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Prove that the set $\mathbb{R}-\{1/n|n \in\mathbb{N}\}$ is not open.

OK, so I am having a little trouble. I know that the definition of open set is : iff every point of $A$ is an interior point of $A$. The definition of a closed set: iff its complement, $A^c$ is open.

I have made a number line to list the numbers in the set to find the interior points. I came up with,

$(-\infty, 1)\cup(1,1/2)\cup(1/2, 1/3)\cup(1/3,1/4).........$

From my thoughts, the set $A=Int(A)$. Am I on the right track?

Further, I am not sure how this set is not open (I guess meaning that it's closed), if all the points of $A$ is in the interior of $A$.

Pretty sure I am missing something, so any guidance would be appreciated.

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HINT: Show that $\{\frac1n\mid n\in\Bbb N\}$ is not closed. Recall that a set $A$ is closed if and only if whenever $x_n\in A$ is a convergent sequence, then $\lim x_n\in A$ as well.

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  • $\begingroup$ I am not familiar with this topic, but as far as I know if for each $x$ in the set there exists a $\epsilon$ such that all points in $(x-\epsilon,x+\epsilon)$ also be in the set then the set is called open. Is this definition correct? because for each point in the set I can imagine such interval. $\endgroup$ – hhsaffar Nov 17 '13 at 23:04
  • $\begingroup$ Yes, that is another definition of open sets (in the real numbers, anyway). $\endgroup$ – Asaf Karagila Nov 17 '13 at 23:13
  • $\begingroup$ Thank you. Now each $x>1$ in the mentioned set is member of a subset of the form $(1/n,1/(n+1))$ for these $x$s I will chose $\epsilon=min\{|x-1/n|,|x-1/(n+1)|\}$. For $x<1$ I will chose $\epsilon = |x-1|$ so the set is open. Would you please point out my mistake? $\endgroup$ – hhsaffar Nov 17 '13 at 23:22
  • $\begingroup$ First of all, you write the intervals backwards. $(a,b)$ is meaningful only when $a<b$ (else it's just an empty set). Secondly, I'm not sure what you mean. $\endgroup$ – Asaf Karagila Nov 17 '13 at 23:25
  • $\begingroup$ For each $x$ in the set I am introducing an $\epsilon$ such that $(x-\epsilon,x+\epsilon)$ is in the set, and thus I am trying to show the set is open. $\endgroup$ – hhsaffar Nov 17 '13 at 23:27
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Note that: $\mathbb{R}-\{1/n|n \in\mathbb{N}\} \neq (-\infty, 1)\cup(1,1/2)\cup(1/2, 1/3)\cup(1/3,1/4).........$ $\mathbb{R}-\{1/n|n \in\mathbb{N}\}=(-\infty,0]\cup(1,\infty)\cup(1/2,1),(1/3,1/2),(1/4,1/3),....$

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  • $\begingroup$ This representation should make it very clear which point should in your set probably is not an interior point. $\endgroup$ – PVAL-inactive Nov 17 '13 at 23:58
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First of all, not every set that is not open is closed. In fact, this set is an excellent example.

Now, you are told to prove that it's not open. You gave a good definition of "open" in $\mathbb{R}$ - a set can be called open iff all its points are interior points (that is, points that every open interval around them is in the set).

Your set has a special point in it. Can you locate it? Can you prove it's not an interior point?

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In this case the problem comes easy. For example $0\in A=\mathbb{R}-\{\frac{1}{n}:n\in\mathbb{N}\}$ and for any $\delta>0$ exist $n$ such that $\frac{1}{n}<\delta$ so it's imposible to find a ball ($B_{\delta}(0)$) around $0$, contained completly in $A$. Therefore we prove the negation of the definition of an open set.

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