# Question on showing quotient space is not Hausdorff

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?

$$X$$ 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.
• @Gitika so you haven’t shown that $X$ is not regular? – Henno Brandsma Dec 24 '20 at 11:00