avewright/proofwiki-math huggingface dataset:
Theorem: Union of Bounded Below Real Subsets is Bounded Below

Let $A$ and $B$ be sets of real numbers.
Let $A$ and $B$ be bounded below.
Then $A \cup B$ is also bounded below.

Proof:
Let $A$ and $B$ both be bounded below.
Then by definition $A$ and $B$ both have a lower bound $L_A$ and $L_B$ respectively.
Suppose $L_A \ge L_B$.
Then:
:$\forall a \in A: a \ge L_B$
and also, by definition:
:$\forall b \in B: b \ge L_B$
and so $L_B$ is a lower bound for $A$.
Otherwise, suppose $L_A < L_B$.
Then:
:$\forall b \in B: b \ge L_A$
and also, by definition:
:$\forall a \in A: a \ge L_A$
Let $x \in A \cup B$.
Then from the above, either $x \ge L_A$ or $x \ge L_B$.
So either $L_A$ or $L_B$ is a lower bound for $A \cup B$.
Hence, by definition, $A \cup B$ is bounded below by either $L_A$ or $L_B$.
{{qed}}
Theorem: Condition for Rational Number to be Square of Rational Number

Let $m$ and $n$ be (strictly) positive integers which are coprime.
Then:
:$\dfrac m n$ is the square of a rational number
{{iff}}:
:both $m$ and $n$ are square numbers.

Proof:
Let $m$ and $n$ be (strictly) positive integers which are coprime.
Theorem: Greatest Lower Bound Property

Let $S \subset \R$ be a non-empty subset of the set of real numbers such that $S$ is bounded below.
Then $S$ admits an infimum in $\R$.
This is known as the '''greatest lower bound property''' of the real numbers.

Proof:
Let $T = \set {x \in \R: -x \in S}$.
By Set of Real Numbers is Bounded Below iff Set of Negatives is Bounded Above:
:$T$ is bounded above.
Thus by the Least Upper Bound Property, $T$ admits a supremum in $\R$.
From Negative of Supremum is Infimum of Negatives:
:$\ds -\sup_{x \mathop \in T} x = \map {\inf_{x \mathop \in T} } {-x}$
That is, by definition of $T$:
:$\ds -\sup_{x \mathop \in T} x = \inf_{x \mathop \in S} x$
and so $S$ admits an infimum in $\R$.
{{qed}}
Theorem: Greatest Lower Bound Property

Let $S \subset \R$ be a non-empty subset of the set of real numbers such that $S$ is bounded below.
Then $S$ admits an infimum in $\R$.
This is known as the '''greatest lower bound property''' of the real numbers.

Proof:
Let $T$ be the set of lower bounds of $S$:
:$T = \set {x \in \R: x \le \forall y \in S}$
Since $S$ is bounded below, $T$ is non-empty.
Now, every $x \in T$ and $y \in S$ satisfy $x \leq y$.
That is, $T$ is bounded above by every element of $S$.
By the Least Upper Bound Property, $T$ admits a supremum in $\R$.
Let $B = \sup T$.
Since every element of $S$ is an upper bound of $T$ it follows from the definition of supremum that:
:$\forall y \in S : y \ge B$
Thus $B$ is a lower bound of $S$.
Because $B$ is an upper bound of $T$:
:$\forall x \in T : x \le B$
and so $B$ is the infimum of $S$.
{{qed}}
Theorem: Set of Real Numbers is Bounded Below iff Set of Negatives is Bounded Above

Let $S$ be a subset of the real numbers $\R$.
Let $T = \set {x \in \R: -x \in S}$ be the set of negatives of the elements of $S$.
Then:
:$S$ is bounded below
{{iff}}:
:$T$ is bounded above.

Proof:
=== Sufficient Condition ===
Let $S$ be bounded below.
Then $S$ has a lower bound.
Let $B$ be a lower bound for $S$.
From Negative of Lower Bound of Set of Real Numbers is Upper Bound of Negatives:
:$B$ is a lower bound for $S$
{{iff}}
:$-B$ is an upper bound for $T$.
It follows that $T$ is bounded above.
{{qed|lemma}}
Theorem: Negative of Lower Bound of Set of Real Numbers is Upper Bound of Negatives

Let $S$ be a subset of the real numbers $\R$.
Let $T = \set {x \in \R: -x \in S}$ be the set of negatives of the elements of $S$.
Then:
:$B$ is a lower bound of $S$
{{iff}}:
:$-B$ is an upper bound of $T$.

