# Closure of Ball = Closed Ball in Normed Space

Show that, in any normed space $(X, \Vert \cdot \Vert)$, $\overline{B_{\Vert \cdot \Vert} (x, r)}$ = $\hat{B}_{\Vert \cdot \Vert} (x, r)$, i.e. the closure of a ball is the closed ball, for all $x \in X$ and $r > 0$.

$\subseteq$ seems to be trivially true since the closure $\overline{B}$ is the smallest closed set containing $B$, so it has to be inside $\hat B$. The other way doesn't seem to be quite clear though.

$\hat B \subseteq \overline{B}$ means the elements of $\hat B$ has to be in $B$ or the set of limit points $B'$, since by definition $\overline{B} = B \bigcup B'$. But how does the norm come into play here? I don't think I can say "$\hat B$ is closed, so it contains all elements of $B$, $b \in B$'s accumulation points, and perhaps more. From what I understand, a norm is much more strict than a metric, so I don't think what I'm thinking can really be this general for normed spaces. How should I approach the $\hat B \subseteq \overline{B}$ part of the proof? Thank you for your help.

• Can you approximate elements of norm exactly $r$ by elements with norm less than $r$? – fredgoodman Mar 5 '18 at 0:04
• Not in a normed space, right? – Max Mar 5 '18 at 0:07
• Certainly, in a normed space. – fredgoodman Mar 5 '18 at 0:09
• If we can approximate elements of norm $r$ by elements with norms less than $r$, then does that mean we can essentially approximate everything in $\hat B$? – Max Mar 5 '18 at 0:57
• Yes, that;s right. – fredgoodman Mar 5 '18 at 1:01

Let $D(x,r) = \{ y | \|x-y\| \le r \}$ and $B(x,r) = \{ y | \|x-y\| < r \}$.

We have $B(x,r) \subset D(x,r)$ for all $r$. Since the norm is continuous we see that $D(x,r)$ is closed and hence $\overline{B}(x,r) \subset D(x,r)$.

Now suppose $y \notin \overline{B}(x,r)$. Since $\overline{B}(x,r)$ is closed there is some $\epsilon>0$ such that $B(y,\epsilon)$ does not intersect $\overline{B}(x,r)$ (and hence does not intersect $B(x,r)$). It follows that $\|x-y\| \ge r + \epsilon$ and so $y \notin D(x,r)$. Hence $D(x,r) \subset \overline{B}(x,r)$.

• Sir, please can you elaborate "since the norm is continuous we see that...... that is first part of your answer. – Akash Patalwanshi Jul 29 '19 at 3:32
• The inverse image of a closed set ($(-\infty, r]$) under a continuous map (the norm) is a closed set. – copper.hat Jul 29 '19 at 5:14
• I can intuitively see why $||x-y|| \ge r + \epsilon$ follows from the preceding steps, but how would you go about making that part more formal? – MMFTSBU Sep 3 '19 at 12:56
• @jwl: Let $p(t) = y+t (x-y)$. Let $t_1$ be such that $p(t_1)$ lies on the boundary of $B(y,\epsilon)$ and let $t_2$ be such that $p(t_2)$ lies on the boundary of $B(x,r)$. We must have $0 < t_1 \le t_2 < 1$ and we have $\|x-y\| = \|p(1)-p(t_2)\| + \|p(t_2)-p(t_1)\| + \|p(t_1)-p(0)\|$. – copper.hat Sep 3 '19 at 13:18
• I am in thinking as the inverse image of the closed set $[0,r]$ is better than $(-\infty,r]$ in every sense. – user519955 Jul 5 at 14:16

It's sufficient to show that $\hat{B}(x,r)\setminus B(x,r) \subseteq \overline{B(x,r)}$.

Let $y\in\hat{B}(x,r)\setminus B(x,r)$ and notice that $\|y-x\| = r$.

For $n\in\mathbb{N}$, let $y_n\in B(y,1/n)\cap B(x,r)$ (why does such $y_n$ exist? Can you define it?). Clearly $y_n\to y$, which shows that $y\in B(x,r)'$. Thus $y\in\overline{B(x,r)}$.

Hint for existence of $y_n$: Since $B(x,r)$ is convex, $(1-t)x+ty\in B(x,r)$ for every $t\in[0,1)$. You can choose $t$ to be arbitrarily close to $1$.

. If $r>0$ and $\|y-x\|\leq r$ then $y\in Cl(\{x+s(y-x): s\in [0,1)\})\subset Cl (B(x,r)).$

Because if $U$ is a nbhd of $y$ then $B(y,t)\subset U$ for some $t>0,$ and there exists $s\in [0,1)$ such that $\|y-x\|\cdot (1-s)<t.$ Then $$\|y-(\;x+s(y-x)\;)\|=\|y-x)\|\cdot (1-s)<t$$ so $x+s(y-x)\in B(y,t)\subset U.$

So $y\in Cl(B(x,r)).$

. If $\|y-x\| >r$ then $B(y, \|y-x\|-r)\cap B(x,r)=\emptyset.$ Because if $z\in B(y,\|y-x\|-r)\cap B(x,r)$ then

$\|y-x\|\leq \|y-z\|+\|z-x\|<(\|y-x\|-r)+r=\|y-x\|,$ an absurdity.

So the open set $B(y,\|y-x\|-r)$ contains $y$ and is disjoint from $B(x,r).$

So $y\not \in Cl(B(x,r)).$

• For many basic properties of normed linear spaces, just consider the space $\Bbb R^2$ (with the Cartesian norm) geometrically rather than analytically. and you may see a method that applies to all such spaces.In \Bbb R^2, for example $B(y, \|x-y\|-r)$ is an open disc disjoint from the open disc $B(x,r)$, and $\{x+s(y-s): 0\leq s<1\}$ is the half-open line segment from $x$ to $y.$ – DanielWainfleet Mar 5 '18 at 7:08