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This is a homework question. I am asked to show that if $A \subset X$ is closed, then $X \setminus A$ is homeomorphic to $(X/A) \setminus (A/A)$. I have done this, and I now have to show by example that this is false if we do not require that $A$ is closed. Could someone point me in the right direction in finding such an example?

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  • $\begingroup$ what is $X/A$ or $A/A$? $\endgroup$
    – Simone
    Mar 9 '16 at 21:58
  • $\begingroup$ @Simone quotient space $\endgroup$ Mar 9 '16 at 22:00
  • $\begingroup$ @Simone $X$ is an arbitrary topological space, $A$ is a subset of $X$, and $X/A$ and $A/A$ are quotient spaces. $\endgroup$
    – G Pace
    Mar 9 '16 at 22:03
  • $\begingroup$ sorry for the ignorance, but how do you quotient a space by a subset? I mean, I can quotient by a relation (so I can quotient by particular subsets of $X\times X$), but I do not know what $X/A$ means, can you point me to a reference? $\endgroup$
    – Simone
    Mar 9 '16 at 22:07
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    $\begingroup$ @Simone: You identify $A$ to a point. The relation is the one whose equivalence classes are the set $A$ and the sets $\{x\}$ for $x\in X\setminus A$. $\endgroup$ Mar 9 '16 at 22:11
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HINT: Take $X=\Bbb R$ and $A=\Bbb Q$. (On further thought: it may be a little easier to take $A=\Bbb R\setminus\Bbb Q$.)

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  • $\begingroup$ I ended up using another example, which was $X = \Bbb R$ and $A = [0,1)$, but I'm curious how you could show that your example works. I tried looking at it, but found it difficult to say much about $\Bbb R/(R\Q)$. Can you expand on how you might show that $\Bbb Q$ is not homeomorphic to $\Bbb (R/(R\Q))\((R\Q)/(R\Q))$? My text formatting is a little off but hopefully that's still clear enough. $\endgroup$
    – G Pace
    Mar 10 '16 at 0:07
  • $\begingroup$ @GPace: Let $\Bbb P=\Bbb R\setminus\Bbb Q$, and let $p$ be the point of $\Bbb R/\Bbb P$ corresponding to $\Bbb P$. Every non-empty open set in $\Bbb R$ contains in irrational, so every non-empty open set in $\Bbb R/\Bbb P$ contains $p$. From this it follows that the open sets in $(\Bbb R/\Bbb P)\setminus\{p\}$ are essentially the subsets of $\Bbb Q$ of the form $U\setminus\Bbb P$, where $U$ is any open nbhd of $\Bbb P$ in $\Bbb R$. The closed sets in $(\Bbb R/\Bbb P)\setminus\{p\}$ are then essentially the subsets of $\Bbb Q$ that are closed in $\Bbb R$. $[\sqrt2,\sqrt3]\cap\Bbb Q$ is then ... $\endgroup$ Mar 10 '16 at 0:18
  • $\begingroup$ ... closed in $\Bbb Q$ but not in $(\Bbb R/\Bbb P)\setminus\{p\}$. $\endgroup$ Mar 10 '16 at 0:19

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