Proof:
Let $B$ be a lower bound of $S$.
That is:
:$\forall x \in S: x \ge B$
Let $x \in T$ be arbitrary.
{{begin-eqn}}
{{eqn | l = x
      | o = \in
      | r = T
}}
{{eqn | ll= \leadsto
      | l = -x
      | o = \in
      | r = S
      | c = 
}}
{{eqn | ll= \leadsto
      | l = -x
      | o = \ge
      | r = B
      | c = as $B$ is a lower bound for $S$
}}
{{eqn | ll= \leadsto
      | l = x
      | o = \le
      | r = -B
      | c = Ordering of Real Numbers is Reversed by Negation
}}
{{end-eqn}}
As $x$ is arbitrary it follows that:
:$\forall x \in T: x \le -B$
That is, $-B$ is an upper bound for $T$.
{{qed|lemma}}
Theorem: Ordering of Real Numbers is Reversed by Negation

Let $x$ and $y$ in $\R$ be real numbers such that:
:$x \le y$
Then:
:$-y \le -x$
where $-y$ and $-x$ are the negatives of $y$ and $x$ respectively.

Proof:
By definition of ordering:
:$x \le y$
{{iff}}:
:$x < y \text { or } x = y$
From Order of Real Numbers is Dual of Order of their Negatives:
:$x < y \iff \paren {-x} > \paren {-y}$
Hence the result.
{{qed}}
Category:Real Numbers
trgpq86ye6amupvisdvcnjcweln53xy
Theorem: Negative of Upper Bound of Set of Real Numbers is Lower Bound of Negatives

Let $S$ be a subset of the real numbers $\R$.
Let $T = \set {x \in \R: -x \in S}$ be the set of negatives of the elements of $S$.
Then:
:$U$ is an upper bound of $S$
{{iff}}:
:$-U$ is a lower bound of $T$.

Proof:
Let $V$ be the set defined as:
:$V = \set {x \in \R: -x \in T}$
From Negative of Lower Bound of Set of Real Numbers is Upper Bound of Negatives:
:$B$ is a lower bound of $T$
{{iff}}:
:$-B$ is an upper bound of $V$
Then we have:
{{begin-eqn}}
{{eqn | l = V
      | r = \set {x \in \R: -x \in T}
      | c = 
}}
{{eqn | r = \set {x \in \R: -\paren {-x} \in S}
      | c = 
}}
{{eqn | r = \set {x \in \R: x \in S}
      | c = 
}}
{{eqn | r = S
      | c = 
}}
{{end-eqn}}
That is:
:$B$ is a lower bound of $T$
{{iff}}:
:$-B$ is an upper bound of $S$
The result follows by setting $U = -B$.
{{qed}}
Theorem: Combination Theorem for Continuous Functions/Real/Difference Rule

:$f - g$ is continuous on $S$.

Proof:
We have that:
:$\map {\paren {f - g} } x = \map {\paren {f + \paren {-g} } } x$
From Multiple Rule for Continuous Real Functions:
:$-g$ is continuous on $S$.
From Sum Rule for Continuous Real Functions:
:$f + \paren {-g}$ is continuous on $S$.
The result follows.
{{qed}}
Theorem: Dirichlet Function is Discontinuous

Let $D: \R \to \R$ denote the Dirichlet function:
:$\forall x \in \R: \map D x = \begin {cases} c & : x \in \Q \\ d & : x \notin \Q \end {cases}$
where $\Q$ denotes the set of rational numbers.
Then $D$ is discontinuous at every $x \in \R$.

