.
Let $\ds \MM = \prod_{i \mathop = 1}^n M_i$ be the cartesian product of $A_1, A_2, \ldots, A_n$.
Let $\ds \NN = \prod_{i \mathop = 1}^n N_i$ be the cartesian product of $B_1, B_2, \ldots, B_n$.
Let $d_\infty$ be the Chebyshev distance on $\ds \AA = \prod_{i \mathop = 1}^n A_i$, and $\ds \BB = \prod_{i \mathop = 1}^n B_i$, defined as:
:$\ds \map {d_\infty} {x, y} = \max_{i \mathop = 1}^n \set {\map {d_i} {x_i, y_i} }$
:$\ds \map {d_\infty} {x, y} = \max_{i \mathop = 1}^n \set {\map {d'_i} {x_i, y_i} }$
where $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {y_1, y_2, \ldots, y_n} \in \AA$ or $\BB$.
Let $F: M \to N$ be the mapping defined as:
:$\forall x \in \AA: \map F {x_1, x_2, \ldots, x_n} = \tuple {\map f {x_1}, \map f {x_2}, \ldots, \map f {x_n} }$
Then $F$ is continuous.

Proof:
Let $\epsilon \in \R_{>0}$.
Let $x \in \AA$.
Let $k \in \left\{{1, 2, \ldots, n}\right\}$.
Then as $f_k$ is continuous:
:$(1): \quad \exists \delta_k \in \R_{>0}: \forall y_k \in A_k: \map {d_k} {x_k, y_k} < \delta_k \implies \map {d'} {\map {f_k} {x_k}, \map {f_k} {y_k} } < \epsilon$
Let $\delta = \max \set {\delta_k: k \in \set {1, 2, \ldots, n} }$.
Then:
{{begin-eqn}}
{{eqn | q = \forall y \in \AA
      | l = \map {d_\infty} {x, y}
      | o = <
      | r = \delta
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \max_{i \mathop = 1}^n \set {\map {d_i} {x_i, y_i} }
      | o = <
      | r = \delta
      | c = {{Defof|Chebyshev Distance|subdef = General Definition}}
}}
{{eqn | ll= \leadsto
      | q = \forall k \in \set {1, 2, \ldots, n}
      | l = \map {d_k} {x_k, y_k}
      | o = <
      | r = \delta
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map {d'_k} {\map {f_k} {x_k}, \map {f_k} {y_k} }
      | o = <
      | r = \epsilon
      | c = from $(1)$
}}
{{eqn | ll= \leadsto
      | l = \max_{i \mathop = 1}^n \set {\map {d'_i} {\map {f_i} {x_i}, \map {f_i} {y_i} } }
      | o = <
      | r = \epsilon
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map {d_\infty} {\map F x, \map F y}
      | o = <
      | r = \epsilon
      | c = {{Defof|Chebyshev Distance|subdef = General Definition}}
}}
{{end-eqn}}
Thus it has been demonstrated that:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall y \in X: \map {d_\infty} {x, y} < \delta \implies \map {d_\infty} {\map F x, \map F y} < \epsilon$
Hence by definition of continuity at a point, $F$ is continuous at $x$.
As $x$ is chosen arbitrarily, it follows that $F$ is continuous for all $x \in X$.
The result follows by definition of continuous mapping.
{{qed}}
Theorem: Addition of Coordinates on Cartesian Plane under Chebyshev Distance is Continuous Function

Let $\R^2$ be the real number plane.
Let $d_\infty$ be the Chebyshev distance on $\R^2$.
Let $f: \R^2 \to \R$ be the real-valued function defined as:
:$\forall \tuple {x_1, x_2} \in \R^2: \map f {x_1, x_2} = x_1 + x_2$
Then $f$ is continuous.

