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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

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  • 2
    $\begingroup$ In bigger ponds, there are more large fish and also more small fish. $\endgroup$ – Prototank Nov 3 '18 at 11:29
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This is true, your proof is correct.

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