What do you get with this equivalence relationship for all $\mathbb{Q}$ sequences

Consider all $\mathbb{Q}$ Cauchy sequences with this equivalence relationship

$\{x_n\} \sim \{y_n\} \iff \{x_n-y_n\} \rightarrow 0$

Then you get all real numbers as an equivalence class with this relationship.

Let's consider now all $\mathbb{Q}$ sequences with this equivalence relationship. We get a set which contains all real numbers and many more.

What can we say about those new elements? Does that set represent some known set? Can we get some interesting topological or analytic results about it?

Solution 1:

I've spent all night working on this, and figured some good stuff out, as well as some not so good. And I got stuck on completeness. It's a long post.

I'll start with the bad. The reasonable definition of multiplication doesn't work... that is $[x_n]\cdot [y_n] = [x_n y_n]$. Consider the sequence $y_n = n$ and $y'_n = n+1/n$. Notice $\{y_n\} \sim \{y'_n\}$. Let $x_n = n^2$. We want $\{x_n y_n\} \sim \{x_n y'_n\}$. But we sadly compute $$ |x_n y_n - x_n y'_n| = |n^3-n^3 + n| = n $$ which decidedly does not converge to 0. However, if we assume the sequences are bounded, then this is a well defined operation, so that's nice.

Division doesn't work at all, as $[(0,1,0,1,...)] \neq 0$ but there's no way to divide by it. Having any sub sequence approaching $0$ is bad news.

Now for the good news. Addition and scalar multiplication does work, so we are in a $\mathbb{Q}$ vector space. And the real good news is that this space has a reasonable metric! Call our quotient space $X$.

Let $[x_n],[y_n] \in X$. Define $$ d([x_n],[y_n]) = \min\big(1,\limsup|x_n-y_n| \big) $$

We have a bunch of things to check. We prove this is well defined. Let $[y'_n]=[y_n]$ and $[x'_n]=[x_n]$. Then \begin{align*} \limsup|x'_n-y'_n| &=\limsup|(x'_n-x_n) + (x_n - y_n) +(y_n-y'_n)|\\ &\leq \limsup|x'_n-x_n| + \limsup|x_n - y_n| +\limsup|y_n-y'_n|\\ &= \limsup|x_n - y_n| \end{align*}

Then repeating the proof exchanging $x'_n$ with $x_n$ and $y'_n$ with $y_n$ we see that $\limsup|x_n-y_n|= \limsup|x'_n-y'_n|$. Applying the min with $1$ will complete the proof.

The rest of the proofs that $d$ is a metric are pretty straightforward, but they seem to work. It is slightly surprising that this does separate points, but $$ \limsup|x_n-y_n| = 0 \implies \lim|x_n-y_n| = 0 \implies [x_n]=[y_n] $$ I won't write down the rest of the proofs here. I'll leave them as exercises.

Now, we have a metric on a vector space. So it is complete? I got stuck trying to prove it. Does anyone know whether it's true or not? What about if we restrict ourselves to bounded sequences?

Now Looking at what this metric does, it seems clear that @Joel Cohen's comment is pretty accurate. Given $x \in X$, a small neighborhood around $x$ looks like all of the sequences which have the same behavior at infinity modulo a small constant. In fact, this space is badly path disconnected. Each behavior at infinity in in its own component. Suppose we had a path $\gamma:[0,1]\to X$. At each $\alpha \in [0,1]$, we can pick an $\varepsilon$ so $\gamma([\max(0,\alpha-\varepsilon),\min(\alpha+\varepsilon,1)]$ contains only elements of the same asymptotic behavior. Then compactness tells us we can choose finitely many such balls, and we see that $\gamma(0)$ has the same asymptotic behavior as $\gamma(1)$.