# Confused about quotient topology 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.

• The notation $[(0,1)]$ stands for the equivalence class under the relation $\sim$ of the point $(0,1)\in X$. – Javi Oct 5 '18 at 21:07
• Thank you. So it looks like the equivalence classes of $X$ are singletons except for the equivalence class $E_0 = \{(0,1), (0,2)\}$. So then I want to consider neighborhoods $O_1, O_2$ in the induced topology, which would be, for example, $O_1 = (0- \epsilon, 0+\epsilon) \times \{1\}$, and $O_2 = (0- \epsilon, 0+\epsilon) \times \{2\}$. I'm not seeing how I'll then show that some $(x,i) \in O_1 \cap O_2$. For if $(x,i) \in O_1$, it's equivalence class is itself, so it's not anywhere on the line $(-1,1) \times \{2\}$ (in particular, it's not in $O_2$). – Pawnee Oct 5 '18 at 21:31
• Oh wait, I had it backwards. The equivalence classes are all basically pairs $\{(x,1),(x,2)\}$. I'm still unsure how to show it – Pawnee Oct 5 '18 at 21:49
• And $\{0,2)\}$ and $\{0,1)\}$ are one-point classes... – Henno Brandsma Oct 5 '18 at 21:58
• It might help to visualize the quotient space as being almost the same as $(-1, 1)$, just with two "copies" of 0. (Though it's sort of difficult to "draw a picture" of it, since as the exercise shows, the quotient space is not Hausdorff.) – Daniel Schepler Oct 5 '18 at 22:13

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?
• Yeah, I'm really struggling to see why that $a$ exists. That's sort of what my original comment above relates to, although I didn't state it correctly at all because when I posted that I had several additional layers of confusion about this problem – Pawnee Oct 5 '18 at 22:43
• I think something like this: Since $q^{-1}(O_1)$ and $(-1,1) \times \{1\}$ are both neighborhoods of $(0,1)$, so is their intersection which we'll call $U$. By properties of the subspace topology there exists an open interval $I_1$ around $0$ such that $I \times \{1\} \subset U \subset q^{-1}(O_1)$. Likewise we can find an open interval $I_2$ for $(0,2)$. Thus the intersection $I_1 \cap I_2$ is an open interval around $0$ such that $I_1 \cap I_2 \times \{1\} \subset q^{-1}(O_1)$ and likewise $I_1 \cap I_2 \times \{2\} \subset q^{-1}(O_2)$... So we can choose some nonzero $a \in I_1 \cap I_2$. – sfmiller940 Oct 5 '18 at 23:53
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.