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Let $S=(0,1)\cup(1,2).$

I claim that 2 is the supremum of the set. Let $A$ be the set of all upper bounds of $S$. Since we are dealing with real numbers, and $S$ has an upper bound, I claim that it has least upper bound and that l.u.b is 2. Why is it 2? Let $s \in S$ and let it be the l.u.b. We see that $s < 2$. Because we are in the reals there exists an $r \in \mathbb{R}$ such that $s < r < 2$. Hence $r$ is larger than $s$, so $s$ cannot be another upper bound.

Is this proof on the right track?

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You can't say "let $s\in S$ and let it be the l.u.b.", because the least upper bound of a set might not be an element of the set, and indeed, in this case it isn't. –  MJD Sep 13 '12 at 23:02
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up vote 2 down vote accepted

"Hence r is larger than s" is not enough. You need that $r \in S$, which of course you can do.


Let $s = \sup S$. Suppose $s < 2$. Since $1.5 \in S$, you must have that $1.5 \leq s < 2$. Hence there exists a $r \in \mathbb{R}$ such that $1.5 \leq s < r < 2$. Hence $r \in (1,2) \subset S$. So $r \in S$. Thus you have shown that there exists a $r \in S$ such that $r > s$. This contradicts $s = \sup S$.

Of course $s \leq 2$, since $2$ is an upper bound for $s$. You can now conclude that $s = 2$.

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