Consider the family $A_n=\{x \in \mathbb{Z}| -n \leq x \leq n\}, n\in \mathbb{N}$
Let $\tau=\{A_n,\emptyset, \mathbb{Z}\}$ be the topology over $\mathbb{Z}$
Let $f:(\mathbb{N},\varepsilon) \rightarrow (\mathbb{Z},\tau): f(x)=2x $ be a mapping, where $\varepsilon$ is the induced euclidean topology
Say if it is continuous and open
I would like some feedback on my solution:
a)Continuity
The $A_n , n\in \mathbb{N}$ are a basis of the topology, so it is enough to verify that $f^{-1}(A_n) \in \varepsilon$. In fact: I can write $A_n=[-n,n]$
I am unsure as to what should the correct answer be depending if you have to intersect with the subspace to get the open sets of the subspace topology like I did in option 1, or if you just have to check that the preimage is the intersection of the subspace with an open set of the initial topology:
option 1 $f^{-1}(A_n)=f^{-1}([-n,n])=[-n/2,n/2] \cap \mathbb{N}=[1, \lfloor n/2\rfloor]$ which $\in \tau$ so it is continuous I guess this is the correct one
option 2
$f^{-1}(A_n)=f^{-1}([-n,n])=[-n/2,n/2] $ if n happens to be odd, I would have an interval wich endpoints are not integers and therefore it doesn't have the form $U\cap \mathbb{N}$ , with $U$ an open set of $\mathbb{R}$ then it is not continuous
b) Openness Let $A\in \varepsilon$, and $A=\mathbb{N}\cap B(p,r)=[a,b]$,where $B(p,r)$ is an open ball of $(\mathbb{R},\varepsilon)$, $a,b \in \mathbb{N}$, then $f([a,b])=[2a,2b]\in \tau$, with $2a, 2b \in \mathbb{Z}$ So it is open
Since every subset of $(\Bbb N,\varepsilon)$ is an open set, any map from $(\Bbb N,\varepsilon)$ into any topological space is continuous, because the reverse image of anything is an open set.
But $f$ is not open, since $\{1,2\}$ is an open subset of $(\Bbb N,\varepsilon)$, but $f\bigl(\{1,2\}\bigr)=\{2,4\}$, which is not an open subset of $(\Bbb Z,\tau)$.