Proof:
Let $\epsilon = \dfrac {\size {c - d} } 2$.
Let $x \in \Q$.
Let $\delta \in \R_{>0}$ be arbitrary.
Let $y \in \Q$ such that $\size {x - y} < \delta$.
{{WLOG}}, let $y > x$.
From Between two Rational Numbers exists Irrational Number:
:$\exists z \in \R \setminus \Q: x < z < y$
and so:
:$\size {\map D x - \map D z} = \size {c - d} > \epsilon$
Similarly if $y < x$:
:$\exists z \in \R \setminus \Q: y < z < x: \size {\map D x - \map D z} > \epsilon$
and by definition of continuity $D$ is discontinuous at $x$.
{{qed|lemma}}
Let $x \in \R \setminus \Q$.
Let $\delta \in \R_{>0}$ be arbitrary.
Let $y \in \R \setminus \Q$ such that $\size {x - y} < \delta$.
{{WLOG}}, let $y > x$.
From Between two Real Numbers exists Rational Number:
:$\exists z \in \Q: x < z < y$
and so:
:$\size {\map D x - \map D z} = \size {c - d} > \epsilon$
Similarly if $y < x$:
:$\exists z \in \Q: y < z < x: \size {\map D x - \map D z} > \epsilon$
and by definition of continuity $D$ is discontinuous at $x$.
{{qed|lemma}}
$D$ has been shown to be discontinuous at all $x \in \R$ whether $x$ is rational or irrational.
Hence the result.
{{qed}}
Theorem: Thomae Function is Continuous at Irrational Numbers

Let $D_M: \R \to \R$ denote the Thomae function:
:$\forall x \in \R: \map {D_M} x = \begin {cases} 0 & : x = 0 \text { or } x \notin \Q \\ \dfrac 1 q & : x = \dfrac p q : p, q \in \Z, p \perp q, q > 0 \end {cases}$
where:
:$\Q$ denotes the set of rational numbers
:$\Z$ denotes the integers
:$p \perp q$ denotes that $p$ and $q$ are coprime (that is, $x$ is a rational number expressed in canonical form)
Then $\map {D_M} x$ is:
:continuous at all irrational $x$ and at $x = 0$
:discontinuous at all rational $x$ such that $x \ne 0$.

Proof:
=== Rational $x$ ===
Let $x = \dfrac p q \in \Q \setminus \set 0$ such that $\dfrac p q$ is the canonical form of $x$.
Then we have:
:$\map {D_M} x = \dfrac 1 q$ 
Let $\epsilon = \dfrac 1 {2 q}$.
Let $\delta \in \R_{>0}$.
Then from Between two Real Numbers exists Irrational Number:
:$\exists z \in \R \setminus \Q: x < z < x + \delta$
Hence there exists $z \in \R$ such that:
:$z: \size {x - z} < \delta: \map {D_M} z = 0$
that is, such that:
:$\size {\map {D_M} x - \map {D_M} z} = \dfrac 1 q > \epsilon$
That is, there exists an $\epsilon \in \R_{>0}$ such that for all $\delta \in \R_{>0}$ it is possible to find $z$ such that:
:$\size {x - z} < \delta$
but such that:
:$\size {\map {D_M} x - \map {D_M} z} > \epsilon$
Hence when $x$ is rational $\map {D_M} x$ is discontinuous.
{{qed|lemma}}
Theorem: Absolute Value of Continuous Real Function is Continuous

Let $f: \R \to \R$ be a real function.
Let $f$ be continuous at a point $a \in \R$.
Then:
:$\size f$ is continuous at $a$
where:
:$\map {\size f} x$ is defined as $\size {\map f x}$.

Proof:
Let $\epsilon \in \R_{>0}$ be a positive real number.
Because $f$ is continuous at $a$:
:$\exists \delta \in \R_{>0}: \size {x - a} < \delta \implies \size {\map f x - \map f a} < \epsilon$
From Reverse Triangle Inequality:
:$\size {\size {\map f x} - \size {\map f a} } \le \size {\map f x - \map f a}$
and so:
:$\exists \delta \in \R_{>0}: \size {x - a} < \delta \implies \size {\size {\map f x} - \size {\map f a} } < \epsilon$
$\epsilon$ is arbitrary, so:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \size {x - a} < \delta \implies \size {\size {\map f x} - \size {\map f a} } < \epsilon$
That is, by definition:
:$\size f$ is continuous at $a$.
{{qed}}
Theorem: Max Operation on Continuous Real Functions is Continuous

Let $f: \R \to \R$ and $g: \R \to \R$ be real functions.
Let $f$ and $g$ be continuous at a point $a \in \R$.
Let $h: \R \to \R$ be the real function defined as:
:$\map h x := \map \max {\map f x, \map g x}$ 
Then $h$ is continuous at $a$.

