Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Homework, again...

A topological space $(X,\tau)$ is called $T_0$-space if these two equivalent statements are true:

i) If $x,y \in X, x \not= y$, then $\overline {\{x\}} \not= \overline {\{y\}}$

ii)If $x,y \in X, x \not= y$, then $\exists \ U \in \tau, \ x\in U, \ y \not\in U$ OR $\ y\in U, \ x \not\in U$

I need to prove these statements are equivalent.

$ii \Rightarrow i $:

$y \not \in U \Rightarrow y \in U^c$, which is closed. Thus $\overline {\{y\}} \subset U^c$. We notice that $x \not \in \overline {\{y\}}$ (for $x$ cannot be in $U$ and $U^c$ at the same time), but $x \in \overline {\{x\}}.$ Thus $\overline {\{x\}} \not= \overline {\{y\}}$.

This, of course, doesn't mean that they need to be completely separate. Like if $X = \{a,b\}, \ \tau = \{\emptyset, \{a\}, X\}$. This is $T_0$, as $\overline {\{a\}} = X$, but $\overline {\{b\}} = \overline {\{b\}}$. They are not same, but they do have shared points.

$i \Rightarrow ii$:

Thus far the best I can do is state the antithesis: $$x \in U \iff y \in U \land \exists \ z \in \overline {\{x\}}, z \not\in \overline {\{y\}}.$$ Is this even a good way to do this? I have about two pages of trying to see if some kind of intersection of closures or complements or combinations thereof produces a contradiction... Any advice would be appreciated.

share|cite|improve this question
The second definition is wrong. It should say "$\exists U\in\tau. (x\in U\land y\notin U)\lor(x\notin U\land y\in U)$". – Asaf Karagila Feb 4 '13 at 17:33
Or I can just name x to by y and y to be x and be done with it... I figured it would be clear enough without the long ream of inclusions and ors and ands. – Valtteri Feb 4 '13 at 17:36
But that is not the same thing. For example take the topology on $\mathbb R$ where all the non-empty open sets include $0$. The result is $T_0$ but it is not satisfying the condition written in your post. – Asaf Karagila Feb 4 '13 at 17:38
@AsafKaragila So an open set necessarily includes 0, so if I select any other point, there can be no open set that only that and no 0...I can see the point. I will add some correction. – Valtteri Feb 4 '13 at 17:45
up vote 1 down vote accepted

Recall that $x\in\overline{A}\setminus A$ if and only if every open set including $x$ intersects both $A$ and its complement.

Suppose that $\overline{\{x\}}\neq\overline{\{y\}}$. If one of the point is closed then we are done, as its complement is open and is the open set witnessing the $T_0$ separation. If neither points is closed then there is some $z\neq x,y$ such that $z\in\overline{\{x\}}$ or $z\in\overline{\{y\}}$, but not in both.

If $z\in\overline{\{x\}}$ then there exists an open set $U$ such that $y\notin U$, witnessing $z\notin\overline{\{y\}}$, but every open set including $z$ must include $x$ as well, and so that particular $U$ is open $x\in U$, $y\notin U$.

In a similar manner when $z\in\overline{\{y\}}$.

share|cite|improve this answer
very much thank you. The truth that boundary must include points from both inside and outside of the set eluded me. – Valtteri Feb 4 '13 at 18:13
Yes, and with that wrong definition in mind you wouldn't have been able to prove that either! :-) – Asaf Karagila Feb 4 '13 at 18:17

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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