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Is the topology on $\mathbb{R}$ generated by the basis $(a,\infty)$ Hausdorff?

I do not think it is Hausdorff.

Proof: Taking unions of the sets of the form $(a,\infty)$ will always produce another set of the same form. So given $a < b$, whatever $2$ open sets you choose, they will never be disjoint.

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Is the topology on R generated by the basis (a,∞ ) Hausdorff?

No, it isn't.

Let $a,b \in \mathbb R$, say $a<b$. Then every open set $C=(c,\infty)$ containing $a$ also contains $b$, since $$a\in C=(c,\infty) \, \Rightarrow \, c<a \, \Rightarrow \, c<b \, \Rightarrow \, b\in C=(c,\infty)$$

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  • $\begingroup$ So this will be an example of a space that is $\mathrm T_0$ but not $\mathrm T_1$ ? $\endgroup$
    – GEdgar
    Oct 24, 2016 at 12:35
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    $\begingroup$ Yes, this space is $\mathrm T_0$ but not $\mathrm T_1$. $\endgroup$
    – Rodrigo Dias
    Oct 24, 2016 at 16:43

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