avewright/proofwiki-math huggingface dataset:
Theorem: Projection from Cartesian Product under Chebyshev Distance is Continuous

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 $\ds \AA = \prod_{i \mathop = 1}^n A_i$ be the cartesian product of $A_1, A_2, \ldots, A_n$.
Let $d_\infty: \AA \times \AA \to \R$ be the Chebyshev distance on $\AA$:
:$\ds \map {d_\infty} {x, y} = \max_{i \mathop = 1}^n \set {\map {d_i} {x_i, y_i} }$
where $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {y_1, y_2, \ldots, y_n} \in \AA$.
For all $i \in \set {1, 2, \ldots, n}$, let $\pr_i: \AA \to A_i$ be the $i$th projection on $\AA$:
:$\forall a \in \AA: \map {\pr_i} a = a_i$
where $a = \tuple {a_1, a_2, \ldots, a_n} \in \AA$.
Then for all $i \in \set {1, 2, \ldots, n}$, $p_i$ is continuous on $\AA$.

Proof:
Let $\epsilon \in \R_{>0}$.
Let $a = \tuple {a_1, a_2, \ldots, a_n} \in \AA$.
Let $\map {B_\epsilon} {a_i; d_i}$ be the open $\epsilon$-ball of $a_i$ in $M_i$.
From Open Ball in Cartesian Product under Chebyshev Distance:
:$\ds \map {B_\epsilon} {a; d_\infty} = \prod_{i \mathop = 1}^n \map {B_\epsilon} {a_i; d_i}$
By definition of $i$th projection:
:$\forall x \in \map {B_\epsilon} {a; d_\infty}: \map {\pr_i} x \in \map {B_\epsilon} {a_i; d_i}$
Thus by definition of image of subset of $\AA$ under $\pr_i$:
:$\pr_i \sqbrk {\map {B_\epsilon} {a; d_\infty} } \subseteq  \map {B_\epsilon} {a_i; d_i}$
Thus by definition of continuity at a point, $\pr_i$ is continuous at $a$.
As $a$ is arbitrary, it follows that $\pr_i$ is continuous at all points of $\AA$.
Hence, by definition, $\pr_i$ is continuous from $\AA$ to $A_i$.
As $i$ is arbitrary, it follows that the result holds for all $i \in \set {1, 2, \ldots, n}$.
{{qed}}
Theorem: Continuity of Mapping to Cartesian Product under Chebyshev Distance

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 $\ds \AA = \prod_{i \mathop = 1}^n A_i$ be the cartesian product of $A_1, A_2, \ldots, A_n$.
Let $d_\infty: \AA \times \AA \to \R$ be the Chebyshev distance on $\AA$:
:$\ds \map {d_\infty} {x, y} = \max_{i \mathop = 1}^n \set {\map {d_i} {x_i, y_i} }$
where $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {y_1, y_2, \ldots, y_n} \in \AA$.
For all $i \in \set {1, 2, \ldots, n}$, let $\pr_i: \AA \to A_i$ be the $i$th projection on $\AA$:
:$\forall a \in \AA: \map {\pr_i} a = a_i$
where $a = \tuple {a_1, a_2, \ldots, a_n} \in \AA$.
Let $M' = \struct {X, d'}$ be a metric space.
Let $f: X \to \AA$ be a mapping.
Then $f$ is continuous on $X$ {{iff}} each of $\pr_i \circ f: X \to A_i$ is continuous on $X$.

