# Union of connected subsets

• Dec 5th 2010, 06:07 PM
okor
Union of connected subsets
Problem: Let $\displaystyle A_{1}$, $\displaystyle A_{2}$, and $\displaystyle A_{3}$ be connected subsets of X, such that $\displaystyle A_{1} \cap A_{2} \neq \emptyset$ and $\displaystyle A_{2} \cap A_{3} \neq \emptyset$. Prove that $\displaystyle A_{1} \cup A_{2} \cup A_{3}$ is connected.

So far this is what I have:

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

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

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

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

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.
• Dec 5th 2010, 06:13 PM
Drexel28
Quote:

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

So far this is what I have:

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

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

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

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

Remark The same methodology (using induction) shows that if $\displaystyle \{A_n\}$ is a sequence of connected subspaces of some topological space for which $\displaystyle A_n\cap A_{n+1}\ne\varnothing$ then $\displaystyle \displaystyle \bigcup_{n\in\mathbb{N}}A_n$ is connnected.
• Dec 5th 2010, 06:34 PM
okor
Quote:

Originally Posted by Drexel28
I think you're misunderstanding the definitions at hand. Let $\displaystyle A_1\cup A_2\cup A_3$ be given the subspace topology. Assume that $\displaystyle A_1\cup A_2\cup A_3=U\cup V$ where $\displaystyle U,V$ are open and disjoint. We may assume WLOG that $\displaystyle A_1\cap U\ne \varnothing$. Argue then that $\displaystyle A_1\subseteq U$ otherwise $\displaystyle \left(A_1\cap U\right)\cup\left(A_1\cap V\right)$ is a disconnnection of $\displaystyle A$. Then, argue that $\displaystyle A_2\subseteq U$ since otherwise $\displaystyle A_2\cap V$ is non-empty and open in $\displaystyle A_2$ and since $\displaystyle A_1\subseteq U$ and $\displaystyle A_1\cap A_2\ne\varnothing$ then $\displaystyle A_1\cap U\ne\varnothing$ and so $\displaystyle \left(A_2\cap U\right)\cup\left(A_2\cap V\right)$ is a disconnection of $\displaystyle A_2$, thus $\displaystyle A_2\subseteq U$. Apply the exact same argument to show that $\displaystyle A_3\subseteq U$ and thus $\displaystyle V=\varnothing$. Conclude that no disconnection of $\displaystyle A_1\cup A_2\cup A_3$ exists.

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

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