Proof:
From Max is Half of Sum Plus Absolute Difference
:$\max \set {x, y} = \dfrac 1 2 \paren {x + y + \size {x - y} }$
Hence:
:$\max \set {\map f x, \map g x} = \dfrac 1 2 \paren {\map f x + \map g x + \size {\map f x - \map g x} }$
From Difference Rule for Continuous Real Functions:
:$\map f x - \map g x$ is continuous at $a$.
From Absolute Value of Continuous Real Function is Continuous:
:$\size {\map f x - \map g x}$ is continuous at $a$.
From Sum Rule for Continuous Real Functions:
:$\map f x + \map g x$ is continuous at $a$
and hence:
:$\map f x + \map g x + \size {\map f x - \map g x}$ is continuous at $a$
From Multiple Rule for Continuous Real Functions:
:$\dfrac 1 2 \paren {\map f x + \map g x + \size {\map f x - \map g x} }$ is continuous at $a$.
{{qed}}
Theorem: Min Operation on Continuous Real Functions is Continuous

Let $f: \R \to \R$ and $g: \R \to \R$ be real functions.
Let $f$ and $g$ be continuous at a point $a \in \R$.
Let $h: \R \to \R$ be the real function defined as:
:$\map h x := \map \min {\map f x, \map g x}$ 
Then $h$ is continuous at $a$.

Proof:
From Min is Half of Sum Less Absolute Difference
:$\min \set{x, y} = \dfrac 1 2 \paren {x + y - \size {x - y} }$
Hence:
:$\min \set {\map f x, \map g x} = \dfrac 1 2 \paren {\map f x + \map g x - \size {\map f x - \map g x} }$
From Difference Rule for Continuous Real Functions:
:$\map f x - \map g x$ is continuous at $a$.
From Absolute Value of Continuous Real Function is Continuous:
:$\size {\map f x - \map g x}$ is continuous at $a$.
From Sum Rule for Continuous Real Functions:
:$\map f x + \map g x$ is continuous at $a$
and hence from Difference Rule for Continuous Real Functions again:
:$\map f x + \map g x - \size {\map f x - \map g x}$ is continuous at $a$
From Multiple Rule for Continuous Real Functions:
:$\dfrac 1 2 \paren {\map f x + \map g x - \size {\map f x - \map g x} }$ is continuous at $a$.
{{qed}}
Theorem: Power of Maximum is not Greater than Sum of Powers

Let $a_1, a_2, \ldots, a_r$ be non-negative real numbers.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $a = \max \set {a_1, a_2, \ldots, a_r}$.
Then:
:$a^n \le a_1^n + a_2^n + \cdots + a_r^n$

Proof:
By definition of the $\max$ operation:
:$\exists k \in \set {1, 2, \ldots, r}: a_k = a$
Hence:
{{begin-eqn}}
{{eqn | l = a^n
      | r = a_k^n
      | c = 
}}
{{eqn | o = \le
      | r = a_1^n + a_2^n + \cdots + a_k^n + \cdots + a_r^n
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Sum of r Powers is not Greater than r times Power of Maximum

Let $a_1, a_2, \ldots, a_r$ be non-negative real numbers.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $a = \max \set {a_1, a_2, \ldots, a_r}$.
Then:
:$a_1^n + a_2^n + \cdots + a_r^n \le r a^n$

Proof:
By definition of the $\max$ operation:
:$\exists k \in \set {1, 2, \ldots, r}: a_k = a$
Then:
:$\forall i \in \set {1, 2, \ldots, r}: a_i \le a_k$
Hence:
{{begin-eqn}}
{{eqn | q = \forall i \in \set {1, 2, \ldots, r}
      | l = a_i
      | r = a_k
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \sum_{i \mathop = 1}^r a_i^n
      | o = \le
      | r = \sum_{i \mathop = 1}^r a_k^n
      | c = 
}}
{{eqn | r = r a_k^n
      | c = 
}}
{{eqn | r = r a^n
      | c = Definition of $a_k$
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Sum of r Powers is between Power of Maximum and r times Power of Maximum

Let $a_1, a_2, \ldots, a_r$ be non-negative real numbers.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $a = \max \set {a_1, a_2, \ldots, a_r}$.
Then:
:$a^n \le a_1^n + a_2^n + \cdots + a_r^n \le r a^n$

Proof:
This proof is divided into $2$ parts:
Theorem: Limit of nth Root of Sum of nth Powers equals Maximum

Let $a_1, a_2, \ldots, a_r$ be non-negative real numbers.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $a = \max \set {a_1, a_2, \ldots, a_r}$.
Then:
:$\ds \lim_{n \mathop \to \infty} \paren {a_1^n + a_2^n + \cdots + a_r^n}^{1 / n} = a$