Proof:
{{WLOG}}, let $i \in \set {1, 2, \ldots, n}$ be arbitrary.
Theorem: Positive Image of Point of Continuous Real Function implies Positive Closed Interval of Domain

Let $f: \R \to \R$ be a continuous real function.
Let $a \in \R$ such that $\map f a > 0$.
Then:
:$\exists k \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in \closedint {a - \delta} {a + \delta}: \map f x \ge k$

Proof:
Let $\map f a = l$ where $l > 0$.
As $f$ is continuous:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \size {y - x} < \delta \implies \size {\map f y - \map f x} < \epsilon$
Let $\epsilon = \dfrac l 2 = k$.
Then:
:$\exists \delta' \in \R_{>0}: \forall y \in \R: \size {y - x} < \delta' \implies \size {\map f y - \map f a} < \dfrac l 2$
Thus:
:$\forall x \in \openint {a - \delta'} {a + \delta'}: \map f x > \dfrac l 2$
Let $\delta = \dfrac {\delta'} 2$
Then $a - \delta \in \openint {a - \delta'} {a + \delta'}$ and $a + \delta \in \openint {a - \delta'} {a + \delta'}$.
Thus:
:$\closedint {a - \delta} {a + \delta} \subseteq \openint {a - \delta'} {a + \delta'}$
and hence the result.
{{qed}}
Theorem: Limit of Sequence in Metric Space in Neighborhood

Let $M = \struct {A, d}$ be a metric space.
Let $\sequence {a_n}$ be a sequence in $A$.
Then $\ds \lim_{n \mathop \to \infty} a_n = a$ {{iff}} for each neighborhood $V$ of $a$:
:$\exists N \in \N: n > N \implies a_n \in V$

Proof:
=== Necessary Condition ===
Let $\ds \lim_{n \mathop \to \infty} a_n = a$.
Let $V$ be a neighborhood of $a$.
By definition of neighborhood:
:$\exists \epsilon \in \R_{>0}: \map {B_\epsilon} a \subseteq V$
where $\map {B_\epsilon} a$ denotes the open $\epsilon$-ball of $a$ in $M$.
By definition of limit:
:$\exists N \in \N: n > N \implies \map d {a, a_n} < \epsilon$
Hence $a_n \in V$.
{{qed|lemma}}
Theorem: Square of Real Number is Non-Negative

Let $x \in \R$.
Then:
: $0 \le x^2$

Proof:
There are two cases to consider:
: $(1): \quad x = 0$
: $(2): \quad x \ne 0$
Let $x = 0$.
Then:
:$x^2 = 0$
and thus
:$0 \le x^2$
{{qed|lemma}}
Let $x \ne 0$.
From Square of Non-Zero Real Number is Strictly Positive it follows that:
:$0 < x^2$
and so by definition:
:$0 \le x^2$
{{Qed}}
Theorem: Real Number between Zero and One is Greater than Square

Let $x \in \R$.
Let $0 < x < 1$.
Then:
:$0 < x^2 < x$

Proof:
We are given that $0 < x < 1$.
By direct application of Real Number Ordering is Compatible with Multiplication, it follows that:
: $0 \times x < x \times x < 1 \times x$
and the result follows.
{{Qed}}
Theorem: Real Number between Zero and One is Greater than Square

Let $x \in \R$.
Let $0 < x < 1$.
Then:
:$0 < x^2 < x$

Proof:
We have that Real Numbers form Ordered Integral Domain.
Thus Square of Element Less than Unity in Ordered Integral Domain applies directly.
{{qed}}
Theorem: Real Number Greater than One is Less than Square

Let $x \in \R$.
Let $x > 1$.
Then:
:$x^2 > x$

Proof:
As $x > 1$ it follows that $x > 0$.
Thus by Real Number Ordering is Compatible with Multiplication:
: $x \times x > 1 \times x$
and the result follows.
{{Qed}}
Theorem: Power of Real Number greater than One is Unbounded Above

Let $x \in \R$ be a real number such that $x > 1$.
Let set $S = \set {x^n: n \in \N}$.
Then $S$ is unbounded above.

