I'm reading Intro to Topology by Mendelson.

The problem at hand is,

Show that $\text{Bdry}(A)=\emptyset$ if and only if $A$ is closed and open.

This was all the problem statement had, but I'm in the chapter covering closure, interior and boundary with respect to topological spaces.

Here is my attempt at the proof,

Suppose that Bdry$(A)=\emptyset$. Then $\bar{A}=A$ $\cup$ Bdry$(A)=A\cup\emptyset=A$. Hence, $\bar{A}=A$, which implies that $A$ is closed. To show that $A$ is also open we will show that $\overline{C(A)}=C(A)$, that is, $C(A)$ is closed and hence $A$ open. We already know that $C(A)\subset\overline{C(A)}$, thus it suffices to only show that $\overline{C(A)}\subset C(A)$. This is the case since we know that Bdry$(A)=\bar{A}\cap\overline{C(A)}=\emptyset$, which implies that $\overline{C(A)}\subset C(\bar{A})=$ Int$(C(A))\subset C(A)$. Thus, $A$ is open. Suppose now that $A$ is both open and closed. Since $A$ is open we know that Int$(A)=A$. Also, since $A$ is closed $\bar{A}=A=$ Int$(A)$. Now we know that Bdry$(A)=\bar{A}\cup\overline{C(A)}=$ Int$(A)$ $\cup$ $C($Int$(A))=\emptyset.$

I used an the identity $\bar{A}=A\cup\text{Bdry}(A)$, which was asked later in the problem set. I'm wondering if I should use this, since I was able to prove it, or attempt the proof assuming I'm unaware of the identity. I have also been trying to write more concise proofs and this one is definitely not one of them. Would removing some of the words help or is there a cleaner approach that can be pointed out?

Thanks for any feedback!


4 Answers 4


As you said $\bar{A} = A \cup \operatorname{Bd}(A)$, but there's a more stronger statement that you can and you should prove: $\bar{A} = \operatorname{Int}(A) \cup \operatorname{Bd}(A)$.

Now it should be fairly easy to prove that $\operatorname{Bd}(A) = \emptyset \iff \text{A is closed and open}$:

(1)If $\operatorname{Bd}(A)=\emptyset$

Then $\bar{A}=\operatorname{Int}(A)$ and since $\operatorname{Int}(A)\subset A \subset \bar{A}$ we conclude that $\operatorname{Int}(A) = A = \bar{A}$.

$\operatorname{Int}(A) = A$ shows that $A$ is open and $A = \bar{A}$ shows that $A$ is closed.

(2)If $A$ is closed and open.

A is open, then $A=\operatorname{Int}(A)$ and since $\bar{A} = \operatorname{Int}(A) \cup \operatorname{Bd}(A)$ we see that $\bar{A} = A \cup \operatorname{Bd}(A)$

But $A$ is also closed!, this means that $\bar{A} = A$ therefore $A = A \cup \operatorname{Bd}(A)$. Now remember that $\operatorname{Int}(A) \cap \operatorname{Bd}(A) = \emptyset$, then $A \cap \operatorname{Bd}(A) = \emptyset$.

Because of $A = A \cup \operatorname{Bd}(A)$ and $A \cap \operatorname{Bd}(A) = \emptyset$ we conclude $\operatorname{Bd}(A) = \emptyset$

