# 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.

• Let me see if I am following this correctly, – user145551 Nov 8 '16 at 16:34
• Let me see if I am following this correctly, if "a" is a lub for B, then "a" is also a lub for A because A is a subset of B. Since A is a subset of B, then the lub A must be less than or equal to lub B. I think I am understanding this, and I will continue to work to prove the statement through contradiction. – user145551 Nov 8 '16 at 16:40
• @user145551 those are exactly the right claims. A rigorous proof for them should only take manipulation of definitions. – RGS Nov 8 '16 at 16:50
• I'll continue to work on it, thank you. – user145551 Nov 8 '16 at 16:50