Proof:
{{AimForCont}} $S$ were bounded above.
Then $S$ has a supremum $B$.
As $x > 1$, it follows that $\dfrac B x < B$ and so therefore $\dfrac B x$ can not be an upper bound.
Therefore:
:$\exists n \in \N: x^n > \dfrac B x \implies x^{n + 1} > B$
So $B$ can not be an upper bound.
From that contradiction it can be concluded that $S$ can not have an upper bound.
So by Proof by Contradiction it follows that $S$ is unbounded above.
{{qed}}
Theorem: Power of Real Number between Zero and One is Bounded

Let $x \in \R$ be a real number.
Let $0 < x < 1$.
Let set $S = \set {x^n: n \in \N}$.
Then:
:$\inf S = 0$
and:
:$\sup S = 1$
where $\inf S$ and $\sup S$ are the infimum and supremum of $S$ respectively.

Proof:
When $n = 0$ it follows that $x^n = 1$ and so $\sup S \ge 1$.
Let $x^k \in S$.
As $x < 1$, it follows from Real Number Ordering is Compatible with Multiplication that:
:$x^{k + 1} < x^k$
So:
:$\forall x \in S: x \le 1$
Hence it follows that $\sup S = 1$.
As $x > 0$, it follows by Real Number Ordering is Compatible with Multiplication that:
:$\forall x \in S: x \ge 0$
Therefore $x$ is a lower bound of $S$.
Now suppose $h > 0$ is also a lower bound of $S$.
Then:
:$\forall n \in \N: x^n \ge h$
and:
:$\forall n \in \N: \paren {\dfrac 1 x}^n \le \dfrac 1 h$
But as $0 < x < 1$ it follows that:
:$\dfrac 1 x > 1$
Thus $\dfrac 1 h$ is an upper bound for $\set {\paren {\dfrac 1 x}^n: n \in \N}$.
From Power of Real Number greater than One is Unbounded Above, this has been shown to be unbounded above.
Therefore there can be no such lower bound $h > 0$ of $S$.
Hence $\inf S = 0$.
{{qed}}
Theorem: Limit of Function by Convergent Sequences/Real Number Line

Let $f$ be a real function defined on an open interval $\openint a b$, except possibly at the point $c \in \openint a b$.
Then $\ds \lim_{x \mathop \to c} \map f x = l$ {{iff}}:
::for each sequence $\sequence {x_n}$ of points of $\openint a b$ such that $\forall n \in \N_{>0}: x_n \ne c$ and $\ds \lim_{n \mathop \to \infty} x_n = c$
:it is true that:
::$\ds \lim_{n \mathop \to \infty} \map f {x_n} = l$

Proof:
=== Necessary Condition ===
Let $\ds \lim_{x \mathop \to c} \map f x = l$.
Let $\epsilon \in \R_{>0}$.
Then by the definition of the limit of a real function:
:$\exists \delta \in \R_{>0}: \size {\map f x - l} < \epsilon$
provided $0 < \size {x - c} < \delta$.
Now suppose that $\sequence {x_n}$ is a sequence of elements of $\openint a b$ such that:
:$\forall n \in \N_{>0}: x_n \ne c$
and:
:$\ds \lim_{n \mathop \to \infty} x_n = c$
Since $\delta > 0$, from the definition of the limit of a real function:
:$\exists N \in \R_{>0}: \forall n > N: \size {x_n - c} < \delta$
But:
:$\forall n \in \N_{>0}: x_n \ne c$
That means:
:$0 < \size {x_n - c} < \delta$
But that implies:
:$\size {\map f {x_n} - l} < \epsilon$
That is, given a value of $\epsilon > 0$, we have found a value of $N$ such that:
:$\forall n > N: \size {\map f {x_n} - l} < \epsilon$
Thus:
:$\ds \lim_{n \mathop \to \infty} \map f {x_n} = l$
{{qed|lemma}}
Theorem: Continuity of Mapping between Metric Spaces by Convergent Sequence

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 $a \in X$ {{iff}}:
: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:
:$\ds \lim_{n \mathop \to \infty} \map f {x_n} = \map f a$
{{qed|lemma}}
Theorem: Limit of Image of Sequence/Real Number Line

