Why is the $\{\bar 0\}$ is open for this quotient topology? The topological space $(\mathbb{R},T)$ is generated by the $\{[a,b)$ $\vert a < b\}$. Say the partition $\mathbb{R} / E = \{(-\infty,0), \{0\}, (0,2), \{2\}, (2.\infty)\}$ for some equivalence relation $E$ on the $(\mathbb{R}, T)$. For the quotient space $(\mathbb{R} / E, T_E)$, find the number of the open sets in $(\mathbb{R},T_E)$.
In my tutor's textbook, he put the equivalent class $\bar{-1} = (-\infty,0)$,$\bar{0} = \{0\}$, $\bar{1} = (0,2)$, $\bar{2} = \{2\}$ and $\bar{3} = (2, \infty)$. (I.E. $(\mathbb{R} / E, T_E) = \{ \bar{-1}, \bar0,\bar1, \bar2,\bar3\}$)
And next he found the basis like the $\{\{\bar{-1}\}, \{\bar{0}\}, \{\bar{0},\bar{1} \}, \{\bar{3}\},   \{\bar{2},\bar{3} \}  \}$.
My question is why the  $\{\bar{0}\}$ is the open set in  $(\mathbb{R} / E, T_E)$. Because considering the quotient mapping $\pi : \mathbb{R} \to \mathbb{R}/E, \pi^{-1}(\{0\}) = \{0\}$  should be open in $T$ under the hypothesis he's claim is true. But $\{0\}$ is not open in the $T$(or lower limit topology.). Is my thought right? I can't understand his claim.
 A: Yes, I think you are right. The quotient space is rather simple; just $5$ points and it is enough (for a base) to see/reason what the minimal open neighbourhood for each point of it is, as the space is finite. All open sets are then just unions of those minimal neighbourhoods.
$\{\overline{-1\}}$ is open as it corresponds exactly to the open set $(-\infty, 0)$ of the lower limit topology.
$\{\overline{0}\}$ is not open for the reason you indicated (it is a closed set though) but $\{\overline{0},\overline{1}\}$ has inverse image $[0,1) \in T$ so $\{\overline{0},\overline{1}\}$ is the minimal open neighbourhood for $\overline{0}$.
$\{\overline{1}\}$ is open as it corresponds to $(0,1) \in T$ again.
Similarly $\{\overline{2}\}$ is not open but $\{\overline{2},\overline{3}\}$ is (as $[2, \infty) \in T$ again) so that is  the minimal open neighbourhood for $\overline{2}$.
Also, $\{\overline{3}\}$ is open as $(2,\infty) \in  T$.
So the minimal base is $\{\{\overline{-1}\},\{\overline{0},\overline{1}\},\{\overline{1}\},\{\overline{2},\overline{3}\}, \{\overline{3}\}\}$
We can observe that the quotient space is $T_0$ but not $T_1$, and disconnected too.
