I have very little knowledge of set theory and proof writing. This is a problem from Introduction to Topology: Pure and Applied by Colin Adams and Robert Franzosa.
DEFINITION: Digital line topology
For each n ∈ ℤ, define $$ B(n) = \begin{cases} \{n\}, & \text{if $n$ is odd} \\ \{n-1,n,n+1\}, & \text{if $n$ is even} \end{cases} $$
THOUGHTS
(i) For each $n \in ℤ$, $n$ is either in $\{n\}$ or $\{n − 1, n, n + 1\}$.
(ii) Let there be odd numbers $a_n$ and even numbers $b_n$ in $ℤ$.
If $a = a,$ their intersection is ${a}$.
If $a_1 ≠ a_2$, their intersection is $Ø$.
If $b = b$, their intersection is $\{b – 1, b, b + 1\}$.
If $b_1 ≠ b_2$, their intersection is $Ø$.
If $a = b – 1$, then $B(a) ∩ B(b) = \{b – 1, b\}$, which would be contained in $\{n − 1, n, n + 1\}$.
If $a = b + 1$, then $B(a) ∩ B(b) = \{b, b + 1\}$ which would be contained in $\{n − 1, n, n + 1\}$.
If $a < b – 1$, then $\{a\} ∩ \{b\} = Ø$.
If $a > b + 1$, then $\{a\} ∩ \{b\} = Ø$.
Therefore, $\{n\}$ is open if and only if $n$ is odd.
I probably made that way more complicated than I had to, and I'm not even sure if it's right. In addition to that, I don't know how to proceed to answer the question. I know that closed sets are complements of open sets, but if I try to take the complement of a single-point odd set $\{n\}$, it's $ℤ - \{n\}$ which doesn't help me much. Anyway, as always, I appreciate any help.
If $n$ is even, say $n=2k$, then if $m \neq n$ we have two options: $m$ is even too and then $\{m-1,m,m+1\}$ is open and is disjoint from $\{n\}$, as $|n-m| \ge 2$ or $m$ is odd and $\{m\}$ is an open set disjoint from $\{n\}$ too. In both cases we see that $m \notin \overline{\{n\}}$ so $\{n\}$ is closed.
If $n$ is odd then every neighbourhood of $n+1$ contains the open set $B(n+1)= \{n,n+1,n+2\}$ so intersects $\{n\}$ and so $n+1 \in \overline{\{n\}}$ for $n$ odd, so $n$ odd implies $\{n\}$ not closed.
Combined with the previous gives the equivalence asked for.