Problems from Union, intersection and complementary of intervals

Development:

1. We calculate first the intersection, and then we will calculate the complementary of the result given. Observing the endpoints of the given intervals, we have the following order: $-2<1<3<8$

   So, we know that the values between $1$ and $3$ belong to both intervals, and therefore, they belong to the intersection. So the result of the intersection is: $$(1,8)\cap [-2,3]=[1,3)$$ Now let's calculate the complementary of this interval: $$\overline{[1,3)}=(-\mathrm{\infty },1)\cup [3,+\mathrm{\infty })$$

2. We calculate first the complementary: $$\overline{[\sqrt{5},9]}=(-\mathrm{\infty },\sqrt{5})\cup (9,+\mathrm{\infty })$$ $$\stackrel{―}{(-2,{\displaystyle \frac{\sqrt{2}}{3}})}=(-\mathrm{\infty },-2]\cup [{\displaystyle \frac{\sqrt{2}}{3}},+\mathrm{\infty })$$

   Then, as ${\displaystyle \frac{\sqrt{2}}{3}}<\sqrt{5}$, the union is the entire $\mathbb{R}.$

3. We have: $\overline{(-\mathrm{\infty },-{\displaystyle \frac{5}{7}})\cup \stackrel{―}{(-4,+\mathrm{\infty })}}=\overline{(-\mathrm{\infty },-{\displaystyle \frac{5}{7}})}\cap \overline{\stackrel{―}{(-4,+\mathrm{\infty })}}$

   But, as the complementary of the complementary is the same set, we have:

   $\overline{(-\mathrm{\infty },-{\displaystyle \frac{5}{7}})}\cap \overline{\stackrel{―}{(-4,+\mathrm{\infty })}}=\overline{(-\mathrm{\infty },-{\displaystyle \frac{5}{7}})}\cap (-4,+\mathrm{\infty })$

   If we calculate the complementary: $\overline{(-\mathrm{\infty },-{\displaystyle \frac{5}{7}})}=[-{\displaystyle \frac{5}{7}},+\mathrm{\infty })$

   So finally, we calculate the intersection: $$[-{\displaystyle \frac{5}{7}},+\mathrm{\infty })\cap (-4,+\mathrm{\infty })=[-{\displaystyle \frac{5}{7}},+\mathrm{\infty })$$

Solution:

1. $(-\mathrm{\infty },1)\cup [3,+\mathrm{\infty }):$ it is not an interval, since it is the union of two intervals, both unbounded and one open and the other closed.

2. $\mathbb{R}=(-\mathrm{\infty },+\mathrm{\infty }):$ it is an unbounded interval.

3. $[-{\displaystyle \frac{5}{7}},+\mathrm{\infty }):$ it is a closed interval, and unbounded above.

Hide solution and development