Characterization of density through the interior

Question

I'm trying to prove an elemental exercise of general topology but I don't know how: Let $C = B \setminus A$, then $A$ is dense in $B$ if and only if $\operatorname{int}(C) = \varnothing$. Any hints or help are welcomed.

Answer 1

I use the following definition: "$A$ is dense in $B$ if $B = \text{closure}(A)$."

Suppose $\text{int}(C) \ne \varnothing$. Then there exists a point $x$ and an open ball $B(x, r)$ such that $x \in B(x, r) \subseteq C$. Consequently, $x$ cannot be a point in $\text{closure}(A)$ because any sequence of points that approaches $x$ would need to eventually belong to $B(x, r)$ which is outside $A$.

Suppose $\text{int}(C) = \varnothing$. Fix some point $x \in C$; it is not an interior poiont of $C$. For any fixed choice of $r_n > 0$, the ball $B(x, r_n)$ contains a point $x_n$ outside of $C$, i.e. $x_n \in A$. By doing this for $r_n = 1/n$, we obtain a sequence $(x_n)_n$ in $A$ satisfying $|x_n - x| < 1/n$, so $x_n \to x$. This implies $x$ is a limit point of $A$. Since $x$ was arbitrary, all points of $C=B\setminus A$ are limit points of $A$, so $B$ is the closure of $A$.

---

Alternate definition of "$A$ is dense in $B$": for any element $b \in B$ and any radius $r>0$ the intersectioon $B(b, r) \cap A$ is nonempty.

Answer 2

If $A$ is dense in $B$ then $\overline A\supseteq B$ so any nbhd of any $p\in B$ must intersect $A$ so any nbhd of any $p\in C$ must intersect $A$ so $no$ nbhd of any $p\in C$ can be a subset of $B$ \ $A =C$, that is, there are no non-empty nbhds that are subsets of $C$, that is, $int(C)=\emptyset$. Try this method for the reverse implication.

BTW it is standard usage that "$u$ intersects $v$" means $u\cap v\ne\emptyset.$And a nbhd of $p$ is a set $V$ such that there exists an open set $U$ with $p\in U\subseteq V$.