hcju/proofwikiqa huggingface dataset:
=== Proof of $M1$ === {{begin-eqn}} {{eqn | l = \map {d_2} {f, f}       | r = \paren {\int_a^b \paren {\map f t - \map f t}^2 \rd t}^{\frac 1 2}       | c = Definition of $d_2$ }} {{eqn | r = \paren {\int_a^b 0^2 \rd t}^{\frac 1 2}       | c = {{Defof|Absolute Value}} }} {{eqn | r = 0       | c = Definite Integral of Constant }} {{end-eqn}} So axiom $M1$ holds for $d_2$. {{qed|lemma}} === Proof of $M2$ === It is required to be shown: :$\map {d_2} {f, g} + \map {d_2} {g, h} \ge d_2 \map {d_2} {f, h}$ for all $f, g, h \in S$. Let: :$(1): \quad t \in \closedint a b$ :$(2): \quad \map f t - \map g t = \map r t$ :$(3): \quad \map g t - \map h t = \map s t$ Thus we need to show that: :$\displaystyle \paren {\int_a^b \paren {\map f t - \map g t}^2 \rd t}^{\frac 1 2} + \paren {\int_a^b \paren {\map g t - \map h t}^2 \rd t}^{\frac 1 2} \ge \paren {\int_a^b \paren {\map f t - \map h t}^2 \rd t}^{\frac 1 2}$ We have: {{begin-eqn}} {{eqn | l = \map {d_2} {f, g} + \map {d_2} {g, h}       | r = \paren {\int_a^b \paren {\map f t - \map g t}^2 \rd t}^{\frac 1 2} + \paren {\int_a^b \paren {\map g t - \map h t}^2 \rd t}^{\frac 1 2}       | c = Definition of $d_2$ }} {{eqn | r = \paren {\int_a^b \paren {\map r t}^2 \rd t}^{\frac 1 2} + \paren {\int_a^b \paren {\map s t}^2 \rd t}^{\frac 1 2}       | c =  }} {{eqn | o = \ge       | r = \paren {\int_a^b \paren {\map r t + \map s t}^2 \rd t}^{\frac 1 2}       | c = Minkowski's Inequality for Integrals }} {{eqn | r = \paren {\int_a^b \paren {\map f t - \map g t + \map g t - \map h t}^2 \rd t}^{\frac 1 2}       | c = Definition of $\map r t$ and $\map s t$ }} {{eqn | r = \paren {\int_a^b \paren {\map f t - \map h t}^2 \rd t}^{\frac 1 2}       | c =  }} {{eqn | r = \map {d_2} {f, h}       | c = Definition of $d_2$ }} {{end-eqn}} So axiom $M2$ holds for $d_2$. {{qed|lemma}} === Proof of $M3$ === {{begin-eqn}} {{eqn | l = \map {d_2} {f, g}       | r = \paren {\int_a^b \paren {\map f t - \map g t}^2 \rd t}^{\frac 1 2}       | c = Definition of $d_p$ }} {{eqn | r = \paren {\int_a^b \paren {\map g t - \map f t}^2 \rd t}^{\frac 1 2}       | c = {{Defof|Absolute Value}} }} {{eqn | r = \map {d_2} {g, f}       | c = Definition of $d_2$ }} {{end-eqn}} So axiom $M3$ holds for $d_2$. {{qed|lemma}} === Proof of $M4$ === {{begin-eqn}} {{eqn | lo= \forall t \in \closedint a b:       | l = \map f t - \map g t       | o = \ne       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \paren {\map f t - \map g t}^2       | o = \ge       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \paren {\int_a^b \paren {\map f t - \map g t}^2 \rd t}^{\frac 1 2}       | o = \ge       | r = 0       | c = Sign of Function Matches Sign of Definite Integral }} {{end-eqn}} From Zero Definite Integral of Nowhere Negative Function implies Zero Function we have that: :$\map {d_2} {f, g} = 0 \implies f = g$ on $\closedint a b$. So axiom $M4$ holds for $d_2$. {{qed}}
