00:01
In this problem we're asked to find beginning of the intersection of the sets word sets are parametric so they depend on value of i.
00:10
So in part of day we have a set a i represented as i i plus 1, i plus 2 so let's look at a 1.
00:17
When i is 1 we would have 1 2 3 up to infinity.
00:23
When i used 2 we have 2 3 up to infinity.
00:25
When i is 2 we have 2 3 up to infinity.
00:29
When i is 3, we have 3 at the infinity.
00:35
So as you can see, as we increase i as i goes up, we lose an element, right? so a number of elements, number of elements goes down.
00:51
So what does this mean? this means that if you look at the union of all these sets, and if you keep increasing i, then a1 will be the set with the most number of elements right so then the union will be a 1 and i which is 1 2 3 up to infinity now if you now look at the intersection of all these sets then we're actually looking for the set with the least number of elements okay so as we keep again increasing i we lose uh one element so as we keep if we do this um infinite times then we would end up with a set with no elements and it can use that then the intersection would be an empty set.
01:43
So this was a now in part b we have a i defined as zero and i so then a one would look like zero one a two would look like zero two a three would then look like zero three and if you're if you look at the union of all these sets we would have zero one union zero two union zero three up to zero and infinity or yeah so then we could say that union of all these sets will be the natural numbers us then now let's look at the intersection of these sets as you can see, as i increases this number changes, so the second digit changes, but the first one stays the same.
02:43
So then the common element for all the sets is the element zero.
02:50
Now in part c, ai is given as open interval 0i...