Is this sketch work correct way to prove?

Question

If $$S\subseteq T\subseteq R$$ where S is non empty, then show that : if T is bounded above, then $$SupS\leqslant SupT$$.

Sketch: Assume t to be supremum of T. Then $$x\leqslant t$$ for all x belonging to T.

Let p be supremum of S. Then $$s\leqslant p$$ for all s belonging to S.

Since s,p belongs to S, they also belong to T. Now if p>t, then t is not supremum of T(by definition). This implies p<t. If p=t( trivial). Thus $$s\leqslant p\leqslant t$$

score 3 · Accepted Answer · answered Mar 18 '21 at 07:13

3

You are assuming that the supremum of a set must belong to that set. That is not true in general.

If $s\in S$, then $s\in T$, and therefore $s\leqslant\sup T$. So, $\sup T$ is an upper bound of $S$. Since $\sup S$ is, by definition, the least upper bound of $S$, it follows that $\sup S\leqslant\sup T$.

answered Mar 18 '21 at 07:13

José Carlos Santos

427,504

Got it. Thanks. So how can I improve my proof? Thanks again! – Natasha J Mar 18 '21 at 07:54
I am sorry, but my proof is the only way that I see to improve yours. – José Carlos Santos Mar 18 '21 at 08:55
Ok! Thanks you! – Natasha J Mar 18 '21 at 09:14

Is this sketch work correct way to prove?

1 Answers1