# T_1 separation axiom in topology [duplicate]

Possible Duplicate:
Checking separation axiom

Let $R$ be an topology on $\mathbb{R}$ defined by $V$ open if and only if either $0\in V$ or $2\notin V$. Would you help me how to check whether $R$ satisfiying separation axiom $T_1$ .

My work: Let $V$ be open set containing $2$. By definition, $0\in V$. Note that $0\neq 2$. Since for all $V$ open that containing $2$, we have $0\in V$ then $T_1$ is not satisfied. Thanks.

-

## marked as duplicate by Ross Millikan, Henry T. Horton, Alexander Gruber♦, Quixotic, Davide GiraudoDec 24 '12 at 11:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Sounds like you've answered your own question - you should post that answer, which is explicitly allowed here. Also, I don't suppose you're the same user as the one who posted this question? If so, I can merge your accounts if you'd like. – Zev Chonoles Dec 24 '12 at 3:53
What are people's opinions about closing this as a duplicate? – Zev Chonoles Dec 24 '12 at 3:54
@Zev Chonoles : No. I just check if my argument correct or not, and also get a constructive comment. I'm a different user – ask Dec 24 '12 at 3:56
@ZevChonoles: How about my answer? Is it correct? Or i have to add some additional argument? – ask Dec 24 '12 at 3:57
It looks just fine (though I, personally, would want to explicitly justify that it does indeed define a topology). As Zev says, you should post it as an answer. – Cameron Buie Dec 24 '12 at 4:13

## 1 Answer

The space is $T_0$: if $x \neq y$ then at least one of them, say $x$, is unequal to $2$. In that case $\{x\}$ is open and does not contain $y$, and so $T_0$ has been shown.

The space is not $T_1$: let $x = 0$ and $y = 2$. If $X$ were $T_1$ there would be an open set $O$ that contains $2$ but not $0$. But if $O$ contains $2$, by the definition of the topology it must contain $0$, so we have a contradiction.

-