I have the following question:
Let $\tau : = \left\{ {\left( {a,\infty } \right)|a \in \mathbb{R}} \right\} \cup \left\{ {\emptyset ,\mathbb{R}} \right\}{\text{ }}$; $\tau {\text{ is called the right-ray topology of }}\mathbb{R}.$
${\text{Find }}{I^o},\bar I{\text{ and }}\partial I{\text{ where }}I = \mathbb{R} – \mathbb{Q},$ the set of all irrational numbers.
$\begin{gathered}
Note:{\text{ Here }}{I^{\text{o}}}{\text{ is the interior of }}I;{\text{ }}\bar I{\text{ is the closure of }}I,{\text{ }} \hfill \\
{\text{and }}\partial I{\text{ is the boundary/frontier of }}I. \hfill \\
\end{gathered}$
I have the following:
Interior: $\mathbb{R} – \mathbb{Q}{\text{ = }}I \notin \tau$. Now for for all $O \in \tau$, only $\emptyset \subseteq I = \mathbb{R} – \mathbb{Q}$, meaning there are no right-ray subsets in $I$ except $\emptyset$ .
Therefore ${I^{\text{o}}} = {\left( {\mathbb{R} – \mathbb{Q}} \right)^{\text{o}}} = \emptyset$.
Closure:
$\mathbb{R} – \mathbb{Q}{\text{ = }}I \notin \tau$. Now for for all $O \in \tau$, only $\mathbb{R} \supseteq I = \mathbb{R} – \mathbb{Q}$, meaning $\mathbb{R}$ is the only closed superset containing $I$. Therefore:
$\bar I = \overline {\mathbb{R} – \mathbb{Q}} = \mathbb{R}$.
It follows that $\partial I = \bar I – {I^o} = \mathbb{R} – \emptyset = \mathbb{R}$.
Could you please verify my answers?
Best Answer
Suppose $(a, \infty) \subseteq \Bbb P$, where $\Bbb P = \Bbb R\setminus \Bbb Q$ is the set of irrationals. As each such open segment contains (lots of) rationals (e.g. large enough integers, for starters), this cannot hold. The only other candidate open sets inside $\Bbb P$ are $\Bbb R$ (no) or $\emptyset$ (yes) so $\mathbb{P}^\circ=\emptyset$.
The only closed sets of $\tau$ are $(-\infty, a]$, which can never contain the unbounded-above $\Bbb P$, $\emptyset$ and $\Bbb R$, so again by process of elimination, $\overline{\Bbb P}=\Bbb R$.
The boundary being $\Bbb R$ too then follows immediately.
So you're correct.