This question has admittedly been asked numerous times on this website. However, I ask it again because I want to show my proof. The proofs that I have seen as answers on this website are much different than mine. But I've looked over the proof I came up with and I don't see anything wrong with it. It's admittedly a cumbersome attempt compared to the proofs I've seen. But can anyone look over the following proof and tell me if anything's wrong with it?
Theorem: Let $X$ be a topological space and let $A$ be a non-empty connected subset of $X$. Then $\overline{A}$ is also connected.
Proof: If $A=\overline{A}$ then of course $\overline{A}$ is connected. So assume that $A\neq\overline{A}$ and consider an arbitrary $a\in(\overline{A}-A)$. Let $C$ and $D$ be arbitrary non-empty open sets in the subspace with underlying set $A\cup\{a\}$ such that $C\cup D=A\cup\{a\}$. Then there exist open sets $U$ and $V$ of $X$ such that $C=(A\cup\{a\})\cap U$ and $D=(A\cup\{a\})\cap V$. By the distributive law for sets, $C=(A\cap U)\cup(\{a\}\cap U)$ and $D=(A\cap V)\cup(\{a\}\cap V)$. Hence, $C\cup D=(A\cap U)\cup (A\cap V)\cup(\{a\}\cap U)\cup(\{a\}\cap V)$. We can use the distributive law for sets once more to reduce the previous equality to $C\cup D=(A\cap(U\cup V))\cup(\{a\}\cap(U\cup V))$. Since $a$ is not in $A$ it is impossible for $a$ to be in $A\cap(U\cup V)$. Therefore, $a\in(\{a\}\cap(U\cup V))$. It's not hard to see that this implies $\{a\}=\{a\}\cap(U\cup V)$. Consequently, all elements of $A$ must be in $A\cap(U\cup V)$ and we get that $A=A\cap(U\cup V)$. Another use of set distribution yields $A=(A\cap U)\cup(A\cap V)$. But $A\cap U$ and $A\cap V$ are open sets in the subspace with underlying set $A$. Since $A$ is connected, $A\cap U$ and $A\cap V$ can't be disjoint. Thus, $C\cap D\neq\emptyset$ and $A\cup\{a\}$ is connected.
We now consider the union of sets $\bigcup_{a\in\overline{A}-A}(A\cup\{a\})$. Clearly, $A\subseteq\bigcap_{a\in\overline{A}-A}(A\cup\{a\})$ meaning the intersection of all the sets in the union is non-empty. The argument above tells us that each set in this union is connected since our choice of $a$ was arbitrary. It follows that $\bigcup_{a\in\overline{A}-A}(A\cup\{a\})$ is also connected. But $\bigcup_{a\in\overline{A}-A}(A\cup\{a\})=\overline{A}$ and we are done. $Q.E.D$
Is this proof flawed in any way? Thanks.
Best Answer
Your cumbersome argument involving the distributive law can be greatly simplified. Observe that $A\subseteq C\cup D\subseteq U\cup V$. Hence (by definition of intersection) $A\cap(U\cup V)=A$.
There actually is a flaw. You argued that $A\cap U$ and $A\cap V$ are open subsets of $A$, so they cannot be disjoint. There is another possibility: one of them is empty. Say $A$ doesn't intersect $V$. What does that tell you about $D$? How does this contradict $a\in\overline{A}$, which is never used in your arguments (so your argument "works" for all $a\notin A$)?