Let $f$ be a real function which is continuous on the interval $\Bbb I$.
Let $\sequence {x_n}$ be a sequence of points in $\Bbb I$ such that:
:$\ds \lim_{n \mathop \to \infty} x_n = \xi$
where:
:$(1): \quad \xi \in \Bbb I$
:$(2): \quad \ds \lim_{n \mathop \to \infty} x_n$ denotes the limit of $x_n$.
Then:
:$\ds \lim_{n \mathop \to \infty} \map f {x_n} = \map f \xi$
That is:
:$\ds \lim_{n \mathop \to \infty} \map f {x_n} = \map f {\lim_{n \mathop \to \infty} x_n}$

Proof:
From Limit of Real Function by Convergent Sequences, we have:
:$\ds \lim_{x \mathop \to \xi} \map f x = l$
{{iff}}:
:for each sequence $\sequence {x_n}$ of points of $\openint a b$ such that:
::$\forall n \in \N_{>0}: x_n \ne \xi$
:and:
::$\ds \lim_{n \mathop \to \infty} x_n = \xi$
:it is true that:
::$\ds \lim_{n \mathop \to \infty} \map f {x_n} = l$
The result follows directly from this and the definition of continuity on an interval.
{{qed}}
Theorem: Infimum of Subset of Real Numbers is Arbitrarily Close

Let $A \subseteq \R$ be a subset of the real numbers.
Let $b$ be an infimum of $A$.
Let $\epsilon \in \R_{>0}$.
Then:
:$\exists x \in A: x - b < \epsilon$

Proof:
Note that $A$ is non-empty as the empty set does not admit an infimum (in $\R$).
Suppose $\epsilon \in \R_{>0}$ such that:
:$\forall x \in A: x - b \ge \epsilon$
Then:
:$\forall x \in A: b + \epsilon \le x$
and so $b + \epsilon$ would be a lower bound of $A$ which is greater than $b$.
But since $b$ is an infimum of $A$ there can be no such $b + \epsilon$.
From that contradiction it follows that:
:$\exists x \in A: x - b < \epsilon$
{{qed}}
Theorem: Existence of Sequence in Set of Real Numbers whose Limit is Infimum

Let $A \subseteq \R$ be a non-empty subset of the real numbers.
Let $b$ be an infimum of $A$.
Then there exists a sequence $\sequence {a_n}$ in $\R$ such that:
:$(1): \quad \forall n \in \N: a_n \in A$
:$(2): \quad \ds \lim_{n \mathop \to \infty} a_n = b$

Proof:
From Infimum of Subset of Real Numbers is Arbitrarily Close:
For $\epsilon = \dfrac 1 n$ there exists an $a_n \in A$ such that:
:$a_n - b < \dfrac 1 n$
Since $b$ is an infimum of $A$:
:$0 \le a_n - b$
Therefore:
:$\ds \lim_{n \mathop \to \infty} a_n = b$
{{qed}}
Theorem: Existence of Sequence in Subset of Metric Space whose Limit is Infimum

Let $M = \struct {A, d}$ be a metric space.
Let $a \in A$.
Let $S \subseteq A$ be a non-empty subset of $A$.
Then there exists a sequence $\sequence {a_n}$ of points of $S$ such that:
:$\ds \lim_{n \mathop \to \infty} \map d {a, a_n} = \map d {a, S}$

Proof:
From Existence of Sequence in Set of Real Numbers whose Limit is Infimum:
:$\ds \lim_{n \mathop \to \infty} \map d {a, a_n} = b$
where $b$ is an infimum of $\map d {a, a_n}$.
Hence the result by definition of distance to a subset of a metric space.
{{qed}}
Theorem: Limit of Sequence in Product of Metric Spaces under Chebyshev Distance

