avewright/proofwiki-math huggingface dataset:
Theorem: Subset of Bounded Subset of Metric Space is Bounded

Let $M = \struct {A, d}$ be a metric space.
Let $B$ be a bounded subset of $M$.
Let $\map \diam B$ denote the diameter of $B$.
Let $C \subseteq B$ be a subset of $B$.
Then $C$ is a bounded subset of $M$ such that:
:$\map \diam C \le \map \diam B$

Proof:
{{begin-eqn}}
{{eqn | q = \forall x, y \in B
      | l = \map d {x, y}
      | o = \le
      | r = \map \diam B
      | c = {{Defof|Diameter of Subset of Metric Space}}
}}
{{eqn | ll= \leadsto
      | q = \forall x, y \in C
      | l = \map d {x, y}
      | o = \le
      | r = \map \diam B
      | c = {{Defof|Subset}}
}}
{{eqn | ll= \leadsto
      | l = \sup \set {x, y \in C: \map d {x, y} }
      | o = \le
      | r = \map \diam B
      | c = {{Defof|Supremum}}
}}
{{eqn | ll= \leadsto
      | l = \map \diam C
      | o = \le
      | r = \map \diam B
      | c = {{Defof|Diameter of Subset of Metric Space}}
}}
{{end-eqn}}
{{qed}}
Theorem: Union of Non-Disjoint Bounded Subsets of Metric Space is Bounded

Let $M = \struct {A, d}$ be a metric space.
Let $B$ and $C$ be bounded subsets of $M$ such that $B \cap C \ne \O$.
Let $\map \diam B$ and $\map \diam C$ denote the diameters of $B$ and $C$.
Then $B \cup C$ is a bounded subset of $M$ such that:
:$\map \diam {B \cup C} \le \map \diam B + \map \diam C$

Proof:
{{AimForCont}} there exists $x, y \in B \cup C$ such that:
:$\map d {x, y} > \map \diam B + \map \diam C$
$x$ and $y$ cannot both be in $B$ or $C$, otherwise either $\map d {x, y} \le \map \diam B$ or $\map d {x, y} \le \map \diam C$.
{{WLOG}}, let $x \in B$ and $y \in C$.
Let $z \in B \cap C$.
This is possible because $B \cap C \ne \O$ {{hypothesis}}.
Then:
{{begin-eqn}}
{{eqn | l = \map d {x, z}
      | o = \le
      | r = \map \diam B
      | c = {{Defof|Diameter of Subset of Metric Space|Diameter of $B$}}
}}
{{eqn | l = \map d {z, y}
      | o = \le
      | r = \map \diam C
      | c = {{Defof|Diameter of Subset of Metric Space|Diameter of $C$}}
}}
{{eqn | ll= \leadsto
      | l = \map d {x, z} + \map d {z, y}
      | o = \le
      | r = \map \diam B + \map \diam C
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map d {x, y}
      | o = \le
      | r = \map \diam B + \map \diam C
      | c = {{Metric-space-axiom|2}}: $\map d {x, y} \le \map d {x, z} + \map d {z, y}$
}}
{{end-eqn}}
This contradicts our supposition that $\map d {x, y} > \map \diam B + \map \diam C$.
Hence:
{{begin-eqn}}
{{eqn | q = \forall x, y \in B \cup C
      | l = \map d {x, y}
      | o = \le
      | r = \map \diam B + \map \diam C
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \sup \set {\map d {x, y}: x, y \in B \cup C}
      | o = \le
      | r = \map \diam B + \map \diam C
      | c = {{Defof|Supremum}}
}}
{{eqn | ll= \leadsto
      | l = \map \diam {B \cup C}
      | o = \le
      | r = \map \diam B + \map \diam C
      | c = {{Defof|Diameter of Subset of Metric Space}}
}}
{{end-eqn}}
{{qed}}
Theorem: Multiple of Metric forms Metric

Let $M = \struct {A, d}$ be a metric space.
Let $d_1: A^2 \to \R$ be the mapping defined as:
:$\forall \tuple {x, y} \in A^2: \map {d_1} {x, y} = k \map d {x, y}$
for some strictly positive $k \in \R_{>0}$.
Then $d_1$ is a metric for $A$.

