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Prove that: Let $S$ be a nonempty set of real numbers that has a least upper bound, $b$, in $\mathbb{R}$. Let $t$ be any real number such that $t < b$. Then there exists some $s$ in $S$ with $t<s$ and $s\leq b$. I'm can find all sorts of examples, but can't prove this in the general sense. |
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Since $b$ is an upper bound of $S$, then for any $s\in S$, we have $s\le b$, by definition of upper bound. Take any $t<b$, and assume for the moment (just to see what happens) that there aren't any $s\in S$ with $t<s\le b$. In other words, we're supposing that if $s\in S$, then $s\le t$ or $b<s$. We can't have $b<s$ if $s\in S$ (see first paragraph), so that means we're assuming that $s\le t$ for every $s\in S$. But that means that $t$ is an upper bound of $S$. In particular, then, since $t<b$, then $b$ cannot be the least upper bound of $S$. But $b$ was defined to be the least upper bound of $S$, so our assumption led to an impossible situation (contradiction), and so our assumption was false. Thus, the desired conclusion holds. |
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Since $t<b$ and $b$ is the least upper bound of $S$, $t$ is not an upper bound of $S$. Hence there is an $s\in S$ with $s>t$. But since $b$ is an upper bound of $S$, we have $s\leq b$. |
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