Proof:
From Sum of $r$ Powers is between Power of Maximum and $r$ times Power of Maximum:
:$a^n \le a_1^n + a_2^n + \cdots + a_r^n \le r a^n$
and so:
:$a \le \paren {a_1^n + a_2^n + \cdots + a_r^n}^{1 / n} \le r^{1/n} a$
From Limit of Integer to Reciprocal Power:
:$n^{1 / n} \to 1$ as $n \to \infty$
Then for $n > r$:
:$1 < r^{1 / n} < n^{1 / n}$
and so:
:$r^{1 / n} = 1$ as $n \to \infty$
Thus as $n \to \infty$:
:$a \le \paren {a_1^n + a_2^n + \cdots + a_r^n}^{1 / n} \le a$
and the result follows by the Squeeze Theorem.
{{qed}}
Theorem: Convergence of Odd and Even Subsequences to Same Limit

Let $\sequence {s_n}$ be a real sequence.
Let the subsequences $\sequence {s_{2 n} }$ and $\sequence {s_{2 n + 1} }$ both converge to the same limit $l$.
Then $\sequence {s_n}$ also converges to the same limit $l$.

Proof:
Suppose that $\sequence {s_n}$ converge to a limit.
Then from Limit of Subsequence equals Limit of Real Sequence, $\sequence {s_{2 n} }$ and $\sequence {s_{2 n + 1} }$ both converge to the same limit.
They do so converge, and that limit is $l$.
So, if $\sequence {s_n}$ converges, it converges to the limit $l$.
{{AimForCont}} $\sequence {s_n}$ does not actually converge to a limit.
Then it is not the case that:
:$\forall \epsilon \in \R_{>0}: \exists N \in \R_{>0}: k > N \implies \size {s_k - l} < \epsilon$
That is:
:$\exists \epsilon \in \R_{>0}: \forall N \in \R_{>0}: \exists k > N: \size {s_k - l} > \epsilon$
Either:
:$k = 2 n$ for some $n \in \N$
or:
:$k = 2 n + 1$ for some $n \in \N$
In the first case that means $\sequence {s_{2 n} }$ does not converge to $l$.
In the second case that means $\sequence {s_{2 n + 1} }$ does not converge to $l$.
Both cases contradict the supposition that $\sequence {s_{2 n} }$ and $\sequence {s_{2 n + 1} }$ both converge to $l$.
Hence by Proof by Contradiction it must be the case that $\sequence {s_n}$ converges to a limit.
That limit has been shown to be $l$.
{{qed}}
Theorem: Hilbert Cube is Compact/2-Sequence Space with 2-Norm

Let $\ell^2$ be the 2-sequence space.
Let $\norm {\, \cdot \,}_2$ be the 2-norm.
Let $C$ be the Hilbert cube.
Then $C$ is compact in $\struct {\ell^2, \norm {\, \cdot \,}_2}$.

Proof:
=== Construction of subsequence ===
Let $\sequence {\mathbf x_m}_{m \mathop \in \N}$ be a sequence in $C$.
Let $\mathbf x_m = \tuple {x_m^{\paren n}}_{n \mathop \in \N}$.
By definition of Hilbert cube:
:$\ds \forall n,m \in \N : 0 \le x_m^{\paren n} \le \frac 1 n$
Consider the case $n = 1$.
Then:
:$\ds \forall m \in \N : 0 \le x_m^{\paren 1} \le 1$
By Closed Real Interval is Compact Space, $\closedint 0 1$ is compact.
Hence, $\sequence {x^{\paren 1}_m}_{m \mathop \in \N}$ has a subsequence $\sequence {x^{\paren 1}_{\map {m_1} j}}_{j \mathop \in \N}$ that is convergent with limit, say, $\ds x^{\paren 1} \in \closedint 0 1$.
Consider the case $n = 2$.
By analogous arguments, $\sequence {x^{\paren 2}_{\map {m_1} j}}_{j \mathop \ge 2}$ has a subsequence $\sequence {x^{\paren 2}_{\map {m_2} j}}_{j \mathop \ge 2}$ that is convergent with limit, say, $\ds x^{\paren 2} \in \closedint 0 {\frac 1 2}$.
Proceeding in this manner, for all $k \in \N$ a sequence $\sequence {x^{\paren k}_{\map {m_{k - 1}} j}}_{j \mathop \ge k}$ has a convergent subsequence $\sequence {x^{\paren k}_{\map {m_k} j}}_{j \mathop \ge k}$ with a limit $\ds x^{\paren k} \in \closedint 1 {\frac 1 k}$.
From all these sequences we construct a subsequence $\sequence {\mathbf x_{\map {m_j} j}}_{j \mathop \in \N}$.
Let $\mathbf x := \tuple {x^{\paren n}}_{n \mathop \in \N}$.
$\mathbf x \in C$ because $\ds \forall n \in \N : 0 \le x^{\paren n} \le \frac 1 n$.
Theorem: Standard Discrete Metric is not Topologically Equivalent to p-Product Metrics

