Show that if A⊆B -knowing that B is bounded and non-empty set- , then inf B≤ inf A≤ sup A≤ sup B I am currently an analysis 1 student and this is my approach to the answer to the question above, hope that you be very responsive and rectify any errors if they exist, so as, I and other members of the group can learn.
since B is a bounded set we have the following: 
∀x∈B : inf(B)≤x≤sup(B)
and since A⊆B then ∀x∈A : inf(B)≤x≤sup(B)
as a conclusion A is bounded and we have the following: 
(∃ inf(A) ∈ ℝ )(∃ sup(A) ∈ ℝ)/ ∀x∈A : inf(A)≤x≤sup(A)
now we need to prove that inf B≤ inf A≤ sup A≤ sup B
and as a gateway, I chose to prove that (inf(A),sup(A))∈B²
Let m be the set of all minorants of set A and let M be the set of all majorants of set A:
-if m∩B≠∅ then inf(A)∈B (because inf(A) is the greatest minorant so at least we have m∩B={inf(A)})
-if m∩B=∅ then inf(A)=inf(B)
-if M∩B≠∅ then sup(A)∈B (because sup(A) is the smallest majorant so at least we have M∩B={sup(A)})
-if M∩B=∅ then sup(A)=Sup(B)
so, as a conclusion (inf(A),sup(A))∈B²
and if inf(A) and sup(A) are elements of B they verify the following:
inf B≤ inf A≤ sup A≤ sup B 
 A: You also need the assumption that $A$ is non-empty. For, if $A$
is empty, by convention, $\inf\emptyset=+\infty$ while $\sup\emptyset=-\infty$.
Then, we do not have $\inf A\leq\sup A$.
Suppose that $A$ is non-empty, $A\subseteq B$, and $B$ is a bounded
set.


*

*$\inf(B)\leq\inf A$: Let $x\in A$ be arbitrary, then $x\in B$
(because $A\subseteq B$). Now $\inf B$ is a lower bound for $B$,
so $\inf B\leq x$. Since $x\in A$ is arbitrary, this shows that
$\inf B$ is a lower bound for $A$. Therefore $\inf B$ is less than
or equal to the greatest lower bound of $A$, i.e., $\inf B\leq\inf A$.

*$\inf A\leq\sup A$: Since $A$ is non-empty, there exists $x_{0}\in A$.
$\inf A$ is a lower bound for $A$ $\Rightarrow$ $\inf A\leq x_{0}$.
$\sup A$ is an upper bound for $A$ $\Rightarrow$ $x_{0}\leq\sup A$.
Combining, we obtain $\inf A\leq\sup A$.

*$\sup A\leq\sup B$. Let $x\in A$ be arbitrary, then $x\in B$
(because $A\subseteq B$). Since $\sup B$ is an upper bound for $B$,
we have $x\leq\sup B$. Since $x\in A$ is arbitrary, we actually
have proved that $\sup B$ is an upper bound for $A$. By definition,
$\sup A$ is the smallest upper bound for $A$, so $\sup A\leq\sup B$.
A: Let $A \subseteq \Re$. If $\exists$ $sup(A)$ $\in$ $\Re$ and $\alpha \in \Re$ is an upper bound of $A$, then $\alpha=sup(A) \Leftarrow\Rightarrow$ $\forall$ $\epsilon > 0$ $\exists$ $x$ $\in$ $A$ such that $\alpha \ge x > \alpha - \epsilon$. The other definiton of $inf (A)$ is analogous with $\beta \le y < \beta + \epsilon$ for a lower bound of $A$. The proof of this equivalence is pretty simple.
You already have that $sup(B)$ is an upper bound of  $A$ because $A \subseteq B$.
Proof by contradiction: 
Supose that $supA > supB$ $\Rightarrow$ for $\epsilon = supA - supB > 0$, $\exists$ $x\in A$ such that $x > sup(A) - \epsilon = sup(A) - (supA - supB) = sup(B)$ $\Rightarrow$ $\exists$ $x\in A $such that $x>sup(B) $! Because that means that $sup(B)$ isn't  an upper bound of $A$, contrary to our assumption. Therefore, $sup(A) \le sup(B)$ due to tricotomy axioms of the field.
The other proof is analogous.
A: You can't prove that $\inf(A)\in B$, even if $m\cap B \neq \emptyset$ (same reasoning for $\sup(A)$). Take the example : $B = \{0\} \cup (1,2)\cup \{4\}$ and $A  = (1,2)$. If $m\cap B \neq \emptyset$ you can directly conclude that $inf(B) \leq \inf(A)$.
