Let \(A \subset X\). Show that if \(C\) is a connected subspace of \(X\) that intersects both \(A\) and \(X - A\), then \(C\) intersects \(\operatorname{Bd} A\).
Solution
1. \(X-\operatorname{Bd}A=\operatorname{Int}A\cup\operatorname{Int}(X-A)\), a union of disjoint open sets.
A point outside \(\operatorname{Bd}A=\overline A\cap\overline{X-A}\) lies outside \(\overline{X-A}=X-\operatorname{Int}A\) or outside \(\overline A=X-\operatorname{Int}(X-A)\). Conversely \(\operatorname{Int}A\) misses \(\overline{X-A}\) and \(\operatorname{Int}(X-A)\) misses \(\overline A\). The two interiors lie in the disjoint sets \(A\) and \(X-A\).
2. Q.E.D.
Suppose \(C\cap\operatorname{Bd}A=\varnothing\). By step 1, \(C\) is the union of the disjoint relatively open sets \(C\cap\operatorname{Int}A\) and \(C\cap\operatorname{Int}(X-A)\), so one of them is empty because \(C\) is connected. If \(C\cap\operatorname{Int}A=\varnothing\), then \(C\subseteq X-A\), contradicting \(C\cap A\ne\varnothing\); if \(C\cap\operatorname{Int}(X-A)=\varnothing\), then \(C\subseteq A\), contradicting \(C\cap(X-A)\ne\varnothing\).