3
$\begingroup$

I have the following problem that I am stuck on:

Let $T=(X,\mathcal{T})$ be a topological space. Given $H\subset T$, prove that the boundary $\partial(H)$ is closed in $T$.

Here are the definitions that my class is using:

Limit Point: $x\in H\subset T$ is a limit point of $H$ if and only if for any $U\in\mathcal{T}$ with $x\in U$, $(U\setminus\{x\})\cap H\neq \varnothing$.

Closure: The closure of $H$ is the set $\overline{H}=H\cup H'$, where $H'$ is the set of limit points of $H$.

Boundary: The boundary of $H$ is the set $\partial(H)=\overline{H}\cap\overline{T\setminus H}$.

I feel like this should be a really easy problem, but I can't seem to figure it out. Thanks in advance for any help or suggestions!

$\endgroup$
  • $\begingroup$ It is. Look at the defn of boundary as an intersection. $\endgroup$ – Randall Aug 30 '17 at 17:47
  • $\begingroup$ The boundary consists solely of the limit points of H and the limit points of the compliment of H. Prove the any limit point of the boundary is a point of the boundary. Then the boundary contains all its limit points and is thus closed. $\endgroup$ – fleablood Aug 30 '17 at 17:48
  • $\begingroup$ The hypothesis $x \in H$ is not a part of the definition of a limit point. It is $x \in X$. $\endgroup$ – Gribouillis Aug 30 '17 at 17:50
3
$\begingroup$

The fact you wish to use here is:

For any set $A \subseteq X$, its closure $\overline{A}$ is a closed set.

$\partial H=\overline{H}\cap\overline{X\setminus H}$ is an intersection of two closed sets, hence it is closed.

Additional details:

To prove that the closure of any set is always closed it is useful to prove this lemma first:

$x\in\overline{A} \iff$ for all $U \in \mathcal{T}$ such that $x \in U$ is $A\cap U \ne\emptyset$

Proof of lemma:

Assume $x \in \overline{A} = A \cup A'$. If $x \in A$ then for every $U \in \mathcal{T}$ such that $x \in U$ we have $\{x\} \subseteq A \cap U$, hence $A \cap U \ne \emptyset$. If $x \in A'\setminus A$ then for every $U \in \mathcal{T}$ such that $x \in U$ we have $A \cap (U\setminus\{x\})\ne \emptyset$, by the definition of limit point. Then we also have $A \cap U \ne \emptyset$.

For the reverse implication, assume that $\forall U \in \mathcal{T}, x \in U$ is $A \cap U \ne \emptyset$. If $x \in A$, we are done. If $x \notin A$, for any $U \in \mathcal{T}, x \in U$ we additionaly have $A \cap (U\setminus\{x\})\ne\emptyset$. $\tag*{$\blacksquare$}$

Now let's prove that $\overline{A}$ is closed by proving that $X\setminus \overline{A}$ is open:

Let $x \notin \overline{A}$. By the lemma above, there exists $U \in \mathcal{T}$ such that $x \in U$ and $A \cap U = \emptyset$. Thus, to prove $U \cap \overline{A}=\emptyset$ it suffices to prove $U\cap A'=\emptyset$. Let $y \in U\cap A'$. Because $y$ is a limit point of $A$, and $U$ is an open set which contains $y$, we have $A\cap(U\setminus\{y\})\ne\emptyset$, which is a contradiction. This means $U\cup \overline{A} = \emptyset$, or equivalently $U \subseteq X \setminus \overline{A}$.

Thus, for arbitrary element $x\in X\setminus\overline{A}$ we have found an open neighbourhood $U_x \subseteq X\setminus\overline{A}$ of $x$. This implies that $$X\setminus\overline{A} = \bigcup\limits_{x\in X\setminus\overline{A}}U_x$$ which is an open set as a union of open sets, so $\overline{A}$ is closed. $\tag*{$\blacksquare$}$

Remark about your notation: A subset $A$ of a topological space $T=(X,\mathcal{T})$ is usually denoted by $A \subseteq X$, not by $A \subseteq T$. Similarly, for the complement is used $X\setminus A$.

$X$ is the set here, $T$ is a structure.

$\endgroup$

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.