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Analysis (Calculus) proof regarding inequalities, sup/inf

  • Thread starter KiwiBunny
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  • #1
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Hey, my stuff is done in LaTeX if you want to copy into a document and compile it. I guess if you can read TeX you can read it without resorting to that though. My TeX is not the neatest, if you have any problems please tell me.

Homework Statement


1. Let $S$ be a nonempty set of real numbers that is bounded from below. Define $T=\{-s: s \in S\}$. Prove that $T$ is nonempty and bounded from above. Then prove that $\sup(T)=-\inf(S)$.\\




The Attempt at a Solution



$S$ is nonempty
so there is some $s \in S$ and also $s \in R$. \\
All the elements in R have an additive inverse so with $s \in R$ there is $-s \in R$ and $-s \in T$. Thus T is nonempty. \\
S is bounded from below so there is an $inf(S)$ so for $\forall s \in S$, $s \geq inf(S)$
$\Rightarrow$ $-s \leq -inf(S) \Rightarrow$ \\ $\forall t \in T$ $t \leq -inf(S)$. So $-inf(S)$ is an upper bound for T.
\\
How do you prove it's the lowest upper bound?

Say that $sup T < -inf S$ $\Rightarrow$ $-sup T > inf S$ $\Rightarrow$ there's an element smaller than -sup T also in S, call it n.
$n \in S \Rightarrow -n \in T$
$n < -sup T \Rightarrow -n > sup T$
$-n \in T$ and $-n >sup T$
Contradiction, therefore $sup T=-inf S$
 

Answers and Replies

  • #2
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For the second part where you are trying to prove that T has an upper bound for S is bounded above,
let c be a lower bound, then [tex]c \leq x[/tex] for [tex]\forall x \epsilon S[/tex]. From here, you have to use algebra to get -c in the inequality somehow*. Use this to show that -c is an upper bound for S.
Then, use Axiom of Completeness, there exists a b=supS such that for [tex]b \leq c[/tex] for [tex]\forall x \epsilon S[/tex] (same as from *) to continue the proof.
 
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