Suppose $C \subseteq B$ and $B \subseteq A$; prove that $$ \sup_{A} C \leq \sup_{B}C$$ That's the problem, let $s= \sup_{A} C$ and $t= \sup_{B}C$, by definition $(x \leq s)( \forall x \in C) \:$ and $ \: (y \leq t)( \forall y \in B) $
Since $C \subseteq B$, then $(x \leq t)(\forall x \in C)$, so, here my question arises, can I deduce that $ s \leq t $?
this exercise is from the book of Pinter?
If you suppose that $ C \subseteq B $ and $ B \subseteq A $, by transitivity $ C \subseteq B \subseteq A $. Let $ s $ be the minimum upper bound of $ C $ in $ A $ $ (s = \sup_ {A} C )$ i.e $ s \geq x $, $ \forall x \in C $, with $ s \in A $.
Let $ t \geq y $, $ \forall y \in C $, with $ t \in B $, i.e $ t $ is the $ \sup $ of $ C $ in $ B $ $ (t = \sup_ {B} C )$.
Since $ t \in B $ and $ B \subseteq A $, then $ t \in A $, but $ s $ is the minimum upper bound of $ C $ in $ A $, so $ s \leq t $. $$ \therefore \: \: \: \sup_ {A} C \leq \sup_ {B} C $$