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I got this question for homework..

Let $\{E_{\alpha}\}$ be a family of connected subsets of a metric space $X$ such that any two of them have a non-empty intersection: $E_{\alpha} \cap E_{\beta}\ne \emptyset$. Prove that the union $\cup_{\alpha}E_{\alpha}$ is connected.

I'm relatively new to the concept of connected sets. I know that a set is connected iff it is the union of 2 separated sets and that one of the two sets are empty.

I don't understand how the fact that some of these sets share elements have anything to do with the union of all these sets...

Please help!

Thanks

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    $\begingroup$ Draw a picture first. It's easy to see why the union has to be connected. When you are convinced, try to prove it by contradiction. There is a lemma which states that if $X=A\cup B$ and $C \subset X$ is a connected subspace, then $C\subset A$ or $C\subset B$ $\endgroup$ – FormerMath Nov 5 '16 at 12:34
  • $\begingroup$ Your understanding of "connected" is not OK. A set $Y$ is disconnected if there exist disjoint, non-empty open sets $U$ and $V$ such that $Y = U \cup V$. Otherwise, $Y$ is connected. Thus: $Y$ is connected provided that if $U$ and $V$ are disjoint open sets such that $Y = U \cup V$, then either $U \cap Y = \varnothing$ or $V \cap Y = \varnothing$. That's not the same thing as "[$Y$] is the union of 2 separated sets and that one of the two sets are empty": Every set $Y$ can be written $Y \cup \varnothing$. $\endgroup$ – Andrew D. Hwang Nov 5 '16 at 12:35
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Let $Y = \bigcup_{\alpha \in \mathcal A} E_\alpha$. Suppose $Y = A \cup B$, where $A$ and $B$ are disjoint open subsets of $Y$. Let $a$ be a point in $X$ such that all $E_\alpha$ contain it. Then $a \in A$ or $a \in B$. Say that $a \in A$. For every $\alpha$, $A \cap E_\alpha$ and $B \cap E_\alpha$ are open subsets of $E_\alpha$. Thus, we may write the disjoint union $E_\alpha = (A \cap E_\alpha)\cup (B \cap E_\alpha)$. Since $E_\alpha$ is connected and $a \in A \cap E_\alpha$, we conclude that $B \cap E_\alpha = \emptyset$. Thus $B = \bigcup (B \cap E_\alpha) = \emptyset$. Therefore, $Y$ is connected.

Edit: This is incorrect. I need the op to untag accepted answer to take it down.

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    $\begingroup$ There is not always a "a point $a$ in $X$ such that all $E_\alpha$ contain it". In my humble opinion, this is not a correct proof. $\endgroup$ – Helmut Sep 15 '18 at 10:51
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We will argue by contradiction. Assume that $\cup_{\alpha}E_{\alpha}$ is not connected. Then there exist $U,V$ non-empty, open and disjoint such that $\cup_{\alpha}E_{\alpha}=U\cup V.$ Now define $f:U\cup V\to \{0,1\}$ by $$f(x)=\begin{cases}0 ,& x\in U, \\ 1,& x\in V.\end{cases}$$ It is clear that $f$ is continuous. Let $x_0\in U.$ There exist $\alpha$ such that $x\in E_{\alpha}.$ Since $f$ is continuous and $E_{\alpha}$ is connected then $f(E_{\alpha})$ must be connected. That is $f(E_{\alpha})=\{0\}.$ Now, for any $\beta$ we have $E_{\alpha} \cap E_{\beta}\ne \emptyset.$ So we get that $f(E_{\beta})=\{0\}.$ That is, $f$ must be constant. But this contradicts our assumption that $V\ne \emptyset.$ So we are done.

Note that the assumption $E_{\alpha} \cap E_{\beta}\ne \emptyset$ is crucial. In other case think of $(0,1)$ and $(1,2).$ Is it their union connected?

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  • $\begingroup$ Why are $U$ and $V$ open? The set $\{1, 2\}$ is closed and disconnected in $\mathbb{R}$, but there are no two nonempty sets so that $U \cup V = \{1, 2\}$. $\endgroup$ – Björn Lindqvist Jun 27 '18 at 6:13
  • $\begingroup$ @BjörnLindqvist, they are open because he/she is assuming that the union is not connected, so there should be a separation, that is, two open disjoint sets. $\endgroup$ – Sigur Sep 20 at 0:26
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In my answer a separation of $Y\subseteq X$ is by definition a pair $\{A,B\}$ of non-empty sets with $A\cup B=Y$ and $\bar{A}\cap B=\emptyset=A\cap\bar{B}$. Subset $Y$ is connected if no separation of $Y$ exists.


Suppose that $\{A,B\}$ is a separation of the union. Then $A\cap B=\varnothing$ and : $$\forall\alpha [E_{\alpha}\subseteq A\vee E_{\alpha}\subseteq B]$$

This because the negation of it for some $\alpha_0$ would lead to the conclusion that this $E_{\alpha_0}$ is not connected: pair $\{A\cap E_{\alpha_0},B\cap E_{\alpha_0}\}$ would be a separation of $E_{\alpha_0}$.

Then the condition that $\beta\neq\alpha\implies E_{\alpha}\cap E_{\beta}\neq\varnothing$ can be used to show that:$$\forall\alpha\; E_{\alpha}\subseteq A\vee\forall\alpha\; E_{\alpha}\subseteq B$$

However then $B=\varnothing$ or $A=\varnothing$, contradicting that $\{A,B\}$ is a separation.

We conclude that no separation exists, i.e. that the union is connected.

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