Union of connected subsets

**okor** · December 5th 2010, 06:07 PM

Problem: Let $A_{1}$ , $A_{2}$ , and $A_{3}$ be connected subsets of X, such that $A_{1} \cap A_{2} \neq \emptyset$ and $A_{2} \cap A_{3} \neq \emptyset$ . Prove that $A_{1} \cup A_{2} \cup A_{3}$ is connected.

So far this is what I have:

Assume $A_{1} \cup A_{2} \cup A_{3}$ is disconnected, then $(A_{1} \cup A_{2} \cup A_{3}) \cap U \neq \emptyset, (A_{1} \cup A_{2} \cup A_{3}) \cap V \neq \emptyset, (A_{1} \cup A_{2} \cup A_{3}) \cap U \cap V = \emptyset$ .

Edit: Forgot to include $(A_{1} \cup A_{2} \cup A_{3}) \subseteq (U \cup V)$

Therefore $(A_{1} \cap U \cap V) \cup (A_{2} \cap U \cap V) \cup (A_{3} \cap U \cap V) = \emptyset$ , $(A_{1} \cap U) \cup (A_{2} \cap U) \cup (A_{3} \cap U) \neq \emptyset$ , $(A_{1} \cap V) \cup (A_{2} \cap V) \cup (A_{3} \cap V) \neq \emptyset$

Thus $(A_{1} \cap U \cap V) = (A_{2} \cap U \cap V) = (A_{3} \cap U \cap V) = \emptyset$

Now, I'm not 100% sure if I have been approaching this correctly, but I think from here I have to show for one of the 3 connected subsets its intersection with U and Y is nonempty and thus it is disconnected meaning there is a contradiction, but I am not sure how to prove this part.

**Drexel28** · December 5th 2010, 06:13 PM

Originally Posted by okor

Problem: Let $A_{1}$ , $A_{2}$ , and $A_{3}$ be connected subsets of X, such that $A_{1} \cap A_{2} \neq \emptyset$ and $A_{2} \cap A_{3} \neq \emptyset$ . Prove that $A_{1} \cup A_{2} \cup A_{3}$ is connected.

So far this is what I have:

Assume $A_{1} \cup A_{2} \cup A_{3}$ is disconnected, then $(A_{1} \cup A_{2} \cup A_{3}) \cap U \neq \emptyset, (A_{1} \cup A_{2} \cup A_{3}) \cap V \neq \emptyset, (A_{1} \cup A_{2} \cup A_{3}) \cap U \cap V = \emptyset$ .

Therefore $(A_{1} \cap U \cap V) \cup (A_{2} \cap U \cap V) \cup (A_{3} \cap U \cap V) = \emptyset$ , $(A_{1} \cap U) \cup (A_{2} \cap U) \cup (A_{3} \cap U) \neq \emptyset$ , $(A_{1} \cap V) \cup (A_{2} \cap V) \cup (A_{3} \cap V) \neq \emptyset$

Thus $(A_{1} \cap U \cap V) = (A_{2} \cap U \cap V) = (A_{3} \cap U \cap V) = \emptyset$

Now, I'm not 100% sure if I have been approaching this correctly, but I think from here I have to show for one of the 3 connected subsets its intersection with U and Y is nonempty and thus it is disconnected meaning there is a contradiction, but I am not sure how to prove this part.

I think you're misunderstanding the definitions at hand. Let $A_1\cup A_2\cup A_3$ be given the subspace topology. Assume that $A_1\cup A_2\cup A_3=U\cup V$ where $U,V$ are open and disjoint. We may assume WLOG that $A_1\cap U\ne \varnothing$ . Argue then that $A_1\subseteq U$ otherwise $\left(A_1\cap U\right)\cup\left(A_1\cap V\right)$ is a disconnnection of $A$ . Then, argue that $A_2\subseteq U$ since otherwise $A_2\cap V$ is non-empty and open in $A_2$ and since $A_1\subseteq U$ and $A_1\cap A_2\ne\varnothing$ then $A_1\cap U\ne\varnothing$ and so $\left(A_2\cap U\right)\cup\left(A_2\cap V\right)$ is a disconnection of $A_2$ , thus $A_2\subseteq U$ . Apply the exact same argument to show that $A_3\subseteq U$ and thus $V=\varnothing$ . Conclude that no disconnection of $A_1\cup A_2\cup A_3$ exists.

Remark The same methodology (using induction) shows that if $\{A_n\}$ is a sequence of connected subspaces of some topological space for which $A_n\cap A_{n+1}\ne\varnothing$ then $\displaystyle \bigcup_{n\in\mathbb{N}}A_n$ is connnected.

**okor** · December 5th 2010, 06:34 PM

Originally Posted by Drexel28

I think you're misunderstanding the definitions at hand. Let $A_1\cup A_2\cup A_3$ be given the subspace topology. Assume that $A_1\cup A_2\cup A_3=U\cup V$ where $U,V$ are open and disjoint. We may assume WLOG that $A_1\cap U\ne \varnothing$ . Argue then that $A_1\subseteq U$ otherwise $\left(A_1\cap U\right)\cup\left(A_1\cap V\right)$ is a disconnnection of $A$ . Then, argue that $A_2\subseteq U$ since otherwise $A_2\cap V$ is non-empty and open in $A_2$ and since $A_1\subseteq U$ and $A_1\cap A_2\ne\varnothing$ then $A_1\cap U\ne\varnothing$ and so $\left(A_2\cap U\right)\cup\left(A_2\cap V\right)$ is a disconnection of $A_2$ , thus $A_2\subseteq U$ . Apply the exact same argument to show that $A_3\subseteq U$ and thus $V=\varnothing$ . Conclude that no disconnection of $A_1\cup A_2\cup A_3$ exists.

Remark The same methodology (using induction) shows that if $\{A_n\}$ is a sequence of connected subspaces of some topological space for which $A_n\cap A_{n+1}\ne\varnothing$ then $\displaystyle \bigcup_{n\in\mathbb{N}}A_n$ is connnected.

I understood the definitions, but following through with your reasoning, I see where I approached it wrong. I forgot to include the $(A_{1} \cup A_{2} \cup A_{3}) \subseteq (U \cup V)$ condition, which is from where I should have started. I knew I needed to pull in the intersection between $A_{1}$ and $A_{2}$ and the intersection between $A_{2}$ and $A_{3}$ , but I wasn't sure where, which your explanation clears up.

Thank you for your assistance.

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