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Prove inf(S)=-Sup(-S)?

  • Thread starter SMA_01
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  • #1
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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
 

Answers and Replies

  • #2
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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 [itex]-x\leq y[/itex]

So, in order to prove that -x is an upper bound of -S. Take an element -s from -S and prove [itex]-s\leq -x[/itex]. Why is that true?
 
  • #3
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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?
 
  • #4
jbunniii
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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 [itex]-s \in -S[/itex], then [itex]s \in S[/itex]. And by definition, [itex]x[/itex] is a lower bound of [itex]S[/itex]...
 

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