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

Question

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 $S_3$.

Answer 1

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

Answer 2

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.