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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.

I thought about this:

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.

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    $\begingroup$ You should double-check your typesetting for errors. $\endgroup$ – parsiad Apr 17 '17 at 18:26
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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 $B_2$.

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

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

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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.

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