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KiwiBunny
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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.
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 $\sup(T)=-\inf(S)$.\\
$S$ 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 $sup T=-inf S$
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 $sup T=-inf S$