Proof:
It is to be demonstrated that $d_1$ satisfies all the metric space axioms.
Theorem: Standard Bounded Metric is Metric

Let $M = \struct {A, d}$ be a metric space.
Let $\bar d: A^2 \to \R$ be the standard bounded metric of $d$:
:$\forall \tuple {x, y} \in A^2: \map {\bar d} {x, y} = \min \set {1, \map d {x, y} }$
Then $\bar d$ is a metric for $A$.

Proof:
It is to be demonstrated that $\bar d$ satisfies all the metric space axioms.
Theorem: Metric over 1 plus Metric forms Metric

Let $M = \struct {A, d}$ be a metric space.
Let $d_3: A^2 \to \R$ be the mapping defined as:
:$\forall \tuple {x, y} \in A^2: \map {d_3} {x, y} = \dfrac {\map d {x, y} } {1 + \map d {x, y} }$
Then $d_3$ is a metric for $A$.

Proof:
It is to be demonstrated that $d_3$ satisfies all the metric space axioms.
Theorem: Square of Metric does not necessarily form Metric

Let $M = \struct {A, d}$ be a metric space.
Let $d_4: A^2 \to \R$ be the mapping defined as:
:$\forall \tuple {x, y} \in A^2: \map {d_4} {x, y} = \paren {\map d {x, y} }^2$
Then $d_4$ may or may not be a metric for $A$.

Proof:
Let $d$ be the standard discrete metric on $M$.
Then:
:$\forall \tuple {x, y} \in A^2: \map {d_4} {x, y} = \map d {x, y}$
and indeed in this case $d_4$ is a metric for $A$.
{{qed|lemma}}
Let $M = \struct {\R, d}$ be the real number line with the usual (Euclidean) metric.
Let $x = 1$, $y = \dfrac 1 2$ and $z = 0$.
We have:
{{begin-eqn}}
{{eqn | l = \map {d_4} {x, y} 
      | r = \size {1 - \dfrac 1 2}^2
      | c = 
}}
{{eqn | r = \dfrac 1 4
      | c = 
}}
{{eqn | l = \map {d_4} {y, z} 
      | r = \size {\dfrac 1 2 - 0}^2
      | c = 
}}
{{eqn | r = \dfrac 1 4
      | c = 
}}
{{eqn | l = \map {d_4} {x, z} 
      | r = \size {1 - 0}^2
      | c = 
}}
{{eqn | r = 1
      | c = 
}}
{{end-eqn}}
So we have:
:$\map {d_4} {x, y} + \map {d_4} {y, z} < \map {d_4} {x, z}$
and it is seen that $d_4$ does not satisfy {{Metric-space-axiom|2}}.
{{qed}}
Theorem: Arc Length on Circle forms Metric

Let $A \subseteq \R^2$ be the set defined as:
:$A = \set {\tuple {x_1, x_2}: x_1^2 + y_2^2 = 1}$
Thus from Equation of Unit Circle, $A$ is the unit circle embedded in the Cartesian plane.
Let $d: A^2 \to \R$ be the mapping defined as:
:$\forall \tuple {x, y} \in A^2: \map d {x, y} = \begin {cases} 0 & : x = y \\ \pi & : x = -y \\ l & : \text {otherwise} \end {cases}$
where:
:$x$ and $y$ are of the form $\tuple {x_1, x_2}$
:$l$ denotes the length of the minor arc of $A$ between $x$ and $y$.
Then $d$ is a metric for $A$.

Proof:
It is to be demonstrated that $d$ satisfies all the metric space axioms.
Theorem: Catesian Product of Open Real Intervals is Open in Real Number Plane

Let $\struct {\R^2, d}$ denote the real number plane with the Euclidean metric.
Let $\openint a b$ and $\openint c d$ be open real intervals.
Then their Cartesian product:
:$S = \openint a b \times \openint c d$
is an open set of $\struct {\R^2, d}$.