Proof:
First we note that:
{{begin-eqn}}
{{eqn | l = \size {\paren {x_1 + x_2} - \paren {y_1 + y_2} }
      | r = \size {\paren {x_1 - y_1} + \paren {x_2 - y_2} }
      | c = 
}}
{{eqn | o = \le
      | r = \size {x_1 - y_1} + \size {x_2 - y_2}
      | c = Triangle Inequality for Real Numbers
}}
{{eqn | n = 1
      | o = \le
      | r = 2 \max \set {\size {x_1 - y_1}, \size {x_2 - y_2} }
      | c = 
}}
{{end-eqn}}
Let $\epsilon \in \R_{>0}$.
Let $x = \tuple {x_1, x_2} \in \R^2$.
Let $\delta = \dfrac \epsilon 2$.
Then:
{{begin-eqn}}
{{eqn | q = \forall y = \tuple {y_1, y_2} \in \R^2
      | l = \map {d_\infty} {x, y}
      | o = <
      | r = \delta
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \max \set {\size {x_1 - y_1}, \size {x_2 - y_2} }
      | o = <
      | r = \delta
      | c = {{Defof|Chebyshev Distance on Real Number Plane}}
}}
{{eqn | ll= \leadsto
      | l = 2 \max \set {\size {x_1 - y_1}, \size {x_2 - y_2} }
      | o = <
      | r = \epsilon
      | c = Definition of $\epsilon$
}}
{{eqn | ll= \leadsto
      | l = \size {\paren {x_1 + x_2} - \paren {y_1 + y_2} }
      | o = <
      | r = \epsilon
      | c = from $(1)$
}}
{{eqn | ll= \leadsto
      | l = \size {\map f x - \map f y}
      | o = <
      | r = \epsilon
      | c = Definition of $f$
}}
{{eqn | ll= \leadsto
      | l = \map d {\map f x, \map f y}
      | o = <
      | r = \epsilon
      | c = {{Defof|Usual Metric}} on $\R$
}}
{{end-eqn}}
Thus it has been demonstrated that:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall y \in \R^2: \map {d_\infty} {x, y} < \delta \implies \map d {\map f x, \map f y} < \epsilon$
Hence by definition of continuity at a point, $f$ is continuous at $x$.
As $x$ was chosen arbitrarily, it follows that $f$ is continuous for all $x \in \R^2$.
The result follows by definition of continuous mapping.
{{qed}}
Theorem: Addition of Coordinates on Euclidean Plane is Continuous Function

Let $\struct {\R^2, d_2}$ be the real number plane with the usual (Euclidean) metric.
Let $f: \R^2 \to \R$ be the real-valued function defined as:
:$\forall \tuple {x_1, x_2} \in \R^2: \map f {x_1, x_2} = x_1 + x_2$
Then $f$ is continuous.

Proof:
First we note that:
{{begin-eqn}}
{{eqn | o = 
      | r = \size {\paren {x_1 + x_2} - \paren {y_1 + y_2} }
      | c = 
}}
{{eqn | r = \size {\paren {x_1 - y_1} + \paren {x_2 - y_2} }
      | c = 
}}
{{eqn | o = \le
      | r = \size {x_1 - y_1} + \size {x_2 - y_2}
      | c = Triangle Inequality for Real Numbers
}}
{{eqn | n = 1
      | o = \le
      | r = \sqrt 2 \sqrt {\paren {x_1 - y_1}^2 + \paren {x_2 - y_2}^2}
      | c = $p$-Product Metrics on Real Vector Space are Topologically Equivalent
}}
{{end-eqn}}
Let $\epsilon \in \R_{>0}$.
Let $x = \tuple {x_1, x_2} \in \R^2$.
Let $\delta = \dfrac \epsilon {\sqrt 2}$.
Then:
{{begin-eqn}}
{{eqn | q = \forall y = \tuple {y_1, y_2} \in \R^2
      | l = \map {d_2} {x, y}
      | o = <
      | r = \delta
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \sqrt {\paren {x_1 - y_1}^2 + \paren {x_2 - y_2}^2}
      | o = <
      | r = \delta
      | c = {{Defof|Euclidean Metric on Real Number Plane}}
}}
{{eqn | ll= \leadsto
      | l = \sqrt 2 \sqrt {\paren {x_1 - y_1}^2 + \paren {x_2 - y_2}^2}
      | o = <
      | r = \delta \sqrt 2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \size {\paren {x_1 + x_2} - \paren {y_1 + y_2} }
      | o = <
      | r = \epsilon
      | c = from $(1)$
}}
{{eqn | ll= \leadsto
      | l = \size {\map f x - \map f y}
      | o = <
      | r = \epsilon
      | c = Definition of $f$
}}
{{eqn | ll= \leadsto
      | l = \map d {\map f x, \map f y}
      | o = <
      | r = \epsilon
      | c = {{Defof|Euclidean Metric on Real Number Line}}
}}
{{end-eqn}}
Thus it has been demonstrated that:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall y \in \R^2: \map {d_2} {x, y} < \delta \implies \map d {\map f x, \map f y} < \epsilon$
Hence by definition of continuity at a point, $f$ is continuous at $x$.
As $x$ is chosen arbitrarily, it follows that $f$ is continuous for all $x \in \R^2$.
The result follows by definition of continuous mapping.
{{qed}}
Theorem: Metric Space Continuity by Inverse of Mapping between Open Balls

