The union of non-disjoint connected is connected

**Pinkk** · September 15th 2010, 07:17 PM

Suppose $S_{1},S_{2}$ are connected sets in $\mathbb{R}^{n}$ and $S_{1}\cap S_{2} \ne \emptyset$ . Show that $S_{1}\cup S_{2}$ is connected. Does this hold true for $S_{1} \cap S_{2}$ ?

Okay, so I have seen proofs that rely on topological properties that we have not really discussed or proven in my class (like for instance, if $U,V$ are open in $A$ connected, and $U \cup V = A$ , then $U \cap V \ne \emptyset$ . Assuming that, I understand the proof for this theorem pretty well, but how do I prove that assumption? Or is there another way to solve this question without invoking open sets/topology in general? Thanks.

**Failure** · September 15th 2010, 11:07 PM

Originally Posted by Pinkk

Suppose $S_{1},S_{2}$ are connected sets in $\mathbb{R}^{n}$ and $S_{1}\cap S_{2} \ne \emptyset$ . Show that $S_{1}\cup S_{2}$ is connected. Does this hold true for $S_{1} \cap S_{2}$ ?

Okay, so I have seen proofs that rely on topological properties that we have not really discussed or proven in my class (like for instance, if $U,V$ are open in $A$ connected, and $U \cup V = A$ , then $U \cap V \ne \emptyset$ . Assuming that, I understand the proof for this theorem pretty well, but how do I prove that assumption? Or is there another way to solve this question without invoking open sets/topology in general? Thanks.

Since "being connected" is defined in terms of open sets, I don't think there is a way around talking of open sets and the like in a proof of the above proposition.

To show that the union $S_1\cup S_2$ of connected sets $S_{1,2}$ is connected, if $S_1\cap S_2\neq \emptyset$ , it suffices to show that given any two open sets $U,V$ with $S_1\cup S_2\subseteq U\cup V$ , and $U\cap V=\emptyset$ , it follows that either $S_1\cup S_2\subseteq U$ or $S_1\cup S_2\subseteq V$ .

And why is that? Well, suppose that $x\in S_1\cap S_2$ , then we have either $x\in U$ or $x\in V$ (but not both, because U and V are disjoint).
If $x\in U$ , then, because $S_{1,2}$ are both connected, it follows that $S_1\subseteq U$ and $S_2\subseteq U$ , and therefore $S_1\cup S_2\subseteq U$ .
Similarly, if $x\in V$ it follows that $S_1\cup S_2\subseteq V$ .

Finally, to the question as to whether $S_1\cap S_2$ is connected if $S_{1,2}$ are connected and $S_1\cap S_2\neq \emptyset$ . The answer is no, for consider the intersection of a line and a circle in $\mathbb{R}^2$ that consists of two isolated points. In such a case, the intersection is clearly not connected.

**Plato** · September 16th 2010, 02:51 AM

There is another characterization of connect sets.
A set is connected if it is not the union of two separated sets
Two sets are separated if each is non-empty and neither contains a point nor a limit point of the other.

Suppose that $U~\&~V$ are separated sets such $S_1\cup S_2=U\cup V$ .
We know that $\left( {\exists t \in S_1 \cap S_2 } \right)$ so $t\in U\text{ or }t\in V$ .
In either case, because each of $S_1~\&~S_2$ is connected there follows a contradiction.
So $S_1\cup S_2=U\cup V$ must be connected.

**Pinkk** · September 16th 2010, 08:35 PM

I don't see how the contradiction automatically follows from $t$ being in either $U$ or $V$ . That is, I don't see why if $t\in S_{1}$ and $S_{1}$ means $S_{1} \subset U$ .

**Plato** · September 17th 2010, 03:06 AM

Otherwise $S_1\cap U~\&~ S_1\cap V$ would be a separation of $S_1$ which is connected.

The basic idea underlying this proof is this theorem:
If a connected set is a subset of the union of two separated sets then the set is a subset of one of the two.

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