Show that $y=\sup (C)$

Let $A \subseteq \mathbb{R}$ nonempty and bounded above. Also, let $C = \{x+100 : x \in A\}$

Consider $M$ in $\mathbb{R}$ such that $M < y$. Is $M-100$ an upper bound for $A$? Why?

Use this result to show that $y=\sup (C)$

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What is $y$? ${}{}$ –  Git Gud Feb 6 '13 at 23:11
@Git, if you put {}{}{} between dollar signs, they count as characters. –  Gerry Myerson Feb 6 '13 at 23:13
@GerryMyerson Thanks! ${}{}$ –  Git Gud Feb 6 '13 at 23:14

1 Answer

I don't really understand the question, but I'm going to find $\sup (C)$ and prove it is what it is. Hopefully that will be of some help to the OP.

Since $A$ is bounded above, $\sup (A)$ exists.

Let $\displaystyle s_A =\sup (A)$. Now let $s=s_A+100$.

1. Take $c\in C$ arbitrarily. We have $c=x+100$ for some $x\in A$. By definition of $s_A$ we have $x\leq s_a$, therefore $c=x+100\leq s_A+100=s$. This proves that $s$ is an upper bound for $C$.
2. We've now estabilished that the set of upper bounds of $C$ isn't empty. So you can take an arbitrary upper bound of $C$, $m$. Since $m$ is an upper bound of $C$ we know that for any $c\in C$, it is true that $c\leq m$, which means that for any $x\in A$ we have $x+100\leq m$ and it follows that for any $x\in A$, $x\leq m-100$, therefore, since $x$ was arbitrary, $s_A\leq m-100$, so we get $s=s_A+100\leq m$. Since $m$ was an arbitrary upperbound for $C$, we've proved that $s$ is the smallest upper boundof $C$ and therefore $s=\sup (C)$.
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I poorly worded the question but what you replied with helped me think about the problem in a different way. Thanks! –  Math Student Feb 7 '13 at 0:51