Let $A,B\subseteq \mathbb{R}$ such that $\forall b\in B ~ \exists a_n$ where $a_n \in A$ and $a_n\to b$ as $n\to\infty$. Prove $\sup(B)\leq \sup(A)$

Question

Question: Let A and B be two non-emty bounded sub-sets of $\mathbb{R}$ with the following proposition: $\forall_{b\in B}\exists_{(a_n)}$. $(a_n)$ is made up of elements of A such that $a_n\to b$ as $n\to\infty$

I don't know what to do with the sequence part.

To be proven: $\sup(B)\leq \sup(A)$.

My attempt:

Suppose $a\in A$ and $b\in B$.

Then $a\leq sup(A)$ and $b\leq sup(B)$

If $b<a$ then $b\leq \sup(B)\leq a\leq \sup(A)$

Suppose $\sup(B)>\sup(A)$.

Then $\exists_{a\in A}$ such that $a>\sup(A)-\frac{\epsilon}{2}$.

And $\exists_{b\in B}$ such that $b>\sup(B)-\frac{\epsilon}{2}$.

But if $\sup(B)>\sup(A)$, then $b>a$. But $b<a$. So contradiction. So $\sup(B)\leq \sup(A)$.

Hope someone will take a look, thanks in advance!

Answer 1

The sequence part tells you that every point of $B$ is an adherent point of $A$. Thus $B\subset \overline{A}$. where $\overline A$ stands for the closure of $A$ in $\mathbb R$, i.e. the smallest closed subset of $\mathbb R$ containing $A$.

$B\subset \overline A\Rightarrow \overline B\subset \overline{\overline A}$ but $\overline{\overline A}=\overline {A}$. Hence,

$\overline B\subset \overline A$

And,

$\overline B\subset \overline A\Rightarrow \sup(B)\leq \sup(A)$.

since:

(1) $\sup(X)$ is an increasing function of subset $X\subset E$ of an ordered set $E$. $\overline B\subset \overline A\Rightarrow \sup(\overline B)\leq \sup(\overline A)$.

(2) $\overline A$ is closed and bounded (since $A$ is bounded) (then compact) having $\sup(\overline A)=\max(\overline A)=\sup(A)$, idem for $\overline B$.

Answer 2

I'm afraid your proof is not right. You need to use the assumption given in the problem. Let's have a look at your proof. You write:

Suppose $a∈A$ and $b∈B$.

Then $a≤\sup(A)$ and $b≤\sup(B)$

If $b<a$ then $b≤\sup(B)≤a≤\sup(A)$

Your conclusion that $b\leq a$ doesn't follow from what you write. You only have that $b<a$, and $b\leq\sup(B)$, which does not imply $\sup(B)\leq a$.

You write:

Suppose sup(B)>sup(A).

Then ∃a∈A such that $a>\sup(A)−\frac ϵ2$.

And ∃b∈B such that $b>\sup(B)−\frac ϵ2$.

But if sup(B)>sup(A), then $b>a$. But $b<a$. So contradiction.

I don't know how you obtained a contradiction here. You conclude that $b>a$, but that this contradicts $b<a$, which seem to come out of nowhere.

I would do it as follows: Let $x = \sup(B)$, and let $\epsilon>0$. Take $b\in B$, such that $b > x-\frac\epsilon 2$, and let $(a_n)$ be a sequence of elements from $A$, such that $a_n\to b$. Then the sequence $a_n$ comes within $\frac\epsilon 2$ of $b$ from some point on, meaning that it comes within $\epsilon$ of $x$ from some point on. Use this to argue that $\sup(A)\geq x$.

Answer 3

Its very simple:

Assume that $\beta:=\sup B>\sup A=:\alpha$. Then $\mu:={\alpha+\beta\over2}$ satisfies $\beta>\mu>\alpha$. By definition of $\beta$ there is a $b\in B$ with $b>\mu$, and by the special assumption in the question there is an $a\in A$ with $a>\mu$ as well – contradicting the definition of $\alpha$.