00:01
So in this problem we have to show that the cardinality of the closed interval 2 .5 is the same as the cardinality of the open interval 3 comma 4.
00:12
And to do that, we'll try to find a bijection between these two sets.
00:20
Now, first, we can show that there is a bijection between the closed interval 0 .1 and the open interval 0 .1.
00:31
And then we'll use the fact that the composition of bijections is a bijection itself and we'll use that to show our final result now consider the sequence a n defined as follows we have we define a not equals zero a one equals one a two equals half and the general nth term we have a n equals 1 over 2 raise to n minus 1 and so on that's how we define the sequence a n now let's define a function h from the closed interval 0 comma 1 to the open interval 0 .1 given by so we define h as we defined h of 0 equals half h of 1 equals 1 h of 1 equals 1 of 1 2 squared and for any number of the form 1 over 2 raised to n we define h of 1 over 2 raise to n as 1 over 2 raise to n plus 2 or any positive integer n so whenever we have a number of the form 1 over 2 raise to n we define h of that as 1 over 2 raise to n plus 2 and we define that for any for all other values of x that is, whenever x is not of, whenever x is not 0, 1 or of the form 1 over 2 range to n, we define h of x equals x the identity function.
02:21
Now note that the way we defined h, it means that h of a not equals a 2.
02:29
Why? because a not is 0 and we define h of 0 equals half, and half is a 2.
02:34
So we have h of a not equals a 2.
02:38
And we have h of a 1 equals a 3.
02:41
Why? because a1 is 1 and we defined h as h of 1 equals 1 over 2 squared, which is a3.
02:50
So h of a1 equals a3.
02:52
And if you look closely, you'll see that for the general nth term, we have h of a .n equals a .n plus 2 and so on.
03:02
So when h acts on the terms of the sequence, an, it just takes it to the term that's 2 over from it in the index.
03:15
That is h maps an to an plus 2 for the n greater than equal to 0 because the sequence starts from a0.
03:26
And all the terms of the sequence are different.
03:35
So h sends terms of the sequence to some other terms of the sequence.
03:41
And all the terms of the sequence in the way we defined this, are different...