Closure of Connected Space is Connected

[Bashyboy]

Homework Statement

If $C$ is a connected space in some topological space $X$, then the closure $\overline{C}$ is connected.

Homework Equations

The Attempt at a Solution

Suppose that $\overline{C} = A \cup B$ is separation; hence, $A$ and $B$ are disjoint and do not share limit points, which means $A \cap \overline{B}$ and $\overline{A} \cap B$ are both empty. Since $C$ is connected and is contained in $\overline{C} = A \cup B$, it must be contained in exactly one of the two partitions. WLOG, suppose that $C \subseteq A$. Then $C \subseteq A \subseteq \overline{C}$ implies $\overline{A} = \overline{C}$, and therefore $\emptyset = \overline{A} \cap B = \overline{C} \cap B$ implies $B= \emptyset. Hence,$\overline{C}## is connected.

How does this sound?

 

[andrewkirk]

Since $C$ is connected and is contained in $\overline{C} = A \cup B$, it must be contained in exactly one of the two partitions.

I think a little more is needed here. C is connected in the topology of X but to argue that that means it is in either A or B you need to first show that C is connected in the subspace topology of $\bar C$ (as it is in that topology that A and B are open, not necessarily the topology of X).

 

[Bashyboy]

Okay. Here is an attempt a remedying the solution. First, let $cl_X(S)$ denote the closure of $S$ in $X$'s topology, and likewise $cl_{\overline{C}}(S)$ denotes the closure of $S$ in $\overline{C}$'s topology. By way of contradiction, suppose that $C$ is separated in $\overline{C}$. Then there exist $P, Q \subseteq \overline{C}$ such that $C = P \cup Q$ and $cl_{\overline{C}}(P) \cap Q$ and $P \cap cl_{\overline{C}}(Q)$ are both empty. But since $\overline{C}$ is closed in $X$, $cl_{\overline{C}}(P) = cl_X(P)$ and $cl_{\overline{C}}((Q) = cl_X(Q)$, and therefore $cl_X(P) \cap Q$ and $P \cap cl_X(Q)$ are both empty. This means that $C$ is separated in $X$, contradicting our hypothesis.

Sorry for the funky notation.

 

[andrewkirk]

suppose that $C$ is separated in $\overline{C}$. Then there exist $P, Q \subseteq \overline{C}$ such that $C = P \cup Q$ and $cl_{\overline{C}}(P) \cap Q$ and $P \cap cl_{\overline{C}}(Q)$ are both empty.

Your notation is fine. What you wrote is perfectly clear. Unfortunately, I don't find the above convincing. The 'Then there exist...' does not use the standard definition of a separation, as it does not say anything about open sets. Maybe you are using some theorem about a criterion that is equivalent to the definitional criterion for being a separation, but if so that theorem needs to be quoted.

I think the problem is actually easiest to do just using the basic definition of separation/connectedness, which is that C is connected in X iff for any open subsets A,B of X such that $C=(A\cap C)\cup (B\cap C)$ (which we note is equal to $(A\cup B)\cap C$ so that the condition may be restated as $C\subseteq A\cup B$), the set $(A\cap C)\cap(B\cap C)$ must be non-empty (and we note that is equal to $A\cap B\cap C$).

So to prove that $\bar C$ is connected in X given that C is, suppose that there are open subsets A,B of X such that $\bar C\subseteq A\cup B$. Then it is fairly straightforward to use the above definition of connectedness of C in X to prove the connectedness of $\bar C$.

 

[Bashyboy]

Maybe you are using some theorem about a criterion that is equivalent to the definitional criterion for being a separation, but if so that theorem needs to be quoted.

Yeah...Sorry about that. Here is the theorem I am using:

If $Y$ is a subspace of $X$, a separation of $Y$ is a pair of disjoint nonempty sets $A$ and $B$ whose union is $Y,$ neither of which contains a limit point of the other (i.e., $\overline{A} \cap B$ and $A \cap \overline{B}$ are both empty)

Of course, $\overline{A}$ denotes the closure of $A$ in $X$ which is $cl_X(A)$ using my notation.