## Mendicant_Bias one year ago (Introductory Real Analysis) Still trying to prove this theorem. I now understand a little more about the problem and what the Supremum/Infimum are, but have some additional questions below in addition to the actual prompt.

1. Mendicant_Bias

So, I figured out without question despite some poorly written answer keys, the infimum or supremum of a set cannot be infinity or -infinity, because these are not real numbers. Now: 1.) How would somebody else read the following set, using words? $T=\{x|-x \in \text{S}\}$ When I first read that, I interpreted it that the set consists of the entire real number line, but apparently it's not so; how do I know that? By the very use of specifying S, do I know that it doesn't constitute the whole real number line/isn't the set of real numbers?

2. dan815

3. Mendicant_Bias

2.) In any case, as far as I'm aware the set above is just supposed to be a generic, arbitrary set representing any kind of Subset S in R. Is this true? And if it is, can anybody give me an idea of why x is negative? I mean, from the standpoint of understanding how stuff works, what would be the big deal if x wasn't negative? 3.) Here's the original problem prompt: http://i.imgur.com/iAHsh0W.png (I): http://i.imgur.com/qQzCncx.png 1.1.3): http://i.imgur.com/DU5NkoA.png

4. anonymous

ok you can read it x such that minus x belongs to S.

5. beginnersmind

It's giving the elements of set T in terms of the elements of another set S. So if S = {1,2,e,-10} then T = {-1,-2,-e,10}

6. Mendicant_Bias

So for clarity, how do I know that S doesn't consist of all of R? By using S does that denote that it must be a subset and can't be all the reals?

7. anonymous

2.) In any case, as far as I'm aware the set above is just supposed to be a generic, arbitrary set representing any kind of Subset S in R. Is this true? And if it is, can anybody give me an idea of why x is negative? I mean, from the standpoint of understanding how stuff works, what would be the big deal if x wasn't negative? its true that S can be anything belongs to R, x could be negative also could be positive as S is subset of R, the negation of x should be in S though.

8. beginnersmind

Well, they said S is bounded below, so it can't be R.

9. beginnersmind

But in general where you just write $T=\{x|-x \in \text{S}\}$ S could represent any set.

10. anonymous

So for clarity, how do I know that S doesn't consist of all of R? By using S does that denote that it must be a subset and can't be all the reals? well sense its an example for theorem 1.1.3 then it should satisfy nonempty also bounded above as @beginnersmind said that can't be all real in other way main question should say these things to not confusing you guys :P

11. Mendicant_Bias

"Well, they said S is bounded below, so it can't be R." Well no, they said S is bounded above IF S is bounded below. How do we show that S is bounded below?

12. beginnersmind

Yeah, they said T is bounded above if S is bounded below. So technically they didn't say S is bounded below. But to prove the statement you assume that S is bounded below and go on to show that T is bounded above. If S is not bounded below you don't need to prove anything.

13. Mendicant_Bias

And sorry, here's Theorem 1.1.8. This is what's trying to be proved, just an inversion of 1.1.3. http://i.imgur.com/9wNev9L.png

14. Mendicant_Bias

Oh, alright, cool. Let's see.

15. beginnersmind

Well, it's the same issue. You have a premise (S is bounded below) and a conclusion (the bit after then). You assume the premise is true and then prove the conclusion.

16. Mendicant_Bias

1.) Assume the set S is bounded below. 2.) Therefore, T is bounded from above. (Given in prompt) 3.) T is a nonempty set of real numbers that is bounded from above; therefore, T has a supremum. (Property I) And now applying 1.1.3 to prove 1.1.8...

17. beginnersmind

Yeah, that's the idea.

18. beginnersmind

Basically chose x in 1.1.8 to be supT. Then prove that (a) and (b) are true.

19. beginnersmind

sorry alpha = supT

20. Mendicant_Bias

Alright, cool. I think I'm going to keep it at this now that I have a general idea of what to do; I spent like the last three hours just attempting to understand/figure out what this problem required me to do. Thanks.

21. beginnersmind

No problem.