nyuuzyou/wikis huggingface dataset:
One Hundred Fowls/Bakhshali Version

== Problem ==
$20$ men, women and childen earn $20$ coins between them as follows:
:Each man earns $3$ coins.
:Each woman earns $1 \frac 1 2$ coins.
:Each child earns $\frac 1 2$ a coin.
How many men, women and children are there?
== Solution ==
:$2$ men
:$5$ women
:$13$ children.
== Proof ==
Let $m$, $w$ and $c$ denote the number of men, women and children respectively.
Then we have:
We are of course subject to the condition that each of $m$, $w$ and $c$ is not fewer than $0$.
It is seen that $2 w$ is even.
Hence, in order for $5 m + 2 w$ also to be even, it is necessary for $m$ to be even.
If $m > 4$, then $5 m > 20$.
In that case $w < 0$ and so it must be that $m \le 4$.
This allows for $m$ to be equal to $0$, $2$ or $4$.
This gives the solutions:
However, let it be understood that there was at least one man, at least one woman and at least one child.
Then there is one solution remaining:
:$2$ men
:$5$ women
:$13$ children.
== Historical Note ==
== Sources ==
* 
* : Indian Puzzles: $47$
Category:One Hundred Fowls Problems
One Hundred Fowls/Historical Note

== Historical Note on One Hundred Fowls ==
This is one of the earliest known example of this kind of problem, dating back as it does to the $5$th century.
However, it is predated by the example in the , if that does indeed date from the $3$rd century, as is possible.
== Sources ==
* : Sun Tsu Suan Ching
Category:Historical Notes
One Ninth as Pandigital Fraction

== Theorem ==
There are $3$ ways $\dfrac 1 9$ can be expressed as a pandigital fraction:
:$\dfrac 1 9 = \dfrac {6381} {57429}$
:$\dfrac 1 9 = \dfrac {6471} {58239}$
:$\dfrac 1 9 = \dfrac {8361} {75249}$
== Proof ==
Can be verified by brute force.
== Also see ==
* One Half as Pandigital Fraction
* One Third as Pandigital Fraction
* One Quarter as Pandigital Fraction
* One Fifth as Pandigital Fraction
* One Sixth as Pandigital Fraction
* One Seventh as Pandigital Fraction
* One Eighth as Pandigital Fraction
Category:One Ninth
Category:Pandigital Fractions
One Perfect Number for Each Number of Digits

== Conjecture ==
There is one perfect number for each number of digits:
:$1$ digit: $6$
:$2$ digits: $28$
:$3$ digits: $496$
:$4$ digits: $8128$
== Refutation ==
The $5$th perfect number is $33 \, 550 \, 336$, which clearly does not have $5$ digits.
== Also see ==
* Last Digit of Perfect Numbers Alternates between $6$ and $8$
== Historical Note ==
== Sources ==
* : $28$
* : $28$
Category:Famous False Conjectures
Category:Perfect Numbers
One Perfect Number for Each Number of Digits/Historical Note

== Historical Note on One Perfect Number for Each Number of Digits ==
The conjecture that there is One Perfect Number for Each Number of Digits was made by in his , published some time around the $2$nd century.
It was a simple extrapolation from the knowledge of the perfect numbers at the time.
Some sources suggest that made these conjectures, but, as it is believed that already knew the $5$th perfect number to be $33 \, 550 \, 336$, this cannot be correct.
== Sources ==
* : $28$
* : $28$
Category:Historical Notes
One Plus Reciprocal to the Nth

