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Real Analysis Question (Suprema, Infima and $\mathbb R$)

Question:

Suppose $U$ is a non-empty subset of $\mathbb R$, bounded above, with supremum $s$. If $a$ is any number satisfying $a < s$, explain why there is some $u\in U$ with $a < u$.

Please can someone help with this question?

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    Perhaps i$f$ you write out the de$f$inition o$f$ "supremum" all will become clear.2012-02-21

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Hint: If the set $U$ had no element $>a$, then $a$ would be an upper bound for $U$.

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If not, then $u \leq a$ for all $u\in U$, but this shows that $a$ is the sup, contridiction.

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    That doesn’t show that $a=\sup U$: it shows that $a$ is **an** upper bound for $U$, but not necessarily the smallest one. However, that’s enough to get a contradiction.2012-02-22