Let $M_1 = \struct {A_1, d_1}, M_2 = \struct {A_2, d_2}, \ldots, M_k = \struct {A_n, d_k}$ be metric spaces.
Let $\ds \AA = \prod_{i \mathop = 1}^k A_i$ be the cartesian product of $A_1, A_2, \ldots, A_k$.
Let $d_\infty: \AA \times \AA \to \R$ be the Chebyshev distance on $\AA$:
:$\ds \map {d_\infty} {x, y} = \max_{i \mathop = 1}^k \set {\map {d_i} {x_i, y_i} }$
where $x = \tuple {x_1, x_2, \ldots, x_k}, y = \tuple {y_1, y_2, \ldots, y_k} \in \AA$.
Let $\sequence {a_n}$ be a sequence of points of $\AA$:
:$a_n = \tuple {a_{n 1}, a_{n 2}, \ldots, a_{n k} } \in \AA$
Let $c = \tuple {c_1, c_2, \ldots, c_k} \in \AA$.
Then:
:$\ds \lim_{n \mathop \to \infty} a_n = c \iff \forall i \in \set {1, 2, \ldots, k}: \lim_{n \mathop \to \infty} a_{n i} = c_i$
where $\ds \lim_{n \mathop \to \infty}$ denotes a limit.

Proof:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} a_n
      | r = c
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = \lim_{n \mathop \to \infty} \map {d_\infty} {a_n, c}
      | r = 0
      | c = {{Defof|Convergent Sequence|subdef = Metric Space|index = 3}}
}}
{{eqn | ll= \leadstoandfrom
      | l = \lim_{n \mathop \to \infty} \max_{i \mathop = 1}^k \set {\map {d_i} {a_{n i}, c_i} }
      | r = 0
      | c = {{Defof|Chebyshev Distance}}
}}
{{eqn | ll= \leadstoandfrom
      | l = \max_{i \mathop = 1}^k \set {\lim_{n \mathop \to \infty} \map {d_i} {a_{n i}, c_i} }
      | r = 0
      | c = Limit of Image of Sequence
}}
{{eqn | ll= \leadstoandfrom
      | q = \forall i \in \set {1, 2, \ldots, k}
      | l = \lim_{n \mathop \to \infty} \map {d_i} {a_{n i}, c_i}
      | r = 0
      | c = {{Defof|Max Function}}
}}
{{end-eqn}}
{{qed}}
Theorem: Recurrence Formula for Sum of Sequence of Fibonacci Numbers

Let $g_n$ be the sum of the first $n$ Fibonacci numbers.
Then:
:$\forall n \ge 2: g_n = g_{n - 1} + g_{n - 2} + 1$

Proof:
Let $F_n$ be the $n$th Fibonacci number.
By definition:
:$F_0 = 0, F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3, \ldots$
Hence, $g_n$, the sum of the first $n$ Fibonacci numbers is
:$g_0 = 0, g_1 = 1, g_2 = 2, g_3 = 4, g_4 = 7, \ldots$
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\ds \sum_{j \mathop = 0}^n F_j = g_n = g_{n - 1} + g_{n - 2} + 1$
Theorem: Limit of Subsequence equals Limit of Sequence/Metric Space

Let $M = \struct {A, d}$ be a metric space.
Let $\sequence {x_n}$ be a sequence in $A$.
Let $l \in A$ such that:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
Let $\sequence {x_{n_r} }$ be a subsequence of $\sequence {x_n}$.
Then:
:$\ds \lim_{r \mathop \to \infty} x_{n_r} = l$

Proof:
Let $\epsilon > 0$.
Since $\ds \lim_{n \mathop \to \infty} x_n = l$, it follows from the definition of limit that:
:$\exists N: \forall n > N: \map d {x_n, l} < \epsilon$
Now let $R = N$.
Then from Strictly Increasing Sequence of Natural Numbers:
:$\forall r > R: n_r \ge r$
Thus $n_r > N$ and so:
:$\map d {x_n, l} < \epsilon$
The result follows.
{{qed}}
Theorem: Convergent Real Sequence is Bounded

Let $\sequence {x_n}$ be a sequence in $\R$.
Let $l \in A$ such that $\ds \lim_{n \mathop \to \infty} x_n = l$.
Then $\sequence {x_n}$ is bounded.