== Theorem ==
Let $\sequence {x_n}$ be the sequence in $\R$ defined as $x_n = \paren {1 + \dfrac 1 n}^n$.
Then $\sequence {x_n}$ converges to a limit as $n$ increases without bound.
== Proof ==
First we show that $\sequence {x_n}$ is increasing.
Let $a_1 = a_2 = \cdots = a_{n - 1} = 1 + \dfrac 1 {n - 1}$.
Let $a_n = 1$.
Let:
:$A_n$ be the arithmetic mean of $a_1 \ldots a_n$
:$G_n$ be the geometric mean of $a_1 \ldots a_n$
Thus:
:$A_n = \dfrac {\paren {n - 1} \paren {1 + \dfrac 1 {n - 1} } + 1} n = \dfrac {n + 1} n = 1 + \dfrac 1 n$
:$G_n = \paren {1 + \dfrac 1 {n - 1} }^{\dfrac {n - 1} n}$
By Cauchy's Mean Theorem‎:
: $G_n \le A_n$
Thus:
:$\paren {1 + \dfrac 1 {n - 1} }^{\frac {n - 1} n} \le 1 + \dfrac 1 n$
and so:
:$x_{n - 1} = \paren {1 + \dfrac 1 {n - 1} }^{n - 1} \le \paren {1 + \dfrac 1 n}^n = x_n$
Hence $\sequence {x_n}$ is increasing.
Next, we show that $\sequence {x_n}$ is bounded above.
Using the Binomial Theorem:
So $\sequence {x_n}$ is bounded above by $3$.
From the Monotone Convergence Theorem (Real Analysis), it follows that $\sequence {x_n}$ converges to a limit.
== Also see ==
Note that, although we have proved that this sequence converges to some limit less than $3$ (and incidentally greater than $2$), we have not at this stage determined exactly what this number actually is.
See Euler's number, where this sequence provides a definition of that number (one of several that are often used).
== Sources ==
* : $\S 4$: Convergent Sequences: $\S 4.19$: Example
Category:Limits of Sequences
Category:Reciprocals
One Quarter as Pandigital Fraction

== Theorem ==
There are $4$ ways $\dfrac 1 4$ can be expressed as a pandigital fraction:
:$\dfrac 1 4 = \dfrac {3942} {15768}$
:$\dfrac 1 4 = \dfrac {4392} {17568}$
:$\dfrac 1 4 = \dfrac {5796} {23184}$
:$\dfrac 1 4 = \dfrac {7956} {31824}$
== Proof ==
Can be verified by brute force.
== Also see ==
* One Half as Pandigital Fraction
* One Third as Pandigital Fraction
* One Fifth as Pandigital Fraction
* One Sixth as Pandigital Fraction
* One Seventh as Pandigital Fraction
* One Eighth as Pandigital Fraction
* One Ninth as Pandigital Fraction
Category:One Quarter
Category:Pandigital Fractions
One Seventh as Pandigital Fraction

== Theorem ==
There are $7$ ways $\dfrac 1 7$ can be expressed as a pandigital fraction:
:$\dfrac 1 7 = \dfrac {2394} {16 \, 758}$
:$\dfrac 1 7 = \dfrac {2637} {18 \, 459}$
:$\dfrac 1 7 = \dfrac {4527} {31 \, 689}$
:$\dfrac 1 7 = \dfrac {5274} {36 \, 918}$
:$\dfrac 1 7 = \dfrac {5418} {37 \, 926}$
:$\dfrac 1 7 = \dfrac {5976} {41 \, 832}$
:$\dfrac 1 7 = \dfrac {7614} {53 \, 298}$
== Proof ==
Can be verified by brute force.
== Also see ==
* One Half as Pandigital Fraction
* One Third as Pandigital Fraction
* One Quarter as Pandigital Fraction
* One Fifth as Pandigital Fraction
* One Sixth as Pandigital Fraction
* One Eighth as Pandigital Fraction
* One Ninth as Pandigital Fraction
== Historical Note ==
== Sources ==
* : $0 \cdotp 5$
* : $0 \cdotp 5$
Category:One Seventh
Category:Pandigital Fractions
One Sixth as Pandigital Fraction

== Theorem ==
There are $3$ ways $\dfrac 1 6$ can be expressed as a pandigital fraction:
:$\dfrac 1 6 = \dfrac {2943} {17658}$
:$\dfrac 1 6 = \dfrac {4653} {27918}$
:$\dfrac 1 6 = \dfrac {5697} {34182}$
== Proof ==
Can be verified by brute force.
== Also see ==
* One Half as Pandigital Fraction
* One Third as Pandigital Fraction
* One Quarter as Pandigital Fraction
* One Fifth as Pandigital Fraction
* One Seventh as Pandigital Fraction
* One Eighth as Pandigital Fraction
* One Ninth as Pandigital Fraction
Category:One Sixth
Category:Pandigital Fractions
One Succeeds Zero in Well-Ordered Integral Domain

