Let $T$ be an topology on $\mathbb{R}$ defined by $U\in T$ if and only if either $U$ doesn't contain $1$ or $U$ contains $0$. Would you help me how to check whether $T$ satisfiying separation axiom $T_0,T_1,T_2,T_3,T_4$. Thanks.
Checking separation axiom
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general-topology
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1Note that $0 \neq 1$. Let $V$ be any open set containing $1$, then by definition $0\in V$, hence $V$ always intersect open set that contain $0$. Hence, $T$ is not $T_2$. Since $T_3$ and $T_4$ implies $T_2$ then $T$ is not $T_3,T_4$ – 2012-12-24
1 Answers
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You answered most of your own question correctly in your last comment; only the $T_1$ and $T_0$ properties remain to be checked. Your observation in the comment actually shows that $T$ is not $T_1$; do you see why? Finally, $T$ is $T_0$; to prove this, you must show that for each $x,y\in\Bbb R$ with $x\ne y$ there is a $U\in T$ such that either $x\in U$ and $y\notin U$, or $y\in U$ and $x\notin U$. If $x\ne 1$, $U=\{x\}$ works; why? What can you use for $U$ if $x=1$? (There are two fairly natural choices that work nicely.)