# Boundary of a Subset is Closed in a Space

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!

• It is. Look at the defn of boundary as an intersection. – Randall Aug 30 '17 at 17:47
• 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. – fleablood Aug 30 '17 at 17:48
• The hypothesis $x \in H$ is not a part of the definition of a limit point. It is $x \in X$. – Gribouillis Aug 30 '17 at 17:50

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.

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.