Proving that the map $f:\mathbb R \to \text{Seq}(\mathbb Q)/\sim$ is surjective

Question

I was reading about constructing Real numbers using Cauchy sequences of rational numbers. To be more specific, let $\text{Seq}(\mathbb Q)$ be the set of all Cauchy sequences of rational numbers and define the equivalence relation $ \sim$ on $\text{Seq}(\mathbb Q)$ as: $$(a_n)_n \sim(b_n)_n \iff \lim(a_n-b_n)=0$$ Then we define the real numbers as $\text{Seq}(\mathbb Q)/\sim$, the set of all equivalence classes.

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After reading about it, I tried finding a bijection between $\mathbb R$ and $\text{Seq}(\mathbb Q)/\sim$. I defined a function $f:\mathbb R \to \text{Seq}(\mathbb Q)/\sim$ as:

For each $a\in \mathbb R$, if $a$ has decimal expansion $r_1...r_k,a_1...a_n...$ then $$f(a)=[(r_1...r_k,a_1...a_n)_n]_\sim$$

So, for example, $f(\pi)$ is the equivalence class of the sequence: $3.1,\ 3.14,\ 3.141,\ 3.1415\dots$

Using some basic results from group theory I was easily able to prove that the function $f$ is injective, but I'm having a LOT of trouble proving that it is also surjective. I tried a direct proof and a proof by contradiction but got nowhere.

How can I prove that $f$ is indeed surjective?

Answer 1

If $(a_n)_{n\in\Bbb N}$ be a Cauchy sequence of rational numbers, and let $a=\lim_{n\in\infty}a_n$; it exists, since any Cauchy sequence of real numbers converges. Then $f(a)$ is the equivalence class of $(a_n)_{n\in\Bbb N}$.