# Let $P \subset Q \subset ℝ$, $P\neq\emptyset$ and $P$ and $Q$ are bounded above. Show that $\sup P\leq \sup Q$.

I was wondering how to finish this proof, or if I'm even on the right track.

Let $b=\sup Q$. Then $x\leq b$ $\forall x \in Q$ and $b \leq$ any upper bound of $Q.$

Since $P \subset Q, \\ \forall y \in P, y \in Q$

This means that $y \leq b$

I know that $\sup P \in Q$ since all of the elements in $P$ are in $Q$. Since any element of $P$ must be less than or equal to b, can I say that $\sup P$ must be less than or equal to $\sup Q$?

Thanks

• If $P=Q=(0,1)$, then $\sup P \notin Q$. – Surb Sep 8 '18 at 22:20

Note that $b$ is an upper bound for $P$. In particular since $\sup P$ is the least upper bound of $P$, it follows that $$\sup P\leq b$$ The supremum exists since $P$ is bounded above and non-empty.

Untel the point in which you assert that$$(\forall y\in P):y\leqslant b\tag1$$it's fine. But you can't say that $\sup P\in Q$. This is not true in general.

However, $(1)$ telles us that $b$ is an upper bound of $P$. Therefore, $\sup P\leqslant b=\sup Q$.

Assume $$\sup P>\sup Q=\sup P-\epsilon$$

then, by characterisation of the least upper bound,

$$(\exists p\in P) \; \; : \; \;\sup Q<p\le \sup P$$

$$\sup Q <p \implies p\notin Q$$

but

$$p\in P \text{ and } P\subset Q \implies p\in Q$$