My attempt is $(a, b) = (a, \infty) \setminus \cap_{n=1}^{\infty} (b - 1/n, \infty)$ with $n \in \mathbb{Z}^{+}$; $a,b \in \mathbb{R}$ and $a < b$. Is this correct?

  • 2
    $\begingroup$ Looks good to me $\endgroup$ Commented Dec 14, 2020 at 14:49
  • $\begingroup$ But that just a very complicated way of writing $(a, \infty)$\ $(b, \infty)$! $\endgroup$
    – user247327
    Commented Dec 14, 2020 at 15:02
  • 1
    $\begingroup$ @user247327 But isn't your solution equal to $(a, b]$ ? $\endgroup$
    – Darby Bond
    Commented Dec 14, 2020 at 15:05

1 Answer 1


Yes, that works: the elements of $$\bigcap_{n\in\mathbb{N}} (b-{1\over n},\infty)$$ are exactly those real numbers which are $>b-{1\over n}$ for every $n\in\mathbb{N}$, which is to say $$\bigcap_{n\in\mathbb{N}} (b-{1\over n},\infty)=[b,\infty).$$ And $(a,\infty)\setminus [b,\infty)=(a,b)$ as desired.

The one subtlety here (which of course you avoided, but is worth mentioning for completeness) is that one needs to pay attention to the point $b$: e.g. $(a,\infty)\setminus (b,\infty)$ does not work since it gives $(a,b]$ instead.


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