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Let $A = \{\frac{1}{n}: n \in \Bbb{N}\}$ and $X = \Bbb{R}$ with topology $$\tau = \{U\smallsetminus B: \text{$U$ is open in $\Bbb{R}$ with usual topology and }B\subset A\}.$$

I need to show that $X/A$ is not Hausdorff.

For this, I am trying to show $\{0\}$ and $A$ can't be seperated by open sets in $X/A$.
So I let $U$ and $V$ to be open sets containing $\{0\}$ and $A$.

If $p$ is the projection map, then by definition of quotient topology, $p^{-1}$$(U)$ and $p^{-1}(V)$ are open in $\Bbb{R}$. They will contain $0$ and all points of $A$.

How can I show that these two sets will not be disjoint?

Help, please!

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$X$ is a Hausdorff space that is not regular as $A$ cannot be separated from $0$ by open sets.

Sketch of proof: suppose that $0$ is contained in some $U-A$, with $U$ standard-open, and $V \supseteq A$ is also standard open. Then $U$ contains some $\frac1n \in A$ and for some standard open $W$ we will have $\frac1n \in W \subseteq U \cap V$. This shows that $U-A$ and $V$ intersect.

Now we identify $A$ to a point to make $X{/}A$, and then $A$ becomes a point that still cannot be separated from $0$, hence the quotient is not Hausdorff. We just use the fact from the first paragraph about $X$ to show it.

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  • $\begingroup$ This is what I want to show. That A can't be separated from 0 $\endgroup$ – Gitika Dec 24 '20 at 10:59
  • $\begingroup$ @Gitika so you haven’t shown that $X$ is not regular? $\endgroup$ – Henno Brandsma Dec 24 '20 at 11:00
  • $\begingroup$ If I show that A and 0 can't be seperated(This is what I want to show) then X won't be regular..right? $\endgroup$ – Gitika Dec 24 '20 at 14:49
  • $\begingroup$ @Gitika sure. And that’s it very hard. It’s proved on several places on this site. $\endgroup$ – Henno Brandsma Dec 24 '20 at 14:51
  • $\begingroup$ @Gitika I added a proof sketch with some minor details omitted. $\endgroup$ – Henno Brandsma Dec 24 '20 at 15:11

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