To prove $\sup B \leq \sup A$

Question

Assume $A$ and $B$ are non empty and bounded above and satisfy $B \subseteq A$. Show that $\sup B \leq \sup A$

I am thinking of proving using contradiction, but I am getting nowhere. Someone please give hints

Thanks

Answer 1

Hint:

Every upper bound of $A$ (so also the least) is an upper bound of $B\subseteq A$.

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edit (alternative using contradiction):

If $\sup B>\sup A$ then some $b\in B$ must exist with $b>\sup A$ and consequently $b\notin A$. This contradicts $B\subseteq A$.

Answer 2

An upper bound for A is an upper bound for B. In particular, the least upper bound for A is an uppper bound for B.

Answer 3

For all $x\in B$, we have $x\in A$, so $x\le sup A$. That means $sup A$ is a upper bound of $B$. This leads to $sup B\le sup A$.