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Let $E_\alpha$ be open for all $\alpha \in I$ such that $E_\alpha \cap E_{\beta} \neq \emptyset$, then $\displaystyle\bigcup_{\alpha \in I}E_\alpha$ is also connected. There are similar questions to this theorem, they say the whole intersection $\displaystyle\bigcap_{\alpha \in I}E_\alpha$ is non-empty or say the same assumption as above but starts the proof with taking an element from $\displaystyle\bigcap_{\alpha \in I}E_\alpha$ when we have $E_\alpha \cap E_{\beta} \neq \emptyset$. If the connected subsets has non-empty intersections pairwisely, how can I show that their union is connected? I think that we can not pick an element from $\displaystyle\bigcap_{\alpha \in I}E_\alpha$. Consider for example that $E_1 = \{1,2 \}, $E_2 = {1,3 }, $E_3 = \{2,3 \} $ where $X = \{1,2,3\}$ with the coarsest topology.

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Suppose that $f: S := \cup_{\alpha \in I} E_\alpha$ is disconnected, so let $C$ be a $S$ clopen subset of $S$.

Then for each $\alpha \in I$: $E_\alpha \subset C$ or $E_\alpha \cap C =\emptyset$.

If this were not the case, $E_\alpha \cap C$ would be a non-trivial clopen set in $E_\alpha$, which cannot happen by connectedness of $E_\alpha$.

If we'd have for some $\alpha$: $E_\alpha \subset C$ and for some $\beta \neq \alpha$: $E_\beta \cap C =\emptyset$, a point in $E_\alpha \cap E_\beta$ would be in $C$ and not in $C$ at the same time, which cannot be. So the same inclusion or disjunction is true for all $\alpha$, and in the former case $C = S$ and in the latter case $C = \emptyset$. So $S$ is connected.

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  • $\begingroup$ This was good timing. :) $\endgroup$ – Martins Bruveris Mar 28 '17 at 7:49
  • $\begingroup$ What if $E_\alpha \subset C$ for some $C$? $\endgroup$ – Ninja Mar 28 '17 at 8:31
  • $\begingroup$ @Ninja then all $E_\alpha$ do. $\endgroup$ – Henno Brandsma Mar 28 '17 at 8:40
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Argue by contradiction. Assume $X = \bigcup_{\alpha} U_\alpha$ is not connected. Then $X = V \cup W$ with $V$, $W$ open, disjoint and nonempty. Now, each $U_\alpha$ satisfies either $U_\alpha \subseteq V$ or $U_\alpha \subseteq W$ or $U_\alpha \cap V$, $U_\alpha \cap W$ are two open, disjoint and nonempty subsets covering $U_\alpha$ which is not possible because $U_\alpha$ is connected. Finally there must exist $\alpha$, $\beta$ such that $U_\alpha \subseteq V$ and $U_\beta \subseteq W$, because $V$ and $W$ are both nonemtpy. But this is not possible because then $U_\alpha \cap U_\beta \neq \emptyset$. Hence $X$ is connected.

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We have the following elementary result:

Proposition 1: If a family of connected sets has a nonempty intersection, then the union is connected.

The OP is aware of proposition 1, so it is appropriate to employ it in the answer.

Select any $\lambda_0$ and set $B_\lambda = A_{\lambda_0} \cup A_\lambda$. By our hypothesis and proposition 1, each $B_\lambda$ in connected. Moreover, $A_{\lambda_0} \subset B_\lambda$ for all $\lambda$. But then by proposition 1 (again), $\cup B_\lambda$ must be connected. But this is the same set at $\cup A_\lambda$.

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