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?


6 Answers 6


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 !

  • 1
    $\begingroup$ Elegant: definitely +1. $\endgroup$ Commented Dec 26, 2015 at 15:59
  • 3
    $\begingroup$ But bounded definition does not necessarily involve closed ball, normally it is open ball but I guess you can alter the definition without too much shakeup $\endgroup$
    – Fraïssé
    Commented Feb 24, 2016 at 2:41
  • 2
    $\begingroup$ @SaradominZamorakGuthix You can just use the closure of your open ball which is closed. $\endgroup$
    – mdcq
    Commented May 26, 2018 at 17:02

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 $\epsilon >0.$)


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)

  • $\begingroup$ thanks. Youre right, I was confused while reading about diameter of metric subspaces :). Thanks for your trouble $\endgroup$ Commented Mar 6, 2013 at 22:05

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 $$ d(y,x_0)=d(\lim\limits_{n\to\infty} x_n,x_0)=\lim\limits_{n\to\infty}d(x_n,x_0)\leq K $$


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.


$A\subset X$ 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.


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