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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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    @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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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.