The Munkres book states the following definition:

Recall that if $A$ is a subset of a space $X$ the interior of $A$ is defined as the union of all open sets of $X$ that are contained in $A$. To say that $A$ has empty interior is to say then that $A$ contains no open set of $X$ other than the empty set. Equivalently, $A$ has empty interior if every point of $A$ is a limit point of the complement of $A$, that is, if the complement of $A$ is dense in $X$.

In such definition I don't understand the double implication

$A$ has empty interior $\Leftrightarrow$ every point of $A$ is a limit point of the complement of $A$ $\Leftrightarrow$ the complement of $A$ is dense in $X$.

I tried to formally prove the equivalence using the definition of dense set, interior etc but I just got confused.

Could you help me to understand the equivalence reported in the definition?


2 Answers 2


Suppose $A$ has empty-interior. Then if $x \in A$, and $O$ is open and contains $x$, $O$ is non-empty (as it contains $x$!) so cannot be a subset of $A$ (as the interior is empty, see the remarks). So there is some point $x' \in X \setminus A$ that is in $O$. This $x'$ witnesses that $O$ intersects $X \setminus A$ in a point unequal to $x$, and as $O$ was arbitrary containing $x$, $x$ is a limit point of $X \setminus A$ (i.e. $x \in (X \setminus A)'$). So $A \subseteq (X \setminus A)'$ in this case.

On the other hand, if all $x \in A$ are a limit point of $X \setminus A$ (which Munkres denotes by $X - A$, I believe), then $A$ has a empty interior. Otherwise there would be some non-empty open set $O \subseteq A$ and any $x \in O$ is a member of $A$ which is not a limit point of $X \setminus A$, as witnessed by $O$ (as $O$ contains $x$ but no point of $X \setminus A$!). So these are equivalent.

And $X \setminus A$ is dense iff $\overline{X \setminus A} = X$ and $\overline{X \setminus A} = (X \setminus A) \cup (X \setminus A)'$. So the set is dense iff $A \subseteq (X \setminus A)'$ by simple set theory.

  • $\begingroup$ Could you elaborate more the last sentence? (I know it is probably trivial) $\endgroup$ Commented Jan 8, 2016 at 9:38
  • $\begingroup$ I mean i'm with you when you say "$cl(X - A) = (X - A) \cup (X - A)'$. I got lost with the second part. $\endgroup$ Commented Jan 8, 2016 at 10:05
  • $\begingroup$ @lukkio Does not Munkres define the closure of $B$ to be $B$ together with its limit points? In other words $B \cup B' = \overline{B}$? Or prove this, at any rate? $\endgroup$ Commented Jan 8, 2016 at 10:37
  • $\begingroup$ It does prove that, however i don't entirely get the proof you just gave to me. Why $\overline{X \setminus A} = X \Leftrightarrow A \subseteq (X \setminus A)'$? $\endgroup$ Commented Jan 8, 2016 at 10:42
  • $\begingroup$ I'm probably stupid... give me a sec $\endgroup$ Commented Jan 8, 2016 at 10:48

The closure of $X-A$ is equal to $\cap F$, where $F$ is family of all closed sets that have $X-A$ as a subset. The set of complements of members of $F$ is the family $G$ of all open subsets of $A.$ Hence $$Int (A)=\cup G=X-\cap F=X- \overline {X-A}.$$ $$\text {Therefore }\quad Int (A)=\phi \iff \overline {X-A}=X.$$

  • 1
    $\begingroup$ This is nice! Enlightened! $\endgroup$
    – user368131
    Commented Dec 8, 2018 at 4:02
  • $\begingroup$ This would imply that $X \setminus A$ is dense in $\underline X$. $\endgroup$
    – Aelx
    Commented Mar 10, 2021 at 11:20
  • $\begingroup$ @СССР. The usual definition of "$X-A$ is dense in the space $X$" is just $\overline {X-A}=X.$ $\endgroup$ Commented Mar 11, 2021 at 22:10

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