What qualifies as examples consider as "collapsing a point to a set." 
I am not clear on what precisely qualifies as "collapsing a set to a point".  I know the definition states conceptually when defining an equivalence class on a set say, $B$, The set $B$ is consider to be its own equivalence class and any two elements in $B$ are consider to be identical.  Any elements not in $B$ are consider to be in a class of their own.  However.  After I looked over a few textbooks that have examples that are related to this concept, from such examples, I am wondering if they all qualifies to be consider as collapsing a point to a set.
The first three examples (included attached image) are taken from a textbook titled: Introduction to Topology Pure and Applied by Colin Adams and Robert Franzosa.  
The first example states:

Let $X^{*}$ be a collection of mutually disjoint subsets of $X$ whose union is $X$, and let $p:X\rightarrow X^{*}$ be the surjective map that takes each point in $X$ to the corresponding element of $X^{*}.$ We think of the process of going from the topology on $X$ to the quotient topology on $X^{*}$ as taking each subset $S$ in the partition and identifying all the points in $S$ with one another, thereby collapsing $S$ to a single point in the quotient space.

Here the set $X$ is partition into five different subsets or equivalence classes, consisting of open, closed and neither open nor closed subsets.  Each of these subsets are sent to a singleton set by the quotient mapping $p$
The next example (example 3.15), I am not sure the way the quotient space is defined fulfills the criteria of collapsing a set to a point, since $X$ is being partition into two equivalence classes, while the quotient space $X^*$ which $p$ maps to consist of two elements/points set.
The third example is the classical one of mapping a closed interval $I=[0,1]$ homeomorphically to a circle by identifying the endpoints as being equivalent.  But here, the quotient space $X^*$ are made up of the set [x], where each $x\in (0,1)$ gets map by $p$ to itself and becomes its own equivalence class and the single set/element $D$ consisting of two points $\{0,1\}$.
The last example is taken from the text Topology point set and geometric by Paul Shick pp 102 to 103 example 5.2 and definition for $U$ is taken from pg 55 In the example, the topology $U$ is the usual topology and is defined as $\text{$U=\{V\in \mathbb{R}:$ if $x\in V$, then there exists an open interval $(a,b)$ such that $x\in(a,b)\subset V$\}.}$

Let $R_U$ be the equivalence relation defined on the real line with $x\text{~}1/2$ for all $x\in (0,1)$, with each $x\in \mathbb{R}$ equivalent to itself.
  We have an entire open set (0,1) being map to a the point $1/2$ and any elements in (0,1) is consider to be equivalent to $1/2$.  

Thank you in advance.
 A: Let  $T_X$ be a topology on $X$ and let $\emptyset \ne S\subset X.$ Take $p$ such that $p\in S$ or $p\not\in X$ and let $Y=(X\setminus S)\cup \{p\}.$  For $x\in X$ let $f(x)=x$ if $x \not \in S$ and let $f(x)=p$ if $x\in S.$
The $f$-quotient topology $T_Y$ on $Y$ is defined as the $\supset$-strongest topology on $Y$ such that $f$ is continuous. So 
(i). $T_Y\subset \{f(U): U\in T_X\}.$
(ii). If $U\in T_X$ and $U\cap S=\emptyset$ then $f(U)=U\in T_Y.$
(iii). If $S\subset U\in T_X$ then $f(U)=(U\setminus S)\cup \{p\}\in T_Y.$
(iv). If $U\in T_X$ but $U$ does not meet condition (ii) or (ii) above then $f(U) \not \in T_Y.$
Examples. (1).Let $T_X$ be the standard topology on $X=[0,1]$ with $S=\{0,1\}$ and $p=0.$ Observe that when $0\in V\subset Y=[0,1),$  we have $V\in T_Y$ iff $V\cup \{1\} \in T_X.$ And that $Y$ is homeomorphic to the circle $S^1.$
(2). Let $T_X$ be the standard topology on $X=\Bbb R,$ with $S=\Bbb N$ and $p\not \in \Bbb R.$ Observe that $p\in V\in T_Y$ iff $(V\setminus \{p\})\cup \Bbb N\in T_X,$ that is, iff $V=\cup_{n\in \Bbb N}f(U_n)$ where  $n\in U_n\in T_X$ for each $n\in \Bbb N.$ In this example $T_Y$ is not a first-countable topology.
