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!

  • $\begingroup$ 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

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.


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