Prove that if for $E$ and $F$ connected sets, if $E \cap F \neq \emptyset \rightarrow E \cup F$ is connected

Question

Prove that if $E \cap F \neq \emptyset \rightarrow E \cup F$ is connected.

I am trying to do it by contradiction.

Assume that $E \cup F$ is disconnected. Therefore, $\exists U, V$ open, such that

$$[(E \cup F) \cap U] \cap [(E \cup F) \cap V] = \emptyset$$ $$[(E \cup F) \cap U] \cup [(E \cup F) \cap V] = E\cup F$$ $$(E \cup F) \cap U \neq \emptyset$$ $$(E \cup F) \cap V \neq \emptyset$$

I am stuck with how to proceed.

Answer 1

If you correct your first line to ${}=\emptyset$ then you can proceed as follows: you also have $(E\cap U)\cap(E\cap V)=\emptyset$ and $(E\cap U)\cup(E\cap V)=E$, and similarly for $F$. As $E$ is connected you can deduce something about $E\cap U$ and $E\cap V$.

Answer 2

Clearly $$(E \cap U) \cap (E \cap V) \subseteq ((E \cup F) \cap U)\cap ((E \cup F) \cap V) = \emptyset\tag 1$$ as $E \subseteq E \cup F$. The same holds for $F$.

Also $$(E \cap U) \cup (E \cap V) = E\tag 2$$ (left to right inclusion is clear, and if $x \in E$ it is in $E \cup F$ so in either $(E \cup F) \cap U$ (and thus in $E \cap U$) or in $(E \cup F) \cap V$ (and thus in $E \cap V$) for the other).

The same holds for $F$ instead of $E$.

Now, as $E$ is connected, we cannot have that both $E \cap U \neq \emptyset$ and $E \cap V \neq \emptyset$, (or else we'd have disconnection of $E$) so say for definiteness that we have $U \cap E = \emptyset$. This then implies that $E \cap V = E$ by (2), so that $E \subseteq V$.

If now $p \in E \cap F$ (we have to use that too, of course) we see that $F \cap V \neq \emptyset$, and again connectedness of $F$ implies that $F \cap U = \emptyset$ (or $F \cap U$ and $F \cap V$ would otherwise disconnect $F$), and thus (!) $(E \cup F) \cap U = \emptyset$ contradicting how $U$ and $V$ were given. So $E \cup F$ is not disconnected, and hence must be connected.

Answer 3

First try to prove this small lemma: If $C$ and $D$ form a separation of a topological space $X$, and if $Y$ is a connected subspace of $X$, then $Y$ lies entirely within $C$ or $D$. Now in your case say $ X=E\cup Y$, and if possible, $Y=C\cup D$ is a separation of $X$. Let $x\in E\cap F$, and suppose $x\in C$, then the lemma says both $E$ and $F$ are inside $C$. So, $D$ is empty. Contradiction. So, $X$ is connected. Hope this helps.