Let $a \in X$ such that $\exists K \in \R: \forall x \in Y: d \left({x, a}\right) \le K$. Let $a' \in X$. {{begin-eqn}} {{eqn | l =  \left({x, a'}\right)       | o = \le       | r = d \left({x, a}\right) + d \left({a, a'}\right)       | c = Metric Space Axiom $M2$ }} {{eqn | o = \le       | r = K + d \left({a, a'}\right)       | c = by hypothesis }} {{eqn | r = K'       | c = where $K' = K + d \left({a, a'}\right)$ }} {{end-eqn}} {{qed}}
Let $M = \left({X, d}\right)$ be a metric space. Let $M' = \left({Y, d_Y}\right)$ be a subspace of $M$. === Definition 1 implies Definition 2 === Let $M'$ be bounded according to Definition 1: :$\exists a \in X, K \in \R: \forall x \in Y: d \left({x, a}\right) \le K$ Let $x, y \in Y$. Then: {{begin-eqn}} {{eqn | l =  \left({x, y}\right)       | o = \le       | r = d \left({x, a}\right) + d \left({a, y}\right)       | c = Metric Space Axiom $M2$ }} {{eqn | r = d \left({x, a}\right) + d \left({y, a}\right)       | c = Metric Space Axiom $M3$ }} {{eqn | o = \le       | r = K + K       | c = by hypothesis }} {{eqn | r = 2 K       | c = where $K$ }} {{end-eqn}} {{qed}} Thus: : $\left({x, y}\right) \le 2 K$ Thus, setting $r = 2 K$, $M'$ fulfils the conditions to be bounded according to Definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $M'$ be bounded according to Definition 2: :$\exists K \in \R: \forall x, y \in M': d \left({x, y}\right) \le K$ Let $a = y$. Then: : $d \left({x, a}\right) \le K$ and so: :$\exists a \in Y, K \in \R: \forall x \in Y: d \left({x, a}\right) \le K$ As $X \subseteq Y$ it follows by definition of subset that: :$\exists a \in Y \implies \exists a \in X$ and so: :$\exists a \in X, K \in \R: \forall x \in Y: d \left({x, a}\right) \le K$ Thus $M'$ fulfils the conditions to be bounded according to Definition 1. {{qed}} Category:Bounded Metric Spaces ticeeeyy2rxsusx2yh4vzgzsze8mber
Let $n \in \N$. Then: :$\ds \sum_{i \mathop = 1}^n \paren {\lambda x_i + \mu y_i}^2 = \lambda^2 \sum_{i \mathop = 1}^n x_i^2 + \mu^2 \sum_{i \mathop = 1}^n y_i^2 + 2 \lambda \mu \sum_{i \mathop = 1}^n x_i y_i$ By Cauchy's Inequality: :$\ds \sum_{i \mathop = 1}^n x_i y_i \le \paren {\sum_{i \mathop = 1}^n x_i^2}^{\frac 1 2} \paren {\sum_{i \mathop = 1}^n y_i^2}^{\frac 1 2}$ Hence: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^n \paren {\lambda x_i + \mu y_i}^2       | o = \le       | r = \lambda^2 \sum_{i \mathop = 1}^n x_i^2 + \mu^2 \sum_{i \mathop = 1}^n y_i^2 + 2 \size {\lambda \mu} \sum_{i \mathop = 1}^n x_i y_i       | c =  }} {{eqn | o = \le       | r = \lambda^2 \sum_{i \mathop = 1}^n x_i^2 + \mu^2 \sum_{i \mathop = 1}^n y_i^2 + 2 \size {\lambda \mu} \paren {\sum_{i \mathop = 1}^n x_i^2}^{\frac 1 2} \paren {\sum_{i \mathop = 1}^n y_i^2}^{\frac 1 2} }} {{eqn | r = \paren {\size \lambda \paren {\sum_{i \mathop = 1}^n x_i^2}^{\frac 1 2} + \size \mu \paren {\sum_{i \mathop = 1}^n y_i^2}^{\frac 1 2} }^2       | c =  }} {{eqn | o = \le       | r = \paren {\size \lambda \paren {\sum_{i \mathop \ge 