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My textbook say this set is OBVIOUSLY dense in $\mathbb{R}$ and not closed, but I don’t get it. How to show it is dense and not closed?

abeliangrape
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    It's not obvious. The idea is to work by contradiction. Suppose it is not dense, deduce that this means there is a least positive element $x$ and use that to conclude that $\alpha $ is rational, thereby obtaining the desired contradiction. – lulu Dec 04 '19 at 00:27
  • And once you know that it's dense, you know it's not closed, because a dense closed set is the entire space. But this set is countable, so it can't be all of $\mathbb{R}$. – Justin Barhite Dec 04 '19 at 00:29
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    Somehow the OP asked for showing that it is not "closed". This can be shown once we know it's dense: if it's closed, then it is equal to the whole $\mathbb R$ by density. But the set is obviously countable while $\mathbb R$ is not countable. – WhatsUp Dec 04 '19 at 00:32

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