== Theorem ==
Let $\struct {D, +, \times, \le}$ be a well-ordered integral domain.
Let $0$ and $1$ be the zero and unity respectively of $D$.
Then $0$ is the immediate predecessor of $1$:
:$0 < 1$
:$\neg \exists a \in D: 0 < a < 1$
== Proof ==
 there exists an element $a \in D$ such that $0 < a < 1$.
Let us create the set $S$ of all such elements:
:$S = \set {x \in D: 0 < x < 1}$
We know $S$ is not empty because it has already been asserted that $a$ is in it.
And all the elements in $S$ are strictly positive by definition.
Because $D$ is well-ordered it follows that $S$ has a minimal element, which we will call $m$.
Thus we have $0 < m < 1$.
Then from Square of Element Less than Unity in Ordered Integral Domain we have:
:$0 < m \times m < m$
Thus $m \times m \in S$ is an element of $S$ which is smaller than $m$.
But $m$ was supposed to be the minimal element of $S$.
From this contradiction we deduce that there can be no such element $a$ such that $0 < a < 1$.
== Sources ==
* : Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 8$. Well-Order: Theorem $12$
Category:Well-Ordered Integral Domains
One Third as Pandigital Fraction

== Theorem ==
There are $2$ ways $\dfrac 1 3$ can be expressed as a pandigital fraction:
:$\dfrac 1 3 = \dfrac {5823} {17469}$
:$\dfrac 1 3 = \dfrac {5832} {17496}$
== Proof ==
Can be verified by brute force.
== Also see ==
* One Half as Pandigital Fraction
* One Quarter as Pandigital Fraction
* One Fifth as Pandigital Fraction
* One Sixth as Pandigital Fraction
* One Seventh as Pandigital Fraction
* One Eighth as Pandigital Fraction
* One Ninth as Pandigital Fraction
Category:One Third
Category:Pandigital Fractions
One Third as Quotient of Sequences of Odd Numbers

== Theorem ==
:$\dfrac 1 3 = \dfrac {1 + 3} {5 + 7} = \dfrac {1 + 3 + 5} {7 + 9 + 11} = \dfrac {1 + 3 + 5 + 7} {9 + 11 + 13 + 15} = \cdots$
That is:
:$\forall n \in \Z_{> 0}: \dfrac 1 3 = \dfrac {\ds \sum_{k \mathop = 1}^n \paren {2 k - 1} } {\ds \sum_{k \mathop = n + 1}^{2 n} \paren {2 k - 1} }$
== Proof ==
The result follows.
== Historical Note ==
== Sources ==
* : $1 / 3$
Category:One Third
One and Minus One form Subgroup of Multiplicative Group of Rational Numbers

== Theorem ==
Let $\struct {\Q_{\ne 0}, \times}$ be the multiplicative group of rational numbers.
Let $S \subseteq \Q$ where $S = \set {1, -1}$.
Then $\struct {S, \times}$ is a subgroup of $\struct {\Q_{\ne 0}, \times}$.
== Proof ==
By hypothesis, $S$ is not empty.
As $0 \notin S$, it follows that $S \subseteq \Q_{\ne 0}$.
Recall that $-1 \times -1 = 1$ and also $1 \times 1 = 1$.
Thus:
:$\forall x \in S: x \times y^{-1} \in S$
The result follows from the One-Step Subgroup Test.
== Sources ==
* : $\S 3.2$: Groups; the axioms: Examples of groups $\text{(ii)}$
Category:Rational Numbers
Category:Examples of Subgroups
One is Common Divisor of Integers

== Theorem ==
Let $a, b \in \Z$ be integers.
Then $1$ is a common divisor of $a$ and $b$.
== Proof ==
From One Divides all Integers:
:$1 \divides a$
and:
:$1 \divides b$
where $\divides$ denotes divisibility.
The result follows by definition of common divisor.
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor
Category:Divisors
One is not Prime