0} x_i^2}^{\frac 1 2} + \size \mu \paren {\sum_{i \mathop \ge 0} y_i^2}^{\frac 1 2} }^2       | c =  }} {{end-eqn}} Thus the sequence of partial sums $\ds \sum_{i \mathop = 1}^n \paren {\lambda x_i + \mu y_i}^2$ is bounded above. We also have that $\ds \sum_{i \mathop = 1}^n \paren {\lambda x_i + \mu y_i}^2$ is also increasing. So by the Monotone Convergence Theorem, $\ds \sum_{i \mathop = 1}^n \paren {\lambda x_i + \mu y_i}^2$ is convergent. {{qed}}
$\ell^2$ is a particular instance of the general $p$-sequence space $\ell^p$. Hence $p$-Sequence Space of Real Sequences is Metric Space can be applied directly. {{qed}}
By definition of the Hilbert sequence space on $\R$: Let $A$ be the set of all real sequences $\left\langle{x_i}\right\rangle$ such that the series $\displaystyle \sum_{i \mathop \ge 0} x_i^2$ is convergent. Then $\ell^2 := \left({A, d_2}\right)$ where $d_2: A \times A: \to \R$ is the real-valued function defined as: :$\displaystyle \forall x = \left\langle{x_i}\right\rangle, y = \left\langle{y_i}\right\rangle \in A: d_2 \left({x, y}\right) := \left({\sum_{k \mathop \ge 0} \left({x_k - y_k}\right)^2}\right)^{\frac 1 2}$ From Convergence of Square of Linear Combination of Sequences whose Squares Converge we have that $\displaystyle \sum_{k \mathop \ge 0} \left({x_k - y_k}\right)^2$ does actually converge. === Proof of $M1$ === {{begin-eqn}} {{eqn | l = d_2 \left({x, x}\right)       | r = \left({\sum_{k \mathop \ge 0} \left({x_k - x_k}\right)^2}\right)^{\frac 1 2}       | c = Definition of $d_2$ }} {{eqn | r = \left({\sum_{k \mathop \ge 0} 0^2}\right)^{\frac 1 2}       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} So axiom $M1$ holds for $d_2$. {{qed|lemma}} === Proof of $M2$ === Let $z = \left\langle{z_i}\right\rangle \in A$. {{MissingLinks|Minkowski's inequality currently only applies to finite sums}} {{begin-eqn}} {{eqn | l = d_2 \left({x, y}\right) + d_2 \left({y, z}\right)       | r = \left({\sum_{k \mathop \ge 0} \left({x_k - y_k}\right)^2}\right)^{\frac 1 2} + \left({\sum_{k \mathop \ge 0} \left({y_k - z_k}\right)^2}\right)^{\frac 1 2}       | c = Definition of $d_2$ }} {{eqn | o = \ge       | r = \left({\sum_{k \mathop \ge 0} \left({x_k - z_k}\right)^2}\right)^{\frac 1 2}       | c = Minkowski's Inequality for Sums: index $2$ }} {{eqn | r = d_2 \left({x, z}\right)       | c = Definition of $d_2$ }} {{end-eqn}} So axiom $M2$ holds for $d_2$. {{qed|lemma}} === Proof of $M3$ === {{begin-eqn}} {{eqn | l = d_2 \left({x, y}\right)       | r = \left({\sum_{k \mathop \ge 0} \left({x_k - y_k}\right)^2}\right)^{\frac 1 2}       | c = Definition of $d_2$ }} {{eqn | r = \left({\sum_{k \mathop \ge 0} \left({y_k - x_k}\right)^2}\right)^{\frac 1 2}       | c = Definition of Absolute Value }} {{eqn | r = d_2 \left({y, x}\right)       | c = Definition of $d_2$ }} {{end-eqn}} So axiom $M3$ holds for $d_2$. {{qed|lemma}} === Proof of $M4$ === {{begin-eqn}} {{eqn | l = x       | o = \ne       | r = y       | c =  }} {{eqn | ll= \implies       | lo= \exists k \in \N:       | l = x_k       | o = \ne       | r = y_k       | c =  }} {{eqn | ll= \implies       | l = \left({x_k - y_k}\right)^2       | o = >       | r = 0       | c =  }} {{eqn | ll= \implies       | l = \left({\sum_{k \mathop \ge 0} \left({x_k - y_k}\right)^2}\right)^{\frac 1 2}       | o = >       | r = 0       | c =  }} {{eqn | ll= \implies       | l = d_2 \left({x, y}\right)       | o = >       | r = 0       | c = Definition of $d_2$ }} {{end-eqn}} So axiom $M4$ holds for $d_2$. {{qed}}
We have that the supremum metric on $A \times A$ is defined as: :$\displaystyle \forall f, g \in A: \map d {f, g} := \sup_{x \mathop \in X} \size {\map f x - \map g x}$ where $f$ and $g$ are bounded real-valued functions. So: :$\exists K, L \in \R: \size {\map f x} \le K, \size {\map g x} \le L$ for all $x \in X$. First note that we have: {{begin-eqn}} {{eqn | l = \size {\map f x - \map g x}       | r = \size {\map f x + \paren {-\map g x} }       | c =  }} {{eqn | o = \le       | r = \size {\map f x} + \size {\paren {-\map g x} }       | c = Triangle Inequality for Real Numbers }} {{eqn | r = \size {\map f x} + \size {\map g x}       | c = {{Defof|Absolute Value}} }} {{eqn | o = \le       | r = K + L       | c =  }} {{end-eqn}} and so the {{RHS}} exists. === Proof of $M1$ === {{begin-eqn}} {{eqn | l = \map d {f, f}       | r = \sup_{x \mathop \in X} \size {\map f x - \map f x}       | c = Definition of $d$ }} {{eqn | r = \sup_{x \mathop \in X} \size 0       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} So axiom $M1$ holds for $d$. {{qed|lemma}} === Proof of $M2$ === Let $f, g, h \in A$. Let $c \in X$. {{begin-eqn}} {{eqn | l = c       | o = \in       | r = X       | c =  }} {{eqn | ll= \leadsto       | l = \size {\map f c - \map h c}       | o = \le       | r = \size {\map f c - \map g c} + \size {\map g c - \map h c}       | c = Triangle Inequality for Real Numbers }} {{eqn | o = \le       | r = \sup_{x \mathop \in X} \size {\map f c - \map g c} + \sup_{x \mathop \in X} \size {\map g c - \map h c}       | c = {{Defof|Supremum of Real-Valued Function}} }} {{eqn | r = \map d {f, g} + \map d {g, h}       | c = Definition of $d$ }} {{end-eqn}} Thus $\map d {f, g} + \map d {g, h}$ is an upper bound for: :$S := \set {\size {\map f c - \map h c}: c \in X}$ So: :$d \left({f, g}\right) + d \left({g, h}\right) \ge \sup S = d \left({f, h}\right)$ So axiom $M2$ holds for $d$. {{qed|lemma}} === Proof of $M3$ === {{begin-eqn}} {{eqn | l = \map d {f, g}       | r = \sup_{x \mathop \in X} \size {\map f x - \map g x}       | c = Definition of $d$ }} {{eqn | r = \sup_{x \mathop \in X} \size {\map g x - \map f x}       | c = {{Defof|Absolute Value}} }} {{eqn | r = \map d {g, f}       | c = Definition of $d$ }} {{end-eqn}} So axiom $M3$ holds for $d$. {{qed|lemma}} === Proof of $M4$ === As $d$ is the supremum of the absolute value of the image of the pointwise sum of $f$ and $g$: :$\forall