Let $B,C$ be sets, $B\subseteq C$ such that $\forall c∈C,\exists b\in B: c\le b$ then $\sup B=\sup C$.

Question

I had been reading this.
In the proof, below lemma is used. I don't know how to go for proving it.Notice that I want to prove this theorem for set of ordinals not real numbers

Let $B,C$ be sets, $B\subseteq C$ such that $\forall c∈C,\exists b\in B: c\le b$. Then $\sup B=\sup C$.

Answer 1

First of all $B\subseteq C$, so any upper bound of $C$ is an upper bound of $B$, hence $\sup C$ is an upper bound of $B$. Because $\sup B$ is the least upper bound of $B$ we conclude that $\sup B\leq\sup C$.

Now let any $c\in C$. There is an element $b\in B$ such that $c\leq b\leq \sup B$. Hence $\sup B$ is an upper bound of $C$. But $\sup C$ is the least upper bound of $C$, so we conclude $\sup C\leq\sup B$.

Answer 2

This holds generally in partially ordered sets.

Let $B\subseteq C$ be subsets of the partially ordered set $(X,\le)$. Suppose that, for every $c\in C$ there exists $b\in B$ such that $c\le b$. If one among $B$ and $C$ has the supremum, then both have and they are equal.

Consider $B^*$ the set of upper bounds of $B$ and define similarly $C^*$. From $B\subseteq C$ it follows that $C^*\subseteq B^*$.

Let $x\in B^*$ and suppose $c\in C$. Then there is $b\in B$ such that $c\le b$. As $b\le x$, we have $c\le x$. Hence $x\in C^*$.

Therefore $B^*=C^*$ and the statement follows.

If $X$ is a set of ordinals, then every upper bounded subset of $X$ has the supremum (but a subset may not be upper bounded).