What is the Relationship Between inf(S) and -sup(-S)?

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The discussion revolves around proving the relationship inf(S) = -sup(-S). The user begins by defining x as inf(S) and acknowledges that x is greater than or equal to all lower bounds of S. They express confusion about demonstrating that -x is an upper bound of -S and seek clarification on the inequality -s ≤ -x. The explanation provided indicates that since -s belongs to -S, it implies that s is in S, thus reinforcing that x is indeed a lower bound for S. The conversation highlights the logical steps needed to establish the proof, emphasizing the connection between lower and upper bounds through negation.
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Prove inf(S)=-Sup(-S)??

Homework Statement




Let S,T be subsets of ℝ, where neither T nor S are empty and both Sup(S) and Sup(T) exist.

Prove inf(S)=-sup(-S).

Starting with =>

I let x=inf(S). Then by definition, for all other lower bounds y of S, x≥y.

I'm stuck at this point...

Any help please?

Thanks
 
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You have to show two things:

  • -x is an upper bound of -S
  • If y is another upper bound of -S, then -x\leq y

So, in order to prove that -x is an upper bound of -S. Take an element -s from -S and prove -s\leq -x. Why is that true?
 


I'm not following along, why would -s≤-x? Unless you mean if I multiply both side by (-1), then it would be x≤s? Or am I totally off on a tangent?
 


SMA_01 said:
I'm not following along, why would -s≤-x? Unless you mean if I multiply both side by (-1), then it would be x≤s? Or am I totally off on a tangent?

If -s \in -S, then s \in S. And by definition, x is a lower bound of S...
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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