f, g \in A: \map d {f, g} \ge 0$ Suppose $f, g \in A: \map d {f, g} = 0$. Then: {{begin-eqn}} {{eqn | l = \map d {f, g}       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \sup_{x \mathop \in X} \size {\map f x - \map g x}       | r = 0       | c = Definition of $d$ }} {{eqn | ll= \leadsto       | lo= \forall x \in X:       | l = \map f x        | r = \map g x       | c = {{Defof|Absolute Value}} }} {{eqn | ll= \leadsto       | l = f       | r = g       | c = Equality of Mappings }} {{end-eqn}} So axiom $M4$ holds for $d$. {{qed}}
We have that the supremum metric on $A \times A$ is defined as: :$\displaystyle \forall f, g \in A: d \left({f, g}\right) := \sup_{x \mathop \in X} \left\vert{f \left({x}\right) - g \left({x}\right)}\right\vert$ where $f$ and $g$ are bounded real-valued functions. From Real Number Line is Metric Space, the real numbers $\R$ together with the absolute value function form a metric space. The result follows by Supremum Metric is Metric. {{qed}}
By definition, a real sequence is a mapping from the natural numbers $\N$ to the real numbers $\R$. Thus a bounded real sequence is a bounded real-valued function. The result follows from Supremum Metric on Bounded Real-Valued Functions is Metric. {{qed}}
The set $A$ is a subset of the set $A'$ of all bounded mappings $f: M_1 \to M_2$. Let $d': A' \times A' \to \R$ be the supremum metric on $A'$. From Supremum Metric is Metric, $\left({A', d'}\right)$ is a metric space.  By definition, $A$ is a metric subspace of $A'$. Hence the result. {{qed}}
As $H_0 = 0$ and $H_1 = 1$, they are integers. The claim is that $H_n$ is not an integer for all $n \ge 2$. {{AimForCont}} otherwise: :$(\text P): \quad \exists m \in \N: H_m \in \Z$ By the definition of the harmonic numbers: :$H_m = 1 + \dfrac 1 2 + \dfrac 1 3 + \cdots + \dfrac 1 m$ Let $2^t$ denote the highest power of two in the denominators of the summands. Then: {{begin-eqn}} {{eqn | l = H_m       | r = 1 + \frac 1 2 + \frac 1 3 + \cdots + \frac 1 m       | c =  }} {{eqn | n = 1       | ll= \leadsto       | l = H_m - \frac 1 {2^t}       | r = 1 + \frac 1 2 + \frac 1 3 + \cdots + \frac 1 {2^t - 1} + \frac 1 {2^t + 1} + \cdots + \frac 1 m       | c =  }} {{eqn | ll= \leadsto       | l = 2^{t - 1} H_m - \frac 1 2       | r = 2^{t - 1} + \frac {2^{t - 1} } 2 + \frac {2^{t - 1} } 3 + \frac {2^{t - 1} } 4 + \frac {2^{t - 1} } 5 + \frac {2^{t - 1} } 6 + \cdots + \frac {2^{t - 1} } m       | c = multiplying by $2^{t-1}$ }} {{eqn | n = 2       | r = 2^{t - 1} + 2^{t - 2} + \frac {2^{t - 1} } 3 + 2^{t - 3} + \frac {2^{t - 1} } 5 + \frac {2^{t - 2} } 3 + \cdots + \frac {2^{t - 1} } m       | c = cancelling powers of $2$ }} {{end-eqn}} Let $S$ be the set of denominators on the {{RHS}} of $(2)$. Then no element of $S$ can have $2$ as a factor, as follows. Consider an arbitrary summand: :$\dfrac {2^{t - 1} } {2^j \times k}$ for some $k \in \Z$, where $j \ge 0$ is the highest power of $2$ that divides the denominator. For