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here's what i managed so far

using absolute value's properties; $$3x-1>1 \hspace{5mm} x>\frac{1}{3}$$

$$D= \{ x\in \mathbb{R} | \hspace{2mm} |3x-1|>x \}=$$ $$=\{ x\in \mathbb{R} | \hspace{2mm} 3x-1>x \hspace{7mm} \forall x \geq\frac{1}{3} \} \cup\{ x\in \mathbb{R} | \hspace{2mm} 1-3x>x \hspace{7mm} \forall x<\frac{1}{3} \} $$

$$=\{ x\in \mathbb{R} | \hspace{2mm} x>\frac{1}{2} \} \cup\{ x\in \mathbb{R} \hspace{2mm} x<\frac{1}{4} \} = (-\infty,\frac{1}{4})\cup(\frac{1}{2},+\infty)$$

let for semplicity $$A:=(-\infty,\frac{1}{4} ) \hspace{7mm} \land \hspace{7mm} B:=(\frac{1}{2},+\infty)$$

to find the $ \hspace{5mm}\sup A \hspace{5mm}$ let's suppose $ \hspace{5mm}\sup A = \frac{1}{4}\hspace{5mm}$ therefore $$ \hspace{5mm} \forall \epsilon \ge 0 \hspace{2mm} \exists a\in A \hspace{1mm}: \frac{1}{4}-\epsilon < a$$

i supposed to find this precise element $a$ given any arbitrary $\epsilon $ but i don't understand how to find it ( isn't it trivially a>$\frac{1}{4}-\epsilon ?$)

i'm not sure how to find such an a therefore i'm note able to continuing the proof of the infimum,

while to prove the $\sup A= +\infty$ we can suppose that $$ \exists M : |a|<M \hspace{3mm} \forall a \in A \implies -M<a<M \implies a<-M$$ if M is the minimum of the set, M is also a member of A therefore $$ \implies a<M<1/4 \implies $$ M is not the minimum so we can let the sup be $+\infty$

the other half of the proof should follow the same principle and after i found the sup/inf of B i would just let $$ \sup D = \max \{ a, b\}$$ with a,b being the suprema of A and B and $$\inf D = \min \{ a', b'\} $$ with a', b' being the infima of A and B

i don't know of the part the epsilon proof can be done like that or it's is different

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You have that $D=(-\infty,1/4)\cup(1/2,+\infty)$, good!

From this, we can basically read off that $\inf D=-\infty$ and $\sup D=+\infty$. To prove it formally, let us for instance argue that $\sup D=+\infty$. If this wasn't the case, then $\sup D=a$ for some real $a$. This would in particular mean that $a$ would be an upper bound of $D$, that is, $$x\leq a$$ for every $x\in D$. However, since $x=\max(a,1/2)+1\in D$ is bigger than $a$, we get a contradiction.

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  • $\begingroup$ just out of curiosity ( more for learning ) how do i ussually find the contradiction starting from the epsilon definition? $\endgroup$
    – Massimiliano Messina
    Commented Aug 5, 2023 at 16:57
  • $\begingroup$ just let suppose i want to find the sup/inf of A and B as a self exercise, how do i manage to use the epsilon defintion as i tried for A (1/4 =supA)? $\endgroup$
    – Massimiliano Messina
    Commented Aug 5, 2023 at 17:05
  • $\begingroup$ A direct proof would be simpler: $\forall M\in\Bbb R\;\exists x\in D\;x>M.$ Simply take some $x>\max(M,1/2).$ $\endgroup$
    – Anne Bauval
    Commented Aug 5, 2023 at 17:22
  • $\begingroup$ @MassimilianoMessina To show that $\sup A=a$, you need to prove two things: (i) $a$ is an upper bound for $A$, that is, $x\leq a$ for every $x\in A$. (ii) $a$ is the least upper bound for $A$, that is, for every $\epsilon>0$, there is $x\in A$ with $x>a-\epsilon$. The point in this exercise is that already (i) fails, so there is no need to check (ii). $\endgroup$
    – Zuy
    Commented Aug 5, 2023 at 21:45
  • $\begingroup$ @MassimilianoMessina To show that $\sup A=1/4$ for $A=(-\infty,1/4)$, proceed as follows. (i) Check that $1/4$ is an upper bound for $A$ (this is obvious). (ii) Check that $1/4$ is the least upper bound for $A$. For this, let $\epsilon>0$. Taking $x$ between $1/4$ and $1/4-\epsilon$ (say $x:=\frac{1/2-\epsilon}{2}$) gives us an $x\in A$ with $x>1/4-\epsilon$. $\endgroup$
    – Zuy
    Commented Aug 5, 2023 at 21:49

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