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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.

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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
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1 Answer

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 $z\in\overline{\{y\}}$.

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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
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