Let $A\subset X$. Show that if $C$ is a connected subspace of $X$ that intersects both $A$ and $X-A$ , then $C$ intersects $\operatorname{Bd}A$.
My attempt:
Approach(1): Assume towards contradiction, $C\cap \partial A=\emptyset$. By definition of interior, $ A^{\circ}, (X-A)^\circ \in \mathcal{T}_X$. So $C\cap A^\circ, C\cap (X-A)^\circ \in \mathcal{T}_C$. Since $\emptyset \neq A\cap C\subseteq \overline{A}\cap C$, we have $\overline{A}\cap C \neq \emptyset$. By Exercise 19, Section 17 of Munkres’ Topology, $\overline{A}\cap C=(A^\circ \cup \partial A) \cap C=(C\cap A^\circ)\cup (C\cap \partial A)=(C \cap A^\circ)\cup \emptyset=C \cap A^\circ$. So $C \cap A^\circ \neq \emptyset$. Since $\emptyset \neq (X-A)\cap C\subseteq \overline{X-A}\cap C$, we have $\overline{X-A}\cap C \neq \emptyset$. By exercise 19 section 17, $C\cap \overline{X-A}=C\cap ((X-A)^\circ \cup \partial(X-A))= C\cap ((X-A)^\circ \cup \partial A)= C\cap (X-A)^\circ$. So $C\cap (X-A)^\circ \neq \emptyset$. By elementary set theory, $(C\cap A^\circ) \cap (C\cap (X-A)^\circ)= C\cap (A^\circ \cap (X-A)^\circ)=C\cap \emptyset =\emptyset$. By distributive law, $(C\cap A^\circ) \cup (C\cap (X-A)^\circ)= C \cap (A^\circ \cup (X-A)^\circ)$. Since $X =\operatorname{int}(A) \sqcup \operatorname{int}(X\setminus A) \sqcup \partial (A)$, we have $C=C\cap X= C\cap (A^\circ \cup (X-A)^\circ \cup \partial A)= [C\cap (A^\circ \cup (X-A)^\circ)]\cup (C\cap \partial A)= C\cap (A^\circ \cup (X-A)^\circ)$. So $(C\cap A^\circ) \cup (C\cap (X-A)^\circ)= C \cap (A^\circ \cup (X-A)^\circ)=C$. Hence $C\cap A^\circ$ and $C\cap (X-A)^\circ$ form separation of $C$. Which contradicts our initial assumption of $C$ is connected. Is this proof correct?
Approach(2): Assume towards contradiction, $C\cap \partial A=\emptyset$. By definition of closure, $\overline{A}$ and $\overline{X-A}$ are closed in $X$. By theorem 17.2, $C\cap \overline{A}$ and $C\cap \overline{X-A}$ are closed in $C$. Since $C\cap A, C\cap (X-A)\neq \emptyset$, we have $C\cap \overline{A}, C\cap \overline{X-A}\neq \emptyset$. By definition of boundary,$\emptyset =C\cap \partial A = (C\cap \overline{A})\cap (C\cap \overline{X-A})$. By Exercise 19, Section 17 of Munkres’ Topology, $(C\cap \overline{A})\cup (C\cap \overline{X-A})$$=C\cap (\overline{A}\cup \overline{X-A})$$= C\cap (A^\circ \cup \partial A \cup X-A^\circ)$$=C \cap (X\cup \partial A)$$= C\cup \emptyset=C$. Hence $C\cap \overline{A}$ and $C\cap \overline{X-A}$ form separation of $C$. Which contradicts our initial assumption of $C$ is connected. Is this proof correct?
Best Answer
(2) looks fine. It might be a bit more efficient to use $A^\circ$ and $(A^c)^\circ$ to partition $C$, where $A^c$ means $X\setminus A$. These are clearly disjoint open sets.
In detail: it's not difficult to see from definition that $X=A^\circ\sqcup(A^c)^\circ\sqcup\partial A$ for any $A\subseteq X$ ($\sqcup$ means disjoint union), and also $\partial A^c=\partial A$, $A\subseteq A^\circ\sqcup\partial A$, $A^c\subseteq (A^c)^\circ\sqcup\partial A$. Suppose $C$ does not intersect $\partial A$ and yet intersect $A$, then it has to intersect $A^\circ$. Similarly for $A^c$. Then $A^\circ$ and $(A^c)^\circ$ partition $C$.