text: # Supremum and infimum
  
  • Sep 16th 2010, 11:04 AM
  novice
  Supremum and infimum
  Let $\displaystyle S$ be a nonempty subset of $\displaystyle \mathbb{R},$ and let $\displaystyle -S = \{-s: s \in S\}.$
  Prove that $\displaystyle inf S = -sup (-S).$ [Hint: $\displaystyle -\infty < inf S]$
  
  I wrote the proof as follows:
  
  $\displaystyle s \in S \Longrightarrow -s\in -S, \forall -s \in -S$
  $\displaystyle \Longrightarrow -s \leq sup(-S)$
  $\displaystyle \Longrightarrow -sup(-S) \leq s, \forall s \in S.$
  $\displaystyle \Longrightarrow -sup(-S)$ is a lower bound of $\displaystyle S.$
  
  $\displaystyle s \in S \Longrightarrow \exists m \in \mathbb{R}$ such that $\displaystyle m \leq s.$
  $\displaystyle \Longrightarrow -s \leq -m$, so $\displaystyle -m$ is an upper bound, and $\displaystyle sup(-S) \leq -m$
  $\displaystyle \Longrightarrow m \leq -sup(-S)$
  
  Since $\displaystyle m$ is a lower bound of $\displaystyle S$, $\displaystyle -sup(-S)$ is the greates lower bound of $\displaystyle S$.
  Therefore, $\displaystyle inf(S)=-sup(-S).$
  
  Question: Could someone please show me how to use the hint?
  • Sep 16th 2010, 11:19 AM
  Plato
  • Sep 16th 2010, 11:54 AM
  novice
  Quote:
  
  Originally Posted by Plato
  
  Yes sir, you have, but I would like to know how to prove it using
  $\displaystyle \infty < infinum$ to see the difference in two methods.
  
  I think it's supposed to be easier.
  • Sep 17th 2010, 09:54 AM
  Jhevon
  Quote:
  
  Originally Posted by novice
  Let $\displaystyle S$ be a nonempty subset of $\displaystyle \mathbb{R},$ and let $\displaystyle -S = \{-s: s \in S\}.$
  Prove that $\displaystyle inf S = -sup (-S).$ [Hint: $\displaystyle -\infty < inf S]$
  
  I wrote the proof as follows:
  
  $\displaystyle s \in S \Longrightarrow -s\in -S, \forall -s \in -S$
  $\displaystyle \Longrightarrow -s \leq sup(-S)$
  $\displaystyle \Longrightarrow -sup(-S) \leq s, \forall s \in S.$
  $\displaystyle \Longrightarrow -sup(-S)$ is a lower bound of $\displaystyle S.$
  
  $\displaystyle s \in S \Longrightarrow \exists m \in \mathbb{R}$ such that $\displaystyle m \leq s.$
  $\displaystyle \Longrightarrow -s \leq -m$, so $\displaystyle -m$ is an upper bound, and $\displaystyle sup(-S) \leq -m$
  $\displaystyle \Longrightarrow m \leq -sup(-S)$
  
  Since $\displaystyle m$ is a lower bound of $\displaystyle S$, $\displaystyle -sup(-S)$ is the greates lower bound of $\displaystyle S$.
  Therefore, $\displaystyle inf(S)=-sup(-S).$
  
  Question: Could someone please show me how to use the hint?
  
  You really should post questions in their entirelty to help people help you better. The hint actually was:
  
  For the case $\displaystyle - \infty < \inf S$, simply state that this was proved in Exercise 4.9.
  
  At this point, you could have also stated that exercise 4.9 asked you to complete a proof by showing that $\displaystyle \inf S = - \sup (-S)$. (And, just as importantly, the fact that you were assuming the set $\displaystyle S$ is bounded below, that is, $\displaystyle -\infty < \inf S$)
  
  This is the proof Plato helped you with, and what you are doing here. It seems this problem wants you to tackle the case where $\displaystyle \inf S = - \infty$, meaning the set is NOT bounded below--a notion introduced in section 5 that you would perhaps not have considered when doing the problem in section 4. In other words, you have already been helped with what you're asking help for, you should be doing something entirely different.
  • Sep 17th 2010, 04:34 PM
  novice
  Nothing is stated in the question that the set is unbound. So the message being conveyed has been as clear as mud. It's unfortunate that mathematician are often poor communicators.
  
