1
$\begingroup$

Suppose that the sets $A_{1},A_{2} \subset \mathbb{R}^n$ are connected and that they are not disjoint. Prove that $A_{1} \cup A_{2}$ is connected.

The section including this question contains this theorem:

enter image description here

And the following definition:

A is disconnected if there exists open sets B,C such that $A\cap B$ and $A \cap C$ are nonempty, disjoint sets and $A \subset B\cup C.$ A is connected if it is not disconnected.

but I do not know how to use this information to prove the required, could anyone help me please?

$\endgroup$
1
$\begingroup$

Suppose by contradiction that $A_1 \cup A_2 \subset C\cup D$ where $C,D$ are open and $C\cap (A_1 \cup A_2) \ne \emptyset,D\cap (A_1 \cup A_2) \ne \emptyset$ and $(C\cap D) \cap (A_1 \cup A_2) = \emptyset$.

As $A_1,A_2$ are connected each are either in $C$ or $D$ , if ,say , both in $C$ then their union $A_1\cup A_2$ would also be in $C$ and then $D\cap (A_1 \cup A_2)= \emptyset$ a contradiction. So we have that W.L.O.G $A_1 \subset C$ and $A_2 \subset D$ but then $A_1\cap A_2 =\emptyset$ , a contradiction.

Note that we didn't use the fact that we are in $\Bbb R^n$ which means we proved that any union of connected sets with nonempty intersection is connected.

$\endgroup$
  • $\begingroup$ are you proving by contradiction using the definition of disconnectedness? $\endgroup$ – hopefully Feb 28 at 11:24
  • 1
    $\begingroup$ Yes, assuming $A_1\cup A_2$ is not connected we get those $C,D$ $\endgroup$ – user123 Feb 28 at 11:25
  • $\begingroup$ very nice proof :) $\endgroup$ – hopefully Feb 28 at 11:26
  • 1
    $\begingroup$ @hopefully thanks :) note that i didn't use the fact that the space is $\Bbb R^n$ $\endgroup$ – user123 Feb 28 at 11:27
  • $\begingroup$ but disconnectedness say that $A_{1} \cup A_{2} \subset C \cup D$ not equal $\endgroup$ – hopefully Feb 28 at 11:28

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.