== Theorem ==
The integer $1$ (one) is not a prime number.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\text {2-2}$ Divisibility
* : $\text I$: The Series of Primes: $1.2$ Prime numbers
* : Chapter $2$: Integers and natural numbers: $\S 2.2$: Divisibility and factorization in $\mathbf Z$
* : $1$
* : $1$
* : $1$
* : Chapter $7$: Patterns in Numbers: Primes
Category:One is not Prime
Category:Prime Numbers
Category:1
One is not Prime/Proof 1

== Theorem ==
== Proof ==
By definition, a prime number is a positive integer which has exactly $2$ divisors which are themselves positive integers.
From Divisors of One, the only divisors of $1$ are $1$ and $-1$.
So the only divisor of $1$ which is a positive integer is $1$.
As $1$ has only one such divisor, it is not classified as a prime number.
== Sources ==
* : Appendix $1$: Some Results from the Elementary Theory of Numbers
* : $1$
* : $1$
* 
Category:One is not Prime
One is not Prime/Proof 2

== Theorem ==
== Proof ==
From Divisor Sum of Prime Number, the sum $\map {\sigma_1} p$ of all the positive integer divisors of a prime number $p$ is $p + 1$.
But from Divisor Sum of 1, $\map {\sigma_1} 1 = 1$.
If $1$ were to be classified as prime, then $\map {\sigma_1} 1$ would be an exception to the rule that $\map {\sigma_1} p = p + 1$.
== Historical Note ==
== Sources ==
* : $1$
* : $1$
Category:One is not Prime
One of 4 Consecutive Numbers Greater than 11 is Divisible by Prime Greater than 11

== Theorem ==
Let $n \in \Z$ such that $n > 11$.
Then at least one of the set:
:$\set {n, n + 1, n + 2, n + 3}$
is divisible by a prime number greater than $11$.
== Proof ==
== Sources ==
* : $11$
* : $11$
Category:11
Category:Prime Numbers
Only Number Twice Sum of Digits is 18

== Theorem ==
There exists only one (strictly) positive integer that is exactly twice the sum of its digits. 
== Proof ==
Let $n$ be equal to twice the sum of its digits.
Suppose $n$ has $1$ digit.
Then $n$ equals the sum of its digits.
Thus $n$ does not have only $1$ digit.
Suppose $n$ has $3$ digits.
Then $n > 99$.
But the highest number that can be formed by the sum of $3$ digits is $9 + 9 + 9 = 27$.
Similarly for if $n$ has more than $3$ digits.
Hence, if there exists such a number, it must have exactly $2$ digits.
Let the $2$ digits of $n$ be $x$ and $y$.
Thus:
Since $x$ and $y$ are both single digits, it follows that $x = 1$ and $y = 8$.
Hence the result.
== Sources ==
* : $18$
Category:18
Only Number Twice Sum of Digits is 18/Mistake

== Source Work ==
:
:The Dictionary
::$18$
== Mistake ==
:$18$ is the smallest number that is twice the sum of its digits.
While this is in fact technically a true statement, it misrepresents the fact that $18$ is the only number to equal twice the sum of its digits.
== Sources ==
* : $18$
Category:Mistakes/Curious and Interesting Numbers
Only Number which is Sum of 3 Factors is 6

== Theorem ==
The only positive integer which is the sum of exactly $3$ of its distinct coprime divisors is $6$.
== Corollary ==
== Proof ==
Let $n$ be such a positive integer with corresponding divisors $a, b, c$ such that:
:$a + b + c = n$
We note that the set $\set {k a, k b, k c}$ satisfy the same properties trivially as divisors of $k n$.
Hence the specification that $\set {a, b, c}$ is a coprime set.
, suppose $a < b < c$.
Since $a, b, c$ are strictly positive, $n \ne c$.
Suppose $\dfrac n c \ge 3$.
Then:
which is a contradiction.
Hence:
Since $a, b$ are divisors of $n$:
:$a \divides \paren {2 a + 2 b}$
:$b \divides \paren {2 a + 2 b}$
which reduces to:
:$a \divides 2 b$
:$b \divides 2 a$
Suppose $b$ is odd.
Then by Euclid's Lemma, we would have $b \divides a$.
By Absolute Value of Integer is not less than Divisors, this gives $b \le a$, which is a contradiction.
Thus $b$ is even.
Suppose $a$ is even.
Then $a, b, c$ are all even.
So $\gcd \set {a, b, c} \ne 1$, which is a contradiction.
Therefore it must be the case that $a$ is odd.
Then by Euclid's Lemma, we have:
:$a \divides \dfrac b 2$
and:
:$\dfrac b 2 \divides a$
By Absolute Value of Integer is not less than Divisors, this gives:
:$\dfrac b 2 = a$
Because $\gcd \set {a, b, c} = 1$, we must have $a = 1$.
Hence the set $\set {1, 2, 3}$ is obtained.
== Sources ==
* : $6$
* : $6$
Category:6
Category:Divisors
Only Number which is Sum of 3 Factors is 6/Corollary