Proof:
Let $P = \tuple {x, y} \in S$.
Then by definition:
:$a < x < b$
and:
:$c < y < d$
Let $\epsilon = \min \set {x - a, b - x, y - c, d - y}$
By definition, $\epsilon > 0$.
Consider the open ball $\map {B_\epsilon} P$.
Let $Q = \tuple {x_0, y_0} \in \map {B_\epsilon} P$ be an arbitrary point in $\map {B_\epsilon} P$.
By definition of $\map {B_\epsilon} P$:
:$\size {x - x_0} < \epsilon$
and:
:$\size {y - y_0} < \epsilon$
Hence we have:
:$\size {x - x_0} < x - a$
:$\size {x - x_0} < b - x$
:$\size {y - y_0} < y - c$
:$\size {y - y_0} < d - y$
and so:
:$Q \in S$
Thus we have that:
:$\map {B_\epsilon} P \subseteq S$
Hence the result by definition of open set.
{{qed}}
Theorem: Non-Zero Value of Continuous Real-Valued Function has Neighborhood not including Zero

Let $M = \struct {A, d}$ be a metric space.
Let $f: M \to \R$ be a continuous real-valued function.
Let $\map f a > 0$ for some $a \in M$.
Then there exists $\delta \in \R_{>0}$ such that:
:$\forall x \in \map {B_\delta} a$
where $\map {B_\delta} a$ denotes the open $\delta$-ball of $a$ in $M$.

Proof:
Let $\epsilon \in \R_{>0}$ such that $\epsilon < \map f a$.
Let $\map {N_\epsilon} {\map f a} = \openint {\map f a - \epsilon} {\map f a + \epsilon}$ be the $\epsilon$-neighborhood of $\map f a$.
Then $\map {N_\epsilon} {\map f a}$ is an open interval of $\map f a$ such that:
:$\forall y \in \map {N_\epsilon} {\map f a}: y > 0$
We are given that $f$ is continuous.
So, by definition:
:$\exists U \subseteq A: \forall x \in U: \map f x \in \map {N_\epsilon} {\map f a}$
where $U$ is an open set of $M$.
As $a \map {N_\epsilon} {\map f a}$, it follows that:
:$a \in U$
By definition of open set:
:$\exists \delta \in \R_{>0}: \map {B_\delta} a \subseteq U$
Hence:
:$\forall x \in \map {B_\delta} a: \map f x \in \map {N_\epsilon} {\map f a}$
As we have already noted that:
:$\forall y \in \map {N_\epsilon} {\map f a}: y > 0$
the result follows.
{{qed}}
Theorem: Combination Theorem for Continuous Mappings/Metric Space/Sum Rule

:$f + g$ is continuous on $M$.

Proof:
By definition of continuous:
:$\forall a \in M: \ds \lim_{x \mathop \to a} \map f x = \map f a$
:$\forall a \in M: \ds \lim_{x \mathop \to a} \map g x = \map g a$
Let $f$ and $g$ tend to the following limits:
:$\ds \lim_{x \mathop \to a} \map f x = l$
:$\ds \lim_{x \mathop \to a} \map g x = m$
From the Sum Rule for Limits of Real Functions, we have that:
:$\ds \lim_{x \mathop \to a} \paren {\map f x + \map g x} = l + m$
So, by definition of continuous again, we have that $f + g$ is continuous on $M$.
{{qed}}
Theorem: Combination Theorem for Limits of Mappings/Metric Space/Sum Rule

:$\ds \lim_{x \mathop \to a} \paren {\map f x + \map g x} = l + m$

Proof:
Let $\sequence {x_n}$ be any sequence of elements of $S$ such that:
:$\forall n \in \N_{>0}: x_n \ne a$
:$\ds \lim_{n \mathop \to \infty} \ x_n = a$
By Limit of Function by Convergent Sequences:
:$\ds \lim_{n \mathop \to \infty} \map f {x_n} = l$
:$\ds \lim_{n \mathop \to \infty} \map g {x_n} = m$
By the Sum Rule for Real Sequences:
:$\ds \lim_{n \mathop \to \infty} \paren {\map f {x_n} + \map g {x_n} } = l + m$
Applying Limit of Function by Convergent Sequences again, we get:
:$\ds \lim_{x \mathop \to a} \paren {\map f x + \map g x} = l + m$
{{qed}}</onlyinclude>
Category:Combination Theorem for Limits of Mappings in Metric Spaces
98adhwvrk305ywb4iooepymq0wtugdm
Theorem: Combination Theorem for Limits of Mappings/Metric Space/Multiple Rule

