$S_1 = [ |a -b| : a \in A , b \in B]$ What is the infimum and supremum of this set?

Question

$S_1 = \{ |a -b| : a \in A , b \in B\}$ What is the infimum and supremum of this set?

$S_2 = \{ a -b : a \in A , b \in B\}$ What is the infimum and supremum of this set?

My Try : I think $\sup S_1 = \max ( |\sup A - \inf B| , |\sup B - \inf A| )$

$\inf S_1 = \min ( |\sup A - \inf B| , |\sup B - \inf A| , |\inf A - \inf B|, |\sup A - \sup B|)$. Can anyone please tell me if I went correct ?

But it is becoming tedious when I go to find those for $S_2$. Can anyone please help me ?

EDIT : I think $\sup S_2 =\sup A - \inf B$ and $\inf S_2 = \inf A - \sup B$

Please correct me If I had gone wrong anywhere..

Answer 1

I will assume the correctness of the following exchange in the comments-

Did you mean $A,B⊆ℝ$? – Le Anh Dung

yes...…..……………………………………………………………………... – cmi

For a set $U$, define $U_+ =\sup U $ and $U_- = \inf U$. Your formulas for $(S_2)_\pm$ and $(S_1)_+$ are correct. Here is a proof for $(S_2)_+$: Note that $a-b \le A_+ - B_-$, so

$$(S_2)_+ \le A_+ - B_-$$ further for any $\epsilon>0$, if $a\in A, b\in B$ such that $a > A_+ - \epsilon $ and $b < B_- + \epsilon$, then $$ a-b > a_+ - B_- - 2\epsilon$$ so that $$ (S_2)_+ \ge A_+ - B_- - 2\epsilon, \quad \forall \epsilon >0$$ As $\epsilon>0$ was arbitrary, $$(S_2)_+ = A_+ - B_-$$ Similarly, one proves that $$ (S_2)_- = A_- - B_+$$ and as $S_1 = \{|s| : s \in S_2\}$, one can similarly show $$(S_1)_+ = \max( |(S_2)_-|, |(S_2)|_+ )$$

However, your formula for $(S_1)_-$ is not correct. suppose $A = (-1,1)$ and $B=(0,2)$. Then simply because $A\cap B\neq \emptyset$, $(S_1)_- = 0$, but $\pm A_\pm \pm B_\pm \neq 0$. In fact you don't even need an intersection, another counterexample is $ A =(0,1) \cap \mathbb Q, B = (0,1) \setminus A $. You also don't need a common limit point in $\mathbb R$, e.g. $A = \{ \sqrt{2n}, n\in \mathbb N\}, B = \{\sqrt{2n+1}, n\in\mathbb N\}$. Also consider $$ A = \{ 0,-2n,2n\} $$ $$ B = \{ 1/n,-10n,10n\}$$

Then $|A_\pm - B_\pm | = | \pm 2n - (\pm 10 n)|> n$ can be made arbitrarily large, while also $(S_1)_- = \frac 1n \to 0$.

If $A,B$ are connected sets, i.e. intervals, then after checking for an intersection (which would imply $(S_1)_- = 0$), your formula is correct.

Answer 2

If $A,B$ are not separated then, $\inf S_1=0.$

But, if $A,B$ are separated then I think it's not always possible to express $\inf S_1$ explicitly.

For example, consider $A=(-10,-5)\cup (-\frac{1}{2},2),B=(-2,-1)\cup (10,20)$

Here, $\inf S_1=\frac{1}{2}$ which you will not get by looking at supremum and infimum of $A$ and $B$.