Let $C \subset X$ be a connected space and $A \subset X$ s.t. $C\cap A \neq \emptyset$ and $C\cap (X-A) \neq \emptyset$. Prove that $C\cap \partial A \neq \emptyset$.  I don't know how to prove this since I fail to see how I can use the fact that C is connected. Any hint would be appreciated.
Let $X$ be topological space and $A\subseteq X$.   If $E\subseteq X$ is connected, $E\cap A\neq\emptyset$ and $E\cap(X\setminus A)\neq\emptyset$ then $E\cap\partial A\neq\emptyset$.  Assume that $E\cap\partial A=\emptyset$.  If $E\cap\partial A=\varnothing$ then $E=(E\cap\text{int}(A))\cup(E\cap\text{ext}(A))$, i.e. a union of disjoint sets open in $E$.   Also the sets are nonempty because: $E\cap\text{int}(A)=E\cap A\neq\varnothing$ and $E\cap\text{ext}(A)=E\cap A^c\neq\varnothing$.  This contradicts that $E$ is connected.  Fixed.
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