Find the interior, boundary and closure of $[0,1]∪{r∈[1,2]:r∉Q}$
What I did so far:
Let $S=[0,1]∪{r∈[1,2]:r∉\mathbb{Q}}$
Then i checked the definitions $\dots$
Def open ball
$B(a;r)=\{x∈R^n:|x−a|<r\}$.
Def interior
$S^{\circ}=\{x∈R^n:∃ε>0:B(x;ε)⊆S\}.$
Def boundary
$∂S=\{x∈R^n:∀ε>0:B(x;ε)∩S≠\varnothing \wedge B(x;ε)∩S^c≠\varnothing\}$.
Def closure
$\overline{S}=\{{x∈R^n:∀ε>0,B(x;ε)∩S≠\varnothing}\}$.
Looks like an open ball in $\mathbb{R^1}$ is just the set of all the points on some open line segment, or some open interval on $\mathbb{R}$, if some point in $S^{\circ}$ then we can draw a ball with positive radius that this ball is also in $S$, that can conclude: $$S^{\circ}=(0,1)$$
And I try to understand the defination of that so called boundary: If some point in $∂S$ then all the balls centred at this point with positive radius must have non-empty intersection with both $S$ and $S^c$. Therefore I guess $0$ must in $∂S$, also $r\in[1,2]:r\not\in\mathbb{Q}$ in $∂S$.
$$∂S=\{0\}\cup[1,2]\backslash\mathbb{Q}$$
Finally, if a point is in the closure, that means all the balls centred at this point with positive radius must have non-empty intersection with $S$, since $S^{\circ}$ and $∂S$ both have intersection with $S$, that should have $S^{\circ}\cup ∂S\subseteq \overline S$, and any point in $S$ is either in it's boundary or in it's interior, that implies $\overline S\subseteq S^{\circ}\cup ∂S$, therefore: $$\overline S = S^{\circ}\cup ∂S$$
Is my understanding to the definitions correct?
Thanks for your help.
Let $S=[0, 1]\cup r$ such that $r\in [1, 2]\cap \mathbb{Q}^{c}$, then
$S^{\circ}=(0, 1)$
$∂S= \{0\}\cup[1, 2]$ because if $r\in [1, 2] $ then for all $\epsilon >0$, $B(r, \epsilon)\cap S$ and $B(r, \epsilon)\cap S^{c}$ are nonempty set
$\bar{S}=S^{\circ}\cup ∂S$.