:$\ds \lim_{x \mathop \to a} \lambda \map f x = \lambda l$

Proof:
Let $\sequence {x_n}$ be any sequence of elements of $S$ such that:
:$\forall n \in \N_{>0}: x_n \ne a$
:$\ds \lim_{n \mathop \to \infty} \ x_n = a$
By Limit of Function by Convergent Sequences:
:$\ds \lim_{n \mathop \to \infty} \map f {x_n} = l$
By the Multiple Rule for Real Sequences:
:$\ds \lim_{n \mathop \to \infty} \lambda \map f {x_n} = \lambda l$
Applying Limit of Function by Convergent Sequences again, we get:
:$\ds \lim_{x \mathop \to a} \lambda \map f x = \lambda l$
{{qed}}
</onlyinclude>
Category:Combination Theorem for Limits of Mappings in Metric Spaces
a3bhyd0dvio94259wyotdvwmfvwtubk
Theorem: Combination Theorem for Limits of Mappings/Metric Space/Product Rule

:$\ds \lim_{x \mathop \to a} \paren {\map f x \map g x} = l m$

Proof:
Let $\sequence {x_n}$ be any sequence of elements of $S$ such that:
:$\forall n \in \N_{>0}: x_n \ne a$
:$\ds \lim_{n \mathop \to \infty} \ x_n = a$
By Limit of Function by Convergent Sequences:
:$\ds \lim_{n \mathop \to \infty} \map f {x_n} = l$
:$\ds \lim_{n \mathop \to \infty} \map g {x_n} = m$
By the Product Rule for Real Sequences:
:$\ds \lim_{n \mathop \to \infty} \paren {\map f {x_n} \map g {x_n} } = l m$
Applying Limit of Function by Convergent Sequences again, we get:
:$\ds \lim_{x \mathop \to a} \paren {\map f x \map g x} = l m$
{{qed}}</onlyinclude>
Category:Combination Theorem for Limits of Mappings in Metric Spaces
hst8rj63o2r8netl572eb92qfgnecgh
Theorem: Combination Theorem for Limits of Mappings/Metric Space/Quotient Rule

:$\ds \lim_{x \mathop \to a} \frac {\map f x} {\map g x} = \frac l m$
provided that $m \ne 0$.

Proof:
Let $\sequence {x_n}$ be any sequence of elements of $S$ such that:
:$\forall n \in \N_{>0}: x_n \ne a$
:$\ds \lim_{n \mathop \to \infty} \ x_n = a$
By Limit of Function by Convergent Sequences:
:$\ds \lim_{n \mathop \to \infty} \map f {x_n} = l$
:$\ds \lim_{n \mathop \to \infty} \map g {x_n} = m$
By the Quotient Rule for Real Sequences:
:$\ds \lim_{n \mathop \to \infty} \frac {\map f {x_n} } {\map g {x_n} } = \frac l m$
provided that $m \ne 0$.
Applying Limit of Function by Convergent Sequences again, we get:
:$\ds \lim_{x \mathop \to a} \frac {\map f x} {\map g x} = \frac l m$
provided that $m \ne 0$.
{{qed}}</onlyinclude>
Category:Combination Theorem for Limits of Mappings in Metric Spaces
7ve7va4a9a45nngjd9xm8yiud3gq950
Theorem: Combination Theorem for Continuous Mappings/Metric Space/Difference Rule

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

Proof:
:$\map {\paren {f - g} } x = \map {\paren {f + \paren {-g} } } x$
From Multiple Rule for Continuous Mappings on Metric Space:
:$-g$ is continuous on $M$.
From Sum Rule for Continuous Mappings on Metric Space:
:$f + \paren {-g}$ is continuous on $M$.
The result follows.
{{qed}}
Theorem: Combination Theorem for Continuous Mappings/Metric Space/Multiple Rule

:$\lambda f$ is continuous on $M$.

