Metric spaces and closed sets

Hi, I need some help with the following question:

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

Prove the following statements:

(1) $\displaystyle \forall \epsilon \geq 0: \overline{A} \subseteq A^{(c)}$

(2) $\displaystyle \overline{A} = \bigcap_{\epsilon \geq 0} A^{(c)}$

(3) $\displaystyle A^{(c)} \ \mbox{is closed}$

(4) $\displaystyle \partial A^{(c)} \subseteq \{ x \in X | d(x,A)=\epsilon \}$

(5) Does there exist an example wherefore $\displaystyle \partial A^{(c)} \subset \{ x \in X | d(x,A)=\epsilon \}$

Proofs:

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

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

Re: Metric spaces and closed sets

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**Siron** Hi, I need some help with the following question:

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

Prove the following statements:

(1) $\displaystyle \forall \epsilon \geq 0: \overline{A} \subseteq A^{(c)}$

(2) $\displaystyle \overline{A} = \bigcap_{\epsilon \geq 0} A^{(c)}$

(3) $\displaystyle A^{(c)} \ \mbox{is closed}$

(4) $\displaystyle \partial A^{(c)} \subseteq \{ x \in X | d(x,A)=\epsilon \}$

(5) Does there exist an example wherefore $\displaystyle \partial A^{(c)} \subset \{ x \in X | d(x,A)=\epsilon \}$

Proofs:

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

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

In #3 I would show the complement is open.

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

Re: Metric spaces and closed sets

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

Proof:

Suppose $\displaystyle x \in \bigcap_{\epsilon\geq 0} A^{\epsilon} = A^{\epsilon_1} \cap A^{\epsilon_2} \cap \ldots $

thus $\displaystyle x \in A^{\epsilon_1} \wedge x \in A^{\epsilon_2} \wedge \ldots$ which means

$\displaystyle d(x,A) \leq \epsilon_1 \wedge d(x,A) \leq \epsilon_2 \wegde \ldots$ and therefore

$\displaystyle d(x,A) \leq \max\{\epsilon_1,\epsilon_2,\ldots\}$, write $\displaystyle d(x,A)\leq e'$

But $\displaystyle d(x,A)=\inf\{d(x,a)|a \in A \} \leq e'$, therefore we can choose an element $\displaystyle a \in A$ with $\displaystyle d(x,a) \leq \epsilon' \wedge a \in A$

and so $\displaystyle \forall \epsilon' \geq 0: B(x,\epsilon') \cap A \neq \emptyset \Rightarrow x \in \overline{A}$

What do you think about this proof?

Re: Metric spaces and closed sets

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

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

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

Re: Metric spaces and closed sets

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

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

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

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

Re: Metric spaces and closed sets

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

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