Collapsing the $K$-Topology Here is a quote

"Recall that $\Bbb{R}_K$ denotes the real line in the $K$-topology. Let $Y$ be the quotient space obtained from $\Bbb{R}_K$ by collapsing $K = \{1/n ~|~ n \in \Bbb{N} \}$; let $p : \Bbb{R}_K \to Y$ be the quotient map".

Okay. That seems pretty nondescript...What does the rest of $Y$ look like? It isn't even clear what this "collapsing" amounts to. What exactly does this meant? What do the elements of $Y$ look like?
 A: If $X$ is a space and $A \subseteq X$ then the quotient space from "collapsing $A$" is made with an equivalence relation which has its equivalence classes $A, \{\{x\}: x \notin A\}$; note that these form a partition of $X$, so define a unique equivalence relation. So all points of $A$ are equivalent to each other and to no other point, and all points not in $A$ are only equivalent to themselves.
The set $Y$ of these classes gets the quotient topology in the usual way: define $q(x)$ to be the unique class of $x$ that $x \in X$ belongs to, so $q(x) = \{x\}$ for $x \notin A$ while $q(x) = A$ for $x \in A$. 
A subset of $Y$ (so a set of classes!) is open iff $q^{-1}[O] = \{x \in X: q(x) \in O\}$ is open in $X$.
As $A$ is closed (this is usual), then $q: X \setminus A \to Y\setminus \{A\}$ is an open injective map (a homeomorphism even) between an open subspace of $X$ and of $Y$. 
So we can identify all points outside $A$ with all classes not equal to the class $A$. $A$ itself is identified (collapsed) to a single point in the new space. In the case of the $K$-topology, we get a $T_1$ space that is not $T_2$ because we cannot separate the points $\{0\}$ from $A$ in $Y$ by disjoint open sets (or their inverse images would separate $A$ from $0$ in the $K$-topology which cannot be done).
