# Proving that there exists an s in S is between t and the supremum.

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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How did $b$ get to be the least upper bound? Why isn't there a smaller upper bound? –  Will Jagy Feb 8 '13 at 2:43
it is given that b is the l.u.b. of S –  M. J. Feb 8 '13 at 2:45
Yes. I am trying to get you to concentrate on the meaning of the word "least" in the definition of lub. –  Will Jagy Feb 8 '13 at 2:47

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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I get it. The reason t must be less than s is because if s were less than t, by definition t becomes an upper bound to the set. Thanks a lot. –  M. J. Feb 8 '13 at 3:25
@M.J.: You're close (and may even have it). There are a few addenda I'd like to make to your statement, though: The reason $t$ must be less than some $s$ is because if every $s$ were less than or equal to $t$, by definition...(you've got the rest of it). –  Cameron Buie Feb 8 '13 at 5:06

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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That's almost exactly what I got from the T.A., lol. I guess I just don't understand why there is necessarily an s > t. Thanks, though. –  M. J. Feb 8 '13 at 2:52