I'm studying Intro to Topology by Mendelson.
The problem statement is,
Let $A,B\subset X$, $X$ a topological space. If $A$ is connected, $B$ open and closed, and $A\cap B\neq\emptyset$ then $A\subset B$.
My proof is,
By way of contradiction, suppose that $A$ is not a subset of $B$. Then there exists an $a\in A$ such that $a\in C(B)$. Consider the sets $P=A\cap B$ and $Q=A\cap C(B)$. Note that both $P$ and $Q$ are nonempty and open. Thus, we have that $A\subset A\cap B\cup A\cap C(B)$ and $P\cap Q=A\cap B\cap C(B)=A\cap\emptyset=\emptyset\subset C(A)$. Also, $P\cap A\neq\emptyset$ and $Q\cap A\neq\emptyset$. Therefore, $A$ is disconnected, which is a contradiction, since $A$ is assumed connected.
I'm not sure if this was even the right approach, but it's my best shot so far.
Thanks for any hints or feedback!
Best Answer
Your argument is essentially correct, but the wording needs a bit more care in one place. When you say that the sets $P=A\cap B$ and $Q=A\setminus B$ are open, the default reading is ‘open in $X$’, which isn’t necessarily the case. You want to say that $P$ and $Q$ are open in $A$. Specifically, $P$ and $Q$ are non-empty disjoint relatively open subsets of $A$ whose union is $A$, and therefore $A$ is not connected.
Added: If you’ve not already done so, you might find it useful for future reference to prove that a set $A$ in a space $X$ is disconnected iff it has a non-empty, proper subset that is clopen in the relative topology on $A$. Here you could have applied that result immediately: $B\cap A$ is clearly non-empty and relatively clopen in $A$, so if $A$ is connected ...