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The set is obtained by removing the rational points from the interval $[4,7].$ How do I check to see if this set is closed in $\mathbb R$ ?

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    $\begingroup$ It's not. There is a sequence tending to $7$ through the irrationals, but 7 is not in the set. $\endgroup$ Aug 31, 2015 at 19:24
  • $\begingroup$ you mean 7 is a limit point of the set? $\endgroup$
    – Avinash N
    Aug 31, 2015 at 19:26
  • $\begingroup$ Yes. I do. (Still haven't beaten the character limit.) $\endgroup$ Aug 31, 2015 at 19:27
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    $\begingroup$ Is a limit point - there is no the limit point of that set. Indeed, every point in $[4, 7]$ is a limit point of that set. However, that's much more than you need. You can get away with showing that just one rational in $[4, 7]$ is a limit point of the irrationals in $[4, 7]$. $\endgroup$ Aug 31, 2015 at 19:33
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    $\begingroup$ @PatrickStevens To beat the character limit you can just add a bunch of \$ $\endgroup$
    – Math1000
    Aug 31, 2015 at 23:45

2 Answers 2

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Take , $A=[4,7]\cap \mathbb I$ ; where , $\mathbb I$ is the set of all irrational points.

Then , $\bar A=[4,7]\not= A$.

So ..............?

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  • $\begingroup$ i think that, A is dense in [4,7]. Is my think is correct? $\endgroup$
    – Avinash N
    Aug 31, 2015 at 19:25
  • $\begingroup$ Dense in what ? $\endgroup$
    – Empty
    Aug 31, 2015 at 19:26
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    $\begingroup$ @Avinash: Yes: both $A$ and $[4,7]\setminus A$ are dense in $[4,7]$. $\endgroup$ Aug 31, 2015 at 19:30
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    $\begingroup$ @ Avinash N) Do you know the definition of ' Dense ' ? Here $A$ is NOT dense in $\mathbb R$. $\endgroup$
    – Empty
    Aug 31, 2015 at 19:30
  • $\begingroup$ yes i know. Thank you sir. $\endgroup$
    – Avinash N
    Aug 31, 2015 at 19:31
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As an alternative to using the theorem closed iff contains all limit points as in Panja's Answer, it is instructive to appeal to the definition of closed as a set whose complement is open.

$$A^\sim = (-\infty,\,4) \cup (7,\,\infty) \cup ([4,\,7]\cap\mathbb{Q})$$

Consider $5\in A^\sim$. Any neighbourhood of $5$ contains an open interval of the form $(5-\epsilon_-,\,5+\epsilon_+)$ (with $1>\epsilon_+,\,\epsilon_->0$), which in turn contains $[5-\epsilon_-/2,\,5+\epsilon_+/2]$ which in turn contains irrationals not in $A^\sim$. So, no neighborhood of $5$ is contained in $A^\sim$.

Now use theorem open iff a neighborhood of all its points. Thus $A^\sim$ is not open, so its complement $A$ is not closed.

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