6
$\begingroup$

Suppose that C, D are connected subsets of a topological space T such that $\bar{C} \cap \bar{D} \neq \emptyset$. Is it true that $C \cup D$ is necessarily connected?

I think I have a counter example for this:

Take the intervals $C = (0,1)$ and $D = (1,2)$ their closure is $[0,1]$ and $[1,2]$ which gives $\bar{C} \cap \bar{D} = \{1\} \neq \emptyset$ but clearly $(0,1) \cup (1,2)$ is disconnected. Am I right?

$\endgroup$
  • 4
    $\begingroup$ Yes, you're right. $\endgroup$ – Rudy the Reindeer Mar 4 '12 at 15:10
  • 1
    $\begingroup$ Exactly, great example. $\endgroup$ – rfauffar Mar 4 '12 at 15:14
  • 4
    $\begingroup$ So that this doesn’t languish forever on the Unanswered list, in a few hours, when the system allows you to do so, you should submit that example as your own answer to the question and accept it. $\endgroup$ – Brian M. Scott Mar 4 '12 at 15:16
  • 6
    $\begingroup$ @user26069: You might consider taking the second part of your question (beginning with "I think...") and making it an answer on your own question. This is allowed, and in fact explicitly encouraged, on the StackExchange network. $\endgroup$ – Zev Chonoles Mar 4 '12 at 15:17
2
$\begingroup$

To get this off the Unanswered list, I’m posting the OP’s correct counterexample as CW:

Take the intervals $C = (0,1)$ and $D = (1,2)$ their closure is $[0,1]$ and $[1,2]$ which gives $\bar{C} \cap \bar{D} = \{1\} \neq \emptyset$ but clearly $(0,1) \cup (1,2)$ is disconnected.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy