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I have a problem trying to prove the following theorem.

If $X$ is a $T_{0}$ space with a base $C$ of clopen sets, then X is totally disconnected.

By definition, a space is totally disconnected if and only if $C(x) = {x} \quad \forall x \in X$. I try to prove it by contradiction, suposing that $\exists y \in C(x) , x \neq y $ and trying to use the $T_{0}$ property for finding a contradiction, but I am stucked.

I haven't found any solution to the problem in the web, so I would aprecciate if somebody could help me.

Regards.

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  • $\begingroup$ $T_0$ says we have a neighbourhood $U$ of $x$ that doesn't contain $y$, or a neighbourhood $V$ of $y$ that doesn't contain $x$. Now find a $W \in C$ that contains one but nor the other. $\endgroup$
    – Daniel Fischer
    Nov 23, 2016 at 20:28
  • $\begingroup$ Thanks for the quick reply. I managed to solve it with the answer and the property that any open set contains an element from the base. $\endgroup$
    – sayeg84
    Nov 24, 2016 at 1:32

1 Answer 1

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HINT: If $x\ne y$, then either there is an open $U$ such that $x\in U$ and $y\notin U$, or there is an open $U$ such that $y\in U$ and $x\notin U$. Without loss of generality assume that there is an open $U$ such that $x\in U$ and $y\notin U$. Since $X$ has a clopen base, there is a clopen $B$ such that $x\in B\subseteq U$. Use $B$ to show that $y\notin C(x)$.

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  • $\begingroup$ I made the proof, thanks. I used the property that for a given connected subspace $C \subseteq X$ if $B \subseteq X$ is a clopen set such that $C \cap B \neq \emptyset$ then $C \subseteq B$ $\endgroup$
    – sayeg84
    Nov 24, 2016 at 1:39
  • $\begingroup$ @Aldo: You’re welcome. Yes, that’s one way to do it. $\endgroup$
    – Brian M. Scott
    Nov 24, 2016 at 5:10

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