  If it is as you said that it's unbound, then it's piece of cake. I wonder why the author can't be as clear as a child.
  
  At any rate, thanks bro.
  • Sep 17th 2010, 06:29 PM
  Jhevon
  Quote:
  
  Originally Posted by novice
  Nothing is stated in the question that the set is unbound. So the message being conveyed has been as clear as mud. It's unfortunate that mathematician are often poor communicators.
  
  If it is as you said that it's unbound, then it's piece of cake. I wonder why the author can't be as clear as a child.
  
  At any rate, thanks bro.
  
  Indeed! Unboundedness is not a stated condition in the problem, but it is one that must be considered nonetheless (nothing said you were dealing with a bounded set either...).
  
  Haha, I won't deny that mathematicians communicate poorly sometimes, but i don't think Ross should be accused of such a thing--definitely not in this case at least.
  
  One thing mathematicians try to do is to cover all angles, all perspectives and considerations. We try not to think inside a box, but to consider all possibilities. So when someone makes a general claim about sets, the mathematician wanting to prove/disprove it will automatically consider all kinds of sets. It is not good enough to just consider one case, but all possible cases, unless the claim restricts us otherwise.
  
  With this problem it is no different. When reading the problem, you should automatically wonder, "wow, is this really true for all nonempty sets of real numbers? what if the set is bounded above? below? neither? both? what if...?" Ross decided to be a nice guy here and remind you that you dealt with one case already in a previous section, so there is no need to worry about it now, just prove the remaining case(s). The guy's not communicating poorly, he's just trying to save you from re-inventing the wheel.
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distilabel_metadata.content: # Supremum and infimum
  
  • Sep 16th 2010, 11:04 AM
  novice
  Supremum and infimum
  Let $\displaystyle S$ be a nonempty subset of $\displaystyle \mathbb{R},$ and let $\displaystyle -S = \{-s: s \in S\}.$
  Prove that $\displaystyle inf S = -sup (-S).$ [Hint: $\displaystyle -\infty < inf S]$
  
  I wrote the proof as follows:
  
  $\displaystyle s \in S \Longrightarrow -s\in -S, \forall -s \in -S$
  $\displaystyle \Longrightarrow -s \leq sup(-S)$
  $\displaystyle \Longrightarrow -sup(-S) \leq s, \forall s \in S.$
  $\displaystyle \Longrightarrow -sup(-S)$ is a lower bound of $\displaystyle S.$
  
  $\displaystyle s \in S \Longrightarrow \exists m \in \mathbb{R}$ such that $\displaystyle m \leq s.$
  $\displaystyle \Longrightarrow -s \leq -m$, so $\displaystyle -m$ is an upper bound, and $\displaystyle sup(-S) \leq -m$
  $\displaystyle \Longrightarrow m \leq -sup(-S)$
  
  Since $\displaystyle m$ is a lower bound of $\displaystyle S$, $\displaystyle -sup(-S)$ is the greates lower bound of $\displaystyle S$.
  Therefore, $\displaystyle inf(S)=-sup(-S).$
  
  Question: Could someone please show me how to use the hint?
  • Sep 16th 2010, 11:19 AM
  Plato
  • Sep 16th 2010, 11:54 AM
  novice
  Quote:
  
  Originally Posted by Plato
  
  Yes sir, you have, but I would like to know how to prove it using
  $\displaystyle \infty < infinum$ to see the difference in two methods.
  
