If $A\subseteq B$ then prove that $\inf B \le \inf A \le \sup A \le \sup B$

Question

Prove $\inf B \le \inf A \le \sup A \le \sup B$

Let $a\in A$. Then $a \le \sup B$ because $A$ is a subset of $B$, so $B$ contains all elements of $A$. Therefore, $\sup A \le \sup B$.

Also, similarly, $\inf B \le a$. Therefore, $\inf B \le \inf A$.

Could you check this proof is valid??

Thank you in advance.

Answer 1

Assuming that $A\subseteq B$, your proof is correct.

It will be helpful if you add couple of statements to your proof to make it more accessible to readers.

For example if your mention that supremum is the least upper bound and infimum is the greatest lower bound,the reader will have an easier time to digest the corresponding inequalities.