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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.

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  • $\begingroup$ There is one snag here: you have to know whether these are the interior/boundary/closure of $S$ in itself or in $\mathbb{R}$. They are not the same. Most likely you want to take the operations in $\mathbb{R}$, which will mean that the closure and boundary may contain points which aren't actually in $S$ (and will, in this case). $\endgroup$ – Ian Sep 26 at 2:12
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Let $S=[0, 1]\cup r$ such that $r\in [1, 2]\cap \mathbb{Q}^{c}$, then

  1. $S^{\circ}=(0, 1)$

  2. $∂S= \{0\}\cup[1, 2]$ 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

  3. $\bar{S}=S^{\circ}\cup ∂S$.

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