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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.
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.

This Question is Closed

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0So, 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? \[T=\{xx \in \text{S}\}\] 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?

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.02.) 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok you can read it x such that minus x belongs to S.

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.2It'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}

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0So 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.02.) 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.

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.2Well, they said S is bounded below, so it can't be R.

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.2But in general where you just write \[T=\{xx \in \text{S}\}\] S could represent any set.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So 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

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0"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?

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.2Yeah, 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.

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0And 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

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0Oh, alright, cool. Let's see.

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.2Well, 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.

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.01.) 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...

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.2Yeah, that's the idea.

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.2Basically chose x in 1.1.8 to be supT. Then prove that (a) and (b) are true.

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.2sorry alpha = supT

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0Alright, 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.
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