# Need help with proof verification about supremums and infemums

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?

• This is correct as is. – 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.