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Prove that the set $A =\displaystyle \left \{ \frac{1}{n+1} : n \in \mathbb N \right \} $ is a nowhere dense subset of $\displaystyle{ \mathbb R }$.

I have think two ways but I can't finish it. Here it is

  1. I tried to prove that $ \text{int} (\bar{A}) = \emptyset$. For this it is enough to prove that $ \bar A =\mathbb Q $ but I can't show this.

  2. I tried to prove that every interval of $ \mathbb R$ contains a subinterval whose intersection with $A$ is the empty set.

Any help?

Thank's in advance!

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    Kindly do not use `\displaystyle` in title as it breaks the layout of the list of questions2012-05-11
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    You want to prove that the closure of $A$ is a set that's not closed?2012-05-11
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    The closure of $A$ cannot be $\mathbb{Q}$ because $\mathbb{Q}$ is not closed while the closure of $A$ is.2012-05-13

1 Answers 1

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You couldn't prove that $\operatorname{cl}A=\Bbb Q$ because it's not true. Try proving instead that $\operatorname{cl}A=A\cup\{0\}$.

By the way, it's also possible to use your second approach: the key step is showing that if $0, there is an $n\in\Bbb N$ such that $$\frac1{n+2}

Added: If you've not already done so, draw a picture:

enter image description here

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    You're too fast!2012-05-11
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    Can you give some hint how to prove $\operatorname{cl}A=A\cup\{0\}$ ? And why is this the key using the second approach?2012-05-11
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    @passenger: I assume that you've no trouble proving that $0$ is a limit point of $A$. Let $x\in\Bbb R\setminus (A\cup\{0\})$. If $x<0$ or $x>1$ you should have no trouble finding an open interval around $x$ disjoint from $A$. If $0, show that $$\frac1{n+2} for some $n\in\Bbb N$. (Doing this should answer your other question for you as well.)2012-05-11
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    Done! Thank you very much for your time!2012-05-11