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A space is said to be totally disconnected iff its components (i.e. the maximally connected subsets of the space) are the one-point sets. In other words, a space is totally disconnected iff the one-point sets cannot be decomposed into the union of two or more disjoint open sets.

I am working on the proof that the algebra of clopen subsets of the Cantor space is atomless. The argument, based on the property that the Cantor space has no isolated points, implies that a clopen set of the Cantor space can always be decomposed into two non-empty clopen sets again.

How is this possible. Doesn't this argument contradict the property of total disconnectedness? What do I misunderstand?

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    $\begingroup$ Consider $\mathbb Q$ which is totally disconnected and having no isolated points wrt the standard topology induced from $\mathbb R$. Hence totally disconnected spaces and discrete spaces are different. $\endgroup$ Oct 23, 2017 at 12:16
  • $\begingroup$ $\mathbb R$ with the usual topology is not totally disconnected, so how do you decomposed a one-point set into the union of two or more disjoint open sets? $\endgroup$
    – Mr. T
    Oct 23, 2017 at 12:28
  • $\begingroup$ @Mr.T You can never decompose a one point set into a union of two or more nonempty sets, open or not. Totally disconnected is that any subset with more than one point can be decomposed. $\endgroup$
    – J126
    Oct 23, 2017 at 12:41
  • $\begingroup$ @Joe Johnson I know, I just wanted that OP notice that. $\endgroup$
    – Mr. T
    Oct 23, 2017 at 12:46

1 Answer 1

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I think I see where your problem is stemming from. Consider the following two statements:

Any subset of the Cantor can be decomposed into the union of two disjoint, nonempty open sets if and only if it has more than one element.

Any clopen subset of the Cantor set can be decomposed into the union of two disjoint, nonempty clopen sets.

Using these two statements, what can you conclude about the clopen sets of the Cantor set? Are all subsets clopen? Why or why not?

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