# Problem of topology about connectedness (problem 3 of Munkres second edition, section 3, page 153 )

Let $\lbrace A _{\alpha} \rbrace$ be a collection of connected subspaces of $X$; let $A$ be a connected subspace of $X$. Show that if $A \cap A _{\alpha} \neq \emptyset$ for all $\alpha$, then $A\cup ( \bigcup A_\alpha)$ is connected.

By contradiction, I supposed that $A\cup ( \bigcup A_\alpha)$ is disconnected, i.e., $A\cup ( \bigcup A_\alpha) = C \cup D$, where $C \neq \emptyset$ and $D \neq \emptyset$ are open and disjoint. As by hypothesis we know that $A$ and $A _{\alpha }$ for all $\alpha$ are connected, then

($A \subset C$ and $A _{\alpha} \subset C$ ) or ($A \subset D$ and $A _{\alpha} \subset D$)

Now, I saw the following cases:

1. If $A \subset C$ and $A _{\alpha }$ for all $\alpha$, then $A \subset C$ and $\bigcup A _{\alpha }$, this implies that $A \cup (\bigcup A _{\alpha }) \subset C$ this means that $D \neq \emptyset$ and this is a contradiction.

2. If $A \subset C$ and exist a $\beta$ such that $A _{\beta } \subset D$, then $A \cap A _{\beta } \subset C \cap D$, but $C \cap D = \emptyset$, then $A \cap A _{\beta } = \emptyset$, but this is a contradiction, because $A \cap A _{\alpha }$ for all $\alpha$

The case 3 (where $A \subset D$ and $A _{\alpha } \subset D$ for all $\alpha$ ) and the case 4 (where $A \subset D$ and exist $\beta$ such that $A _{\beta } \subset C$) are analogous to cases 1 and 2 respectively. Hence $A\cup ( \bigcup A_\alpha)$ is connected.

• You should double-check your typesetting for errors. – parsiad Apr 17 '17 at 18:26

Suppose $A \cup \left(\bigcup_\alpha A_\alpha\right)$ is disconnected. Then it can be expressed as $B_1 \cup B_2$ where $B_1$ and $B_2$ are nonempty, open, and disjoint. Then $A$ and each $A_\alpha$ lies either entirely in $B_1$ or entirely in $B_2$, since each subspace is connected. Because $A_\alpha \cap A$ is nonempty for all $\alpha$, that means all of the $A$ and $A_\alpha$ either all lie in $B_1$ or all lie in $B_2$. This contradicts nonemptiness of $B_1$ and $B_2$.

• Your answer helped me a lot, thanks :) – G. P Apr 17 '17 at 20:42

I highly suspect you meant that $A\cap A_\alpha \ne \empty$ instead.

In that case, we have by the pivot theorem that $A\cup A_\alpha$ is connected, and then again by the pivot theorem $\cup_\alpha A\cup A_\alpha$ is connected.

• It's true, It's $A \cap A _{\alpha } \neq \emptyset$ thank you – G. P Apr 17 '17 at 18:45

My preferred way to think about connectedness is as follows. Let $Z=\{0,1\}$ be a two-point space with the discrete topology. Then a space $X$ is connected iff all continuous $f:X\to Z$ are constant.

Here if you have a continuous $f:A\cup\bigcup A_\alpha\to Z$ then $f$ is constant on $A$ and also on each $A_\alpha$. Therefore....

Using simple arguments in point set topology, it can be shown that the following result provides the solution to the OP's question.

Let $X$ be a topological space and assume we have a family $(A_\alpha)_{\alpha \in I}$ of connected subspaces of $X$, such that

$\tag 1 \text{There is an } \alpha_0 \in I \text{ such that } A_{\alpha_0} \cap A_\alpha \ne \emptyset \text{ for every } \alpha \in I$

and

$\tag 2 X = \bigcup_{\alpha \in I} A_{\alpha}$

Then $X$ is connected.

Proof:

Let $U$ and $V$ be any two disjoint open sets of $X$ with $X = U \cup V$. It is easy to show that each of the connected subspaces $A_{\alpha}$ must be wholly contained in one of these two open sets. But since all of these spaces have at least one point in common with $A_{\alpha_0}$, they must all be wholly contained in the same open set that $A_{\alpha_0}$ is contained in. But this means that

$\quad U = X \text{ and } V = \emptyset$

or

$\quad V = X \text{ and } U = \emptyset$.

So $X$ is connected.