The supremum, infimum, maximum and minimum of the set $B:=\{x \in\mathbb{R}, \left||x-1|-|x-2|\right|<1\}$

real-analysissupremum-and-infimum

What is the supremum, infimum, maximum and minimum of the set
$$B:=\{x \in\mathbb{R}, \left|\left|x-1\right|-\left|x-2\right|\right|<1 \}?$$

I am not sure how to find any of them. I thought I have to find for which values is the equation right and then from the values I could have some interval for which I could say what is the min, max, sup and inf.. But whatever I put for $x$ I get that $1 < 1$.

Best Answer

A good place to start, at least for an intuitive grasp of the matter, would be to graph $f(x) = ||x-1| - |x-2||$ and $g(x)=1$, and see whenever $f$ is below $g$. Doing so, we see that, indeed, $f(x) = 1$ for all $x$, except on a particular interval:

enter image description here

For $x\in (1,2)$, $f(x) \ne 1$ (you can show this algebraically if you choose$^{(1)}$), and this gives you an idea of what your set actually looks like. Namely, the graph suggests your set is equivalent to $(1,2)$, and deriving the quantities you desire becomes easier.

$(1)$: Showing This Algebraically:

To show $f(x) = 1$ for all $x$ outside $(1,2)$, consider two cases: $x\le 1$ and $x\ge2$.

  • In the former, then $|x-1| = 1-x$ and $|x-2| = 2-x$. Then $f(x) = |1-x-(2-x)| = 1$.
  • In the latter case, $|x-1| = x-1$ and $|x-2| = x-2$. Then $f(x) = |x-1-(x-2)| = 1$.

On the other hand, if $x \in (1,2)$, i.e. $1 < x < 2$, then $|x-1| = x-1$ but $|x-2|=2-x$. Then $$f(x) = |x-1-(2-x)| = |2x-1|$$ which is clearly not always equal to $1$.

Related Question