Saturated subset of a quotient map

Question

First, here are the two equivalent definitions I was given for a saturated subset in my course:

Definition 1: Let $p : X \longrightarrow X'$ be a quotient map. We say that a subset $U\subseteq X$ is saturated if $p^{-1}(p(U))=U$.

Definition 2: Let $\pi : X \longrightarrow X/\sim$ a quotient map, where $\sim$ denotes a equivalence relation. Then, a subset $U\subseteq X$ is saturated if and only if $U$ is a union of equivalences classes.

Here is the question I am working on:

Exercice: Let $([0,1],\varepsilon_{[0,1]})$, with $\varepsilon_{[0,1]}$ denoting the topology induced by the standard topology on $[0,1]$ and $A=\left\{ 0,\frac{1}{2},1 \right\}$. Graph a visual representation of $[0,1]/A$ and describe the quotient topology.

Intuitively, with this quotient, we're matching the points of $A$ to be the same point. Let's call $p : [0,1] \longrightarrow [0,1]/A$ the quotient map. Thus, we can get the following visual representation:

Now, by definition, we can define the following quotient topology on $[0,1]/A$:

$$\tau'=\left\{ V \subseteq [0,1]/A : p^{-1}(V)\in\varepsilon_{[0,1]} \right\}=\left\{ p(V)\subseteq [0,1]/A : V\in\varepsilon_{[0,1]},V \text{ is saturated} \right\}$$

So, to describe our quotient topology, we need to identify the saturated open subset of $\varepsilon_{[0,1]}$. For any open subset of $\varepsilon_{[0,1]}$ that does not intersect $A$, it is clear that they are indeed open in our quotient topology by definition. But, I have some troubles with open subset of $\varepsilon_{[0,1]}$ that does intersect $A$. Here is my problem illustrated by the following diagram:

You might guess where I am heading to: depending on which orientation you "glue" the interval to get the quotient space, the open set (denoted in blue) might end up in the four following pattern:

That is where I have troubles: are these different sets under the quotient map $p$ the same? The same question applies for open set of the form $[0,a)\cup (b,1]$ where $a<\frac{1}{2}$ and $b>\frac{1}{2}$. One way that I've thought of and that would fix this issue, is to consider the following open subset of $\varepsilon_{[0,1]}$:

$$V=[0,a)\cup(a,b)\cup(b,1]$$

where $a<\frac{1}{2}$ and $b>\frac{1}{2}$. That way, no matter how we "glue" the points of $A$, we would always have the same direct image in our quotient space. More generally, I would like to know how one would use the definition of a saturated set to avoid such issues.

Thanks in advance for the help.

Answer 1

While working on this problem, I think I solved it and I might as well share it here. I'll be using a nice equivalence that I haven't seen (or at least, I could not find it) being shared there.

Notation: $\bar{x}$ denotes the equivalence class of $x$.

I'll first introduce the following proposition and prove it:

Proposition: Let $(X,\tau)$ be any topological space and $A\subseteq X$. Let $p:X\rightarrow X/A$ be the quotient map. Then, $$U\subseteq X \text{ is saturated} \iff A\subseteq U \text{ or } A\cap U = \emptyset$$

Proof: We begin by showing "$\Leftarrow$". Let $U\subseteq X$, then by definition, we have that $$p^{-1}(p(U))=\bigcup_{x\in U}\bar{x}.$$

• If $U\cap A = \emptyset$, then for any $x\in U$, we have that $\bar{x}=\left \{ x \right \}$. It follows directly than $$p^{-1}(p(U))=\bigcup_{x\in U}\bar{x}=\bigcup_{x\in U} \left \{ x \right \} = U.$$ So, $U$ is indeed saturated by the first definition.
• If $A\subseteq U$, we compute $$\begin{eqnarray} p^{-1}(p(U)) & = & \bigcup_{x\in U}\bar{x} \nonumber\\ & = & \left (\bigcup_{x\in U\setminus A}\bar{x} \right )\cup\left ( \bigcup_{x\in A} \bar{x} \right ) \nonumber\\ & = & \left (\bigcup_{x\in U\setminus A} \left \{x \right \} \right )\cup\left ( \bigcup_{x\in A} \bar{x} \right ) \nonumber\\ & = & \left (U \setminus A \right )\cup A \nonumber\\ & = & U . \end{eqnarray}$$ So $U$ is indeed saturated.

Now, we show "$\Rightarrow$". Let $U\subseteq X$ be a saturated subset. Thus, we have $$p^{-1}(p(U)) = U = \bigcup_{x\in U}\bar{x} = \left (\bigcup_{x\in U\setminus A}\ \left \{x \right \} \right )\cup\left ( \bigcup_{x\in A\cap U} \bar{x} \right ). \tag{1}$$ We then suppose by contradiction that $A\nsubseteq U$ and $A\cap U \neq \emptyset$. So, using (1), we get that $$U = \bigcup_{x\in U}\bar{x} = (U\setminus A)\cup A = U\cup A,$$ which is a contradiction. So, we need that $A\subseteq U$ or $A\cap U = \emptyset$. $$\tag*{$\blacksquare$}$$

Using this proposition, we can directly see that the saturated open subsets $V$ of $\varepsilon_{[0,1]}$ are either

1. Open subset of $\varepsilon_{[0,1]}$ that does not intersect $A$, namely $(a,b)\in \varepsilon_{[0,1]}$ such that $0<a<b<1/2$ or $1/2<a<b<1$.
2. Open subset of $\varepsilon_{[0,1]}$ such that $A\subseteq V$, namely $[0,a)\cup (a,b) \cup (b,1] \in \varepsilon_{[0,1]}$ such that $0<a<1/2$ and $1/2<b<1$.