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I've seen two different and perhaps equivalent definitions of connected sets.

set $E$ is connected when.

  1. $\nexists{C, D}$ such that both open and $C \cap D = \emptyset$ & $E = C \cup D$
  2. $\nexists{C, D}$ such that $C \cap \bar{D} = \emptyset$ & $\bar{C} \cap D = \emptyset$ & $E = C \cup D$

I can prove 1 -> 2, but 2->1 is assured too? i failed to prove it. how to find proper open sets generally?

For instances, $E = [0,1] \cup [2,3]$ then by def2, it's cleary seperated but how to find two open sets to use def 1?

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3 Answers 3

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If I understand correctly, the confusion is on what is an open set.

In your example, $[1,2]$ is indeed OPEN in $E$. It is not open in $\mathbb R$, but it is open in $E$ as it is the intersection of $E$ with an open set of $\mathbb R$, for instance $[1,2]=(0.5,2.5)\cap E$.

The definition $1)$ of connectedness, requires $C,D$ to be open in $E$. Similarly, in definition $2)$ the closures $\bar D, \bar C$ are in the closures in $E$. In particular, if $E=C\cup D$, the condition "$C$ is open in $E$" is equivalent to "$D$ is closed in $E$" (because $D$ is the complement of $C$).

Since $\bar D$ always contains $D$, if $E=C\cup D$, then the condition $$C\cap \bar D=\emptyset$$ is equivalent to $\bar D=C^c=D$, that is, $D$ is closed in $E$. So in $2)$ the conditions stated are equivalent to ask that both $C$ and $D$ are closed in $E$, hence their complements, $D,C$ respectively, are open. This is why $1)$ and $2)$ are equivalent.

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the fact is the first definition holds also for closed sets indeed when $E$ is disconnected $C,D$ are clopen. How is this true? Suppose $E$ is not connected so exist $C,D$ open s.t. $C\cap D = \emptyset,\; E = C\cup D$, so who's the complementary of $D$ in $E$? $D^c = E-D = C$ so we show that the complementary of $D$ is open and $D$ has to be closed. You can reiterate the same reasoning for $C$.

In your example you can choose $C = [0,1]$ and $D = [2,3]$


In addition to prove that $2) \implies 1)$ try with contraposition if you need I can elaborate furhter, let me know.

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(2) implies (1). Assume you have $C, D$ with $C \cap {\rm Cl} D = \emptyset$, ${\rm Cl} C \cap D = \emptyset$, and $E = C \cup D$. By the way, there is an additional requirement that neither $C, D$ should be empty (otherwise, choosing $C = E, D = \emptyset$ would trivially "separate" any space $E$, which is not desirable). We claim that $C, D$ satisfy the first definition too. For this, we just need to show that $C, D$ are disjoint and open.

The fact that $C, D$ are disjoint is easy to check, since we know any set is contained in its closure. So $C \cap {\rm Cl} D = \emptyset$ alone implies $C \cap D = \emptyset$.

To see that $C$ is open, consider arbitrary $c \in C$ - we show there is a neighborhood of $c$ contained in $C$. Note that $c \in C$ along with $C \cap {\rm Cl} D = \emptyset$ imply that $c \in X - {\rm Cl} D$, which is an open set and thus contains a neighborhood $U$ of $c$ disjoint from ${\rm Cl} D$. We claim that $U \subseteq C$. Take arbitrary $x \in U$. Since $x \notin {\rm Cl} D$, and $D \subseteq {\rm Cl} D$, then $x \notin D$. Since $E = C \cup D$, this implies $x \in C$, as desired. A similar argument can be applied to show that $D$ is open as well.

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