I'm trying to solve this exercise:

Let $A$ be a non-empty set $\mathbb R$-bounded. Let $B=\{|x-y|: x,y\in A\}$.

  1. Prove that $B$ has a least upper-bound and a greatest lower-bound.
  2. Prove that $\sup B = \sup A - \inf A$.
  3. Find $\inf B$.

I've completed part 1, but how should I tackle parts 2 and 3?

  • $\begingroup$ One can use mathjax for math symbols: meta.math.stackexchange.com/questions/5020/… $\endgroup$ – Cristhian Gz Dec 30 '14 at 14:05
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    $\begingroup$ start with small cases like $A$ is a bounded interval. first try to see what will happen if we start with an interval with only non-negative reals, with only non-positive reals, and an interval that contains both positive and negative reals. e.g. $A = [0, 1], [-1, 0], [-1, 1].$ Try to prove the results in this cases. it will give you an insight about the general case. $\endgroup$ – Krish Dec 30 '14 at 14:17
  • $\begingroup$ For the 3rd question, let $y=x$. $\endgroup$ – Nishant Dec 30 '14 at 16:23

Given $x, y \in A$, $x \le \sup A$ and $y \ge \inf A$, thus $x - y \le \sup A - \inf A$ Similarly $y - x \le \sup A - \inf A$. Therefore $$|x - y| \le \sup A - \inf A.$$ Since this is true for all $x,y \in A$, the set $\{|x - y| : x, y\in A\}$ is bounded above by $\sup A - \inf A$. Consequently, $$\sup B \le \sup A - \inf A.$$ On the other hand, given $\varepsilon > 0$, $\inf A + \frac{\varepsilon}{2}$ is not a lower bound for $A$ and $\sup A - \frac{\varepsilon}{2}$ is not an upper bound for $A$. Hence, there exists $x, y \in A$ for which $\inf A + \frac{\varepsilon}{2} > y$ and $\sup A - \frac{\varepsilon}{2} < x$. Thus $$\sup A - \inf A - \varepsilon = \left(\sup A - \frac{\varepsilon}{2}\right) - \left(\inf A + \frac{\varepsilon}{2}\right) < x - y \le |x - y| \le \sup B.$$ Since $\sup A - \inf A \le \sup B + \varepsilon$ for all $\varepsilon > 0$, we have $$\sup A - \inf A \le \sup B.$$ Hence $$\sup B = \sup A - \inf A.$$

You can see that $\inf B = 0$ by noting that $(1)$ $|x - y| \ge 0$ for all $x,y \in A$ and $(2)$ $0 \in B$ by taking an $x\in A$ and writing $0 = |x - x|$.

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  • $\begingroup$ For the second part (proving the least upper bound property), could we try to prove that for any arbitrary upper-bound d of B, $sup(A)-inf(A)\leq d$ ? Is the following argument right ? We just let $\varepsilon >0$ be arbitrary, and then let $x\in A$ denote the element such that $sup(A)-\varepsilon<x$ . Then, since $\forall y\in A, inf(A) \leq y $, we can let $y \in A$ arbitrary and subtract the second inequation from the first, concluding that $sup(A)-inf(A)-\varepsilon < x-y \leq |x-y| \leq d$. From that, we can conclude that $sup(A)-inf(A)\leq d$ . Is that a correct reasoning ?Thanks $\endgroup$ – nerdy Apr 6 '15 at 6:34
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    $\begingroup$ @nerdy you have some good ideas, but argument fails at the end when you conclude $\sup A - \inf A - \epsilon < x - y$; while $\sup A - x < \epsilon$, $-\inf A \color{red}{\ge} -y$, so your inequality does not follow. This is why I chose $x$ and $y$ such that $\inf A + \epsilon/2 > y$ and $\sup A - \epsilon/2 < x$ in my answer. If you do the same, then your argument will be valid. $\endgroup$ – kobe Apr 6 '15 at 12:39
  • $\begingroup$ Thanks, this was a really basic error of mine. $\endgroup$ – nerdy Apr 8 '15 at 4:56

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