$A\cup B$ is connected when $A$ is connected in $X$ and $B$ clopen in $X-A$

Question

Let $A$ be connected subset of a connected space $X$, and $B\subset X-A$ be an open and closed set in the topology of the subspace $X-A$ of the space $X$. Prove that $A\cup B$ is connected.

I think the idea is to show limit point of $A$ is in $B$ or vice-versa.But I have no idea how to show it. I am preparing for qualifying so please do for me... Thanks for help....

Answer 1

I'm going to use the following characterization (number 3 in Wikipedia): A topological space $X$ is connected iff $\varnothing$ and $X$ are the only clopen (closed and open) subsets of $X$. Also, let's assume $A\neq\varnothing$, otherwise the question is trivial.

Let $C$ be clopen in $A\cup B$. By going to the complement (in $A\cup B$) if necessary, assume that $C\cap A\neq\varnothing$. Since $C\cap A$ is also clopen in $A$, then $C\cap A=A$, that is, $A\subseteq C$ (so, in particular, $C\neq\varnothing$).

Since $B$ is clopen in $X\setminus A$, then there exists $U_B$ open in $X$ and $F_B$ closed in $X$ such that $B=(X\setminus A)\cap U_B=(X\setminus A)\cap F_B$, that is $B=U_B\setminus A=F_B\setminus A$. Similarly, since $C$ is clopen in $A\cup B$, there exists $U_C$ open and $F_C$ closed in $X$ such that $C=(A\cup B)\cap U_C=(A\cup B)\cap F_C$.

Set $D=(A\cup B)\setminus C=B\setminus C$ (because $A\subseteq C$), the complement of $C$ in $A\cup B$. Then \begin{align*} D&=B\setminus C=(U_B\setminus A)\setminus C=U_B\setminus C=U_B\setminus(F_C\cap(A\cup B))\\ &=(U_B\setminus F_C)\cup( U_B\setminus (A\cup B))=U_B\setminus F_C \end{align*} because $U_B\setminus(A\cup B)=(U_B\setminus A)\setminus B=B\setminus B=\varnothing$, so $D=U_B\setminus F_C$ is open in $X$. Similarly, $D=F_B\setminus U_C$ is also closed in $X$, so $D$ is a clopen in $X$ and its complement contains $C$, which is non-empty. Thus, $D=\varnothing$, because $X$ is connected.

Therefore, $A\cup B=C\cup D=C$, and this proves that $A\cup B$ is connected.