Partial ordered sets and supremum.

Question

Let $A,B$ subsets of a partial ordered set $(X,\succeq)$, such that $A\subseteq B$. Suppose that $sup(B)$ exists, Do $sup(A)$ exists?

I could prove that all upper bound of $B$ is an upper bound of $A$ I wanted to prove that $sup(A)$ exists by the supremum axiom, but I don't know if $X$ is complete. Is there a counterexample pf $sup(B)$ exists but $sup(A)$ does not?

Answer 1

Let $X=[0,1]\cap\Bbb Q$ with the usual order. Let $B=X$ and $A=\left[0,\sqrt2-1\right)\cap\Bbb Q$. Then $\sup B=1$, but $A$ has no supremum in $X$, because $\sqrt2-1$ is irrational.