  I think it's supposed to be easier.
  • Sep 17th 2010, 09:54 AM
  Jhevon
  Quote:
  
  Originally Posted by novice
  Let $\displaystyle S$ be a nonempty subset of $\displaystyle \mathbb{R},$ and let $\displaystyle -S = \{-s: s \in S\}.$
  Prove that $\displaystyle inf S = -sup (-S).$ [Hint: $\displaystyle -\infty < inf S]$
  
  I wrote the proof as follows:
  
  $\displaystyle s \in S \Longrightarrow -s\in -S, \forall -s \in -S$
  $\displaystyle \Longrightarrow -s \leq sup(-S)$
  $\displaystyle \Longrightarrow -sup(-S) \leq s, \forall s \in S.$
  $\displaystyle \Longrightarrow -sup(-S)$ is a lower bound of $\displaystyle S.$
  
  $\displaystyle s \in S \Longrightarrow \exists m \in \mathbb{R}$ such that $\displaystyle m \leq s.$
  $\displaystyle \Longrightarrow -s \leq -m$, so $\displaystyle -m$ is an upper bound, and $\displaystyle sup(-S) \leq -m$
  $\displaystyle \Longrightarrow m \leq -sup(-S)$
  
  Since $\displaystyle m$ is a lower bound of $\displaystyle S$, $\displaystyle -sup(-S)$ is the greates lower bound of $\displaystyle S$.
  Therefore, $\displaystyle inf(S)=-sup(-S).$
  
  Question: Could someone please show me how to use the hint?
  
  You really should post questions in their entirelty to help people help you better. The hint actually was:
  
  For the case $\displaystyle - \infty < \inf S$, simply state that this was proved in Exercise 4.9.
  
  At this point, you could have also stated that exercise 4.9 asked you to complete a proof by showing that $\displaystyle \inf S = - \sup (-S)$. (And, just as importantly, the fact that you were assuming the set $\displaystyle S$ is bounded below, that is, $\displaystyle -\infty < \inf S$)
  
  This is the proof Plato helped you with, and what you are doing here. It seems this problem wants you to tackle the case where $\displaystyle \inf S = - \infty$, meaning the set is NOT bounded below--a notion introduced in section 5 that you would perhaps not have considered when doing the problem in section 4. In other words, you have already been helped with what you're asking help for, you should be doing something entirely different.
  • Sep 17th 2010, 04:34 PM
  novice
  Nothing is stated in the question that the set is unbound. So the message being conveyed has been as clear as mud. It's unfortunate that mathematician are often poor communicators.
  
  If it is as you said that it's unbound, then it's piece of cake. I wonder why the author can't be as clear as a child.
  
  At any rate, thanks bro.
  • Sep 17th 2010, 06:29 PM
  Jhevon
  Quote:
  
  Originally Posted by novice
  Nothing is stated in the question that the set is unbound. So the message being conveyed has been as clear as mud. It's unfortunate that mathematician are often poor communicators.
  
  If it is as you said that it's unbound, then it's piece of cake. I wonder why the author can't be as clear as a child.
  
  At any rate, thanks bro.
  
  Indeed! Unboundedness is not a stated condition in the problem, but it is one that must be considered nonetheless (nothing said you were dealing with a bounded set either...).
  
  Haha, I won't deny that mathematicians communicate poorly sometimes, but i don't think Ross should be accused of such a thing--definitely not in this case at least.
  
  One thing mathematicians try to do is to cover all angles, all perspectives and considerations. We try not to think inside a box, but to consider all possibilities. So when someone makes a general claim about sets, the mathematician wanting to prove/disprove it will automatically consider all kinds of sets. It is not good enough to just consider one case, but all possible cases, unless the claim restricts us otherwise.
  
  With this problem it is no different. When reading the problem, you should automatically wonder, "wow, is this really true for all nonempty sets of real numbers? what if the set is bounded above? below? neither? both? what if...?" Ross decided to be a nice guy here and remind you that you dealt with one case already in a previous section, so there is no need to worry about it now, just prove the remaining case(s). The guy's not communicating poorly, he's just trying to save you from re-inventing the wheel.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math question and a detailed attempt at answering it. The question is about proving a relationship between the supremum and infimum of a set and its negation. The user also asks for help on how to use a specific hint provided in the problem.
  
  Label: question-and-answer