Vrftsjtryk

Given two bounded, nonempty sets $A$ and $B$, such that $sup(A) = inf(B)$.

$(a)$ Show that $A cap B$ contains at most $1$ element.

Suppose that two distinct $a, b in A cap B $, we then have:
$$ a leq sup(A) = inf(B)$$
since $a in A$, and also since $b in B$ we can say that that:
$$ a leq sup(A) = inf(B) leq b rightarrow a leq b$$

We now make the same statement, but reverse the roles of $a$ and $b$, since $b in A$, and also since $a in B$
$$ b leq sup(A) = inf(B) leq a rightarrow b leq a$$
We conclude that $a=b$. $square$

I think this proof is fine so far.

(b) Show that:

$$ forall epsilon >0: space{ } exists a in A, b in B: space |a-b|< epsilon $$

I think the point is to use the first exercise,

1 element so suppose we have $c in A cap B$ then also $c in A,B$ , we then have that $|c-c|=0<epsilon$

No elements For the second case we would need to prove that whenever $A cap B = emptyset$, we can still find such an $a$ and $b$. Maybe we could somehow use the fact about the supremum and infimum, but I am not sure how to use this. We know that $a leq sup(A)=inf(B) leq b$, this just tells us how they are ordered.

I think it's unnecessarily complicated to use the first question. It could be useful to use the following property of $c=sup(A)$. For every $varepsilon>0$ there exists $ain A$ such that
$$c-varepsilon<a$$
(otherwise $c$ wouldn't be the lowest upper bound!). Similarly if $d=inf(B)$ then for every $varepsilon>0$ there exists $bin B$ such that
$$b<d+varepsilon$$
Can you finish it from there?

You can use the first statement if you wish. So you have two cases.

Case 1 A$cap$B = {a} You'll want to show that a = inf A = sup B. Then you can prove your claim using definitions.

Case 2 A$cap$ B = $emptyset$ For elements to be within distance $epsilon$ of each other, you can show they are distance $frac{epsilon}{2}$ away from s = inf A = sup B. This can also be done with definitions of sup and inf.

What @Olivier Moschetta has written is also correct. You'll notice you'll use the same arguments in both cases.

I don't think the first statement is necessary for the second.

But the definition (one way or another) of $sup A$ is that if $w < sup A$ then $w$ is not upper bound. So if $w < sup A$ there exists an $ain A$ so that $w < a le sup A$.

Likewise if $v > inf B$ then $v$ is not a lower bound of $B$ and so there is a $b in B$ so so that $inf B le b < v$.

If we let $w = sup A - frac 12epsilon$ and $v = inf B + frac 12 epsilon $ we get that there are elements $a,b$ so that:

$w = sup A - frac 12epsilon < a le sup A = sup B le b < sup B + frac 12 epsilon$.

And so $|b-a| = b- a < (inf B + frac 12 epsilon) - (sup - frac 12epsilon = epsilon$.

Now it could be that $a = b = sup A=inf B$. Or it might not. If $Acap B = {z}$ where $z = inf B=sup A$ that's a perfectly good example that works. But it need not be the only. If $Acap B =emptyset$ then we know that there is another example. But none of this is necessary

Let $c=sup(A)$ and $d=inf(B)$

For every $varepsilon>0$ there exists $ain A$ such that
$$c-frac{varepsilon}{2}<a rightarrow c-a <frac{varepsilon}{2} $$
Similarly, for every $varepsilon>0$ there exists $bin B$ such that
$$b<d+ frac{varepsilon}{2} rightarrow b-d < frac{varepsilon}{2}$$
Also notice that $c=d$, so we have that
$$b-c < frac{varepsilon}{2}$$
Great, now notice that by the triangle inequality:

$$ |a-b|=|a-c + c-b| leq |a-c| + |c-b|=|c-a|+|b-c|<frac{varepsilon}{2} + frac{varepsilon}{2} = varepsilon $$
Booya!