Proof:
From Real Number Line is Metric Space, the set $\R$ under the usual metric is a metric space.
By Convergent Sequence in Metric Space is Bounded it follows that:
:$\exists M > 0: \forall n, m \in \N: \size {x_n - x_m} \le M$
Then for $n \in \N$, by the Triangle Inequality for Real Numbers:
{{begin-eqn}}
{{eqn | l = \size {x_n}
      | r = \size {x_n - x_1 + x_1}
      | c = 
}}
{{eqn | o = \le
      | r = \size {x_n - x_1} + \size {x_1}
      | c = 
}}
{{eqn | o = \le
      | r = M + \size {x_1}
      | c = 
}}
{{end-eqn}}
Hence $\sequence {x_n}$ is bounded.
{{qed}}
Theorem: Convergent Real Sequence is Bounded

Let $\sequence {x_n}$ be a sequence in $\R$.
Let $l \in A$ such that $\ds \lim_{n \mathop \to \infty} x_n = l$.
Then $\sequence {x_n}$ is bounded.

Proof:
Let $\sequence {x_n}$ be a sequence in $\R$.
Let $x_n \to l$ as $n \to \infty$.
To show that $\sequence {x_n}$ is bounded sequence, we need to find $K$ such that:
:$\forall n \in \N: \size {x_n} \le K$
Because $\sequence {x_n}$ converges:
:$\forall \epsilon > 0: \exists N: n > N \implies \size {x_n - l} < \epsilon$
In particular, this is true when $\epsilon = 1$.
That is:
:$\exists N_1: \forall n > N_1: \size {x_n - l} < 1$
By Backwards Form of Triangle Inequality:
:$\forall n > N_1: \size {x_n} - \size l \le \size {x_n - l} < 1$
That is:
:$\size {x_n} < \size l + 1$
So we set:
:$K = \max \set {\size {x_1}, \size {x_2}, \ldots, \size {x_{N_1} }, \size l + 1}$
and the result follows.
{{qed}}
Theorem: Monotone Convergence Theorem (Real Analysis)/Decreasing Sequence

Let $\sequence {x_n}$ be a decreasing real sequence which is bounded below.
Then $\sequence {x_n}$ converges to its infimum.

Proof:
Let $\sequence {x_n}$ be decreasing and bounded below.
Then $\sequence {-x_n}$ is increasing and bounded above.
Thus the Monotone Convergence Theorem for Increasing Sequence applies and the proof follows.
{{Qed}}
Theorem: Monotone Convergence Theorem (Real Analysis)/Increasing Sequence

Let $\sequence {x_n}$ be an increasing real sequence which is bounded above.
Then $\sequence {x_n}$ converges to its supremum.

Proof:
Suppose $\sequence {x_n}$ is increasing and bounded above.
By the Continuum Property, it has a supremum, $B$.
We need to show that $x_n \to B$ as $n \to \infty$.
Let $\epsilon \in \R_{>0}$.
By the definition of supremum, $B - \epsilon$ is not an upper bound.
Thus:
:$\exists N \in \N: x_N > B - \epsilon$
But $\sequence {x_n}$ is increasing.
Hence:
:$\forall n > N: x_n \ge x_N > B - \epsilon$
But $B$ ''is'' still an upper bound for $\sequence {x_n}$.
Then:
{{begin-eqn}}
{{eqn | q = \forall n > N
      | l = B - \epsilon
      | o = <
      | r = x_n \le B
      | c =
}}
{{eqn | ll= \leadsto
      | q = \forall n > N
      | l = B - \epsilon
      | o = <
      | r = x_n < B + \epsilon
      | c = Real Plus Epsilon
}}
{{eqn | ll= \leadsto
      | q = \forall n > N
      | l = \size {x_n - B}
      | o = <
      | r = \epsilon
      | c = Negative of Absolute Value: Corollary 1
}}
{{end-eqn}}
Hence the result.
{{Qed}}
Theorem: Open Ball in Euclidean Plane is Interior of Circle

