# The Closure of a Connected Set is Connected

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.

• To typeset $\{a\}$, use $\{a\}$, with a single backslash before each curly brace. The double backslash causes a line break, which makes your post pretty difficult to read! Commented Oct 25, 2023 at 2:18
• Thanks for reminding me, haha. Commented Oct 25, 2023 at 2:52
• You wrote “consider an arbitrary $a\in(\overline{A}-\{a\})$” but I think you meant “consider an arbitrary $a\in(\overline{A}-A)$”.
– MJD
Commented Oct 25, 2023 at 2:55
• I did. Let me fix that. Commented Oct 25, 2023 at 3:07
• There. I'm sorry that my mistake upset you. Commented Oct 25, 2023 at 3:13

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 $$a\notin A$$)?
But If $$A \cap V = \varnothing$$, as $$\{a\} \notin U \cap V$$, then, If assume $$a \in V$$ without loss of generality, $$C = (A \cap U) \cup (\{a\} \cap U) = \varnothing$$. Therefore, we have contradict.