# verification of my proof proving the set of boundary points are closed and set of interior points is open

Let $$\overline{A} = \{x : x \text{ is a boundary point of A}\}$$ and $$\mathring{A} = \{x : x \text{ is a interior point of A}\}$$ we will show that $$\overline{A}$$ is closed and $$\mathring{A}$$ is open.

In order to show that $$\mathring{A}$$ is open we have to show that every element of $$\mathring{A}$$ is an interior point. Since any $$x\in\mathring{A}$$ is an interior point of $$A$$ we now that there exist an $$r>0$$ s.t. the open ball $$B(x,r)\subseteq A$$. But since $$B(x,r)$$ is open we now that for any $$x^*$$ we can find $$r^*>0$$ s.t. $$B(x^*, r^*) \subseteq B(x,r) \subseteq A$$. But this shows that every element of $$B(x,r)$$ is an interior point of A and thus the open ball $$B(x,r)$$ is also contained in $$\mathring{A}$$, i.e. $$B(x,r) \subseteq \mathring{A}$$. Thus, every element of $$\mathring{A}$$ is an interior point and we conclude that $$\mathring{A}$$ is open.

In order to show that $$\overline{A}$$ is closed we have to show that $$\overline{A}$$ contain all its boundary points. In order for this to happen it must be the case that $$\overline{A}$$ and $$A$$ share the same boundary points.

Thus, let $$x_0$$ be a boundary point of $$\overline{A}$$, we will show that $$x_0$$ is also a boundary point of $$A$$. Since $$x_0$$ is a boundary point of $$\overline{A}$$ we now that for every $$r>0$$ we can find $$x,y\in B(x_0, r)$$ s.t. $$x\in\overline{A}$$ and $$y\in \overline{A}^\mathrm{C}$$. We know that $$x$$ is a boundary point of $$A$$ and thus for every $$r^*>0$$ we can fins $$x^*,y^*\in B(x, r^*)$$ s.t. $$x^*\in{A}$$ and $$y^*\in{A}^\mathrm{C}$$. If we can show that their exist an $$r^*>0$$ s.t. $$B(x,r^*) \subseteq B(x_0, r)$$ then we now that $$x^*,y^* \in B(x_0, r)$$ where $$x^*\in{A}$$ and $$y^*\in{A}^\mathrm{C}$$ we have shown that $$x_0$$ is a boundary point of $$A$$ and thus $$\overline{A}$$ contain all its boundary points and is closed.

Set $$r^* = r - d(x, x_0)$$ it then follows that $$d(y^*,x_0)\leq d(y^*,x) + d(x, x_0)< r-d(x, x_0) + d(x,x_0) = r$$ which shows that $$B(x,r^*) \subseteq B(x_0, r)$$ and hence $$B(x_0, r)$$ contains at least one $$x^*\in A$$ and $$y^*\in A ^\mathrm{C}$$ $$\implies$$ $$x_0$$ is a boundary point of A $$\implies$$ $$x_0\in{\overline{A}}$$ $$\implies$$ $$\overline{A}$$ is closed.

It is not true that $$\overline{A}$$ and $$A$$ must share the same boundary points. For example, the boundary of the rational numbers is the the real numbers, while the boundary of the real numbers is empty.
However, your argument doesn’t use this claim and is essentially correct as written, but it needs some editing for clarity. For example, the last paragraph is a bit confusing to read since you use $$y^*$$ is multiple places with different meanings.
• @Philips Hoskins - So, I agree with your comment above. A more correct statement would be that any boundary point of $\bar{A}$ is also an boundary point of A. For your example with the set of irrational numbers this would be a correct statement. Apr 26, 2022 at 20:02
• @xxtensionxx It’s an understandable misconception. At this point, you should be able to prove that any closed set admits the following decomposition: $A=\partial A \cup \mathring{A}$. I like to use the $\partial$ symbol for set boundaries. It follows then that a closed set will equal its boundary if and only if its interior is empty. Most of the familiar geometric examples of boundary sets satisfy this condition. Apr 27, 2022 at 4:37