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If $A\subset B$ and $B$ is bounded above, show that $\operatorname{lub}A \leq \operatorname{lub}B$.

This seems very obvious but I am not able to write a proper solution for this. Does it really require a proof or we can logically conclude this? Kindly help.

Ottavio
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V2002
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  • By "lub" you mean supremum, i.e. least-upper-bound? Are we talking of real numbers? ā€“ Ottavio Nov 25 '20 at 18:04
  • Yes, it requires a proof. To show that the least upper bound of a set $S$ is less than $x$ all you need to do is to show that $sā‰¤x$ for all $s\in S$. ā€“ lulu Nov 25 '20 at 18:14
  • Yes @OttavioBartenor , by lub I mean supremum of a set ā€“ V2002 Nov 25 '20 at 18:16

1 Answers1

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Yes it requires a proof, by logically concluding.

Let $x$ be the least upper bound of $B$, then it's an upper bound for $A\subseteq B$, hence $\sup A\le x$.

Berci
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