TITLE: Empty interior, equivalent definitions from Munkres.
QUESTION [5 upvotes]: 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?

REPLY [3 votes]: 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 $$Int (A)=\cup G=X-\cap F=X- \overline {X-A}.$$  $$\text {Therefore }\quad  Int (A)=\phi \iff \overline {X-A}=X.$$