# How do you construct interval $(a, b)$ using just intervals of the form $(b, \infty)$?

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?

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

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.