# Union of connected sets which have pairwisely nonempty intersections

If connected subsets have non-empty intersections pairwisely, how can I show that their union is connected? Formally, let $$E_\alpha$$ be connected for every $$\alpha \in I$$, and suppose that $$E_\alpha \cap E_{\beta} \neq \emptyset$$ for every distinct pair of indices $$\alpha$$ and $$\beta$$ in $$I$$. How can I show that $$\displaystyle\bigcup_{\alpha \in I}E_\alpha$$ is 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 $$E_\alpha \cap E_{\beta} \neq \emptyset$$. 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.

Define $$S := \cup_{\alpha \in I} E_\alpha$$, and 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.

• This was good timing. :) Mar 28, 2017 at 7:49
• What if $E_\alpha \subset C$ for some $C$? Mar 28, 2017 at 8:31
• @Ninja then all $E_\alpha$ do. Mar 28, 2017 at 8:40

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.

We have the following elementary result:

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 $\cup A_\lambda$.