Let $(X, d)$ be a metric space, $x_{0}$ be a point in $X$, and $r>0 .$ Let $B$ be the open ball $B:=B\left(x_{0}, r\right)=\left\{x \in X: d\left(x, x_{0}\right) 0$ such that $\overline{B}$ is not equal to $C$.

\begin{proof} Let $x \in \bar{B}=\operatorname{int}(B) \cup \partial B$. Then we have two cases (since by definition a point cannot be both a interior point and boundary point of the same set).\par Case 1: $(x \in \operatorname{int}(B))$. This means that $x \in B$ so $d\left(x, x_{0}\right)<r$ but then we also have $d\left(x, x_{0}\right) \leq r$ so $x \in C .$\par Case 2: $(x \in \partial B)$. Here is a rough outline of what we will do: We will assume $x \notin \subset$. Then we will show that this leads to $x \in \operatorname{ext}(B)$ (which we will do by finding $r^{\prime}$ such that $\left.B\left(x, r^{\prime}\right) \cap B\left(x_{0}, r\right)=\emptyset\right)$ which contradicts that $x \in \partial B$\par Assume, for sake of contradiction, that $x \notin C$. That is, $d\left(x, x_{0}\right)>r$.\par Now, let $r^{\prime}=d\left(x, x_{0}\right)-r>0$.\par Assume $y \in B\left(x, r^{\prime}\right) \cap B\left(x_{0}, r\right)$. That is, $y \in B\left(x, r^{\prime}\right)$ and $y \in B\left(x_{0}, r\right)$. This means that $d(y, x)<r^{\prime}$ and $d\left(y, x_{0}\right)<r$.\par From the triangle inequality we have: $$d\left(x, x_{0}\right) \leq d(x, y)+d\left(y, x_{0}\right)<r^{\prime}+r=r-d\left(x, x_{0}\right)+r=d\left(x, x_{0}\right) .$$ So we have $d\left(x, x_{0}\right)<d\left(x, x_{0}\right)$. Thus there is no element $y \in B\left(x, r^{\prime}\right) \cap B\left(x_{0}, r\right)$. Hence, $B\left(x, r^{\prime}\right) \cap B\left(x_{0}, r\right)=\emptyset$. Thus $x$ is an exterior point of $B$, but this contradicts that $x \in \partial B$, so our original assumption that $x \notin C$ was wrong. And therefore $x \in C$.\par In both cases we get that $x \in C$. Thus $x \in \bar{B} \Rightarrow x \in C$, hence $\bar{B} \subset C$\par \end{proof} For (b), consider the following \begin{example*} Let $X=\mathbb{R}^{2}$ and $d$ be the discrete metric. Let $x_{0}=(0,0)$ and take $r=1$. Then we have $B_{\left(\mathbb{R}^{2}, d_{\text {disc }}\right)}((0,0), 1)=\{(0,0)\}$ And so $\bar{B}=\{(0,0)\}$ But $C=\mathbb{R}^{2}$ since $\forall x, y \in \mathbb{R}^{2}, d(x, y) \leq 1$. Thus $\bar{B} \neq C$. \end{example*}

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Analysis II

Exercise 1.2.4

Let $(X, d)$ be a metric space, $x_{0}$ be a point in $X$ , and $r > 0 .$ Let $B$ be the open ball $B : = B (x_{0}, r) = {x \in X : d (x, x_{0}) < r}$ and let $C$ be the closed ball $C : = {x \in X : d (x, x_{0}) \leq r} .$

(a) Show that $B \subset C$ .
(b) Give an example of a metric space $(X, d)$ , a point $x_{0}$ , and a radius $r > 0$ such that $\bar{B}$ is not equal to $C$ .

Answers

Proof. Let $x \in \bar{B} = int (B) \cup ∂B$ . Then we have two cases (since by definition a point cannot be both a interior point and boundary point of the same set).

Case 1: $(x \in int (B))$ . This means that $x \in B$ so $d (x, x_{0}) < r$ but then we also have $d (x, x_{0}) \leq r$ so $x \in C .$

Case 2: $(x \in ∂B)$ . Here is a rough outline of what we will do: We will assume $x \notin \subset$ . Then we will show that this leads to $x \in ext (B)$ (which we will do by finding $r^{'}$ such that $B (x, r^{'}) \cap B (x_{0}, r) = \emptyset)$ which contradicts that $x \in ∂B$

Assume, for sake of contradiction, that $x \notin C$ . That is, $d (x, x_{0}) > r$ .

Now, let $r^{'} = d (x, x_{0}) - r > 0$ .

Assume $y \in B (x, r^{'}) \cap B (x_{0}, r)$ . That is, $y \in B (x, r^{'})$ and $y \in B (x_{0}, r)$ . This means that $d (y, x) < r^{'}$ and $d (y, x_{0}) < r$ .

From the triangle inequality we have:

d (x, x_{0}) \leq d (x, y) + d (y, x_{0}) < r^{'} + r = r - d (x, x_{0}) + r = d (x, x_{0}) .

So we have $d (x, x_{0}) < d (x, x_{0})$ . Thus there is no element $y \in B (x, r^{'}) \cap B (x_{0}, r)$ . Hence, $B (x, r^{'}) \cap B (x_{0}, r) = \emptyset$ . Thus $x$ is an exterior point of $B$ , but this contradicts that $x \in ∂B$ , so our original assumption that $x \notin C$ was wrong. And therefore $x \in C$ .

In both cases we get that $x \in C$ . Thus $x \in \bar{B} \Rightarrow x \in C$ , hence $\bar{B} \subset C$ □

For (b), consider the following

Let $X = ℝ^{2}$ and $d$ be the discrete metric. Let $x_{0} = (0, 0)$ and take $r = 1$ .

Then we have $B_{(ℝ^{2}, d_{disc})} ((0, 0), 1) = {(0, 0)}$ And so $\bar{B} = {(0, 0)}$

But $C = ℝ^{2}$ since $\forall x, y \in ℝ^{2}, d (x, y) \leq 1$ .

Thus $\bar{B} \neq C$ .

mathchakra

2021-12-10 20:31

Exercise 1.2.4

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