Will a “continuous” Cauchy sequence construction generate the real numbers?

There’s two common ways to construct the real number system: Dedekind cuts and equivalence classes of Cauchy sequences. I’d like to try a different way.

Let $X$ be the set of all functions $f:Q\rightarrow Q$ such that for any positive rational number $\epsilon$ there exists a positive rational number $\delta$ such that for all rational numbers $x,y>\delta$, we have $|f(x)-f(y)|<\epsilon$. (Intuitively, it means that its limit as $x$ goes $\infty$ is a real number.) And let’s define an equivalence relation $\sim$ on $X$ by saying that $f\sim g$ if the limit of $f(x)-g(x)$ as $x\rightarrow\infty$ is $0$.

My question is, is the set of equivalence classes of elements of $X$ under $\sim$ isomorphic to the set of real numbers?

• One should not write $f$~$g$. Proper usage is $f\sim g.$ It's all between just one pair of dollar signs and all within MathJax. I edited accordingly. – Michael Hardy Aug 2 '18 at 2:13

Yes. Specifically, let $Y=X/{\sim}$ and define a bijection $\varphi:Y\to \mathbb{R}$ as follows. Given any $f\in X$, define $\varphi([f])$ to be the limit of $f(x)$ as $x\to \infty$. It is clear that this limit exists (for instance, take the limit of the Cauchy sequence $(f(n))$, and then the condition on $f$ implies that actually this limit is the limit of $f(x)$ where $x\to\infty$ ranges over all rationals) and that it is independent of the choice of $f$ in a given equivalence class. To see that $\varphi$ is an injection, just observe that if $f$ and $g$ have the same limit then clearly $f\sim g$. To see that $\varphi$ is a surjection, note that given any $r\in\mathbb{R}$ we can choose a sequence $(q_n)$ of rationals approaching $r$ and then define $f(x)=q_n$ where $n$ is the least nonnegative integer greater than $x$, and then $\varphi([f])=r$.