1
$\begingroup$

I would like a proof verification, alternative proofs and in general some backround to this problem or related problems:

A closed subset of $\mathbb R^2$ with the euclidean topology is the frontier of a subset of $\mathbb R^2$.


I came up with the following proof: Suppose we want to prove the closed set $C$ is the frontier of a set. Consider the set $A=C\setminus C^\circ \cup S$ where $S=\{(x,y)\in C^\circ | x+y\in \mathbb Q\}$. It is clear the interior of $A$ is empty. Suppose the interior is not empty, then pick a point $x$ in the interior of $A$. There is an open set around $x$ in $A$. This open set must necessarly contain a point of $C^\circ$, so the intersection of the open set with $C^\circ$ is an open set of $S$, but $S$ contains no open sets since in general any non-empty open set of $\mathbb R^2$ has a point with rational sum of coordinates. Therefore $A^\circ=\emptyset$

It is also clear that $\overline A=C$ since every point $x$ that is in $C$ but not in $A$ satisfies that every neighbourhood of $x$ intersects $A$. And so a closed set containing $A$ must contain every such element.

Therefore $Fr(A)=\overline A\setminus A^\circ=C\setminus \emptyset =C$ as desired.

$\endgroup$
1
$\begingroup$

I think the ideas of your proof are better summarized by this way of putting it:

Let $C$ be a closed subset of $\mathbb{R}^2$. If $C$ has empty interior then we are done, because $C$ is the boundary of itself. Otherwise, construct a dense subset of $C$ with empty interior, and call it $D$. Given such a $D$, the closure of $D$ is $C$, and the interior of $D$ is empty, so the boundary of $D$ is $C \setminus \emptyset = C$.

I think this better focuses on the key step, which is the construction of $D$. It also uses some routine topology to smooth out the rest of the proof.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.