What topologies on $\mathbb{Z}$ make the floor function continous?

Question

I've been wondering about the floor function $f: \mathbb{R}\to \mathbb{Z}$ that maps $x \mapsto [x]$, where $[x]$ is the greatest $n\in \mathbb{Z}$ such that $n \leq x$.

The function is discontinous at each integer, and I'm asking myself what topologies in $\mathbb{Z}$ make the floor function continous when $\mathbb{R}$ has the usual topology $\tau_u$, and furthermore, if $\tau$ is the final topology on $\mathbb{Z}$, what separation axioms does $(\mathbb{Z} , \tau)$ satisfy?

I'd have shown my work, but I'm having troubles in just picturing how such topology should be to make $[x]$ contiuous first so I've not been able to work much further. Thanks in advance!

Answer 1

Suppose that $\varnothing\ne A\subseteq\Bbb Z$, and suppose that $f^{-1}[A]$ is open in $\Bbb R$. Let $n\in A$; then $n-1$ must belong to $A$. Thus, $A$ must be an initial segment of $\Bbb Z$ (which can of course be $\Bbb Z$ itself). If $U_n=\{k\in\Bbb Z:k\le n\}$ for $n\in\Bbb Z$, then

$$\tau=\{\varnothing,\Bbb Z\}\cup\{U_n:n\in\Bbb Z\}\;.$$

This topology is $T_0$ but not $T_1$.

Any topology that is a subset of $\tau$ will of course also make $f$ continuous. No proper subset of $\tau$ is $T_0$, however.

Answer 2

Note that for each $n\in\mathbb Z$, $f^{-1}(n)=[n,n+1)$. Note that no such set is open in $\mathbb R$; moreover, no union of such sets is open unless it is empty, can be written as $(-\infty, n)$, or is $\mathbb R$ itself. Otherwise, it would contain a boundary point.

So the only candidates for nontrivial open sets are those of the form $L_n\equiv \{\ldots, n-2, n-1, n\}$. Any such collection, together with $\varnothing$ and $\mathbb Z$, forms a topology on $\mathbb Z$.