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The example is as follows:

$\mathbb{R}\times \{0,1\}$ is clearly a Hausdorff space. But the quotient space of $\mathbb{R}\times \{0,1\}$ defined by the equivalence relation $\langle x,0\rangle$~$\langle x,1\rangle \iff x\neq 0$ is not Hausdorff.

Why is that?

Is the quotient space the set $Q=\mathbb{R}$, such that $U\subset Q$ is open if and only if $U\times\{0\}\cup U\times\{1\}$ is open in $\mathbb{R}\times\{0,1\}$?

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  • $\begingroup$ The quotient space is the real line with a double point. You can't separate the double point by open sets. $\endgroup$ – user302982 Aug 16 '17 at 9:24
  • $\begingroup$ what do you mean by a double point? $\endgroup$ – Sid Caroline Aug 16 '17 at 9:29
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    $\begingroup$ If you identify each $\left< x, 0 \right>$ with $x$, you get $\mathbb{R}$ and there is one more point in the space: $\left< 0, 1 \right>$ (which is not identified with $0$ because $\left< 0, 0 \right> \not \sim \left< 0, 1 \right>$). So the quotient space can be thought of as $\mathbb{R} \cup \{ 0' \}$ with appropriate topology. $\endgroup$ – Adayah Aug 16 '17 at 9:31
  • $\begingroup$ But I thought the set of equivalence classes is just $(\mathbb{R}\setminus\{0\}) \cup\{0\}$. which is just $\mathbb{R}$? Oh I see, the set of equivalence classes is $(\mathbb{R}\setminus\{0\}) \cup\{0,0'\}$, since $\langle 0,0\rangle$ is not equivalent to $\langle 0,1\rangle$. $\endgroup$ – Sid Caroline Aug 16 '17 at 9:33
  • $\begingroup$ For $r \in \mathbb{R}$ denote $r = [ \left< r, 0 \right> ]_{\sim}$ and $r' = [ \left< r, 1 \right> ]_{\sim}$ where $[x]_{\sim}$ means equivalence class of $x$. Now $Q = \{ r : r \in \mathbb{R} \} \cup \{ r' : r \in \mathbb{R} \}$. But for each $r \neq 0$ we have $r = r'$ from the definition of $\sim$, so we are left with $\{ r : r \in \mathbb{R} \} \cup \{ 0' \}$ and the first set is just $\mathbb{R}$, so $Q = \mathbb{R} \cup \{ 0' \}$. $\endgroup$ – Adayah Aug 16 '17 at 9:39
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Is the quotient space the set $Q=\mathbb{R}$, such that $U\subset Q$ is open if and only if $U\times\{0\}\cup U\times\{1\}$ is open in $\mathbb{R}\times\{0,1\}$?

No. The quotient space, by definition, is the set $X/_\sim$ where $X$ is the original space (in your case, $X=\mathbb R\times\{0,1\}$. This means $Q=\{[(x,y)]| (x,y)\in X\}$.

In $Q$, a set $U\subset Q$ is open if $q^{-1}(U)$ is open in $X$.


Now, for $x\neq 0$, you have $[(x,0)] = [(x,1)] = \{(x,0),(x,1)\}$, and you can certainly see that the local neighborhood around the point $[(x,0)]\in Q$ is homeomorphic to the local neighborhood around the point $x\in R$, but that does not mean that the two sets are equal! In fact, you get into trouble around $x=0$, since $[(0,0)] = \{(0,0)\}$ and $[(0,1)]=\{(0,1)\}$


So, the idea is to prove that those two points do not have disjoint neighborhoods.


Take any neighborhood $O_0$ containing $[(0,0)]$. Then, $q^{-1}(O)$ is an open set around $(0,0)$, which means that it contains some "interval" (i.e. there exists some $\epsilon_0 > 0$ such that $(-\epsilon_0, \epsilon_0)\times\{0\}\subset q^{-1}(O_0)$.

From this, it should be easy to show that $I_0 = \{[(x, 0])| |x|<\epsilon_0\}$ must be a subset of $O_0$

If you now repeat the process on a neighborhood $O_1$ containing $[(0,1)]$, you will find that $I_1 = \{[(x,0)]; |x<\epsilon_1\}$ is a subset of $O_1$, and thus $O_0\cap O_1\neq\emptyset$

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