00:01
Hello students, let s0 equal to infimum of s and let t0 equal to minus supremum of minus s.
00:15
We want to show that s0 is less than equal to t0 by the definition of infimum of s.
00:24
By the definition of infimum of s is for any epsilon greater than 0, there exists small n s in capital s such that s0 is less than equal to s, s0 is less than equal to s0 plus epsilon.
01:02
Also by the definition of by the definition of supremum of s supremum of minus s, we can say that for any epsilon greater than 0, there exists t in minus s such that t0 minus epsilon is less than equal to t0.
01:35
So, let us choose epsilon equal to minus t0 which is positive since t0 is negative.
01:51
For this choice of epsilon, we have for this choice of epsilon, we have s0 less than equal to s less than s0 minus t0 which is equal to s0 plus epsilon and t0 minus epsilon less than equal to t, t0 minus epsilon less than t less than equal to t0.
02:27
Adding this, adding this, we have s plus t0 minus epsilon less than equal to s plus t less than equal to s0 plus t0.
02:51
So, notice that s plus t0 minus epsilon is a sum of number from s and a number from minus s which means that it is non -negative.
03:06
Similarly, s plus t is a sum of a number from s and a number from minus s which means it is non -positive.
03:17
This shows that s plus t0 minus epsilon is non -negative and s plus t is non -positive.
03:36
Therefore, s plus t0 minus epsilon less than equal to 0 less than equal to s0 plus t0.
03:48
This implies that t0 minus epsilon less than equal to s0.
03:55
Since epsilon equal to minus t0, we have t0 plus t0 less than equal to s0 which simplifies that to t0 less than equal to s0 or equivalently t0 less than equal to s0 by 2...