I think the answer is: yes, it is!. Due to all the points of an open set are interior points, the set contains a ball of radius r>0 , I tried to translate the question to demonstrate that any ball is an infinity uncountable set, but I don't see how to do it. I would like to receive some suggestion.
Is any open subset $A≠∅$ of $R^n $ with the usual topology uncountable?
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0any ball contains an interval. Then prove this interval is uncountable. – 2017-02-14
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0Ok I see, there is a set in bijection with some interval – 2017-02-14
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Yes !!. If a non-empty finite set A, with cardinality 'n', was open. Take an element x$\in$A. Then, there must be a ball around 'x', say B$_{\epsilon}$(x) completely in the parent set.
i.e. B$_{\epsilon}$(x)$\subseteq$A which will imply $|$ B$_{\epsilon}$(x)$|$$\leq$$|$A$|$ i.e. $\infty$$\leq$n. A contradiction.
Similarly, you can do the same for any countably infinite open set.
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1I don't understand this answer. It already assumes that $B_\varepsilon(x)$ is uncountable. If you "only assume it is infinite", then I don't see how "Similarly, you can do the same for any countably infinite open set". – 2017-02-14
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0@AsafKaragila I think you are correct. Best way is to show that open ball, having an interval, is uncountable and hence every open set is uncountable, because each open set has an open ball – 2017-02-15