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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.

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    Sounds like you've answered your own question - [you should post that answer, which is explicitly allowed here](http://blog.stackoverflow.com/2011/07/its-ok-to-ask-and-answer-your-own-questions/). Also, I don't suppose you're the same user as the one who [posted this question](http://math.stackexchange.com/q/264366/264)? If so, I can merge your accounts if you'd like.2012-12-24
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    What are people's opinions about closing this as a duplicate?2012-12-24
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    @Zev Chonoles : No. I just check if my argument correct or not, and also get a constructive comment. I'm a different user2012-12-24
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    @ZevChonoles: How about my answer? Is it correct? Or i have to add some additional argument?2012-12-24
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    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.2012-12-24
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    @ZevChonoles: I would close it as a duplicate. The other one asked more questions, but this one is included.2012-12-24
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    @Zev: I’d leave it open long enough to give *ask* a reasonable chance to write up and accept an answer.2012-12-24

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