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I was trying to go through one of my undergrad calculus textbooks as I am planning to apply for masters in the next two years. I was trying to prove that ∂(∂)=∂ iff E is open. I was having a hard time proving this with just the definition of an open set and the boundary definition. I understand that it is easy to prove if I show that the interior is empty but I also wanted to prove it with just the original definitions.

The boundary point (x) of a set A is a point such that a ball centered at a point x the points in this ball belong to both A and its complement.

mathfanatic
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  • The definition of boundary point is more clearly expressed here: https://math.stackexchange.com/questions/78873 – 311411 Oct 04 '22 at 02:56
  • Did you try to prove $\partial E \subset \partial\ (\partial E)$ under the assumption $E$ open? That is a good exercise in the definitions of "element of an open set" and "boundary point of a set". You should show some kind of attempt above. – 311411 Oct 04 '22 at 03:13
  • @311411 Sorry I should have stated what I tried. I started by writing down the definition of ∂ and ∂(∂). After that, for the definition of ∂(∂), I know that the points in a ball are in ∂ and (∂)^c and I also know ∂ = ∂^c but I wasn't sure if I could just take the complement inside. Thanks! – mathfanatic Oct 06 '22 at 04:00
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    Wait, how is this true? Doesn't $[0,1]\subseteq\Bbb R$ satisfy this? – Akiva Weinberger Oct 06 '22 at 07:14
  • mathfanatic, do you not want to update your question, in consideration of Akiva Weinberger's comment? (Also you should include your attempt in the body of your question, not place it in a comment.) – 311411 Oct 08 '22 at 17:15

4 Answers4

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A correct statement is: $\partial E = \partial(\partial E) \iff \partial E$ has no inner points, the non-trivial part being $\partial E \subseteq \partial(\partial E)$. This is clear from the original definition, and is true if $E$ is open, but also if $E$ is, for example, closed, so open subsets cannot be characterised that way.

Lisa
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If the original problem seems confusing, substituting the $\partial E$ which repeatedly appeared in the problem with any arbitrary set with some certain property, say $A$, thus we need to prove $\partial A=A$, and according to the definition that $\partial A = A\backslash \mathring{A}$, we need to show that $\mathring{A}=\emptyset$, which can be shown by the fact that there is no inner point of $A$ (which is a border of some open set $E$).

The fact mentioned above can be rigorously proven as follows (and substituting back):

Suppose the opposite: there is an $x\in\partial E$ and a neighbourhood $U$ satisfying $U\subseteq \partial E$. But we also know that $U\cap E$ is not empty, but because $E$ is open, $\partial E\cap E$ is empty, which leads to contradiction.

Note: this is also true for a closed set $E$ by substituting the $E$ with $E^{C}$ in the proof.

erpxyr2001
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Unless I am mistaken somewhere, this iff statement is not true.

$$\partial E=Cl(E)\setminus Int(E)$$

Consider the plane $\mathbb{R}^2$ with the standard topology and let $E=S^1$. Since $S^1$ is closed and has no interior then $$Cl(S^1)=S^1 \text{ and } Int(S^1)=\emptyset$$ $$\Rightarrow \partial S^1=S^1$$ $$\Rightarrow \partial (\partial S^1)=\partial S^1=S^1$$ and $S^1$ is not open.

Sam
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We want to prove $\partial\partial E=\partial E$ when $E$ is open set. Since, $\partial\partial E=\overline{\partial E}-\partial E^{\circ}$, we must prove two things.

  1. Boundry is always closed, i.e. $\overline{\partial E}=\partial E$. Proof: By definition, ${\partial E}=\overline{E}\cap\overline{E'}$, so $\partial{E}$ is closed. (Here $A'$, $\overline{A}$, $A^{\circ}$ denote the complement, the closure, the interior of $A$ respectively.)
  2. $\partial E^{\circ}=\emptyset$ when $E$ is open set. Proof: We have $\partial E=\overline{E}\cap\overline{E'}=\overline{E}\cap (E^{\circ})'=\overline{E}\cap E'$ since $E^{\circ}=E$ by openness of $E$. So, we have $(\partial E)^{\circ}=(\overline{E})^{\circ}\cap (E')^{\circ}=(\overline{E})^{\circ}\cap (\overline{E})'=\emptyset$ because the closure contains its interiror so that $(\overline{E})^{\circ}\cap (\overline{E})'\subset \overline{E}\cap (\overline{E})'=\emptyset$.

In a similar manner, we can prove that $\partial\partial\partial E=\partial\partial E$ for any $E$.

Bob Dobbs
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