# set connectedness with two equivalent definition. how to prove they are equal?

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?

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.

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.

(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.