Equip the integers with the topology such that each element is a set of three consecutive integers Define S = { {2k-1, 2k, 2k+1} | k $\in$ $\mathbb{Z}$ }
(a) Show that S is a subasis for a topology on $\mathbb{Z}$. Call the topological space generated by this subasis X.
I'm fairly certain I answered this correctly, but here's my working:
I show that the union of all subsets in the form of S is equal to $\mathbb{Z}$.
k = ..., -1, 0, 1, ...
... $\cup$ {2(-1)-1, 2(-1), 2(-1)+1} $\cup$ {2(0)-1, 2(0), 2(0)+1} $\cup$ {2(1)-1, 2(1), 2(1)+1} $\cup$ ...
= ... $\cup$ {-3, -2, -1} $\cup$ {-1, 0, 1} $\cup$ {1, 2, 3} $\cup$ ...
= { ..., -3, -2, -1, 0, 1, 2, 3, ... } = $\mathbb{Z}$
Hence S is a subasis for $\mathbb{Z}$
(b) For any k $\in$ $\mathbb{Z}$, calculate the closure of {k} in X.
I feel a little hazy about calculating closures. I think each set of the form {2k-1, 2k, 2k+1} is clopen, since for any k, {2k-1, 2k, 2k+1} is open in X and X - {2k-1, 2k, 2k+1} is the union of open sets of the form {2a-1, 2a, 2a+1}, where a is such that 2a + 1 < 2k - 1 or 2a - 1 > 2k + 1
The closure of {k} would depend on it's parity. If k is odd, then it is it's own closure because it is the intersection of two closed sets:
{ 2($\frac{k-1}{2}$)-1, 2($\frac{k-1}{2}$), 2($\frac{k-1}{2}$)+1 } $\cap$  { 2($\frac{k+1}{2}$)-1, 2($\frac{k+1}{2}$), 2($\frac{k+1}{2}$)+1 } = { k }
If k is even, then k is only contained in one set, and hence the closure of {k} will be the set that it is contained in:
{ 2($\frac{k}{2}$)-1, 2($\frac{k}{2}$), 2($\frac{k}{2}$)+1 }
Is this the right train of thought to approach this problem?
(c) Show that X fails this condition: For A and B disjoint sets that are both closed in X, there exist disjoint open sets U and V in X such that A $\subseteq$ U and B $\subseteq$ V.
I'm confused by this question. A does not have to be a proper subset of U and B does not have to be a proper subset of V according to the condition, so if we choose A = U and B = V, then the condition is satisfied. 
 A: The first part  can be done much more clearly and succinctly. Let $k\in\Bbb Z$. If $k$ is even, then $k\in\{k-1,k,k+1\}\in S$, and if $k$ is odd, then $k\in\{k,k+1,k+2\}\in S$, so $\Bbb Z\subseteq\bigcup S$. On the other hand, every member of $S$ is a subset of $\Bbb Z$, so $\bigcup S\subseteq\Bbb Z$, and therefore $\bigcup S=\Bbb Z$.
It’s not true that every member of $S$ is clopen, but it’s not hard to calculate $\operatorname{cl}\{k\}$ directly. You’re right in thinking that it depends on the parity of $k$, though the details are quite different from what you propose. First note that if $k$ is odd, then $k-1$ and $k+1$ are even, so
$$\{k\}=\{k-2,k-1,k\}\cap\{k,k+1,k+2\}$$
is the intersection of two members of $S$ and is therefore open. In other words, each odd integer is an isolated point. Now suppose that $k$ is even, and that $U$ is an open nbhd of $k$. Then by the definition of subbase there is a finite subset $S_0$ of $S$ such that $k\in\bigcap S_0\subseteq U$. But the only member of $S$ that contains $k$ is $\{k-1,k,k+1\}$, so $S_0=\big\{\{k-1,k,k+1\}\big\}$, and therefore $\{k-1,k,k+1\}\subseteq U$. In other words, every open nbhd of the even integer $k$ contains the set $\{k-1,k,k+1\}$: this set is the smallest open set containing $k$. Using these observations we can show that 


*

*$\operatorname{cl}\{k\}=\{k\}$ if $k$ is even, and  

*$\operatorname{cl}\{k\}=\{k-1,k,k+1\}$ if $k$ is odd.


Suppose first that $k$ is even. If $\ell\in\Bbb Z$ is odd, then $\{\ell\}$ is an open nbhd of $\ell$ that is disjoint from $\{k\}$, so $\ell\notin\operatorname{cl}\{k\}$. And if $\ell\in\Bbb Z\setminus\{k\}$ is even, then $\{\ell-1,\ell,\ell+1\}$ is an open nbhd of $\ell$ disjoint from $\{k\}$, so again $\ell\notin\operatorname{cl}\{k\}$, and it follows that $\{k\}=\operatorname{cl}\{k\}$ is closed.
Now suppose that $k$ is odd. If $\ell\in\Bbb Z\setminus\{k\}$ is odd, then $\{\ell\}$ is an open nbhd of $\ell$ disjoint from $\{k\}$, and $\ell\notin\operatorname{cl}\{k\}$. If $\ell\in\Bbb Z$ is even, and $\ell\notin\{k-1,k+1\}$, then $\{\ell-1,\ell,\ell+1\}$ is an open nbhd of $\ell$ disjoint from $\{k\}$, so again $\ell\notin\operatorname{cl}\{k\}$. However, we just say that every open nbhd of the even integer $k-1$ contains the set $\{k-2,k-1,k\}$, and every open nbhd of $k+1$ contains the set $\{k,k+1,k+2\}$. In particular, every open nbhd of $k-1$ contains $k$, and every open nbhd of $k+1$ contains $k$, so $k-1,k+1\in\operatorname{cl}\{k\}$. Thus, $\operatorname{cl}\{k\}=\{k-1,k,k+1\}$ when $k$ is odd.
For the final part of the problem, note that we just showed that the sets $\{0\}$ and $\{2\}$ are closed (because $0$ and $2$ are even). Take $A=\{0\}$ and $B=\{2\}$, and suppose that $U$ and $V$ are open sets such that $A\subseteq U$ and $B\subseteq V$. Bearing in mind what we’ve shown about open nbhds of even integers in this topology, show that $U\cap V\ne\varnothing$ by actually finding a specific element of $U\cap V$.
