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Question : Let $A$ and $B$ $\subset \mathbb{R}$ nonempty and bounded above. Prove that $A \subset B$ implies $\sup A \le \sup B$.

This is a pretty basic analysis question. I am presenting my solution here so that anybody who has more experience in mathematics can help me to check if there is any flaw in the argument. So here it goes:

From the hypothesis, $A$ and $B$ nonempty and bounded above, $\alpha := \sup A$ and $\gamma := \sup B$ exist by the least upper bound property of $\mathbb{R}$. Goal is to prove $\alpha \le \gamma$.

$\alpha \ge a \quad \forall a \in A$. Since if $a \in A \Rightarrow a \in B$, we have $a \le \gamma \quad \forall a \in A$. $\gamma$ is an upper bound for $A$ by definition of upper bound. Again by definition of supremum, $\alpha \le \gamma$ since $\alpha$ is the least upper bound of A. $\Box$

Just a side note: The definition of supremum I am using has the following component: If $s$ is $\sup A$ for some subset $A$ in $\mathbb{R}$, $s \le u \quad \forall u$ the upper bounds of $A$.

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    Thanks a lot for the help!2012-06-20

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If $s$ is $\sup A$ for some subset $A$ in $\mathbb R$, $s\le u$ for any upper bound $u$ of $A$.

This makes the proof quick (summarized here to not leave the question unanswered):

  1. Since $\gamma$ is an upper bound for $B$ and $A\subseteq B$, it follows that $\gamma$ is an upper bound for $A$
  2. By the definition of supremum of $A$, $\alpha\le \gamma$.