what is $U$ here and how it look like?

Question

Quotient topology : let $X$ be a topological space and $\sim$ an equivalence relation on $X$. For every $x \in X$ , denote by $[x]$ its equivalence class .The quotient space of $X$ modulo $ \sim$ is given by the set

$X/\sim~ =\{[x] : x \in X \}$

we have the projection map $p: X \to /\sim ,x \to [x]$ and we equip $X/\sim$ with the topology

$U\subseteq$ $X/ \sim~$ is open if and only if $p^{-1}(U)$ is an open subset of $X$

From my previous post I got example

But I'm confused that how can we colloborate the example with definition of quotient topology

My attempt : Here $X=[0,1]$ then projection map $p : [0,1] \to [0,1]/\sim~$

$p(x) =[x]$ where $[0,1]/\sim~ \cong S^1$

$U\subseteq [0,1]/\sim$ mean $U\subseteq S^1$ and $p^{-1}(U)$ is open

according to the definition I'm confused about $U$ . I mean what is $U$ here and how it look like ?

Answer 1

I imagine you're considering the euclidean topology on $[0,1]$. In that case, since open intervals generate the euclidean topology on $\Bbb R$, every open subset of $[0,1]$ is generated by intersections of open intervals with $[0,1]$ (by their union).

There are five types of intersections: $(a,b)\cap[0,1]=\begin{cases}(a,b)&\text{if }0\le a,b\le 1\\ [0,b)&\text{if }a<0,b\le1\\(a,1]&\text{if }0\le a,1<b\\ [0,1]&\text{if }a<0,1<b\\\emptyset&\text{otherwise}\end{cases}$.

If $0\le a,b\le1$ then $p^{-1}((a,b))=(a,b)$, so every $(a,b)$ with $0\le a,b\le1$ is open in $[0,1]/\sim$

Let $m$ denote $m=[0]=[1]=\{0,1\}$. If we have $U$ a nonempty open subset of $[0,1]/\sim$, then we have two options: $U$ contains $m$ or $U$ doesn't contain $m$. In the second case we have that $p^{-1}(U)=U$, so $U$ must be the union of intervals $(a,b)$ with $0\le a,b\le1$ (since $p^{-1}(U)$ is open in $[0,1]$ and it contains neither $0$ nor $1$) $\Rightarrow U=\bigcup_i(a_i,b_i)$ with $0\le a_i,b_i\le1\,\forall i$, so (by the previous paragraph) we have $U$ expressed as a union of open sets in $[0,1]/\sim$

If $U$ contains $m$ then $p^{-1}(U)$ contains $0$ and $1$, so $p^{-1}(U)$ must contain some intervals $(c,1]$ and $[0,d)$ (which are open in $[0,1]$) with $0\le c<1,0<d\le1$, so in consequence $[0,d)\cup(c,1]\subset p^{-1}(U)$. This implies that $(0,d)\cup(c,1)\cup\{m\}\subset U$, since that's the only way we can obtain $p^{-1}((0,d)\cup(c,1)\cup\{m\})=[0,d)\cup(c,1]$. Note that this means $(0,d)\cup(c,1)\cup\{m\}$ is open in $[0,1]/\sim$

Note also $\{m\}$ is not open, since $p^{-1}(\{m\})=\{0,1\}$.

Of course, now we can say unions of intervals $(a,b)$ with $0\le a,b\le1$ and/or subsets of the form $(0,d)\cup(c,1)\cup\{m\}$ with $0\le c<1,0<d\le1$ are open in $[0,1]/\sim$

And actually that's how every nonempty open subset $U$ of $[0,1]/\sim$ is constructed:

1. If $U$ doesn't contain $m$ we've shown $U=\bigcup_i(a_i,b_i)$ for some intervals $(a_i,b_i)$.

2. If it does contain $m$ then $U=\{m\}\cup(U\setminus\{m\})=\{m\}\cup\bigcup_i(a_i,b_i)$, since $U\setminus\{m\}=U\cap\{m\}^\mathsf c=U\cap(0,1)$, which is an open subset not containing $m$; and we also know $U$ contains a set $(0,d)\cup(c,1)\cup\{m\}$, so $U=\big((0,d)\cup(c,1)\cup\{m\}\big)\cup\bigcup_i(a_i,b_i)$. In fact $(0,d)$ and $(c,1)$ must be in $\bigcup_i(a_i,b_i)$, so we are kind of writing them down twice in that expression, but I think it shows that, effectively, $U$ is union of the kind of open subsets we talked about.