I was giving some thought to this answer and I was curious about the sentence
$Y$ is connected
How close is $Y$ is connected to a $\Sigma^0_1$, or $\Pi^0_1$ sentence?
For a little more context... Let a subset $X$ of the positive dyadic and ternary rationals $Y$ be closed if and only if for all $x∈X$, the set also contains $\dfrac{3x+2^{\nu_2(x)}}2$. Then unless I've made a topological error (which is sadly much more likely than I would like), if $Y$ is connected the Collatz conjecture is true. I was curious how close "Y is connected" is, to a "provably true, if true" statement (in your favourite base theory).
Noah writes
Note that checking whether a sentence is $\Sigma^0_1$, or $\Pi^0_1$, is decidable: it's a purely syntactic property. The issue is the "equivalent" bit.
But I wasn't able to associate a statement of the form I wrote above, with any of the cases in the Wikipedia article.
Connectedness is in general a very complicated property: to say that a space $\mathcal{S}$ is connected, we need to quantify over the open sets of that space. Generally a space has many more open sets than it does points; consider Cantor space, for example. So a priori the connectedness of even a "simply definable" topological space (such as the one in the OP) is a $\Pi^1_1$ property. This is galactically far from $\Pi^0_1$ or $\Sigma^0_1$ - it's beyond the entire arithmetical, or even hyperarithmetical, hierarchy.
A bit more precisely, conflating $Y$ and $\mathbb{N}$ via a simple bijection appropriately what you have here is a "monotone $\Sigma^0_1$" formula defining a closure operation which gives the closed sets in your topology: $x\in cl(X)$ iff there is some $y\in X$ such that [simple condition]. Note that every monotone $\Sigma^0_1$ formula (and we can go well beyond $\Sigma^0_1$ here, but never mind) gives rise to a closure operation which in turn generates a topology, so this perspective shift is appropriate. Unfortunately, the set of monotone $\Sigma^0_1$ closure conditions giving rise to connected topological spaces is in fact $\Pi^1_1$ complete (this isn't hard but it is a bit tedious), so in general you won't find a way to express connectedness better than $\Pi^1_1$.
However, particular examples - including that in the OP - may be much nicer. It's totally possible that there is some much simpler way to express connectedness in this case. But I don't see one at the moment.