Prove that a subset $A$ of $\mathbb{R}$ is bounded if and only if there is $M∈\mathbb{R}$ such that $|x|≤M$ for all $x∈A$. 
Prove that a subset $A$ of $\mathbb{R}$ is bounded if and only if there is $M∈\mathbb{R}$ such that $|x|≤M$ for all $x∈A$.

Help me understand this question I don't have any idea where to start. 
 A: Proofs like this depend crucially on how things are defined.  In this case, the problem comes from a book (see the link in my comment below the OP) in which a set $A\subset\mathbb{R}$ is said to be bounded if it bounded both above and below, where $A$ is bounded above if there is a number $M$ such that $x\le M$ for all $x\in A$, and bounded below if there is a number $L$ such that $L\le x$ for all $x\in A$.
The proof now amounts to putting these definitions together with a property of the absolute value function, namely $-|x|\le x\le |x|$, along with a property of inequalties, namely that if $|x|\le M$, then $-M\le-|x|$, so that
$$|x|\le M\implies -M\le-|x|\le x\le|x|\le M$$
Thus $A$ is bounded above (by $M$) because $|x|\le M$ for all $x\in A$ implies $x\le M$ for all $x\in A$, and likewise bounded below (by $L=-M$) because $|x|\le M$ for all $x\in A$ implies $-M\le x$ for all $x\in A$.  Finally it follows that $A$ is bounded by the book's definition of "bounded."
Remark: One subtlety in all this is that the proof given above depends not just on knowing the book's definition of "bounded" but also on knowing certain facts about inequalities and the absolute value function.  Presumably, though, the relevant properties were established in earlier sections of the book.
A: That what you stated is literally the definition of a bounded set.
I mean you could try and assume the opposite: that for every x from A  there exists y from A such that $y>x$ but then you easily achieve contradiction, that A is in that case unbounded.
And the other way around if A is bounded that means that there exists a number M such that it is greater or equal than all of the members of the set, which I guess completes it.
A: Note that $|x|\leq M$ means $$-M \leq x \leq M$$ that means every $x$ in $A$ is smaller than $M$ (i.e. $A$ is bounded from above by $M$) and every $x$ in $A$ greater than $-M$ (i.e. $A$ is bounded from below by $-M$).
A: Let's start with another definition of boundedness: $A\subset\Bbb R$ is bounded iff $A$ is a subset of some interval, i.e. $a\le x\le b$ for some $a,b$ and for any $x\in A$. Now it is rather easy to find $M$ to fulfil $|x|\le M$. Try to play with $|a|$, $|b|$, some maximum etc. It could happen that both $a,b$ are negative and so on.
The bound $|x|\le M$ could be strongly unoptimal from one side. See $A=(2,3)$. Then $|x|\le3$ for any $x\in A$, but saying the truth, $|x|\le 3\iff -3\le x\le 3$.
A: 
1) $ \emptyset \neq A \subseteq \Bbb{R}$, $A$ is bounded $\leftrightarrow$ $\exists M \in \Bbb{R}:\forall x \in A: |x| \leq M$

proof 1)

"$\leftarrow$" $|x| \leq M \to -M\leq x\leq M$ with $x \in A$, then
  $A$ is bounded
"$\rightarrow$" $A$ is bounded then $\exists S\in \Bbb{R}: \forall u
 \in A: u \leq S$ and $\exists s \in \Bbb{R}: \forall w\in A: s\leq w$.
  I define $T:=\max_\leq\{|S|,|s|\}$ and we have:
  
  
*
  
*$s\leq x \leq S$ with $x \in A$
  
*$s\leq |s|$ and $-s\leq |s|$ (then $-|s| \leq s$)
  
*$S \leq |S|$ and $-S\leq |S|$
  
*$|S| \leq T$ and $|s| \leq T$ (then $-T\leq -|s|$)
  
  
  then $$-T\leq-|s|\leq s \leq x \leq S \leq |S| \leq T \to -T \leq x
 \leq T \to x \leq |T| \text{, with } x \in A$$

