Exercise On Quotient Topology Let $X = \mathbb R$ (real numbers) in the standard topology. Take the partition $X' = \{\dots, (-1 , 0] , (0 , 1] , (1 , 2] ,\dots\}$.
My question is this: 
Describe the open sets in the resulting quotient topology on $X'$?
 A: Hint: The open subsets of $X'$ are those sets $\mathcal O\subseteq X'$ such that $\bigcup\mathcal O$ is open in $X=\Bbb R$. What can a subset of $X'$ look like in order for the union of its members to be a non-trivial open subset of $X=\Bbb R$?
A: For $n\in\Bbb Z$ let $p_n=(n-1,n]$, so that $X'=\{p_n:n\in\Bbb Z\}$. Let $q:X\to X'$ be the quotient map: $q(x)=p_n$ if and only if $n-1<x\le n$, i.e., if and only if $\lceil x\rceil=n$. By definition a set $U\subseteq X'$ is open in $X'$ if and only if $q^{-1}[U]$ is open in $X$. 
Now $q^{-1}[U]$ is always a union of intervals of the form $(n-1,n]$; under what circumstances is such a union open in $X$? Suppose, for instance, that $p_1\in U$, so that one of the intervals is $(0,1]$. In order for $q^{-1}[U]$ to be open in $X$, it must contain an ordinary open interval around $1$, and the only way to get that is to have $p_2\in U$, so that $(1,2]\subseteq q^{-1}[U]$. Now we know that $q^{-1}[U]\supseteq(0,2]$, so in order for it to be open, it must contain an open interval around $2$. The only way to get that is to have $p_3\in U$, so that $(2,3]\subseteq q^{-1}[U]$. Keep going: by induction you can show that if $p_1\in U$, and $U$ is open in $X'$, then $p_k\in U$ for each $k\ge 1$, and $q^{-1}[U]$ contains all of the positive reals. In fact, we can now see that $\{p_n:n\ge 1\}$ is open in $X'$: its inverse under $q$ is just the positive reals.
Now try to generalize from that example to find the rest of the open sets in $X'$.
A: I keep  the notation defined by Brian M.Scott.
If $J$ a non empty subset of $\Bbb Z$, let us put: $$O_J=\{p_k / k \in J\}$$
We can show that  $O_J$ is an open of $X'$ if and only if $J$ has the form : $$J=]\!]m,+\infty [\![=\{ k \in \Bbb Z / k >  m\}$$ for some $m \in \Bbb Z \cup \{-\infty\}$ where $]\!]-\infty,+\infty[\![ = \Bbb Z$
Suppose that $O_J$ is an open set of $X'$,
First steep : we have : $\forall k \in \Bbb Z  \quad k \in J \Rightarrow  k+1 \in J$. if not , let $k \in \Bbb Z$ such that : $k \in J$ and  $k+1 \notin J$. then $p_k \in O_J$ and $p_{k+1} \notin O_J$. this gives : $(k-1,k] \subset q^{-1}(O_J)$ and $]k,k+1[ \cap q^{-1}(O_J) = \emptyset$, who contredise that $ q^{-1}(O_J)$  is  one open set of $\Bbb R$
Second steep: there are two cases: 
$\bullet$ first case : if $J$ has a smaller element (minimum),then by first steep $J=]\!]m,+\infty[\![$ where $m+1=\min(J)$ 
$\bullet$ second case : if not, we have by first steep: $J=\Bbb Z$ 
Conversly, if $J=]\!]m,+\infty[\![$ for certain $m \in \Bbb Z$, then $q^{-1}[O_J]= ]m,+\infty[$ who is an open of $\Bbb R$. if $m=-\infty$ then $q^{-1}[O_J]=\Bbb R$.
That gives the topology of $X'$; it's:$$\mathscr T=\{U_m / m \in \Bbb Z \cup \{-\infty ,+\infty\}\}$$  where :$$\left\{\begin{array}{lcl}U_m=\{p_k / k \in ]\!]m,+\infty[\![ \}& \text{if}& m \in \Bbb Z \\U_{-\infty} = X' &&  \\U_{+\infty} = \emptyset  &&  \end{array} \right.$$
