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.


$E\cap \partial{E}$ 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.

  • 1
    $\begingroup$ 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. $\endgroup$ – Kavi Rama Murthy Jun 4 '19 at 23:58
  • $\begingroup$ You are confusing subspace and subset. $\endgroup$ – William Elliot Jun 5 '19 at 0:58
  • $\begingroup$ @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). $\endgroup$ – monoidaltransform Jun 5 '19 at 1:04
  • $\begingroup$ The boundary of any subspace is empty. The boundary of the subset is what you claimed to be the boundary of the subspace, $\endgroup$ – William Elliot Jun 5 '19 at 1:08
  • $\begingroup$ @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. $\endgroup$ – 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 $E=int(E).$


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