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Prove that the boundary set of any non-empty subset of the real numbers is a closed set, and also that a non-empty subset of the real numbers union its boundary set is a closed set.

$\textbf{Prove that the boundary set of any non-empty subset of the real numbers is a closed set.}$

$\textbf{Proof:}$ Let the boundary set be denoted as $\delta X$ and the interior set be denoted as $X'$. So $$\delta X=\{ x \in \mathbb{R} | \forall \mathscr{O} \,with x \in \mathscr{O}, \mathscr{O} \cap X \neq \phi, \mathscr{O} \cap X^c \neq \phi\}$$ The definition of the closure of X, denoted as $\bar{X}$, is $\bar{X}=\delta X \cup X'$ where $\delta X \cap X'=\phi$ and we know that $\bar{X}$ is closed.

$X' \in \mathbb{R}$ is a union of all open intervals inside X. So $\bar{X}-X'$ is a collection of points. Thus $\bar{X}-X'$ is closed. $$\bar{X}=\delta X \cup X' \Rightarrow \delta X=(\bar{X}-X') \cup (\delta X \cap X')$$ But $\delta X \cap X' = \phi$. So $\delta X$ is a collection of points. Thus, $\delta X$ is closed in $\mathbb{R}$.

$\textbf{Prove that a non-empty subset of the real numbers union its boundary set is a closed set.}$

How am I suppose to prove this? Do I just say it because of the first proof?

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  • $\begingroup$ What do you mean by $\bar{X}-X'$ is a collection of points? Are you saying they are isolated points? $\endgroup$ – Pedro Sep 30 '14 at 13:25
  • $\begingroup$ Yes I am saying they are isolated points $\endgroup$ – Username Unknown Sep 30 '14 at 14:28
  • $\begingroup$ Then the affirmation $\bar{X}-X'$ is a collection of points is false, take for example $X=\mathbb{Q}$. Look that $\bar{X}-X'$ don't need to be a collection of points to be closed. $\endgroup$ – Pedro Sep 30 '14 at 14:30
  • $\begingroup$ For the second question you are right since $X\subset\bar{X}=\delta{X}\cup X'$ and $X'\subset X$ implies $\bar{X}=\delta X\cup X'$, which is closed. $\endgroup$ – Pedro Sep 30 '14 at 14:43
  • $\begingroup$ So I can delete that line about $\bar{X}-X'$ is closed from my proof? $\endgroup$ – Username Unknown Sep 30 '14 at 14:58
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Thanks to @Pedro for the assistance. Here is the final answer.

$\textbf{Prove that the boundary set of any non-empty subset of the real numbers is a closed set.}$

$\textbf{Proof:}$ Let the boundary set be denoted as $\delta X$ and the interior set be denoted as $X'$. So $$\delta X=\{ x \in \mathbb{R} | \forall \mathscr{O} \,with x \in \mathscr{O}, \mathscr{O} \cap X \neq \phi, \mathscr{O} \cap X^c \neq \phi\}$$ The definition of the closure of X, denoted as $\bar{X}$, is $\bar{X}=\delta X \cup X'$ where $\delta X \cap X'=\phi$ and we know that $\bar{X}$ is closed.

$X' \in \mathbb{R}$ is a union of all open intervals inside X. Thus $\bar{X}-X'$ is closed. $$\bar{X}=\delta X \cup X' \Rightarrow \delta X=(\bar{X}-X') \cup (\delta X \cap X')$$ But $\delta X \cap X' = \phi$. So $\delta X$ is a collection of points. Thus, $\delta X$ is closed in $\mathbb{R}$.

$\textbf{Prove that a non-empty subset of the real numbers union its boundary set is a closed set.}$

$\textbf{Proof:}$ Since $X \subset \bar{X}=\delta X \cup X'$ and $X' \subset X$ implies $\bar{X}=\delta X \cup X'$, which is closed. Hence a non-empty subset of the real numbers union its boundary set is a closed set.

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