any $2$ to remain after simplification, we would need $j > t - 1$. Were this to be so, then $2^j \times k$ would have $2^t$ as a factor, and some denominator would be a multiple of $2^t$. By Greatest Power of Two not Divisor, the set of denominators of the {{RHS}} of $(1)$: :$\set {1, 2, 3, \ldots, 2^t - 1, 2^t + 1, \ldots, m}$ contains no multiple of $2^t$. Therefore there can be no multiple of $2$ in the denominators of the {{RHS}} of $(2)$. Let: :$\ell = \map \lcm S$ be the lowest common multiple of the elements of $S$. Because $2$ is not a divisor of any of the elements of $S$, it will not be a divisor of $\ell$. Hence $\ell$ is likewise odd. We have: {{begin-eqn}} {{eqn | l = 2^{t - 1} H_m - \frac 1 2       | r = 2^{t - 1} + 2^{t - 2} + \frac {2^{t - 1} } 3 + 2^{t - 3} + \frac {2^{t - 1} } 5 + \frac {2^{t - 2} } 3 + \ldots + \frac {2^{t - 1} } m       | c = from $(2)$ }} {{eqn | ll= \leadsto       | l = \frac {2^t H_m - 1} 2       | r = \frac {2^{t - 1} \ell + 2^{t - 2} \ell + 2^{t - 1} \paren {\ell / 3} + 2^{t - 3} \ell + 2^{t - 1} \paren {\ell / 5} + \ldots + 2^{t - 1} \paren {\ell / m} } \ell       | c = multiplying top and bottom by $\ell$ }} {{eqn | ll= \leadsto       | l = \ell \paren {2^t H_m - 1}       | r = 2 \paren {2^{t - 1} \ell + 2^{t - 2} \ell + 2^{t - 1} \paren {\ell / 3} + 2^{t - 3} \ell + 2^{t - 1} \paren {\ell / 5} + \ldots + 2^{t - 1} \paren {\ell / m} }       | c = multiplying both sides by $2 \ell$ }} {{end-eqn}} But the {{LHS}} of that last equation is odd, while its {{RHS}} is even. As this is a contradiction, it follows that our assumption $(\text P)$ that such an $m$ exists is false. That is, there are no harmonic numbers apart from $0$ and $1$ which are integers. {{qed}}
{{AimForCont}}: :$(\text P): \quad \exists m \in \N: H_m \in \Z$ By the definition of the harmonic numbers: :$1 + \dfrac 1 2 + \dfrac 1 3 + \cdots +\dfrac 1 m = H_m$ $m$ is either prime or composite. If $m$ is prime, we have that: {{begin-eqn}} {{eqn | l = 1 + \frac 1 2 + \frac 1 3 + \dots + \frac 1 m       | r = H_m }} {{eqn | l = m! + \frac {m!} 2 + \frac {m!} 3 + \cdots + \paren {m - 1}!       | r = m!\cdot H_m       | c = multiplying by $m!$ }} {{end-eqn}} If $k \le n$, then $k$ divides $n!$, so all terms are integers. Every term on the {{LHS}} is divisible by $m$ except for one, namely $\paren {m - 1}!$. Hence, the {{LHS}} is not divisible by $m$. But the {{RHS}} is divisible by $m$, so this is a contradiction. If $m$ is composite, let $p$ be the largest prime which is less than $m$. Then we have: {{begin-eqn}} {{eqn | l = 1 + \frac 1 2 + \frac 1 3 + \dots + \frac 1 p + \cdots + \frac 1 m       | r = H_m }} {{eqn | l = m! + \frac {m!} 2 + \frac {m!} 3 + \cdots \frac {m!} p + \cdots + \paren {m - 1}!       | r = m! \cdot H_m       | c = multiplying by $m!$ }} {{end-eqn}} For $k < p$, we have that $m!$ is divisible by $p$ and $k$ is not. So from Euclid's Lemma, every term $\dfrac {m!