00:01
Hello, let's have a look at the question.
00:04
So the question state, suppose that there are a function in which f is defined in the interval a to b and the sets a1, a2 is a subset of a and b1, b2 is a subset of b.
00:24
Now let's recall that the pre -image of a subset b ' of b through the given function which is defined from a to b.
00:34
Now here in the a part we have to prove the following identities.
00:38
So let us begin the solution.
00:41
So in the a part we have to prove that f a1 union a2 is equals to f a1 union f a2.
00:54
So now here let us take y belongs to f a1 union a2.
01:01
So then we can say y will be equals to f a or we can say y is equals to f b where a belongs to a1 and b belongs to a2.
01:19
Hence here we can say that y belongs to f a1 and union f a2.
01:30
So that is from here we can say that f a1 union a2 is a subset of f a1 union f a2.
01:43
Let it be the first equation.
01:46
Now conversely here we can say that if y belongs to f a1 union f a2.
01:58
So then we can say y will be equals to f at a or y is equals to f at b where a belongs to a and b belongs to b.
02:13
Therefore here we will have that y belongs to f a1 union a2.
02:21
That is here we can say f a1 union f a2 is a subset of f a1 union a2.
02:32
Let it be the second equation.
02:35
Now from the first and second equation we have that f a1 union a2 is equals to f a1 union f a2.
02:49
Next we have f inverse a1 union a2 is equals to f inverse a1 union f inverse a2.
03:01
So now to prove this here first of all let us take that x belongs to f inverse a1 union a2.
03:11
We can say that fx will belongs to a1 union a2.
03:17
So now here let us consider the two cases as first of all we have that if fx belongs to a1 then we can say that x belongs to f inverse a1.
03:33
So therefore here we can say that x belongs to f inverse a1 union f inverse a2.
03:43
Next we have that if fx does not belongs to a1 so this implies that fx belongs to a2.
03:53
So this means that x will belongs to f inverse a2.
03:58
So from here also we can imply that x belongs to f inverse a1 union f inverse a2.
04:08
Now from the above we have proved that f inverse a1 union a2 is a subset of f inverse a1 union f inverse a2.
04:23
Now here also let us take this as the first equation.
04:27
Now conversely here we can say that since f inverse a1 is a subset of f inverse a1 union a2 and f inverse a2 is a subset of f inverse a1 union a2 then we will write that f inverse a1 union f inverse a2 will be a subset of f inverse a1 union a2.
05:03
Let it be the second equation.
05:05
Now from the first and second equation we can say f inverse a1 union a2 is equals to f inverse a1 union f inverse a2.
05:23
So we have proved the second identity also.
05:26
Now the third identity that we have we have f inverse a1 intersection a2 is equals to f inverse a1 intersection f inverse a2.
05:43
Now to prove this identity here let us take we have x belongs to f inverse a1 intersection f inverse a2.
05:54
Then we can say that fx belongs to a1 and fx belongs to a2...