# Let $X$ be a topological space. Prove that for any $x$ in the intersection of all opens sets $=\{x \}$, the space $X$ need not be Hausdorff.

Let$X$ be a topological space. Prove that for any $x$ in the intersection of opens sets $=\{x \}$, the space $X$ need not be Hausdorff.

My thoughts / strategies.

I want to choose some other topology and show that if $x$ is in the intersection of open sets where the intersection contains only the set {x}, there exists some x' that is not in the intersection of open sets. My initial thoughts are to use the discrete metric. But, then I think that the discrete metric is Hausdorff. Our hint is to use the Zariski topology, but it seems impossible to find an intersection containing just one point. So, I tried looking at closed sets in the Zariski topology instead. I do not want a solution, but more of a point in the right direction if possible.

• Can you rephrase the assumption? Do you mean that if $x$ is in the intersection of two open sets then this intersection must be equal to the singleton of $x$? – T. Eskin Mar 26 '15 at 20:01
• Yes, @ThomasE. that is exactly how I have been going about it. – Zeta10 Mar 26 '15 at 20:03
• Ok. So the assumption is that if two open sets intersect, then the intersection is a singleton. – T. Eskin Mar 26 '15 at 20:05
• Any explanation for the down vote? – Zeta10 Mar 26 '15 at 20:28

Hint: The set $\{0,1\}$ has $3$ different non-trivial topologies, $2$ of them are not Hausdorff. Take either one and it should do the job.
• I see. So, this topology is not Hausdorff, but if I intersect ${0}$ and ${0,1}$ I get the set ${0}$, which is the condition I am attempting to achieve. – Zeta10 Mar 26 '15 at 20:17
• @Zeta10: Well the Zariski topology on $\mathbb{R}$ is obviously not going to work, because every open set has a finite complement. So any two open sets have an uncountable intersection. – T. Eskin Mar 27 '15 at 1:17
Let $V = \displaystyle\bigcap_{x \in U;\ U \subset \mathbb{R}}= \{x \}$. We want to show $y \ne x , y \notin V.$ Let $U = R - \{ y \}.$ Then we have $x \in U$ and clearly $y \notin U$ Taking the intersection of all open sets around $x$ gives us $V = \displaystyle\bigcap_{x \in U; U \subset \mathbb{R}} = \{x \}$, with $y \notin V$.