Show that the closure of a subset is bounded if the subset is bounded Let $A$ be a subset of $X$, and let $A$ be bounded. I.e.: $\exists x_0\in X : d(x,x_0)\le K,   \forall x\in A.$
I want to show that $\overline{A}$, the closure of $A$ is bounded as well, but as simple it may seem, I have some trouble with proving it. Should I use the definition that the $\overline{A}=A\cup\{\text{All limit points of }  A\}$ , and then show that if you pick an element from the limit points, that $d(x,x_0) \le Q$ for some $Q\in \mathbb{R}, x_0 \in X$?
Or should I use the defintion of closure that $x\in \overline{A}$ if for any $\epsilon>0$ we have $B_{\epsilon}(x)\cap A \neq \emptyset$?
What should be my strategy? 
The writer  states that $\inf A $ and $\sup A$ are both in $\overline{A}$. How can that be shown?
 A: Either definition works:
If $a$ is a limit point of $A$, there is some sequence $a_n \in A $such that $a_n \rightarrow a$. It then follows that $d(a,x_0) \leq d(a,a_n) + d(a_n,x_0) \leq d(a,a_n)+K$ for any $n$. Letting $n \rightarrow \infty$ we get that $d(a_n,a) \rightarrow 0$ so $d(a,x_0) \leq K$. (Since the left hand side is independent of $n$.)
Otherwise, for any $a \in \bar A$, $B_1(a)\cap A \not = \emptyset$, so we can pick a point $a'$ in the intersection. Hence $d(a,x_0) \leq d(a,a') + d(a',x_0) \leq 1 + K$. (You can even show that $\bar A$ is bounded by the same $K$ as before by noting that the above is true with $1$ replaced by any $\epsilon >0.$)
A: If $A$ is bounded then there exists some closed ball $B$ such that $A \subset B$.
Now by definition of $\bar A$ as the smallest closed set that contains $A$, it must be that $\bar A \subset B$.
This proves that $\bar A$ is bounded.
No computation involved in this proof !
A: Let $y\in\mathrm{cl}A$, then $y=\lim\limits_{n\to\infty}x_n$ for some $\{x_n:n\in\mathbb{N}\}\subset A$, since $d:X\times X\to\mathbb{R}_+$ is continuous then
$$
d(y,x_0)=d(\lim\limits_{n\to\infty} x_n,x_0)=\lim\limits_{n\to\infty}d(x_n,x_0)\leq K
$$ 
A: Not much more than the definitions of closure and metric and bounded are needed.
Suppose $d(x,x_0)<Q$ for all $x\in A$.
If $y\in \bar A$, there exists $x\in A$ with $d(x,y)<1$. 
By triangle inequality $d(x_0,y)<Q+1$.
Remark: Working with $\inf A$ and $\sup A$ works of ocurse only if $A$ is ordered ($\subseteq \mathbb R$, say)
A: Since $A$ is bounded, $\exists \, \,M > 0$ and $x_0 \in X$ such that $A \subset B_M(x_0)$.
$\Rightarrow$ diam $A$ < $M$  
$\Rightarrow$ diam cl $(A)$ < $M$
$\Rightarrow$ cl $A \subset B_M(y_0)$ $\qquad $ for some $y_0 \in X$ 
Therefore cl $A$ is bounded.
A: $A\subset X$ is a bounded set w.r.t $d$, if $\exists M \in \mathbb R^+:d(a,b)<M$. for every $a,b\in A.$
We want to prove $x,y \in \overline A, \exists x_n\in A,$ $x_n \to x$ as $n\to \infty$ and $\exists y_n\in A,$ $y_n \to y$ as $n\to \infty$.
By Triangle inequality, we have $d(x,y)\leq d(x,x_n)+d(x_n,y_n)+d(y_n,y)$. From the convergence of $x_n$, $y_n$ and boundedness of $A$. We get $\overline A$ is bounded.
