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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

mfl
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mathinthecity
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    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 $C\subset B$ – FormerMath Nov 05 '16 at 12:34
  • 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 $Y \cup \varnothing$. – Andrew D. Hwang Nov 05 '16 at 12:35
  • @user26857 The supposedly duplicate question you linked to is not really a duplicate. In the present question OP asks about pairwise intersecting sets, whereas the linked question deals with sets that have a non-empty intersection. This question is a true duplicate. – Evan Aad Jun 10 '22 at 07:36

2 Answers2

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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?

mfl
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  • 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 $U \cup V = {1, 2}$. – Gaslight Deceive Subvert Jun 27 '18 at 06:13
  • @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. – Sigur Sep 20 '19 at 00:26
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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.

drhab
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