} k$ is an integer divisible by $p$. {{AimForCont}} there exists $k \in \Z$, with $p< k< m$ such that $\dfrac {m!} k$ is not divisible by $p$. Because $m!$ is divisible by $p$ and $\dfrac{m!} k$ is not, it follows from Euclid's Lemma that $k$ is a multiple of $p$. Since $k$ is a multiple of $p$ which is greater than $p$: :$2 p \le k$ Thus: :$p < 2 p \le k < m$ From Bertrand's Postulate, there exists a prime $q$ such that $p < q < 2 p$. Hence: :$p < q <m $ This contradicts the fact that $p$ is the largest prime less than $m$. Hence, the assumption that there exists $k \in \Z$ with $p < k < m$ such that $\dfrac {m!} k$ is not divisible by $p$ is false. Therefore, every term of the {{LHS}}, except perhaps $\dfrac{m!} p$, is a multiple of $p$. From above: :$p < m < 2 p$ so $p$ is the only positive integer less than $m$ which is divisible by $p$. Since $m!$ is the product of positive integer less than or equal to $m$, it follows that $m!$ is divisible by $p$ only once. Therefore, $\dfrac {m!} p$ is not divisible by $p$. Because every term of the {{LHS}} is divisible by $p$ except for one, it follows that the {{LHS}} is not divisible by $p$. Because the {{RHS}} is a multiple of $m!$, and hence a multiple of $p$, this is a contradiction. So the assumption that there exists $m \in \N: H_m \in \Z$ is false. {{qed}}
Let $\mathscr B \left({\left[{a \,.\,.\, b}\right], \R}\right)$ be the set of all bounded real functions $f: \left[{a \,.\,.\, b}\right] \to \R$. From Supremum Metric on Continuous Real Functions is Subspace of Bounded, $\left({\mathscr C \left[{a \,.\,.\, b}\right], d_{\mathscr C} }\right)$ is a (metric) subspace of $\left({\mathscr B \left({\left[{a \,.\,.\, b}\right], \R}\right), d}\right)$. The result follows from Subspace of Metric Space is Metric Space. {{qed}}
Let $A := \mathscr D^r \left[{a \,.\,.\, b}\right]$ be the set of all continuous functions $f: \left[{a \,.\,.\, b}\right] \to \R$ which are of differentiability class $r$. Let $d_r: A \times A \to \R$ be the supremum metric on $A$, defined as: :$\displaystyle \forall f, g \in A: d \left({f, g}\right) := \sup_{\substack {x \mathop \in \left[{a \,.\,.\, b}\right] \\ i \mathop \in \left\{ {0, 1, 2, \ldots, r}\right\} } } \left\vert{f^{\left({i}\right)} \left({x}\right) - g^{\left({i}\right)} \left({x}\right)}\right\vert$ where: : $f$ and $g$ are continuous functions on $\left[{a \,.\,.\, b}\right]$ which are $r$ times differentiable : $r \in \N$ is a natural number. From Supremum Metric on Differentiability Class is Metric, $d_r$ is a metric on $A$. By definition, real functions which are continuous on $\left[{a \,.\,.\, b}\right]$ are of differentiability class $0$. Thus setting $r = 0$ it follows that $\mathscr C \left[{a \,.\,.\, b}\right]$ is $\mathscr D^0 \left[{a \,.\,.\, b}\right]$. The result follows. {{qed}}
