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After having proven that:

• every Cauchy sequence is bounded
• that for any positive real numbers $x$ and $y$ there is some natural number $k\in {\mathbb N}$ such that $kx > y$
• that for any positive real number $x$ there is some rational number $b\in {\mathbb Q}$ such that $0 < b < x$
• and the theorem that every Cauchy sequence of rational numbers $\{a_n\}_{n\in {\mathbb N}}$ is convergent in $\mathbb R$ with the limit $\lim_{n \rightarrow \infty} a_n = [(a_n)]$ where $[a_n]$ has been defined as the equivalence class of equivalent Cauchy sequences in the set of rational numbers i.e. $[(a_n)] = \{{b_n} \in A│{{b_n}} \sim {{a_n}}\}$, where $A$ is the set of all Cauchy sequences in $\mathbb Q$ and $a_n \in A$,

I am trying to next prove as a corollary that for a real number $[(a_n)]$ and a positive real number $\epsilon$ , there is a rational number $a$ such that $|[(a_n)] - a|<\epsilon$. Equivalent sequences are defined by $[{{a_n}}\sim {{b_n}}] \equiv [\lim(a_n - b_n) = 0]$

After having proven that:

• every Cauchy sequence is bounded
• that for any positive real numbers $x$ and $y$ there is some natural number $k\in {\mathbb N}$ such that $kx > y$
• that for any positive real number $x$ there is some rational number $b\in {\mathbb Q}$ such that $0 < b < x$
• and the theorem that every Cauchy sequence of rational numbers $\{a_n\}_{n\in {\mathbb N}}$ is convergent in $\mathbb R$ with the limit $\lim_{n \rightarrow \infty} a_n = [(a_n)]$ where $[a_n]$ has been defined as the equivalence class of equivalent Cauchy sequences in the set of rational numbers i.e. $[(a_n)] = \{{b_n} \in A│{{b_n}} \sim {{a_n}}\}$, where $A$ is the set of all Cauchy sequences in $\mathbb Q$ and $a_n \in A$,

I am trying to next prove as a corollary that for a real number $[(a_n)]$ and a positive real number $\epsilon$ , there is a rational number $a$ such that $|[(a_n)] - a|<\epsilon$. Equivalent sequences are defined by $[{{a_n}}\sim {{b_n}}] \equiv [\lim(a_n - b_n) = 0]$ How do I prove this corollary from the previous propositions given?

After having proven that:

• every Cauchy sequence is bounded
• that for any positive real numbers $x$ and $y$ there is some natural number $k\in {\mathbb N}$ such that $kx > y$
• that for any positive real number $x$ there is some rational number $b\in {\mathbb Q}$ such that $0 < b < x$
• and the theorem that every Cauchy sequence of rational numbers $\{a_n\}_{n\in {\mathbb N}}$ is convergent in $\mathbb R$ with the limit $\lim_{n \rightarrow \infty} a_n = [a_n]$ where $[a_n]$ has been defined as the equivalence class of equivalent Cauchy sequences in the set of rational numbers i.e. $[a_n] = \{{b_n} \in A│{b_n} \sim {a_n}\}$, where $A$ is the set of all Cauchy sequences in $\mathbb Q$ and $a_n \in A$,

I am trying to next prove as a corollary that for a real number $[a_n]$ and a positive real number $\epsilon$ , there is a rational number $a$ such that $|[a_n] - a|<\epsilon$. Equivalent sequences are defined by $[{a_n}\sim {b_n}] \equiv [\lim(a_n - b_n) = 0]$

After having proven that:

• every Cauchy sequence is bounded
• that for any positive real numbers $x$ and $y$ there is some natural number $k\in {\mathbb N}$ such that $kx > y$
• that for any positive real number $x$ there is some rational number $b\in {\mathbb Q}$ such that $0 < b < x$
• and the theorem that every Cauchy sequence of rational numbers $\{a_n\}_{n\in {\mathbb N}}$ is convergent in $\mathbb R$ with the limit $\lim_{n \rightarrow \infty} a_n = [(a_n)]$ where $[a_n]$ has been defined as the equivalence class of equivalent Cauchy sequences in the set of rational numbers i.e. $[(a_n)] = \{{b_n} \in A│{{b_n}} \sim {{a_n}}\}$, where $A$ is the set of all Cauchy sequences in $\mathbb Q$ and $a_n \in A$,

I am trying to next prove as a corollary that for a real number $[(a_n)]$ and a positive real number $\epsilon$ , there is a rational number $a$ such that $|[(a_n)] - a|<\epsilon$. Equivalent sequences are defined by $[{{a_n}}\sim {{b_n}}] \equiv [\lim(a_n - b_n) = 0]$

After having proven that every Cauchy sequence is bounded, that for any positive real numbers x and y there is some natural number k such that kx > y, that for any positive real number x there is some rational number b such that 0 < b < x, and the theorem that every Cauchy sequence of rational numbers {an} is convergent in R with the limit lim an = [(an)] where [(an)] has been defined as the equivalence class of equivalent Cauchy sequences in the set of rational numbers i.e. [(an)] = {{bn}∈ A│{bn}~{an}} where A is the set of all Cauchy sequences in Q and {an}∈ A. I am trying to next prove as a corollary that for a real number [(an)] and a positive real number ϵ , there is a rational number a such that |[(an)]- a|<ϵ. Equivalent sequences are defined by [{an}~{bn}] ≡ [lim(an - bn) = 0]

After having proven that:

• every Cauchy sequence is bounded
• that for any positive real numbers $x$ and $y$ there is some natural number $k\in {\mathbb N}$ such that $kx > y$
• that for any positive real number $x$ there is some rational number $b\in {\mathbb Q}$ such that $0 < b < x$
• and the theorem that every Cauchy sequence of rational numbers $\{a_n\}_{n\in {\mathbb N}}$ is convergent in $\mathbb R$ with the limit $\lim_{n \rightarrow \infty} a_n = [a_n]$ where $[a_n]$ has been defined as the equivalence class of equivalent Cauchy sequences in the set of rational numbers i.e. $[a_n] = \{{b_n} \in A│{b_n} \sim {a_n}\}$, where $A$ is the set of all Cauchy sequences in $\mathbb Q$ and $a_n \in A$,

I am trying to next prove as a corollary that for a real number $[a_n]$ and a positive real number $\epsilon$ , there is a rational number $a$ such that $|[a_n] - a|<\epsilon$. Equivalent sequences are defined by $[{a_n}\sim {b_n}] \equiv [\lim(a_n - b_n) = 0]$