Let's call a space zero-dimensional if it has a basis of clopen sets, and is $T_0$. Is there a zero-dimensional space that is not Hausdorff?
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asked Apr 21 '15 at 6:43
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No. Let $x\neq y$ and let be $ U$ a clopen neighbourhood of $x$ not containing $y$. Then the complement of $U$ is a clopen neighbourhood of $y$, disjoint from $ U$.
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$\begingroup$ I'm not sure I agree. How do you know such a $U$ exists? What if the only neighborhood of $x$ is the entire space? $\endgroup$ – manthanomen Apr 21 '15 at 7:02
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$\begingroup$ @manthanomen Such a $U$ is guaranteed by the definition of a basis: for any open set $U$ and $x \in U$, there is a basis element $V$ such that $x \in V \subset U$. $\endgroup$ – user98602 Apr 21 '15 at 7:05
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$\begingroup$ @MikeMiller But the containment $V \subset U$ isn't necessarily proper. But I realized the argument works because of the $T_0$ condition. If we can't find a neighborhood of $x$ not containing $y$, we can find a neighborhood of $y$ not containing $x$ and proceed similarly. $\endgroup$ – manthanomen Apr 21 '15 at 7:29
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$\begingroup$ Oh, I misread your complaint. Yes, I agree that $x$ needs to be the one that actually has a neighorhood (otherwise, as you said, just swap them). Sorry for the confusion. $\endgroup$ – user98602 Apr 21 '15 at 7:32
