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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.

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  • $\begingroup$ The notation $[(0,1)]$ stands for the equivalence class under the relation $\sim$ of the point $(0,1)\in X$. $\endgroup$ – Javi Oct 5 '18 at 21:07
  • $\begingroup$ 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 $O_2$). $\endgroup$ – Pawnee Oct 5 '18 at 21:31
  • $\begingroup$ 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 $\endgroup$ – Pawnee Oct 5 '18 at 21:49
  • $\begingroup$ And $\{0,2)\}$ and $\{0,1)\}$ are one-point classes... $\endgroup$ – Henno Brandsma Oct 5 '18 at 21:58
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    $\begingroup$ 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.) $\endgroup$ – Daniel Schepler Oct 5 '18 at 22:13
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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?

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  • $\begingroup$ 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 $\endgroup$ – Pawnee Oct 5 '18 at 22:43
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    $\begingroup$ 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 $a \in I_1 \cap I_2$. $\endgroup$ – sfmiller940 Oct 5 '18 at 23:53
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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.

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