Closure of Connected Space is Connected

In summary: Also, this is not the standard definition of a separation, but it is equivalent. I think the proof I gave is still valid, but I can give a proof using the definition you gave if you prefer.
  • #1
Bashyboy
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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?
 
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  • #2
Bashyboy said:
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).
 
  • #3
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.
 
  • #4
Bashyboy said:
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##.
 
  • #5
andrewkirk said:
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.
 

FAQ: Closure of Connected Space is Connected

1. What is the definition of "Closure of Connected Space is Connected"?

The closure of a connected space is connected if every limit point of the space is also a limit point of the connected space.

2. How is the closure of a connected space related to its connectedness?

The closure of a connected space is an important property that guarantees the connectedness of the space. It ensures that the space remains connected even when additional points are added to it.

3. Can a space be connected but not have a connected closure?

Yes, it is possible for a space to be connected but have a closure that is not connected. This can occur if there are isolated points in the space that do not have any nearby points.

4. What is the significance of the closure of connected space being connected?

The closure of a connected space being connected is a useful property in topology, as it allows for the construction of continuous functions on the space. It also helps in proving the connectedness of more complex spaces.

5. How is the closure of connected space different from the closure of a disconnected space?

The closure of a connected space guarantees the connectedness of the space, while the closure of a disconnected space may not necessarily be connected. Additionally, a disconnected space may have multiple connected components in its closure, while a connected space will have only one connected component in its closure.

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