Let $A, B \subset \Bbb{R}$ be two bounded non-empty sets. Proof that – $$A \subseteq B \implies \inf(B) \le \inf(A) \le \sup(A) \le \sup(B)$$
I've done it in this way:
Since $A \; and \; B$ are bounded and non-empty sets, they surely admit $\inf$ and $\sup$ in $\Bbb{R}$.
In particular $\inf(A) \le \sup(A)$ and $\inf(B) \le \sup(B)$.
Again, since $A \subseteq B,\;$ if $\; \inf(B) \le b; \forall b \in B \implies \inf(B) \le a; \forall a \in A$.
Being $\inf(A)$ the greatest lower bound of $A \implies \inf(B) \le \inf(A)$
Finally, since $A \subseteq B,\;$ if $\; \sup(B) \ge b; \forall b \in B \implies \sup(B) \ge a; \forall a \in A$.
Being $\sup(A)$ the least upper bound of $A \implies \sup(A) \le \sup(B)$
$\inf(B) \le \inf(A) \le \sup(A) \le \sup(B)$
I think it's right but I'm not sure. Can you please tell me what you think?
Thanks,
Lorenzo
Best Answer
This is true, your proof is correct.