# Quotient spaces homeomorphic to the reals

Let $Z$ be a topological space. Given a subspace $A$ of $Z$, define an equivalence relation $R_A$ such that its equivalence classes are $\{x\}$ for $x\in Z\setminus A$ and $A$. Let $Z/A$ be the appropriate quotient space.

Let $Z=\mathbb{R}$, and let $A_1 = [0,1]$, $A_2=(0,1)$, $A_3=[0,1)$. I need to find for which $i=1,2,3$ does the quotient space $\mathbb{R}/A_i$ is homeomorphic to $\mathbb{R}$.

I found out that $\mathbb{R}/A_1$ is homeomorphic since the open interval $(-\infty,0)$ is homeomorphic to itself, the open interval $(1,\infty)$ is homeomorphic to $(0,\infty)$, and $\{A_1\}$ is homeomorphic to $\{0\}$.

$\mathbb{R}/A_2$ is not homeomorphic since the singleton $\{A_2\}$ is an open interval since $A_2$ is an open interval.

I am trying to come up for a solution for $A_3$, and also, to come up with a better explanation for $A_1$.

For (1), one can always show this directly, i.e., by constructing a bijection $\phi: \Bbb R \to \Bbb R / [0, 1]$ and showing that $\phi$ and its inverse are both continuous. (Of course, $\phi$ here cannot be the canonical quotient map.)
For (3), one option is to show that the quotient is not Hausdorff: Since $\Bbb R$ is Hausdorff, this would show that the two sets are not homeomorphic. Can you find two points in $\Bbb R / [0, 1)$ that cannot be separated by open sets?