Proof:
By definition of continuous:
:$\forall a \in M: \ds \lim_{x \mathop \to a} \map f x = \map f a$
Let $f$ tend to the following limit:
:$\ds \lim_{x \mathop \to a} \map f x = l$
From the Multiple Rule for Limits of Real Functions, we have that:
:$\ds \lim_{x \mathop \to a} \paren {\lambda \map f x} = \lambda l$
So, by definition of continuous again, we have that $\lambda \map f x$ is continuous on $M$.
{{qed}}
Theorem: Combination Theorem for Continuous Mappings/Metric Space/Absolute Value Rule

:$\size f$ 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 $a$:
:$\exists \delta \in \R_{>0}: \map d {x, a} < \delta \implies \size {\map f x - \map f a} < \epsilon$
From Reverse Triangle Inequality:
:$\size {\size {\map f x} - \size {\map f a} } \le \size {\map f x - \map f a}$
and so:
:$\exists \delta \in \R_{>0}: \map d {x, a} < \delta \implies \size {\size {\map f x} - \size {\map f a} } < \epsilon$
$\epsilon$ is arbitrary, so:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \map d {x, a} < \delta \implies \size {\size {\map f x} - \size {\map f a} } < \epsilon$
That is, by definition:
:$\size f$ is continuous at $a$.
{{qed}}
Theorem: Combination Theorem for Continuous Mappings/Metric Space/Maximum Rule

:$\max \set {f, g}$ is continuous on $M$.

Proof:
Let $a \in M$ be arbitrary.
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 Mappings on Metric Space:
:$\map f x - \map g x$ is continuous at $a$.
From Absolute Value Rule for Continuous Mappings on Metric Space:
:$\size {\map f x - \map g x}$ is continuous at $a$.
From Sum Rule for Continuous Mappings on Metric Space:
:$\map f x + \map g x$ 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 Mappings on Metric Space:
:$\dfrac 1 2 \paren {\map f x + \map g x + \size {\map f x - \map g x} }$ is continuous at $a$.
As $a$ is arbitrary:
:$\max \set {f, g}$ is continuous on $M$.
{{qed}}
Theorem: Combination Theorem for Continuous Mappings/Metric Space/Minimum Rule

:$\min \set {f, g}$ is continuous on $M$.

Proof:
Let $a \in M$ be arbitrary.
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 Mappings on Metric Space:
:$\map f x - \map g x$ is continuous at $a$.
From Absolute Value Rule for Continuous Mappings on Metric Space:
:$\size {\map f x - \map g x}$ is continuous at $a$.
From Sum Rule for Continuous Mappings on Metric Space:
:$\map f x + \map g x$ is continuous at $a$
From Difference Rule for Continuous Mappings on Metric Space:
:$\map f x + \map g x - \size {\map f x - \map g x}$ is continuous at $a$
From Multiple Rule for Continuous Mappings on Metric Space:
:$\dfrac 1 2 \paren {\map f x + \map g x - \size {\map f x - \map g x} }$ is continuous at $a$.
As $a$ is arbitrary:
:$\min \set {f, g}$ is continuous on $M$.
{{qed}}
Theorem: Combination Theorem for Continuous Mappings/Metric Space/Combined Sum Rule

:$\lambda f + \mu g$ is continuous on $M$.

Proof:
From the Multiple Rule for Continuous Mappings on Metric Space, we have that:
:$\lambda \map f x$ and $\mu \map g x$ are continuous.
From the Sum Rule for Continuous Mappings on Metric Space, we have that:
:$\lambda \map f x + \mu \map g x$ is continuous.
So, by definition of continuous again, we have that $\lambda f + \mu g$ is continuous on $M$.
{{qed}}
Category:Combination Theorem for Continuous Mappings
eszx0n9ll44pa13x7ttstq2m5862b0d
Theorem: Combination Theorem for Continuous Mappings/Metric Space/Product Rule

:$f g$ is continuous on $M$.

Proof:
By definition of continuous:
:$\forall a \in M: \ds \lim_{x \mathop \to a} \map f x = \map f a$
:$\forall a \in M: \ds \lim_{x \mathop \to a} \map g x = \map g a$
Let $f$ and $g$ tend to the following limits:
:$\ds \lim_{x \mathop \to a} \map f x = l$
:$\ds \lim_{x \mathop \to a} \map g x = m$
From the Product Rule for Limits of Real Functions, we have that:
:$\ds \lim_{x \mathop \to a} \paren {\map f x \map g x} = l m$
So, by definition of continuous again, we have that $f g$ is continuous on $M$.
{{qed}}
Theorem: Combination Theorem for Continuous Mappings/Metric Space/Quotient Rule

:$\dfrac f g$ is continuous on $M \setminus \set {x \in A: \map g x = 0}$.
that is, on all the points $x$ of $A$ where $\map g x \ne 0$.