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 $d_2$ {{iff}}:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \map {B_\delta} {a; d_1} \subseteq f^{-1} \sqbrk {\map {B_\epsilon} {\map f a; d_2} }$
where $\map {B_\epsilon} {\map f a; d_2}$ denotes the open $\epsilon$-ball of $\map f a$ with respect to the metric $d_2$, and similarly for $\map {B_\delta} {a; d_1}$.

Proof:
By definition, $f$ is continuous at $a$ with respect to the metrics $d_1$ and $d_2$ {{iff}}:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: f \sqbrk {\map {B_\delta} {a; d_1} } \subseteq \map {B_\epsilon} {\map f a; d_2}$
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.
{{qed}}
Theorem: Open Ball is Neighborhood of all Points Inside

Let $M = \struct {A, d}$ be a metric space.
Let $a \in A$.
Let $\map {B_\epsilon} a$ be an open $\epsilon$-ball of $a$ in $M$.
Let $x \in \map {B_\epsilon} a$.
Then $\map {B_\epsilon} a$ is a neighborhood of $x$ in $M$.

Proof:
From Open Ball of Point Inside Open Ball:
:$\exists \delta \in \R: \map {B_\delta} x \subseteq \map {B_\epsilon} a$
Thus by definition $\map {B_\epsilon} a$ is a neighborhood of $x$ in $M$.
{{qed}}
Theorem: Metric Space Continuity by Neighborhood

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:

Proof:
=== $\epsilon$-Ball Definition implies Definition by Neighborhood ===
Suppose that $f$ is $\tuple {d_1, d_2}$-continuous at $a$ in the sense that:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: f \sqbrk {\map {B_\delta} {a; d_1} } \subseteq \map {B_\epsilon} {\map f a; d_2}$
where $\map {B_\epsilon} {\map f a; d_2}$ denotes the open $\epsilon$-ball of $\map f a$ with respect to the metric $d_2$, and similarly for $\map {B_\delta} {a; d_1}$.
Let $N'$ be a neighborhood of $\map f a$ in $M_2$.
Then:
{{begin-eqn}}
{{eqn | q = \exists \epsilon \in \R_{>0}
      | l = \map {B_\epsilon} {\map f a; d_2}
      | o = \subseteq
      | r = N'
      | c = {{Defof|Neighborhood (Metric Space)}}
}}
{{eqn | ll= \leadsto
      | q = \exists \delta \in \R_{>0}
      | l = f \sqbrk {\map {B_\delta} {a; d_1} }
      | o = \subseteq
      | r = \map {B_\epsilon} {\map f a; d_2}
      | c = as $f$ is $\tuple {d_1, d_2}$-continuous at $a$
}}
{{end-eqn}}
But by Open Ball is Neighborhood of all Points Inside, $\map {B_\delta} {a; d_1}$ is a neighborhood of $a$ in $M_1$.
Thus letting $N := \map {B_\delta} {a; d_1}$:
:$f \sqbrk N \subseteq \map {B_\epsilon} {\map f a; d_2} \subseteq N'$
and so the condition for $f$ to be $\tuple {d_1, d_2}$-continuous at $a$ by Neighborhood is fulfilled.
{{qed|lemma}}