For $n \in \N$, let $\R^n$ be an Euclidean space.
Let $p \in \R_{\ge 1}$.
Let $d_p$ be the $p$-product metric on $\R^n$.
Let $d_0$ be the standard discrete metric on $\R^n$.
Then $d_p$ and $d_0$ are not topologically equivalent.

Proof:
From Open Ball in Standard Discrete Metric Space it is seen that singletons are open sets in $\struct {\R^n, d_0}$.
However, this is not the case in the $\struct {\R^n, d_p}$.
{{qed}}
Theorem: Supremum Metric and L1 Metric on Closed Real Intervals are not Topologically Equivalent

Let $S$ be the set of all real functions which are continuous on the closed interval $\closedint a b$.
Let $d_1$ be the $L^1$ metric on $S$:
:$\ds \forall f, g \in S: \map {d_1} {f, g} := \int_a^b \size {\map f t - \map g t} \rd t$
Let $d_\infty$ be the supremum metric on $S$:
:$\ds \forall f, g \in S: \map {d_\infty} {f, g} := \sup_{x \mathop \in \closedint a b} \size {\map f x - \map g x}$
Then $d_1$ and $d_\infty$ are not topologically equivalent.

Proof:
Let $f, g \in S$.
Then by definition of supremum metric:
:$\forall x \in \closedint a b: \size {\map f x - \map g x} \le \map {d_\infty} {f, g}$
Hence by ...
:$\map {d_1} {f, g} = \ds \int_a^b \size {\map f x - \map g x} \rd x \le \paren {b - a} \map {d_\infty} {f, g}$
{{finish|Find that link}}
and so:
:$\map {B_\epsilon} {f; d_\infty} \subseteq \map {b_{\paren {b - a} } } {f; d_1}$
Now let $f_0$ denote the constant mapping:
:$\forall x: \map {f_0} x = 0$
We show that $\map {B_1} {f_0; d_\infty}$ is not $d_1$-open.
{{AimForCont}} instead that $\map {B_1} {f_0; d_\infty}$ is $d_1$-open.
Then we should have:
:$\map {B_1} {f_0; d_1} \subseteq \map {B_1} {f_0; d_\infty}$
for some $\epsilon \in \R_{>0}$.
But for $\epsilon > 0$ there exists a continuous function on $\closedint a b$ such that:
:$\map {d_1} {f, f_0} < \epsilon$
yet:
:$\map {d_\infty} {f, f_0} = 1$
So:
:$f \in \map {B_1} {f_0; d_1}$
but:
:$f \notin \map {B_1} {f_0; d_\infty}$
which contradicts our deduction that $\map {B_1} {f_0; d_1} \subseteq \map {B_1} {f_0; d_\infty}$.
Hence it cannot be the case that $\map {B_1} {f_0; d_\infty}$ is $d_1$-open.
The result follows.
{{qed}}
Theorem: Metrics on Space are Topologically Equivalent iff Identity Mapping is Homemorphism

Let $M_1 = \struct {A, d_1}$ and $M_2 = \struct {A, d_2}$ be metric spaces on the same underlying set $A$.
Let $I_A$ denote the identity mapping on $A$.
Then:
:$d_1$ and $d_2$ are topologically equivalent
{{iff}}:
:$I_A: M_1 \to M_2$ is a homeomorphism.