  • $\begingroup$ Thanks for the response, I definitely will try and prove the identity you gave. I liked how it was used in your proof and even yielded the identity I used. $\endgroup$
    – S.D.
    Jul 21, 2013 at 2:26
  • $\begingroup$ I tried to show the identity you gave and was able to show that $\text{Int}(A)\cup\text{Bdry}(A)\subset\bar{A}$. I'm having trouble showing the other direction. I know that if $a\in\bar{A}$ then every neighborhood of $N$ of $a$, $N\cap A\neq\emptyset$. Would you have any hints on how I can proceed from here? $\endgroup$
    – S.D.
    Jul 21, 2013 at 5:36
  • 1
    $\begingroup$ If you already proved the first inclusion and you're working with intro to topology by Mendelson there's a easier way than prove the other direction. Mendelson's Theorem 4.13 shows that $\bar{A}$ is the smallest closed subset containing $A$, then the proof could be finished as follows: (1) Prove that $A \subset \operatorname{Int}(A)\cup\operatorname{Bdry}(A)$. (2) Prove that $\operatorname{Int}(A)\cup\operatorname{Bdry}(A)$ is closed. (3) Because of (1) and what you proved $A \subset \operatorname{Int}(A)\cup\operatorname{Bdry}(A) \subset \bar{A}$, the identity follows from (2) and Th. 4.13. $\endgroup$
    – Cure
    Jul 21, 2013 at 20:02

$\emptyset=\partial A=\overline{A}\setminus \mathring{A}$ (the first is the hypothesis, the second a well-known property) together with $\mathring{A} \subset A\subset\overline{A} $ gives you $\mathring{A}=A=\overline{A}$ so that A is clopen (and from the first inequalities the other implication is obvious).

  • $\begingroup$ Thank you for the response. I never thought about using that property and it was even shown in the text. This definitely is a more consice approach than mine, appreciate the insight. $\endgroup$
    – S.D.
    Jul 21, 2013 at 2:29
  • $\begingroup$ I was able to show the well-known property you mentioned above, but that was only when we had that $\text{Bdry}(A)=\emptyset$. Does $\text{Bdry}(A)=\bar{A}-\text{Int}(A)$ in general? $\endgroup$
    – S.D.
    Jul 21, 2013 at 5:55
  • $\begingroup$ Yes, it's a general property: a point is in the boundary of A iff each of its neighborhood intersects both A and X\A; thus iff it's in the closure of A but not in its interior. $\endgroup$ Jul 21, 2013 at 8:30
  • $\begingroup$ I see, appreciate the help. $\endgroup$
    – S.D.
    Jul 21, 2013 at 11:07

We know $\operatorname{Bdry}(A)=\operatorname{cl}(A)\cap \operatorname{cl}(X-A)$. So $\operatorname{Bdry}(A)=\operatorname{cl}(A)\cap (X-\operatorname{int}(A))$. Hence $\operatorname{Bdry}(A)=\operatorname{cl}(A)-\operatorname{int}(A)$. Now use this identity.


(1) If $Bdry(A) = \emptyset$.

Then, $$\bar A = A \cup Bdry(A) = A \cup \emptyset = A$$ Hence $\bar A = A$, which implies A is closed.

Since the $Bdry(A) = Bdry(A^c)$ this implies $$\overline {A^c} = A^c \cup Bdry(\overline {A^c}) = A^c \cup Bdry(A) = A ^c \cup \emptyset.$$ Hence $\overline {A^c} = A^c$, which implies $A^c$ is closed, which implies A is open.

(2) If $A$ is closed

Then $A = \bar A$. So $$Bdry(A) = \bar A \cap \overline {A^c} = A \cap \overline {A^c} = A \cap (Int(A))^c$$

Since $A$ is open, $A$ is a neighborhood of all its points so $A = Int(A)$,

which implies $A \cap C(Int(A)) = A \cap C(A) = \emptyset$

  • 1
    $\begingroup$ I know this post is 5 years old but I just wanted to put up my take on the question and I feel it requires less moving parts than previous answers. $\endgroup$
    – Ross
    Mar 6, 2019 at 4:07
  • 3
    $\begingroup$ Please use MathJax to format the mathematics in your post. $\endgroup$ Mar 6, 2019 at 4:33
  • $\begingroup$ There is noting wrong in answering old questions: however, as @AlexProvost writes, it is better to use Math Jax to type mathematical text. Here you'll find a tutorial. $\endgroup$ Mar 6, 2019 at 9:12

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