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Let $C \subseteq \mathbb{R}$ be a set that has uncountably many connected components. What can be said about the openness/ closedness of $C$ in $\mathbb{R}$?

(One such example is the Cantor set. It is totally disconnected so its connected subsets are only singletons, of which there are uncountably many since the Cantor set is uncountable as it can be put in bijection with the set of all binary sequences.)

Question: is my proof of the following claim true?

Claim: $C$ is not open.

Proof: Suppose $C$ was open. Then $C$ can be written as the union of countably many disjoint open intervals. Recalling the proof of this result, one knows that each interval is the largest interval containing every point in that interval and hence the largest connected set containing every point in it. Hence $C$ has only countably many connected components, a contradiction. Hence $C$ is not open.

I don't think anything can be said about the closedness of $C$ since the irrationals are also totally disconnected and uncountable and hence constitute another example of $C$ but they are neither open nor closed.

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  • $\begingroup$ You are right about your claim. $\endgroup$ – pisco Jun 3 '18 at 7:11
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In a locally connected space (like $\mathbb{R}$ is) the connected components of an open set are open.

In a separable space (like $\mathbb{R}$ is) there cannot be an uncountable family of pairwise disjoint open sets (every one of these open sets must contain a necessarily different member of a countable dense set).

These facts together imply that an open subset of a locally connected separable space has at most countably many components. So your hypothesis is right.

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As pisco said, you are right. But the follwing argument is even easier.

If $C$ is open, you can write it as the union of countably many (not necessarily disjoint) open intervals $J_n$. Each is $J_n$ is connected, therefore contained in a component $C_n$ of $C$. The $C_n$ cover $C$. Hence $C$ is covered by countably many components.

Note that you may have $C_n = C_m$ for some pairs $(n,m)$ with $n \ne m$, but that is irrelevant for the countability argument.

The argument can be generalized to open subsets of any locally connected separable space.

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