Let $A\subset X$. Show that if $C$ is a connected subspace of $X$ that intersects both $A$ and $X-A$ , then $C$ intersects $\operatorname{Bd}A$

Question

Problem Let $A\subset X$. Show that if $C$ is a connected subspace of $X$ that intersects both $A$ and $X-A$ , then $C$ intersects $\operatorname{Bd}A$.

Attempt- Suppose $C\cap \operatorname{Bd}A=\varnothing$ then $$C\cap (\overline{A}\cap \overline{X-A})=\varnothing$$

therefore, $$(C\cap \overline{A})\cap (C\cap \overline{X-A})=\varnothing$$

Edit: If $C\cap A\neq \varnothing$ and $C\cap (X-A)\neq\varnothing$ but $C\cap \operatorname{Bd}A=\varnothing$. Then $C\cap (A-\operatorname{Bd}{A})=C\cap A^0\neq \varnothing$ and $C\cap ((X-A)-\operatorname{Bd}(X-A))=C\cap (X-A) ^0\neq \varnothing$

This will form a separation for $C$ if we show that $C\cap \overline{A}$ and $C\cap \overline{X-A}$ are open.Clearly, both sets are non-empty but How will it be open? Am I on right track? Thanks !

Answer 1

They are not closed subsets of $C$ (in the subspace topology). It follows that they are also open in this topology because they are each other's complements.

Alternatively, you can argue as follows: The complement of the boundary of $A$ is the union of $A^{0}$ (the interior of $A$) and $(A^{c})^{0}$. Since $C \subseteq A^{0} \cup (A^{c})^{0}$ we get a contradiction to connectedness of $C$. ( $A^{0}$ and $(A^{c})^{0}$ are disjoint open sets).