Find supremum and infimum of the set $\{ x \in \mathbb R : |x| +|x+1|<2 \}. $

Question

Find supremum and infimum of the set $\{ x \in \mathbb R : |x| +|x+1|<2 \}. $

Please help someone how can I solve it?

If we take the different cases $(i) x >0, (ii) x<0$ the we are getting some region in $\mathbb R$ but finally will we have to take the union of this solutions or intersection?

Answer 1

You have to separate three cases: $x\geq 0$, $-1\leq x\leq 0$ and $x\leq -1$

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The entire set is then the union of the three regions. But it's importand that you understand why you take the union, don't just memorize you have to take a uniuon:

The point is that separating the cases, you say

• "OK, if $x\geq 0$, then $|x|+|x+1| = 2x+1<2\iff x<\frac12$"
• "And if $-1\leq x\leq 0$, then $|x|+|x+1| = 1 <2$ always"

Now, you know that $x$ is either greater than $0$ OR it is smaller than $0$. So, the OR is what makes you construct a UNION, because an element is in a union $A\cup B$ if it is in $A$ OR in $B$.

Answer 2

Hint: I like to use an array to cut $\mathbb{R}$ in subintervals where I can remove the absolute values.

$$\renewcommand{\arraystretch}{1.5} \begin{array}{|c|ccc|c|ccc|}\hline x&{-\infty }&&{-1}&&0&&{+\infty }\\ \hline \left|x\right|&&{-x}&&{-x}&&x&\\ \hline \left|x+1\right|&&{-(x+1)}&&x+1&&x+1&\\ \hline \left|x\right|+\left|x+1\right|-2&&&&&&&\\ \hline \end{array}$$

You only need to complete the array's last line and think.