== Corollary to Only Number which is Sum of 3 Factors is 6 ==
$1$ can be expressed uniquely as the sum of $3$ distinct unit fractions:
:$1 = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 6$
== Proof ==
Suppose we have $1 = \dfrac 1 a + \dfrac 1 b + \dfrac 1 c$, where $a, b, c$ are distinct positive integers in ascending order.
Multiplying both sides by $m = \lcm \set {a, b, c}$:
:$m = \dfrac m a + \dfrac m b + \dfrac m c$
Each fraction on the right is a factor of $m$.
Moreover, they must be comprime:
Suppose not. Then:
:$\exists d > 1: d \divides \dfrac m a, \dfrac m b, \dfrac m c$
Then:
:$\dfrac {m / d} a, \dfrac {m / d} b, \dfrac {m / d} c \in \Z$
Which shows that $\dfrac m d$ is a common multiple of $a, b, c$ less than $m$, a contradiction.
From Only Number which is Sum of 3 Factors is 6 we have:
:$\dfrac m a = 3, \dfrac m b = 2, \dfrac m c = 1$
Their sum is:
:$m = 3 + 2 + 1 = 6$
Thus:
:$m = 3 a = 2 b = c = 6$
with the unique solution:
:$\tuple {a, b, c} = {2, 3, 6}$
== Sources ==
* : $6$
* : $6$
Category:6
Category:Divisors
Only Number which is Sum of 3 Factors is 6/Mistake

== Source Work ==
:
:The Dictionary
:: $6$
:
:The Dictionary
:: $6$
== Mistake ==
:It is the only number that is the sum of exactly $3$ of its factors, ...
== Correction ==
The statement should be:
:It is the only number that is the sum of exactly $3$ of its distinct coprime factors, ...
Let $n$ be such a positive integer with corresponding divisors $a, b, c$ such that:
:$a + b + c = n$
We note that the set $\set {k a, k b, k c}$ satisfy the same properties trivially as divisors of $k n$.
Hence it is necessary definitely to require the set to be coprime.
Also note the trivial observation that $1 + 1 + 1 = 3$, which explains the need to specify that the factors be distinct.
== Sources ==
* : $6$
* : $6$
Category:Mistakes/Curious and Interesting Numbers
Open Ball Centred at Origin in Normed Vector Space is Symmetric

== Theorem ==
Let $\struct {X, \norm \cdot}$ be a normed vector space.
Let $\map B {0, r}$ be the open ball in $X$ centered at $0$ with radius $r$.
Then $\map B {0, r}$ is symmetric.
== Proof ==
Let $x \in \map B {0, r}$.
Then: 
:$\norm x < r$
We then have: 
:$\norm {-x} = \cmod {-1} \norm x = \norm x < r$
So $-x \in \map B {0, r}$.
So $\map B {0, r}$ is symmetric.
Category:Symmetric Subsets of Vector Spaces
Category:Open Balls
Open Ball Contains Open Ball Less Than Half Its Radius

== Therorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $a, b \in A$.
Let $\epsilon, \delta \in \R_{>0}$ such that $\delta \le \dfrac \epsilon 2$.
Let $\map {B_\epsilon} a$ be the open $\epsilon$-ball on $a$.
Let $\map {B_\delta} b$ be the open $\delta$-ball on $b$.
Let $a \in \map {B_\delta} b$.
Then:
:$\map {B_\delta} b \subseteq \map {B_\epsilon} a$
== Proof ==
We have:
Category:Open Balls
Open Ball contains Smaller Open Ball

