Closure of Ball = Closed Ball in Normed Space

Question

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.

Answer 1

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 $D(x,r) \subset \overline{B}(x,r)$.

Addendum: To show why $\|x-y\| \ge r + \epsilon$:

Let $\phi(t) = t y +(1-t)x$. For $0 \le t_1 \le t_2 \le 1$ we have $\|\phi(t_2)-\phi(t_1)\| = (t_2-t_1) \|x-y\|$.

Note that $\phi(t) \in B(x,r)$ if $t \in [0,{r \over \|x-y\|})$ and $\phi(t) \in B(y,\epsilon)$ if $t \in (1-{\epsilon \over \|x-y\|}), 1]$, and we must have $t_1={r \over \|x-y\|} \le t_2=1-{\epsilon \over \|x-y\|}$ since the balls do not overlap.

Finally $\|x-y\|= \|\phi(0)-\phi(1)\| \ge \|\phi(0)-\phi(t_1)\| + \|\phi(t_2)-\phi(1)\| = r+ \epsilon$.

Another take:

Consider $\phi(t)$ for $t \in [0,1]$. For the $t \in [0,{r \over \|x-y\|})$ part we have $\phi(t) \in B(x,r)$ and for $t \in (1-{\epsilon \over \|x-y\|}), 1]$ we have $\phi(t) \in B(y,\epsilon)$. The balls do not overlap, so the intervals $[0,{r \over \|x-y\|}), (1-{\epsilon \over \|x-y\|}), 1]$ are disjoint and the combined length of the intervals is ${r+ \epsilon \over \|x-y\|}$.

Since $\|\phi(t_2)-\phi(t_1)\| = (t_2-t_1) \|x-y\|$ for $0 \le t_1\le t_2 \le 1$ we see that $\|x-y\|=\|\phi(1)-\phi(0)\| = (1-t_2)+(t_2-t_1)+(t_1-0)) \|x-y\| \ge (1-t_2)+(t_1-0) \|x-y\|$, and since $(1-t_2)+(t_1-0) = {r+ \epsilon \over \|x-y\|}$ we have the desired result.

Answer 2

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 $y\in\overline{B(x,r)}$.

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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 $1$.

Answer 3

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

[2]. 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)). $