Show that if A⊆B -knowing that B is bounded and non-empty set- , then inf B≤ inf A≤ sup A≤ sup B

Question

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

Answer 1

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.

1. $\inf(B)\leq\inf A$: 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., $\inf B\leq\inf A$.

2. $\inf A\leq\sup A$: 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 $\inf A\leq\sup A$.

3. $\sup A\leq\sup B$. 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 $\sup A\leq\sup B$.

Answer 2

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 $inf(B) \leq \inf(A)$.

Answer 3

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.