# About definition of interior, boundary and closure

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|.

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?

• 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). – Ian Sep 26 at 2:12
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. $$\bar{S}=S^{\circ}\cup ∂S$$.