00:01
This question we are required to compute the infinite union and infinite section for some specific set, for sequence of set a i.
00:10
Okay, first, the first a i, the first sequence a i is equal to, okay, minus i, minus i plus 1, blah, blah, blah, minus 1, 0, 1, 2, i.
00:29
Or a i can be written as n is some integers and the absolute of n is less or equal to i.
00:43
Okay, so it's easy for us to see a i, a 1 is a subset of a 2, is a subset of a 3, is a subset of a i, and blah, blah, blah.
00:55
So it's a fact that we can observe.
01:00
And that means the infinite union will be just equal to this one, right? because if we use this expression, we know the infinite union will be actually equal to the union of all sets, such that n is the integer, the absolute value of n is less or equal to i.
01:33
You can see i approaches infinite, and the range of our absolute value also approaches positive infinity.
01:41
So this is actually equal to the integer.
01:46
And once we notice this fact, we find the intersection of a 1 and a 2 is equal to a 1.
01:54
And if we take a 3, we still get a 1.
01:58
So this infinite intersection is just equal to a 1.
02:05
Which is actually equal to minus 1, 0, 1.
02:11
Okay, this is the first sequence.
02:15
The second sequence, a i is equal to minus i and i.
02:23
Then we can see a i and a j are some disjoint set if i is not equal to j.
02:36
Because, for example, a 1 contains minus 1 and 1.
02:40
A 3 contains minus 3 and a 3.
02:42
They are disjoint with each other.
02:45
That means this infinite intersection must be equal to the empty set.
02:52
Because the intersection of any two of them must be empty.
02:57
That means if we intersect them all, we must get an empty set.
03:02
And the infinite union is the infinite union of a lot of disjoint sets.
03:11
So it will be just equal to z minus 0.
03:15
Because 0 is not contained in any a i.
03:23
So this guy contains any integers just as we've discussed here.
03:32
But 0 is not included.
03:36
Okay, the third sequence, now a i is then closed interval from minus i to i.
03:46
Now, again, we can find a 1 is a subset of a 2.
03:53
Because a 1 is a subset from minus 1 to 1.
03:56
While a 2 is from minus 2 to 2.
04:00
A 3.
04:03
That means the infinite intersection will be just equal to a 1...