-1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $x$? $\endgroup$ – T. Eskin Mar 26 '15 at 20:01
  • $\begingroup$ Yes, @ThomasE. that is exactly how I have been going about it. $\endgroup$ – Zeta10 Mar 26 '15 at 20:03
  • $\begingroup$ Ok. So the assumption is that if two open sets intersect, then the intersection is a singleton. $\endgroup$ – T. Eskin Mar 26 '15 at 20:05
  • $\begingroup$ Any explanation for the down vote? $\endgroup$ – Zeta10 Mar 26 '15 at 20:28
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Zeta10 Mar 26 '15 at 20:17
  • 1
    $\begingroup$ @Zeta10: Yes. Correct. $\endgroup$ – T. Eskin Mar 26 '15 at 20:41
  • $\begingroup$ I was working with a partner who said that the Zariski topology was the hint, but later they mentioned that was NOT actually the hint! This is just one of the costs of doing mathematics! Thanks for your help! $\endgroup$ – Zeta10 Mar 26 '15 at 20:45
  • $\begingroup$ @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. $\endgroup$ – T. Eskin Mar 27 '15 at 1:17
0
$\begingroup$

It turns out the Zariski topology does work. Here is the proof.

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$.

$\endgroup$
  • $\begingroup$ This is answering a different question than what you formulated in the problem statement in the comments. Saying that the intersection of all open sets around a point will yield the singleton, is not the same as saying that any two open sets containing a point will yield the singleton. With Zariski topology the intersection of two open sets is always uncountably infinite, while with uncountably infinite intersections you can get the singleton. $\endgroup$ – T. Eskin Mar 29 '15 at 0:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.