Ordering the reals (from Analysis I by T. Tao) (Part 3)

In this part we solve Exercise 5.4.1, Exercise 5.4.2, Exercise 5.4.3, and Exercise 5.4.4. In the next part we will solve the remaining exercises.

[vichg]

Hello friend, I would like to know if you can help me with a step of the proof of proposition 5.4.4. Specifically with the following: At minute 40:29 of this video you say: "N cannot possibly be greater than all the elements of S1". What I would like to see is a proof that the following statement leads to a contradiction: for all n in S1then N>n. This would justify what you said at minute 40:29 and would allow us to choose an m' in S1 such that m' is greater than or equal to N. (What I am trying to do is understand how you took m' in S1 such that m' is greater than or equal to N) Thank you very much for your attention and for the work you are doing with these videos, it is excellent.

[vichg]

Hello friend @A Mathematical Room. I hope you are well, let me ask you a question please. At minute 1:29:39 of this video you say "...that's why A is finite and because it is finite it will have a smallest element, it will have a largest element also..." The question is: What theorem or result allows you to conclude that A has a smallest and a largest element from the fact that A is a nonempty finite set of integers?