# Proving that infimum and supremum of a set lie between those of its superset

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

• In bigger ponds, there are more large fish and also more small fish. – Prototank Nov 3 '18 at 11:29