By definition of restriction: :$\forall x, y \in H: d_H \left({x, y}\right) = d \left({x, y}\right)$ As $d$ is a metric, the metric space axioms are all fulfilled by all $x, y \in A$ under $d$. As $H \subseteq A$, by definition of subset, all $x, y \in H$ are also elements of $A$. Therefore the metric space axioms are all fulfilled by all $x, y \in H$ under $d_H$. {{qed}}
By definition of the $p$-sequence metric on $\R$: Let $A$ be the set of all real sequences $\left\langle{x_i}\right\rangle$ such that the series $\displaystyle \sum_{i \mathop \ge 0} x_i^2$ is convergent. Then $\ell^p := \left({A, d_2}\right)$ where $d_p: A \times A: \to \R$ is the real-valued function defined as: :$\displaystyle \forall x = \left\langle{x_i}\right\rangle, y = \left\langle{y_i}\right\rangle \in A: d_p \left({x, y}\right) := \left({\sum_{k \mathop \ge 0} \left\vert{x_k - y_k}\right\vert^p}\right)^{\frac 1 p}$ === Proof of $M1$ === {{begin-eqn}} {{eqn | l = d_p \left({x, x}\right)       | r = \left({\sum_{k \mathop \ge 0} \left\vert{x_k - x_k}\right\vert^p}\right)^{\frac 1 p}       | c = Definition of $d_p$ }} {{eqn | r = \left({\sum_{k \mathop \ge 0} 0^p}\right)^{\frac 1 p}       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} So axiom $M1$ holds for $d_p$. {{qed|lemma}} === Proof of $M2$ === Let $z = \left\langle{z_i}\right\rangle \in A$. {{MissingLinks|Minkowski's inequality currently only applies to finite sums}} {{begin-eqn}} {{eqn | l = d_p \left({x, y}\right) + d_p \left({y, z}\right)       | r = \left({\sum_{k \mathop \ge 0} \left\vert{x_k - y_k}\right\vert^p}\right)^{\frac 1 p} + \left({\sum_{k \mathop \ge 0} \left\vert{y_k - z_k}\right\vert^p}\right)^{\frac 1 p}       | c = Definition of $d_p$ }} {{eqn | o = \ge       | r = \left({\sum_{k \mathop \ge 0} \left\vert{x_k - z_k}\right\vert^p}\right)^{\frac 1 p}       | c = Minkowski's Inequality for Sums }} {{eqn | r = d_p \left({x, z}\right)       | c = Definition of $d_p$ }} {{end-eqn}} So axiom $M2$ holds for $d_p$. {{qed|lemma}} === Proof of $M3$ === {{begin-eqn}} {{eqn | l = d_p \left({x, y}\right)       | r = \left({\sum_{k \mathop \ge 0} \left\vert{x_k - y_k}\right\vert^p}\right)^{\frac 1 p}       | c = Definition of $d_2$ }} {{eqn | r = \left({\sum_{k \mathop \ge 0} \left\vert{y_k - x_k}\right\vert^p}\right)^{\frac 1 p}       | c = {{Defof|Absolute Value}} }} {{eqn | r = d_p \left({y, x}\right)       | c = Definition of $d_p$ }} {{end-eqn}} So axiom $M3$ holds for $d_p$. {{qed|lemma}} === Proof of $M4$ === {{begin-eqn}} {{eqn | l = x       | o = \ne       | r = y       | c =  }} {{eqn | ll= \implies       | lo= \exists k \in \N:       | l = x_k       | o = \ne       | r = y_k       | c =  }} {{eqn | ll= \implies       | l = \left\vert{x_k - y_k}\right\vert^p       | o = >       | r = 0       | c =  }} {{eqn | ll= \implies       | l = \left({\sum_{k \mathop \ge 0} \left\vert{x_k - y_k}\right\vert^p}\right)^{\frac 1 p}       | o = >       | r = 0       | c =  }} {{eqn | ll= \implies       | l = d_p \left({x, y}\right)       | o = >       | r = 0       | c = Definition of $d_p$ }} {{end-eqn}} So axiom $M4$ holds for $d_p$. {{qed}}