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Let $S$ and $T$ be nonempty bounded subsets of $\mathbb R$ with $S \subseteq T$. Prove that $$\inf T \le \inf S \le \sup S \le \sup T.$$

Proof:

Since $S$ is bounded, then $S$ is bounded above and bounded below.

So $m=\sup S$ and $n=\inf S$ where $m \ge n$.

By the same argument above, $a = \sup T$ and $b= \inf T$ where $a \ge b$.

Since $S \subseteq T$, $\sup T$ is an upper bound for $S$ and $\inf T$ is a lower bound for $S$, so $a \ge m \ge n \ge b$.


Is my proof correct? And if it is, can it be written better?

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  • $\begingroup$ This is correct as is. $\endgroup$ – rubikscube09 Mar 12 '19 at 18:48
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I think you pretty much have it. I think if I had written the proof, I would have talked about $T$ first and then transitioned into the definition of a subset since $S$ is contained in $T$, but that is just personal preference.

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