Let $\R^2$ be the real number plane with the usual (Euclidean) metric.
Let $x = \tuple {x_1, x_2} \in \R^2$ be a point in $\R^2$.
Let $\map {B_\epsilon} x$ be the open $\epsilon$-ball at $x$.
Then $\map {B_\epsilon} x$ is the interior of the circle whose center is $x$ and whose radius is $\epsilon$.

Proof:
Let $S = \map {B_\epsilon} x$ be an open $\epsilon$-ball at $x$.
Let $y = \tuple {y_1, y_2} \in \map {B_\epsilon} x$.
Then:
{{begin-eqn}}
{{eqn | l = y
      | o = \in
      | m = \map {B_\epsilon} x
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = \map d {y, x}
      | o = <
      | m = \epsilon
      | c = {{Defof|Open Ball of Metric Space|Open $\epsilon$-Ball}}
}}
{{eqn | ll= \leadstoandfrom
      | l = \sqrt {\paren {y_1 - x_1}^2 + \paren {y_2 - x_2}^2}
      | o = <
      | m = \epsilon
      | c = {{Defof|Real Number Plane with Euclidean Metric}}
}}
{{eqn | ll= \leadstoandfrom
      | l = \paren {y_1 - x_1}^2 + \paren {y_2 - x_2}^2
      | o = <
      | m = \epsilon^2
}}
{{end-eqn}}
But from Equation of Circle:
:$\paren {y_1 - x_1}^2 + \paren {y_2 - x_2}^2 = \epsilon^2$
is the equation of a circle whose center is $\tuple {x_1, x_2}$ and whose radius is $\epsilon$.
The result follows by definition of interior and Open Ball of Point Inside Open Ball.
{{qed}}
Theorem: Open Ball in Euclidean 3-Space is Interior of Sphere

Let $\R^3$ be the real Euclidean $3$-space considered as a metric space under the usual metric.
Let $x = \tuple {x_1, x_2, x_3} \in \R^3$ be a point in $\R^3$.
Let $\map {B_\epsilon} x$ be the open $\epsilon$-ball at $x$.
Then $\map {B_\epsilon} x$ is the interior of the sphere whose center is $x$ and whose radius is $\epsilon$.

Proof:
Let $S = \map {B_\epsilon} x$ be an open $\epsilon$-ball at $x$.
Let $y = \tuple {y_1, y_2, y_3} \in \map {B_\epsilon} x$.
Then:
{{begin-eqn}}
{{eqn | l = y
      | o = \in
      | m = \map {B_\epsilon} x
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = \map d {y, x}
      | o = <
      | m = \epsilon
      | c = {{Defof|Open Ball of Metric Space|Open $\epsilon$-Ball}}
}}
{{eqn | ll= \leadstoandfrom
      | l = \sqrt {\paren {y_1 - x_1}^2 + \paren {y_2 - x_2}^2 + \paren {y_3 - x_3}^2}
      | o = <
      | m = \epsilon
      | c = {{Defof|Euclidean Metric|subdef = Ordinary Space}}
}}
{{eqn | ll= \leadstoandfrom
      | l = \paren {y_1 - x_1}^2 + \paren {y_2 - x_2}^2 + \paren {y_3 - x_3}^2
      | o = <
      | m = \epsilon^2
}}
{{end-eqn}}
But from Equation of Sphere in Rectangular Coordinates:
:$\paren {y_1 - x_1}^2 + \paren {y_2 - x_2}^2 + \paren {y_3 - x_3}^2 = \epsilon^2$
is the equation of a sphere whose center is $x$ and whose radius is $\epsilon$.
The result follows by definition of interior and Open Ball of Point Inside Open Ball.
{{qed}}
Theorem: Open Ball in Real Number Plane under Chebyshev Distance