Proof:
By definition of continuous:
:$\forall a \in M: \ds \lim_{x \mathop \to a} \map f x = \map f a$
:$\forall a \in M: \ds \lim_{x \mathop \to a} \map g x = \map g a$
Let $f$ and $g$ tend to the following limits:
:$\ds \lim_{x \mathop \to a} \map f x = l$
:$\ds \lim_{x \mathop \to a} \map g x = m$
From the Product Rule for Limits of Real Functions, we have that:
:$\ds \lim_{x \mathop \to a} \paren {\dfrac {\map f x} {\map g x} = \dfrac l m$
whenever $m \ne 0$.
So, by definition of continuous again, we have that $\dfrac f g$ is continuous on $M \setminus \set {x \in A: \map g x = 0}$.
{{qed}}
Theorem: Canonical Injection into Metric Space Product with P-Product Metric is Continuous

Let $M_1 = \struct {A_1, d}$ and $M_2 = \struct {A_2, d'}$ be metric spaces.
Let $\AA := A_1 \times A_2$ be the cartesian product of $A_1$ and $A_2$.
Let $\MM = \struct {\AA, d_p}$ denote the metric space on $\AA$ where $d_p: \AA \to \R$ is one of the $p$-product metrics on $\AA$:
:$\map {d_p} {x, y} := \begin {cases}
\paren {\paren {\map d {x_1, y_1} }^p + \paren {\map {d'} {x_2, y_2} }^p}^{1/p} & : p \in \Z_{>0} \\
\ds \max_{i \mathop = 1}^n \set {\map d {x_1, y_1}, \map {d'} {x_2, y_2} } & : p = \infty
\end {cases}$
where:
:$x = \tuple {x_1, x_2}$
:$y = \tuple {y_1, y_2}$
Let $a \in A_1$ and $b \in A_2$ be fixed and arbitrary.
Let:
:$i_b: A_1 \to \AA$ be the mapping defined as:
::$\forall x \in A_1: \map {i_b} x = \tuple {x, b}$
:$i_a: A_2 \to \AA$ be the mapping defined as:
::$\forall y \in A_2: \map {i_a} y = \tuple {a, y}$
Then $i_b$ and $i_a$ are continuous in $M_1$ and $M_2$ respectively.

Proof:
Let $\pr_1: \MM \to M_1$ and $\pr_2: \MM \to T_2$ be the first and second projections from $\MM$ onto its factors.
From Projection from Metric Space Product with P-Product Metric is Continuous, both $\pr_1$ and $\pr_2$ are continuous
The result follows from Continuous Mapping to Product Space.
{{qed}}
Theorem: Metric is Continous Mapping

Let $M = \struct {A, d}$ be a metric space.
Consider the distance function:
:$d: A \times A \to \R$
Then $\R$ is a continuous function.

Proof:
Let $\epsilon > 0$.
Let $\tuple {x_1, x_2} \in A \times A$.
Let $\delta = \dfrac \epsilon 2$.
Then $U = \map {B_\delta} {x_1} \times \map {B_\delta} {x_2}$ is a neighborhood of $\tuple {x_1, x_2}$ in $X \times X$ such that:
:$d \sqbrk U \subseteq \openint {\map d {x_1, x_2} - \epsilon} {\map d {x_1, x_2} + \epsilon}$
Therefore $d: X \times X \to \mathbb R$ is a continuous function.
{{qed}}
Category:Metric Spaces
Category:Continuous Mappings
nh6o7an4iq0nqexk3gsgb9xkcjyoslm
Theorem: Taxicab Metric on Metric Space Product is Continuous

Let $M = \struct {A, d}$ be a metric space.
Let $\AA$ be the Cartesian product $A \times A$.
Let $d_1$ be the taxicab metric on $\AA$:
:$\ds \map {d_1} {x, y} = \sum_{i \mathop = 1}^n \map {d_{i'} } {x_{i'}, y_{i'} }$
for $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {y_1, y_2, \ldots, y_n} \in \AA$.
Then $d_1: \AA \to \AA$ is a continuous function.

