What are the open sets in the quotient space formed when the endpoints of a closed interval are "glued together"?

Question

I understand the definition of the quotient topology, but can't intuitively understand how the open sets are formed.

Answer 1

The gluing process is made precise by following this procedure, which will be familiar to you, but is worth writing out in detail, to see what the open sets in the quotient topology are:

$1).\ $ if $[a,b]$ is a closed interval, define an equivalence relation $\sim$ by declaring the classes to be: the singleton sets $\{\{x\}:x\in (a,b)\}$ and the set $\{a,b\}.$ All this means is that we have defined a new set $[a,b]/\sim$ which consists of points in the interior of the interval $[a,b]$ and the "point" $\{a,b\}$.

$2).\ $ To topologize $[a,b]/\sim$, we define a function $q:[a,b]\to [a,b]/\sim$ by sending each $x\in [a,b]$ to its equivalence class. So, for example, if $a<x<b,\ q$ sends $x$ to $\{x\}$, which we may as well just take to be $x$ itself. But if $x=a$ or $x=b,$ then $q$ sends $x$ to the class $\{a,b\}$ which is a new "point", which we call $[a]\ (\text{or}\ [b])$.

Now, we define a set $V$ in $[a,b]/\sim$ to be open if and only if $q^{-1}(V)$ is open in $[a,b].$

To see what the open sets are, first note that if $V$ is an open interval $(x,y)$ with $x,y\neq a$ or $b$, then $q^{-1}(V)=V$ itself, which is open in $[a,b]$, and so $V$ is open in $[a,b]/\sim$. This is no surprise, since $q$ sends all interior points of $[a,b]$ to themselves (identifying $x$ with $\{x\}$). Now check some other possibilities for $V$:

If $V=[a]$, with then $q^{-1}(V)$ is $\{a,b\}$ which is not open in $[a,b]$ and so the point $[a]$ is not open in $[a,b]/\sim$.

If $V=[a]\cup (a,x),\ x<b$, then $q^{-1}(V)=[a,x)\cup\{b\}$ which is not open in $[a,b]$ so $V$ is not open in $[a,b]/\sim$.

If $a<x<y<b$ and $V=[a]\cup (a,x)\cup (y,b)$, then $q^{-1}(V)=[a,x)\cup (y,b]$, which is open in $[a,b]$ so $V$ is open in $[a,b]/\sim$.

Let $a<x<y<b$. It should be becoming clear that the basic open sets in the quotient are those sets $V$ for which $q^{-1}(V)$ is of the form $(x,y)$ or $[a,x)\cup (y,b]$. And since $a$ and $b$ are identified in the quotient, this latter set may be considered to be a "gluing" of $[a,x)$ and $(y,b]$ into a "single" open set, which after all, is what the quotient was supposed to do.

Answer 2

If $q:[a,b]\to [a,b]/\sim$ is the quotient map, then through the inverse image map $q^{-1}[\bullet]:\mathcal P([a,b]/\sim)\to \mathcal P([a,b])$ they are naturally in bijection with the open sets which are either subsets of $]a,b[$, or the whole $[a,b]$, or which may be written as a disjoint union $[a,\delta)\cup U\cup (\varepsilon, b]$ with $U$ open in $]a,b[$.