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Let $\{E_\alpha \}$ be a collection of connected sets in $\mathbb{R}^d$ such that $\cap \ E_\alpha\ne \emptyset.$ Show that $E=\cup \ E_\alpha$ is connected.

Attempt at a solution

Assume $E=\cup \ E_\alpha$ is disconnected, then there exists two open sets $A,B$ such that $A\cap E$ and $B\cap E$ are disjoint, nonempty and there union $(A\cap E) \cup (B\cap E)=E$. Let $x\in E$, thus $x$ is either in $A$ or $B$. Suppose $x\in A$, then since $A$ and $B$ are open and the intersections of all $E$, which is denoted by $E_{\alpha}$, is nonempty then $x$ is also an element of $B$. Thus $x\in B$, but that is a contradiction since $A$ and $B$ are disjoint.

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  • $\begingroup$ You may have the right idea, but it’s hard to tell, since you’ve a major error when you write $E=\bigcap E_\alpha$. Are you sure that that’s what you mean? $\endgroup$ – Brian M. Scott Apr 3 '13 at 2:46
  • $\begingroup$ @BrianM.Scott I will now edit. $\endgroup$ – user60514 Apr 3 '13 at 2:49
  • $\begingroup$ @BrianM.Scott is that better or still wrong? $\endgroup$ – user60514 Apr 3 '13 at 2:53
  • $\begingroup$ @user60514: How did you conclude that "$x$ is also an element of $B$"? $\endgroup$ – wj32 Apr 3 '13 at 2:58
  • $\begingroup$ @wj32 Because they are both open and from the hypothesis, which states $\cap \ E_\alpha\ne \emptyset.$ Am I wrong for doing that? $\endgroup$ – user60514 Apr 3 '13 at 3:00
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First, a notational problem: since $E_\alpha$ is a name for one of the sets in the collection, you should not use it as a name for the intersection of those sets. Let’s let $F=\bigcap_\alpha E_\alpha$ instead.

Starting with your $A$ and $B$ is fine, but then you go a bit astray. Let $x\in E$; then $x$ belongs to exactly one of $A$ and $B$. Without loss of generality suppose that $x\in A$. Since $x\in E$, there is some index $\alpha$ such that $x\in E_\alpha$, so $x\in A\cap E_\alpha$. But $E_\alpha$ is connected, and $(A\cap E_\alpha)\cap(B\cap E_\alpha)=\varnothing$, so $B\cap E_\alpha$ must be empty. In other words, $E_\alpha\subseteq A$.

Now use the fact that $F\ne\varnothing$, and pick any point $y\in F$. Then $y\in E_\alpha\subseteq A$.

Now let $\beta$ be any index; $y\in A\cap E_\beta$, and $E_\beta$ is connected, so $E_\beta\subseteq\ldots~$? Can you see the contradiction here?

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  • $\begingroup$ I see my error now. Thank you! If you were to rate my answer 1-15, what would you give it? Where 15 is the highest. $\endgroup$ – user60514 Apr 3 '13 at 3:08
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    $\begingroup$ @user60514: You’re welcome. May I change to a $0$-$10$ scale? That’s what I’ve used for the last $40$ years or so. On that scale probably $3$, meaning roughly that you’ve made a useful start, but most of the work remains to be done. Crucially, you’ve not actually used anywhere the fact that the sets $E_\alpha$ are connected, and the result depends on that fact. $\endgroup$ – Brian M. Scott Apr 3 '13 at 3:12
  • $\begingroup$ I got a $1$ out of $15$ for that answer on my quiz. I see what I did wrong now, but do you think $1$ is a bit low or justifiable? $\endgroup$ – user60514 Apr 3 '13 at 3:15
  • $\begingroup$ @user60514: It’s lower than I’d have given, but I don’t think that it’s indefensible; depending on an instructor’s marking habits, I think that out of $15$ one could probably defend anything from $1$ to about $5$. $\endgroup$ – Brian M. Scott Apr 3 '13 at 3:22
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    $\begingroup$ @user60514: You’re welcome! $\endgroup$ – Brian M. Scott Apr 3 '13 at 3:26

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