# Prove that $\sup(A)<\sup(B)$ implies there is an upper bound of $A$ in $B$

I am trying to prove the proposition:

$\sup A <\sup B$ implies there is an upper bound of $A$ in $B$ ($A ,B \subseteq \mathbb{R}$).

Is my proof sensible?

If there was no $b \in B$ with $\sup{A} < b$:

$\Rightarrow \sup A \geq b$ for all $b \in B$

$\Rightarrow \sup A$ is an upper bound of B

as $\sup B \leq$ all upper bounds of $B$ this means we have a contradiction as $\sup A <\sup B$ by the proposition.

Can we conclude from this there must be a $b \in B$ such that $\sup{A} < b$. Making it an upper bound of $A$?

• The phrase “upper bound” is more common than “upper limit”. – egreg May 7 '17 at 21:31
• Thanks, the word limit is certanly used in a different way in analysis anyways. – pindakaas May 8 '17 at 9:51

Your proof is quite correct though I would write it a bit differently:

Let $A,B$ be non-empty sets with $\sup(A) < \sup(B)$.

Suppose for a contradiction that $\forall b \in B: \lnot(\sup(A) < b)$

Then $\forall b \in B: b \le \sup(A)$, as we have a linear order. Which makes $\sup(A)$ an upperbound for $B$ and by minimality of $\sup(B)$ among all upperbounds of $B$, we get $\sup(B) \le \sup(A) (< \sup(B))$, a contradiction.

So $\exists b \in B: \sup(A) < b$ and as all $a \in A$ obey $a \le \sup(A)$, this newly found $b$ is indeed an upperbound for $A$.

Note also that the reverse almost but not quite holds: if $B$ contains a strict upperbound for $A$, so $\exists b \in B :\forall a \in A: a < b$ , then $\sup(A) \le b \le \sup(B)$, and we can do no better (not $\sup(A)< \sup(B)$) as $A = (0,1), B= (0,1]$ shows.

Some detail has to be fixed. You can avoid contradiction by proving the contrapositive.

Suppose no element of $B$ is an upper bound of $A$ (better than “suppose $\sup A<b$ for no $b\in B$”).

Then, for every $b\in B$, there is $a\in A$ with $a\ge b$. Since $\sup A$ is an upper bound for $A$, $a\le\sup A$.

Hence, for every $b\in B$, we have $b\le\sup A$. In particular $\sup A$ is an upper bound for $B$ and therefore $\sup B\le\sup A$.

On the other hand, you can do a direct proof.

Let $\varepsilon=\sup B-\sup A>0$. Then there exists $b\in B$ such that $b>\sup B-\varepsilon$. This means $b>\sup B-(\sup B-\sup A)$, that is, $b>\sup A$. Since $\sup A$ is an upper bound for $A$, then $b$ is an upper bound as well.

Since $\sup B$ is the least upper bound of $B,$ no number less that $\sup B$ can be an upper bound of $B.$ In particular, since $\sup A\lt\sup B,$ it follows that $\sup A$ is not an upper bound of $B,$ Therefore, there is an element $b\in B$ such that $b\gt\sup A.$ Since $b\gt\sup A,$ it follows that $b$ is an upper bound of $A.$

It is wrong for instance take $A=[0,\sqrt{2})$ and $B=[0,2) \cap \mathbb Q$. What would be an upper limit for $A$ in $B$?

• What is it that you are saying is wrong? One answer to your question is "1.5". That is a member of B that is an upper limit for A. – user247327 May 7 '17 at 21:16
• $\frac{3}{2} > \sqrt{2}$ last time I checked. So $B$ has an upper bound for $A$, not its least upper bound, but that was not the claim. – Henno Brandsma May 7 '17 at 21:17
• I think you are misinterpreting "upper limit." An upper limit of $A$ seems to be any number which is larger than all members of $A$. In your example, 1.5 would be an upper limit for $A$ in $B$. – ForgotALot May 7 '17 at 21:17
• Indeed, I thought upper limit meant supremum. English is not my native language, I am sorry. – Régis May 8 '17 at 9:40