[Math] Topology exercises – closure, frontier

general-topology

Are my proofs correct? The topology is $\mathbb{R^n}$

Exercises.

  1. Prove that a set $A$ is closed iff $Fr(A)\subseteq A$
  2. A set $A$ is closed iff $A=Cl(A)$
  3. For any set, $A$, $Fr(A)$ is closed
  4. For any set $A\subseteq \mathbb{R^n}$, $Fr(A)=Fr(\mathbb{R^n}-A)$
  5. For any set $A\subseteq \mathbb{R^n}$, $Fr(A)=Cl(A)\cap Cl(\mathbb{R^n}-A)$
  6. A set $A$ is open iff $A=Int(A)$

Proofs. The closure of a set $A$ is defined by $Cl(A)=A\cup Fr(A)$

  1. Suppose $Fr(A)\subseteq \mathbb{R^n}-A$. As $A$ is closed, $\forall x\in \mathbb{R^n}-A, \exists N_x$ such that $N_x\cap A=\emptyset$. This is true for every $x\in Fr(A)$ (by assumption). But $x\in Fr(A) \implies$ every neighborhood of $x$ intersect $A$ non-trivially; a contradiction. So $Fr(A)\subseteq A$. Conversely, suppose $Fr(A)\subseteq A$ and let $x\notin A$, then $\exists N$ neighborhood of $x$ such that $N\cap A=\emptyset$ thus $A$ is closed.
  2. Suppose $A$ is closed, then $Fr(A)\subseteq A \implies Fr(A)\cup A \subseteq A \implies Cl(A)\subseteq A$. As $A\subseteq Cl(A)$ for any $A$, it follows that $A=Cl(A)$. Conversely, suppose $A=Cl(A)=A\cup Fr(A)$, then $Fr(A)\subseteq A$ thus $A$ is closed (by the previous result).
  3. As $Fr(Fr(A))\subseteq Fr(A)$, it follows (by 1.) that $Fr(A)$ is closed.
  4. Let $x\in Fr(A)$, then $\exists$ neighborhood $N$ such that $N\cap A\neq \emptyset$ and $N\cap (\mathbb{R^n}-A)\neq \emptyset$. As $A=\mathbb{R^n}-(\mathbb{R^n}-A)$, it follows that $N\cap (\mathbb{R^n}-(\mathbb{R^n}-A))\neq \emptyset$ and $N\cap (\mathbb{R^n}-)\neq \emptyset$ thus $x\in Fr(\mathbb{R^n}-A)$. The converse can be proved similarily.
  5. $Cl(A)\cap Cl(\mathbb{R^n}-A) = (A\cup Fr(A))\cap ((\mathbb{R^n}-A)\cup Fr(A))=Fr(A)\cup (A\cap (\mathbb{R^n}-A))=Fr(A)$
  6. Suppose $A=Int(A)$. As $Int(A)$ is a union of open sets it's open thus $A$ is open. Conversely, suppose that $A$ is open, then there is an open neighborhood of every point $x\in A$ such that $x\in N\subseteq A$. As $x\in N$, $x$ is in any union of open sets containing $N$ so $x\in Int(A)$. As $Int(A)\subseteq A$ for any $A$, it follows that $A=Int(A)$

Best Answer

$\newcommand{\fr}{\operatorname{Fr}}\newcommand{\cl}{\operatorname{cl}}$You get off on the wrong foot in the proof of (1): the negation of $\fr(A)\subseteq A$ is $\fr(A)\nsubseteq A$, not the much stronger statement $\fr(A)\subseteq\Bbb R^n\setminus A$. For that direction of the equivalence you can argue as follows:

Suppose that $A$ is closed and $x\notin A$; then $x$ has a nbhd $N$ such that $N\cap A=\varnothing$. This clearly implies that $x\notin\fr(A)$, and it follows at once that $\fr(A)\subseteq A$.

For (3) I take it that you’ve already shown elsewhere that $\fr(\fr(A))\subseteq\fr(A)$ for any set $A$.

In (4) you have the wrong quantifier: if $x\in\fr(A)$, then for every nbhd $N$ of $x$ it’s true that $N\cap A\ne\varnothing\ne N\cap(\Bbb R^n\setminus A)$, and you need this fact for every nbhd of $x$, not for just one, in order for your argument actually to yield the conclusion that $x\in\fr(\Bbb R^n\setminus A)$.

The rest looks fine, however.

Related Question