I'd like a check about this exercise.
Let $\tau$ be the topology so defined $\tau=\bigl\{A \subset \mathbb{R}\,|\, (\forall x \in A \cap \mathbb{Z})( \exists \epsilon >0): (x-\epsilon,x+\epsilon) \subset A\bigr\}$
(i) Is $\tau$ finer or coarser than the euclidian topology?
(ii) Find $\operatorname{Int}[0,+\infty), \operatorname{Int}[0,1], \operatorname{Int}\mathbb{Z}$
(iii) Find $\overline{[0,1]}$
(iv) Is $\mathbb{R},\tau$ an Hausdorff space?
(v) Is $\mathbb{R}$ compact
(vi) Let $f\colon(\mathbb{R},\tau_e) \rightarrow (\mathbb{R},\tau_e)$ be a continuous function such that $f(\mathbb{Z}) \subset \mathbb{Z}$.
Show that $f\colon(\mathbb{R},\tau) \rightarrow (\mathbb{R},\tau)$ is continuous.
(i) $\tau_e$ is finer than $\tau$ because $\tau \subset \tau_e$. In fact, every open set in $\tau$ is open in the standard euclidian topology. I also have that $\tau_e \subset \tau$, so the two topologies are equivalent.
(ii)$\operatorname{Int}[0,1]=\emptyset$, since the only integers in this set are {$0,1$}, but if $0$ is included, then $(0-\epsilon,0+\epsilon) \notin [0,1]$
With the same argument, $\operatorname{Int}[0,+\infty)=(0,+\infty)$
$\operatorname{Int}\mathbb{Z}=\emptyset$ because every open set which contains integer points, automatically contains also real points.
(iii) $[0,1]$ is closed since $(\infty,0) \cup (1,+\infty)$ is an open set. So $\overline{[0,1]}=[0,1]$.
(iv)Yes, it's an Hausdorff space since for every different points $x,y \in \mathbb{R}$ I can find two open neighborhood $U,V$ such that $U \cap V = \emptyset$.
(v)It's not compact: if I take the open cover $\mathcal{R}= \cup_{i \in \mathbb{N}}A_i$, $A_i=(i,i+2)$, I can't take a finite subcover.
(vi) Since $f$ is continuous (with the euclidean topology), then for every open set $U \in \tau_e$: $f^{-1}(U) \in \tau_e \subset \tau$. So $f:(\mathbb{R},\tau) \rightarrow (\mathbb{R},\tau)$ is continuous.