Proof of a set union and intersection

[Guest]

Hello!

Lately, I've been struggling with this assignment. (angle brackets represent closed interval)

Screenshot_20211116_184448(1).png (40.15 KiB) Viewed 445 times

I figured out that:

a)
union = R
intersection = {0}

b)
union = (0, 2)

intersection = {1}


I asked my prof about this and she explained to me that it should be shown that if a set is an intersection of sets, then it belongs to each of those sets and, conversely, nothing else belongs to the intersection, so every other element does not belong to at least one of those sets. But I don't really know how to interpret this or where to even start. (normally, when proving the equality of two sets, I would try to prove that A⊆B and B⊆A, but I don't see how that's applicable here).


Thank you for your help!

[Guest]

x is in [tex]A\cup B[/tex] if and only if x is in A or in B.
x is in [tex]A\cap B[/tex] if and only if x is in A and in B.

(a) x is in [tex]\cap A_{t\in T}[/tex] if and only if x is in all of the sets [tex]A_t[/tex]. Given any non-zero number, x, there exist some t> 0 such that |t|< x. x is NOT in [tex]A_t[/tex] for that t so is NOT in the intersection of such sets.

x is in [tex]\cup A_{t\in T}[/tex] if and only if x is in any one of the sets [tex]A_t[/tex]. Given any number, x, there exist some t> 0 such that x< |t| so -t< x< t. x is in [tex]A_t[/tex] for that t so is in the union of such sets.

(b) Now, T is the set of all numbers between 0 and 1 (but does not include 0 or 1). For t just slightly larger than 0, t+ 1 is just slightly larger than 1, (0, 1). For t just slightly less than 1, t + 1 is just slightly less than 2, (1, 2). The union of all such sets includes numbers from slightly larger than 0 to slightly less than 2, (0, 2). The intersection of all such sets is empty.