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 $1 < 1$.

  • $\begingroup$ $B=(1,2)$. It has no max nor min. $\endgroup$ Jul 26, 2020 at 9:00

2 Answers 2


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:

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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 $1$.

  • 2
    $\begingroup$ Nice answer (+1). $\endgroup$
    – S.H.W
    Jul 26, 2020 at 8:21

If $x \le 1$ then $$|1-x + x - 2|\lt 1 \implies 1 \lt 1$$Which is a contradiction. If $1\lt x \le 2$ then $$|x-1 + x - 2| \lt 1 \implies 1 \lt x \lt 2$$And if $x\gt 2$ then $$|x - 1 - x + 2| \lt 1 \implies 1 \lt 1$$ So the answer is $A = \{x \in \mathbb{R} | 1\lt x \lt2\}$. I think you can find the values that you've mentioned then.


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