== Theorem ==
Let $M = \struct{A, d}$ be a metric space.
Let $a \in A$.
Let $\epsilon, \delta \in \R_{>0}$ such that $\epsilon \le \delta$.
Let $\map {B_\epsilon} a$ be the open $\epsilon$-ball on $a$.
Let $\map {B_\delta} a$ be the open $\delta$-ball on $a$.
Then:
:$\map {B_\epsilon} a \subseteq \map {B_\delta} a$
== Proof ==
By definition of subset:
:$\map {B_\epsilon} a \subseteq \map {B_\delta} a$
Category:Open Balls
Open Ball contains Strictly Smaller Closed Ball

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $a \in A$.
Let $\epsilon, \delta \in \R_{>0}$ such that $\epsilon < \delta$.
Let $\map {B^-_\epsilon} a$ be the closed $\epsilon$-ball on $a$.
Let $\map {B_\delta} a$ be the open $\delta$-ball on $a$.
Then:
:$\map {B^-_\epsilon} a \subseteq \map {B_\delta} a$
== Proof ==
By definition of subset:
:$\map {B^-_\epsilon} a \subseteq \map {B_\delta} a$
Category:Open Balls
Category:Closed Balls
Open Ball in Cartesian Product under Chebyshev Distance

== Theorem ==
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 $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {y_1, y_2, \ldots, y_n} \in \AA$.
Let $a = \tuple {a_1, a_2, \ldots, a_n} \in \AA$.
Let $\epsilon \in \R_{>0}$.
Let $\map {B_\epsilon} {a; d_\infty}$ be the open $\epsilon$-ball of $a$ in $M = \struct {\AA, d_\infty}$.
Then:
:$\ds \map {B_\epsilon} {a; d_\infty} = \prod_{i \mathop = 1}^n \map {B_\epsilon} {a_i; d_i}$
== Proof ==
Let $\epsilon \in \R_{>0}$.
Let $x = \tuple {x_1, x_2, \ldots, x_n} \in \AA$.
Then:
And then:
The result follows by definition of set equality.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 4$: Open Balls and Neighborhoods: Exercise $7$
Category:Chebyshev Distance
Category:Open Balls
Open Ball in Euclidean 3-Space is Interior of Sphere

== Theorem ==
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:
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.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.3$: Open sets in metric spaces: Example $2.3.2$
Category:Open Balls
Category:Euclidean Space
Open Ball in Euclidean Plane is Interior of Circle

== Theorem ==
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:
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.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.3$: Open sets in metric spaces: Example $2.3.2$
Category:Open Balls
Category:Real Number Plane with Euclidean Metric
Open Ball in Euclidean Plus Metric is Subset of Equivalent Ball in Euclidean Metric

== Theorem ==
Let $\R$ be the set of real numbers.
Let $d: \R \times \R \to \R$ be the Euclidean plus metric:
:$\map d {x, y} := \size {x - y} + \ds \sum_{i \mathop = 1}^\infty 2^{-i} \map \inf {1, \size {\max_{j \mathop \le i} \frac 1 {\size {x - r_j} } - \max_{j \mathop \le i} \frac 1 {\size {y - r_j} } } }$
Let $d': \R \times \R \to \R$ be the Euclidean metric.
Let $\epsilon \in \R_{>0}$ be a (strictly) positive real number.
Let $\map {B_\epsilon} {p; d}$ be an open $\epsilon$-ball of $p$ in $\R$ on $d$.
Let $\map {B'_\epsilon} {p; d'}$ be an open $\epsilon$-ball of $p$ in $\R$ on $d'$.
Then:
:$\map {B_\epsilon} {p; d} \subseteq \map {B'_\epsilon} {p; d'}$
== Proof ==
Let $p \in \R$.
Let $\epsilon \in \R_{>0}$.
Let $x \in \map {B_\epsilon} {p; d}$.
Then:
Thus:
:$x \in \map {B_\epsilon} {p; d} \implies x \in \map {B_\epsilon} {p; d'}$
and the result follows by .
== Sources ==
* : Part $\text {II}$: Counterexamples: $30$. The Rational Numbers: $5$
Category:Euclidean Plus Metric
Category:Open Balls