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Hi everyone: Is a null set of $\mathbb{R}^n$, $(n>0)$, necessarily closed? Give a counter example. Thanks for your reply.

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I up-voted this question and its total is $0$, which means (1) someone down-voted it and (2) Why are those who've answered this and who've voted on the answers neglecting to up-vote it? –  Michael Hardy Jul 5 at 18:30
Dear @MichaelHardy : the solutions have merit, but the question is only a PSQ, so this distribution seems natural. Regards –  rschwieb Jul 6 at 0:08
@rschwieb : I'd rather not spend vast amounts of effort trying to decrypt your comment, especially when they would probably not succeed. Can you translate it into English? This web page doesn't help: acronyms.thefreedictionary.com/PSQ –  Michael Hardy Jul 6 at 15:09
Dear @MichaelHardy : Sorry, I thought you were aware of this abbreviation, which we use regularly in meta. From the context, I believe you can deduce the meaning of the sentence is "The solutions have merit, but the question does not." Hope this makes things clear. Regards –  rschwieb Jul 6 at 16:51
Could someone explain why my second answer below was down-voted? –  Michael Hardy Jul 6 at 18:22

4 Answers 4

up vote 7 down vote accepted

Any countable set is a null set, there are many nonclosed countable sets. eg $\mathbb{Q}^n$

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The set $\{\frac1n\}_{n\geq 1}$ is a nice countable (and therefore zero-measure) set in $\mathbb{R}$ that isn't closed, because it doesn't contain its limit point $0$.

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$$ \left\{ \frac 1 n : n=1,2,3,\ldots \right\} $$

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The set of members of the Cantor set that are not endpoints of the deleted intervals is an uncountable set of measure $0$ that is not closed.

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Is there a reason why this was down-voted? –  Michael Hardy Jul 6 at 15:08
Dunno. I thought it was a good answer. –  G Tony Jacobs Jul 6 at 16:02

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