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

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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
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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.

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