Proof:
First we establish the following:
From Identity Mapping is Bijection, we have that $I_A$ is a bijection.
From Inverse of Identity Mapping, $\paren {I_A}^{-1} = I_A$.
By Identity Mapping on Metric Space is Continuous:
:$(1): \quad I_A$ is a $\tuple {d_1, d_1}$-continuous mapping.
:$(2): \quad I_A^{-1} = I_A$ is a $\tuple {d_2, d_2}$-continuous mapping.
Theorem: Metrics on Space are Lipschitz Equivalent iff Identity Mapping is Lipschitz Equivalence

Let $M_1 = \struct {A, d_1}$ and $M_2 = \struct {A, d_2}$ be metric spaces on the same underlying set $A$.
Let $I_A$ denote the identity mapping on $A$.
Then:
:$d_1$ and $d_2$ are Lipschitz equivalent
{{iff}}:
:$I_A: M_1 \to M_2$ is a Lipschitz equivalence.

Proof:
By definition of identity mapping:
:$\forall x \in A: \map {I_A} x = x$
Theorem: Homeomorphism yields Topologically Equivalent Metric

Let $M_1 = \struct {A, d_1}$ and $M_2 = \struct {A, d_2}$ be metric spaces on the same underlying set $A$.
Let $f: M_1 \to M_2$ be a homeomorphism.
Let $d_3: A^2 \to \R$ be a mapping defined as:
:$\map {d_3} {x, y} := \map {d_2} {\map f x, \map f y}$
Then $d_3$ yields a metric on $M_1$ which is topologically equivalent to $d_1$.

Proof:
{{ProofWanted|Easy when you understand what you're doing}}
Theorem: Lipschitz Equivalence yields Lipschitz Equivalent Metric

Let $M_1 = \struct {A, d_1}$ and $M_2 = \struct {A, d_2}$ be metric spaces on the same underlying set $A$.
Let $f: M_1 \to M_2$ be a Lipschitz equivalence.
Let $d_3: A^2 \to \R$ be a mapping defined as:
:$\map {d_3} {x, y} := \map {d_2} {\map f x, \map f y}$
Then $d_3$ yields a metric on $M_1$ which is Lipschitz equivalent to $d_1$.

Proof:
{{ProofWanted|Easy when you understand what you're doing}}
Theorem: Open Balls whose Distance between Centers is Twice Radius are Disjoint

Let $M = \struct {A, d}$ be a metric space.
Let $x, y \in A$ such that $\map d {x, y} = 2 r > 0$.
Let $\map {B_r} x$ denote the open $r$-ball of $x$ in $M$.
Then $\map {B_r} x$ and $\map {B_r} y$ are disjoint.

Proof:
{{AimForCont}} $\map {B_r} x \cap \map {B_r} y \ne \O$.
Then:
{{begin-eqn}}
{{eqn | q = \exists z \in A
      | l = z \in \map {B_r} x
      | o = \text { and }
      | r =  z \in \map {B_r} y
      | c = {{Defof|Set Intersection}}
}}
{{eqn | ll= \leadsto
      | l = \map d {x, z} < r
      | o = \text { and }
      | r = \map d {z, y} < r
      | c = {{Defof|Open Ball of Metric Space}}
}}
{{eqn | ll= \leadsto
      | l = \map d {x, z} + \map d {z, y}
      | o = <
      | r = 2 r
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map d {x, y}
      | o = <
      | r = 2 r
      | c = {{Metric-space-axiom|2}}: $\map d {x, y} \le \map d {x, z} + \map d {z, y}$
}}
{{end-eqn}}
But this contradicts our initial assertion that $\map d {x, y} = 2 r$.
The result follows by Proof by Contradiction.
{{qed}}
Theorem: Triangle Inequality/Examples/4 Points

Let $M = \struct {A, d}$ be a metric space.
Let $x, y, z, t \in A$.
Then:
:$\map d {x, z} + \map d {y, t} \ge \size {\map d {x, y} - \map d {z, t} }$

Proof:
We have that $\map d {x, z}$, $\map d {y, t}$, $\map d {x, y}$, $\map d {z, t}$ are themselves all real numbers.
Hence the Euclidean metric on the real number line can be applied:
{{begin-eqn}}
{{eqn | l = \size {\map d {x, y} - \map d {z, t} }
      | o = \le
      | r = \size {\map d {x, y} - \map d {y, z} } + \size {\map d {y, z} - \map d {z, t} }
      | c = 
}}
{{eqn | o = \le
      | r = \map d {x, z} + \map d {y, t}
      | c = Reverse Triangle Inequality twice
}}
{{end-eqn}}
{{qed}}