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


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?

  • $\begingroup$ This is correct as is. $\endgroup$ – rubikscube09 Mar 12 '19 at 18:48

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.