Quotient Topology Lets $X$ is a topological space and $Y$ is some subset of $X$. How we define topology of quotient space
$$X/Y=\{x\in X~|~x\sim y\Leftrightarrow x, y\in Y\}.$$
Thanks.
 A: For a set $X$ and an equivalence relation $R$, the quotient set $X/R$ is the set of equivalence classes $[x]$ of $R$.  The quotient set comes with a quotient map $\pi:X\rightarrow X/R$, naturally defined by sending $x$ to its equivalence class $[x]$.  The "quotient topology" on $X/R$ is defined by saying "$U\subset X/R$ is open iff $\pi^{-1}(U)$ is open in $X$."  (the quotient topology itself is a special case of two things: 1) the weak topology induced by a famaily of maps too/from your set, and 2) a pushout diagram.  You should check these things out, they're cool.)
Your example, quotienting by a subset, is a special case of a quotient set.  To make your relation an equivalence relation, just add the diagonal $\Delta_X= \{ (x,x)\in X\times X\ |\ x\in X\}$.  An element $[x]\in$"$X/Y$" is either a singleton $[x]=\{x\}$ (where $x\not\in Y$), or $[x]=Y$.  Hence the quotient map is a bijection on $X-Y$.
If $U\subset X/Y$, then $U$ is open iff $\pi^{-1}(U)=\{x\in X\ |\ [x]\in U\}$ is open in $X$.  We can write any $U$ as $(U-\{Y\})\ \dot{\cup}\ (\{Y\}\cap U)$, so in general $U$ is open iff $\pi^{-1}(U)=\pi^{-1}(U-\{Y\})\ \dot{\cup}\ \pi^{-1}(\{Y\}\cap U)$ is open in X.  Thus for $U\subset X/Y$: if $Y\not\in U$ then $\pi^{-1}(U)$ is disjoint from $Y\subset X$ so the quotient map is a bijection; if $Y\in U$, then we must check that $\{x\in X-Y\ |\ [x]\in U\}\cup Y$ is open in X.  
So if you want to think of $X/Y$ as "$X$, except $Y$ has collapsed" you can think "$U$ is open iff $U$ is open in $X$ and doesn't intersect $Y$, or $U\cup Y$ is open in $X$."   
A: Just as in the case of groups, as a set the quotient $X/Y$ is the set where $Y$ has been collapsed to a point. So define an equivalence relation on $X$ by $x_1 \sim x_2$ iff $ x_1$ and $x_2$ belong to $Y$. So now all the points of $Y$ are identified with each other, and all the points disjoint from $Y$ contain only themselves in their corresponding classes. Then $X/Y:= X/\sim$.
Now, we have an obvious surjective map $q: X \to X/\sim$ which sends each point of $X$ to its corresponding class. We want to give $X/\sim$ a topology so that $q$ is continuous. The obvious choice is the final topology: that is, the finest topology such that $q$ is still continuous. 
The reason we make this choice is because then the quotient satisfies the universal property that any continuous function $g: X \to Z$ which makes the same identifications as $q$ passes to a unique continuous map $\tilde{g} : X/\sim \to Z$ such that $\tilde{g} \circ q = g$.
