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Let $\mathbb{Q}$ denotes the set of rational numbers. Find sets $E \subset S_1 \subset S_2 \subset S_3 \subset \mathbb{Q}$ such that $E$ has a least upper bound in $S_1$, but does not have a least upper bound in $S_2$, yet does have a least upper bound in $S_3$.

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Please show us what you've thought of so far so that we might be able to guide you through it. Also, please don't ask anything in the imperative - we don't like being told what to do. – mixedmath May 12 '11 at 1:07

Take $E_0 = [0,1)$, $E_1=E\cup \{2\}$, $E_2=E\cup(1,2]$, $E_3=E\cup[1,2]$. Now take $E=E_0\cap \mathbb Q$, $S_i=E_i\cap \mathbb Q$.

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Take $E = [0,1[$, $S_1 = [0,1[ \, \cup \, {2}$, $S_2 = [0,1[ \, \cup \, ]1,3[$ and $S_3 = [0,3]$. The inclusions are trivial, in $S_1$ the l.u.b. is $2$ because it is the only upper bound, in $S_2$ there is no l.u.b. because supposing $b$ was a l.u.b. for $E$ in $S_2$, then $b \in ]1,3[$ and $(1+b)/2$ is also an upper bound of $E$ in $S_2$, contradicting the choice of $b$ since $(1+b)/2 < b$. Similarly you can prove that $1$ is the l.u.b. in $S_3$.

P.S. I wrote intervals as if they were subsets of the reals but it is understood that $[a,b[$ denoted intervals containing rationals.

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There are maybe more trivial examples to come up with, but the idea is to work out the fact that the l.u.b. of $E$ is not in $E$. Then you put $E$ and add a big upper bound to it (hence my construction of $S_1$), then you remove the upper bound property by taking an "open" interval (in the ][ sense, not in the topology one). Adding the endpoints to my "open" interval in $S_2$ brings back the l.u.b. in $S_3$. – Patrick Da Silva May 12 '11 at 1:21
you forgot to intersect the sets with $\mathbb Q$. – lhf May 12 '11 at 1:21
... Read my P.S. The sets were supposed to be included in $\mathbb Q$ anyway, I think everyone got the point. – Patrick Da Silva May 12 '11 at 9:10

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