Bounded sets are relatively compact in $\mathbb{R}$ I know that every closed and bounded set in $\mathbb{R}$ is compact (like $[a,b]$)
so i can conclude that every bounded set in $\mathbb{R}$ is relatively compact, by contradiction i say that let $A\subset\mathbb{R}$ be a bounded set but not relatively compact it means that $\overline{A}$ is not compact, but $\overline{A}$ is closed and bounded in $\mathbb{R}$ so it is compact, contradiction. 
My question is if $A$ is bounded, why $\overline{A}$ still bounded ?
Thank you.
 A: If $A$ is bounded this means there is some $R > 0$ such that $d(x,y) \le R$ for all $x,y \in  A$. But the same then holds for $\overline{A}$ as well. This can for instance be seen as follows: pick $p \in A$, then all members of $A$ are in the closed ball $D(p,R) = \{x \in X: d(x,p) \le R \}$. So $\overline{A} \subseteq \overline{D(p,R)} = D(p,R)$ as well.
So If $A$ is bounded, $\overline{A}$ is bounded (see above) and closed (by definition), so if we are in $\mathbb{R}$ or $\mathbb{R}^n$, $\overline{A}$ is compact by Heine-Borel.
A: Since A is bounded, it has a supremum and infimum. But  $\sup A=\sup \bar {A} $ and $\inf A=\inf \bar {A}$. Hence $\bar A $ is bounded.  
To see why this claim is true. Suppose $\sup \bar {A} $ is bigger than $\sup A $ by $ \epsilon$. Since $\bar {A} $ is closed,  $\sup\bar {A}\in \bar {A} $. Now the ball of radius $\epsilon/2 $ around this point will intersect $ A $, which is a contradiction, since then $ A $ contains an element bigger than its supremum. The argument for the infimum is similar.
A: As $A$
  is bounded, we have $A\subset\bar{\mathcal{B}}\left(0,M\right)$
  for a certain $0<M<+\infty$
 . Since $\bar{A}$
  is the smallest closed set that contains $A$
 , thus we must have $\bar{A}\subseteq\bar{\mathcal{B}}\left(0,M\right)$
  and then $\bar{A}$
  is bounded.
A: Recall what bounded means: When $A$ is bounded, there exists an open interval $(x_0-\delta;x_0+\delta)$ that contains $A$. Then the corresponding closed interval $[x_0-\delta;x_0+\delta]$ also contains $A$, and since the closure of $A$ is the smallest closed set containing $A$, the closure of $A$ will be contained in the interval $[x_0-\delta;x_0+\delta]$. But then the closure of $A$ is also contained in the open interval $(x_0-2\delta;x_0+2\delta)$, which makes the closure of $A$ bounded. 
A: Suppose $A$ is bounded so $|x-y|\leq M$ for $x,y \in A$. If $\overline A$ is not bounded then there is a $z\in \overline A$ such that $dist(z,A)>0$. But $z\in \overline A$ means $dist(z,A)=0$. Contradiction.