Let $\R^2$ be the real number plane.
Let $d_\infty: \R^2 \times \R^2 \to \R$ be the Chebyshev Distance on $\R^2$:
:$\ds \map {d_\infty} {x, y} := \max \set {\size {x_1 - y_1}, \size {x_2 - y_2} }$
where $x = \tuple {x_1, x_2}, y = \tuple {y_1, y_2} \in \R^2$.
For $a \in \R^2$, let $\map {B_\epsilon} a$ be the open $\epsilon$-ball at $a$.
Then $\map {B_\epsilon} a$ is the interior of the square centered at $a$ and whose sides are of length $2 \epsilon$ parallel to the coordinate axes.

Proof:
Let $a = \tuple {a_1, a_2}$.
From Open Ball in Cartesian Product under Chebyshev Distance:
:$\map {B_\epsilon} {a; d_\infty} = \map {B_\epsilon} {a_1; d} \times \map {B_\epsilon} {a_2; d}$
where $d$ is the usual (Euclidean) topology.
From Open Ball in Real Number Line is Open Interval:
:$\map {B_\epsilon} {a_1; d} \times \map {B_\epsilon} {a_2; d} = \openint {a_1 - \epsilon} {a_1 + \epsilon} \times \openint {a_2 - \epsilon} {a_2 + \epsilon}$
That is:
:$x \in \map {B_\epsilon} {a; d_\infty} \iff \paren {a_2 - \epsilon < x_2 < a_2 + \epsilon} \land \paren {a_2 - \epsilon < x_2 < a_2 + \epsilon}$
from which the result follows.
{{qed}}
Theorem: Open Ball in Standard Discrete Metric Space

Let $M = \struct {A, d}$ be a metric space.
Let $d$ be the standard discrete metric on $M$.
Let $a \in A$.
Let $\map {B_\epsilon} {a; d}$ be an open $\epsilon$-ball of $a$ in $M$.
Then:
:$\map {B_\epsilon} {a; d} = \begin {cases} \set a & : \epsilon \le 1 \\ A & : \epsilon > 1 \end {cases}$

Proof:
Let $\epsilon \in \R_{>0}: \epsilon \le 1$.
Then:
{{begin-eqn}}
{{eqn | q = \forall x \in A
      | l = x
      | o = \ne
      | r = a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map d {x, a}
      | o = \ge
      | r = \epsilon
      | c = {{Defof|Standard Discrete Metric}}
}}
{{eqn | ll= \leadsto
      | l = x
      | o = \notin
      | r = \map {B_\epsilon} {a; d}
      | c = {{Defof|Open Ball of Metric Space|Open $\epsilon$-Ball}}
}}
{{eqn | ll= \leadsto
      | l = \map {B_\epsilon} {a; d}
      | r = \set a
      | c = 
}}
{{end-eqn}}
{{qed|lemma}}
Let $\epsilon \in \R_{>0}: \epsilon > 1$.
Then:
{{begin-eqn}}
{{eqn | q = \forall x \in A
      | l = \map d {x, a}
      | o = >
      | r = \epsilon
      | c = {{Defof|Standard Discrete Metric}}
}}
{{eqn | ll= \leadsto
      | l = x
      | o = \in
      | r = \map {B_\epsilon} {a; d}
      | c = {{Defof|Open Ball of Metric Space|Open $\epsilon$-Ball}}
}}
{{eqn | ll= \leadsto
      | l = \map {B_\epsilon} {a; d}
      | r = A
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}