# 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?

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 $\epsilon >0.$)

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 !

• Elegant: definitely +1. – Tom Collinge Dec 26 '15 at 15:59
• 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 – Carlos - the Mongoose - Danger Feb 24 '16 at 2:41
• @SaradominZamorakGuthix You can just use the closure of your open ball which is closed. – philmcole May 26 '18 at 17:02

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)

• thanks. Youre right, I was confused while reading about diameter of metric subspaces :). Thanks for your trouble – Applied mathematician Mar 6 '13 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 $$\exists M \in \mathbb R^+:d(a,b). for every $$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.