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Let $X$ be two disjoint copies of $\mathbb{R}$, that is say $X = (\{a\} \times \mathbb{R}) \cup (\{b\} \times \mathbb{R})$ for real numbers $a<b$ and consider X as a subspace of $\mathbb{R} \times \mathbb{R}$. Define an equivalence relation by $a \times t \sim b \times t$ for all $t \neq 0$. Show that the quotient space $X^{*}$ determined by this equivalence relation is locally Hausdorff, but not Hausdorff.

Definition of Locally Hausdorff: For any $x$ in $X$ there is a neighborhood $U$ of $x$ such that $U$, with the subspace topology, is Hausdorff.

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  • $\begingroup$ Well we know all open subsets of X* are not necessairly disjoint. I just can't find a way to prove that their intersections are nonempty. $\endgroup$ – KangHoon You Oct 15 '13 at 4:45
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HINT: $X^*$ is almost homeomorphic to $\Bbb R$; there is exactly one pair of points in $X^*$ that cannot be separated by disjoint open sets, and knowing that should make the pair easy to find.

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  • $\begingroup$ Is the pair when a and b are equal? $\endgroup$ – KangHoon You Oct 15 '13 at 5:17
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    $\begingroup$ @KangHoonYou: I don’t know what you mean: $a$ and $b$ aren’t equal. $\endgroup$ – Brian M. Scott Oct 15 '13 at 5:19
  • $\begingroup$ Is the pair (a,0) and (b,0)? $\endgroup$ – KangHoon You Oct 16 '13 at 3:34
  • $\begingroup$ @KangHoonYou: Yes, it is. Can you see why they don’t have disjoint nbhds? $\endgroup$ – Brian M. Scott Oct 16 '13 at 3:37
  • $\begingroup$ Yes, is it because these points are not part of the equivalence class? $\endgroup$ – KangHoon You Oct 16 '13 at 3:42

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