# Boundary Points and Metric space

Definition:The boundary of a subset of a metric space X is defined to be the set $$\partial{E}$$ $$=$$ $$\bar{E} \cap \overline{X\setminus E}$$

Definition: A subset E of X is closed if it is equal to its closure, $$\bar{E}$$.

Theorem: Let C be a subset of a metric space X. C is closed iff $$C^c$$ is open.

Definition: A subset of a metric space X is open if for each point in the space there exists a ball contained within the space

Show that if $$E \cap \partial{E}$$ $$=$$ $$\emptyset$$ then $$E$$ is open.

Proof:

$$E\cap \partial{E}$$ being empty means that $$E\subseteq (\bar{E}^c \cup \overline{X\setminus E}^c)$$. Since $$E \subseteq \bar{E}$$ it follows that $$E \subseteq \overline{X\setminus E}^c$$ which implies that $$E \cap \overline{X\setminus E}$$ is empty. Since every subset is a subset of its closure, it follows that $$X\setminus E$$ $$=$$ $$\overline{X\setminus E}$$ and so $$X\setminus E$$ is closed, and therefore $$E$$ is open.

Is the proof correct? I would really love feedback.

• Yes it is correct. After saying that $E \cap \overset{-} {(X\setminus E)}$ is empty you can add: $\overset{-} {(X\setminus E)} \subset X\setminus E$ for clarity. – Kavi Rama Murthy Jun 4 '19 at 23:58
• You are confusing subspace and subset. – William Elliot Jun 5 '19 at 0:58
• @WilliamElliot Every subset of a metric space is also a metric space wrt the same metric. May I know where I confused the term? A subspace is a subset, by definition and every subset of a metric space is a subspace (a metric space in its own right). – monoidaltransform Jun 5 '19 at 1:04
• The boundary of any subspace is empty. The boundary of the subset is what you claimed to be the boundary of the subspace, – William Elliot Jun 5 '19 at 1:08
• @WilliamElliot What do you mean the boundary of any subspace is empty? Clearly not, (0,1) is a subset\subspace of the reals and 1 is an element of the boundary. – monoidaltransform Jun 5 '19 at 1:13

After William Elliot's feedback on your proof and this comment of yours, I don't think there is much that needs to be clarified. Still if you have anything specific regarding your proof to ask me, I welcome you to come here.

In any case, let me try to write a proof that I believe is in line with your attempt.

\begin{align*}E\cap \partial{E}=\emptyset&\implies E\cap(\overline{E}\cap \overline{X\setminus E})=\emptyset\\&\implies (E\cap\overline{E})\cap \overline{X\setminus E}=\emptyset\\&\implies E\cap \overline{X\setminus E}=\emptyset\\&\implies \overline{X\setminus E}\subseteq X\setminus E\\&\implies \overline{X\setminus E}=X\setminus E\end{align*}This shows that $$X\setminus E$$ is closed and hence $$E$$ is open.

In any topological space $$X$$ and any $$E\subset X,$$ the 3 sets $$int(E),\, int(X\setminus E),\, \partial E)$$ are pair-wise disjoint and their union is $$X.$$

So if $$E\cap \partial E=\emptyset$$ then $$E=E\cap X=E\cap (\,int (E) \cup int (X\setminus E)\cup \partial E\,)=$$ $$=(E\cap int E)\,\cup\, (E\cap int (X\setminus E))\,\cup\, (E\cap \partial E)\subset$$ $$\subset (E\cap int(E)\,\cup \,( E\cap (X\setminus E)\,\cup\, (E\cap \partial E)=$$ $$=int (E)\,\cup \, ( \emptyset)\,\cup \,(\emptyset)=$$ $$=int (E)\subset E$$ so $$E=int(E).$$

OR, from the first sentence above, for any $$E\subset X$$ we have $$int(E)\subset E\subset \overline E=int(E)\cup \partial E.$$

So if $$E\cap \partial E=\emptyset$$ then $$E=E\cap \overline E=E\cap (int (E) \cup \partial E)=$$ $$=(E\cap int (E))\,\cup \,(E\cap \partial E)=$$ $$=(E\cap int (E))\cup(\emptyset)=$$ $$=int(E)\subset E$$ so $$E=int(E).$$