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If $A \subset B$, then $\sup_B \geq \sup_A$.

$\textbf{Proof:}$

Let $k = \sup_B$. If $\sup_A$ were greater than $\sup_B$, then $\sup_A=k+\delta$ where $\delta$ is some positive number. Since, however, $k$ is the maximum value in $B$, $k+\delta$ cannot exists within $B$. Since we stipulated that $A \subset B$, $k+\delta$ cannot exist within $A$ either, proving that $\sup_B \geq \sup_A$.

Is this a valid proof?

Izzy M.
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  • Out of curiosity... If I were to say "Since for all values $b \in B$, $b \leq \sup_B$," would the statement be more correct? – Izzy M. Jun 02 '16 at 18:41

1 Answers1

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An easier approach might be the following: by definition, for all $x \in A$, $x \leq \sup_A$. But because $A \subset B$, it follows that $x \in B$ as well, so also by definition $x \leq \sup_B$ for all $x \in A$. So $\sup_B$ is an upper bound for all the elements in $A$. But by definition, $\sup_A$ is the LEAST upper bound for the elements of $A$, so it must be that $\sup_A\leq \sup_B$, as desired.

As for your proof, $k$ need not be the "maximum" value in $B$; the supremum need not be in the set. I think this could probably be worked around, but would take some extra work.

J.G
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