Proof:
Recall the definition of continuous mapping in this context.
Given metric spaces $M_X = \struct {X, d_X}$ and $M_Y = \struct {Y, d_Y}$, and a mapping $f : X \to Y$, we say that $f$ is $\struct {X, d_X} \to \struct {Y, d_Y}$-continuous {{iff}}:
:$\forall x_0 \in X: \forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in X: \map {d_X} {x, x_0} < \delta \implies \map d {\map f x, \map f {x_0} } < \epsilon$
Hence it is necessary to prove that $d: \struct {A \times A, d_1} \to \struct {\R, \size {\,\cdot \,} }$ is continuous.
That is, given $\tuple {x_0, y_0} \in A \times A$ and $\epsilon>0$, it is necessary to find $\delta \in \R_{>0}$ such that:
:$\forall x, y \in A: \map {d_1} {\tuple {x, y}, \tuple {x_0, y_0} } < \delta \implies \size {\map d {x, y} - \map d {x_0, y_0} } < \epsilon$
Hence, let $\tuple {x_0, y_0} \in A \times A$.
If $\map {d_1} {\tuple {x, y}, \tuple {x_0, y_0} } = \map d {x, x_0} + \map d {y, y_0} < \epsilon$, then:
:$\size {\map d {x, y} - \map d {x_0, y_0} } \le \size {\map d {x, y} - \map d {x, y_0} } + \size {\map d {x, y_0} - \map d {x_0, y_0} } \le 2 \epsilon$
using {{Metric-space-axiom|2}}.
The result follows.
{{qed}}
Theorem: Space of Continuous on Closed Interval Real-Valued Functions with Supremum Norm forms Normed Vector Space

Let $I := \closedint a b$ be a closed real interval.
The space of continuous real-valued functions on $I$ with supremum norm forms a normed vector space.

Proof:
We have that:
:Space of Continuous on Closed Interval Real-Valued Functions with Pointwise Addition and Pointwise Scalar Multiplication forms Vector Space
:Supremum Norm is Norm/Continuous on Closed Interval Real-Valued Function
By definition, $\struct {\map C I, \norm {\, \cdot \,}_\infty}$ is a normed vector space.
{{qed}}
Theorem: Neighborhood Filter of Point is Filter

Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in S$ be a point of $T$.
Let $\NN_x$ denote the neighborhood filter of $x$ in $T$.
Then $\NN_x$ is a filter on $S$.

Proof:
By definition, $\NN_x$ is the set of all neighborhoods of $x$ in $T$.
It is to be demonstrated that all the conditions are satisfied for $\NN_x$ to be a filter.
Theorem: Intersection of Compact and Closed Subsets of Normed Finite-Dimensional Real Vector Space with Euclidean Norm is Compact

Let $\struct {\R^d, \norm {\, \cdot \,}_2}$ be the normed finite-dimensional real vector space with Euclidean norm.
Let $K$ be a compact subset of $\struct {\R^d, \norm {\, \cdot \,}_2}$.
Let $F$ be a closed subset of $\struct {\R^d, \norm {\, \cdot \,}_2}$.
Then $F \cap K$ is compact in $\struct {\R^d, \norm {\, \cdot \,}_2}$.

Proof:
By assumption, $K$ is compact.
We have that a compact subset of normed vector space is closed and bounded.
Hence, $K$ is closed and bounded.
Since $K$ is bounded:
:$\exists C \in \R_{> 0} : \forall \mathbf x \in K : \norm {\mathbf x}_2 \le C$.
Then:
:$\forall \mathbf x \in K \cap F : \norm {\mathbf x}_2 \le C$
Hence, $K \cap F$ is bounded.
By assumption, $K$ and $F$ are closed.
We have that intersection of closed sets is closed in normed vector space.
Therefore, $K \cap F$ is closed.
By Heine-Borel theorem, $K \cap F$ is compact.
{{qed}}