# Proving least upper bounds within the set of real numbers

If A is a subset of B and A,B are non-empty sets that are subsets of the set of Real Numbers R, and if B has a least upper bound, then how can I prove that A has a least upper bound and that the lub A is less than or equal to lub B

• What you want to show is that if $\alpha$ is an upper bound of $B$, it must also be an upper bound of $A$. Really, the whole answer comes down to shuffling definitions – Omnomnomnom Nov 8 '16 at 16:27

First suppose that $A$ has no least upper bound to arrive at a contradiction, proving it has. Then suppose its least upper bound is greater than $B$'s to arrive at another contraction, showing it isn't greater i.e. it is less than or equal to $B$ least upper bound.