Confused about quotient topology question

Question

Let $X = (-1,1) \times \{1,2\}$ be endowed with the topology induced from $(\mathbb{R}^2, \tau_{standard})$. Let relation $\sim$ be defined as follows: $(x,i) \sim (y,j)$ if $x=y$ and ($i=j$ or $x \neq 0$). Then $\sim$ is an equivalence relation. Let $Y = X/\sim.$ Show that for any neighborhood $O_2$ of $[(0,2)]$ and any neighborhood $O_1$ of $[(0,1)]$, it holds that $O_2 \cap O_1 \neq \emptyset$.

I'm confused about what I'm being asked to show here. First off, I don't understand what the notation $[(0,1)]$ and $[(0,2)]$ means. If someone can explain this problem and offer a few hints to get started, that would be very helpful.

Answer 1

A neighbourhood $O_1$ of $[(0,1)]$ (the class of $(0,1)$, which is just a singleton) is any set of classes such that $q^{-1}[O_1]$ is open in $(0,1) \times \{1,2\}$ (where $q: (x,i) \to [(x,i)]$ is the standard quotient map). The topology of that set is generated by all sets of the form $U_1 \times \{1\}$ and $U_2 \times \{2\}$ where $U_i$ are open in $(0,1)$ (usual topology), basically two loose copies of the unit interval. Convince yourself that $O_1$ must contain some set $(0,r) \times \{1,2\}$ and $O_2$ as well.

Answer 2

Let $q:X \to Y$ be the quotient map. Let $O_1, O_2$ be neighborhoods in $Y$ containing $[(0,1)],[(0,2)]$ respectively. Then by the definition of quotient topology the preimages $q^{-1}(O_1), q^{-1}(O_2)$ are neighborhoods in $X$ containing $(0,1), (0,2)$ respectively.

Suppose $q^{-1}(O_1) \cap q^{-1}(O_2) \neq \emptyset$, then $O_1 \bigcap O_2 \neq \emptyset$ and we're done.

Suppose $q^{-1}(O_1) \cap q^{-1}(O_2) = \emptyset$. Choose a nonzero $a \in (-1,1)$ such that $(a,1) \in q^{-1}(O_1)$ and $(a,2) \in q^{-1}(O_2)$. Since $a$ is nonzero, $(a,1) \sim (a,2)$. Thus $q(a,1) = q(a,2) \in O_1 \cap O_2$ and we're done.

This is an outline of a proof, but the question remains... Why can we choose that nonzero $a \in (-1,1)$ as above?