Metric spaces and closed sets

**Siron** · April 8th 2012, 11:48 AM

Hi, I need some help with the following question:

Let $(X,d)$ be a metric space and $A \subseteq X$ . Define the set $A^{(c)} \subseteq X$ for $\epsilon>0$ as $\{ x \in X | d(x,A) \leq \epsilon \}$

Prove the following statements:
(1) $\forall \epsilon \geq 0: \overline{A} \subseteq A^{(c)}$
(2) $\overline{A} = \bigcap_{\epsilon \geq 0} A^{(c)}$
(3) $A^{(c)} \ \mbox{is closed}$
(4) $\partial A^{(c)} \subseteq \{ x \in X | d(x,A)=\epsilon \}$
(5) Does there exist an example wherefore $\partial A^{(c)} \subset \{ x \in X | d(x,A)=\epsilon \}$

Proofs:
(1) Suppose $x \in \overline{A}$ thus by definition $d(x,A)=0 \leq \epsilon$ (with $\epsilon \geq 0$ ) therefore $x \in A^{(c)}$

Can someone give some hints to prove the other 4 statements, because I'm stuck there.

**Plato** · April 8th 2012, 12:11 PM

Originally Posted by Siron

Hi, I need some help with the following question:

Let $(X,d)$ be a metric space and $A \subseteq X$ . Define the set $A^{(c)} \subseteq X$ for $\epsilon>0$ as $\{ x \in X | d(x,A) \leq \epsilon \}$

Prove the following statements:
(1) $\forall \epsilon \geq 0: \overline{A} \subseteq A^{(c)}$
(2) $\overline{A} = \bigcap_{\epsilon \geq 0} A^{(c)}$
(3) $A^{(c)} \ \mbox{is closed}$
(4) $\partial A^{(c)} \subseteq \{ x \in X | d(x,A)=\epsilon \}$
(5) Does there exist an example wherefore $\partial A^{(c)} \subset \{ x \in X | d(x,A)=\epsilon \}$

Proofs:
(1) Suppose $x \in \overline{A}$ thus by definition $d(x,A)=0 \leq \epsilon$ (with $\epsilon \geq 0$ ) therefore $x \in A^{(c)}$

From #1 you already have half of #2. So show that $\overline{A} \supset \bigcap_{\epsilon \geq 0} A^{(c)}$ .

In #3 I would show the complement is open.

#4 should fall right from applying the definition of boundary point.

**Siron** · April 8th 2012, 01:32 PM

To prove: $\bigcap_{\epsilon \geq 0} A^{\epsilon} \subset \overline{A}$
Proof:
Suppose $x \in \bigcap_{\epsilon\geq 0} A^{\epsilon} = A^{\epsilon_1} \cap A^{\epsilon_2} \cap \ldots$
thus $x \in A^{\epsilon_1} \wedge x \in A^{\epsilon_2} \wedge \ldots$ which means
$d(x,A) \leq \epsilon_1 \wedge d(x,A) \leq \epsilon_2 \wegde \ldots$ and therefore
$d(x,A) \leq \max\{\epsilon_1,\epsilon_2,\ldots\}$ , write $d(x,A)\leq e'$
But $d(x,A)=\inf\{d(x,a)|a \in A \} \leq e'$ , therefore we can choose an element $a \in A$ with $d(x,a) \leq \epsilon' \wedge a \in A$
and so $\forall \epsilon' \geq 0: B(x,\epsilon') \cap A \neq \emptyset \Rightarrow x \in \overline{A}$

What do you think about this proof?

**Plato** · April 8th 2012, 01:53 PM

Originally Posted by Siron

To prove: $\bigcap_{\epsilon \geq 0} A^{\epsilon} \subset \overline{A}$

What would it mean if $\exists y\in\left( \bigcap_{\epsilon \geq 0} A^{\epsilon}\right) \setminus \overline{A}$ .

Let $\delta=D(A;y)>0$ . If $c=\tfrac{\delta}{2}$ can $y\in A^{(c)}~?$

**Siron** · April 8th 2012, 11:05 PM

Originally Posted by Plato

What would it mean if $\exists y\in\left( \bigcap_{\epsilon \geq 0} A^{\epsilon}\right) \setminus \overline{A}$ .

It means that $\exists y \in \bigcap_{\epsilon \geq 0} A^{\epsilon} \wedge y \notin \overline{A}$

Originally Posted by Plato

Let $\delta=D(A;y)>0$ . If $c=\tfrac{\delta}{2}$ can $y\in A^{(c)}~?$

I never have seen something like $\delta=D(A;y)$ before, what does it mean?

**Plato** · April 9th 2012, 02:53 AM

Originally Posted by Siron

$\delta=D(A;y)$ before, what does it mean?

$\delta=D(A;y)$ denotes ther distance from $y$ to $A$ .

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