# $\mathbb{Q}$ does not have least upper bound property

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 $S_3$.

-
If this is homework, please add the homework tag. Also please show us what you've thought of so far so that we might be able to guide you through it. Finally, 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 $E=E_0\cap \mathbb Q$, $S_i=E_i\cap \mathbb Q$.

-
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.
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