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

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.

• I think you need to rethink the start of your proof. If you assume $E\cup F$ is not connected, you have open sets $U$ and $V$ in $E\cup F$ such that $U\cap V=\emptyset$ while $U\cup V=E\cup F$. – Tartaglia's Stutter Nov 2 '18 at 11:36
• Both are the definitions of a disconnected set (in this case $E \cup F)$. I just took it from the book. – The Bosco Nov 2 '18 at 11:37
• I'm saying I don't think your first line should state that the intersection is not empty, I think it should say that it IS empty. Not connected means two disjoint open sets partition your space. The disjoint is an important part of that. – Tartaglia's Stutter Nov 2 '18 at 11:38
• Then how do I proceed to the definition of disconected? – The Bosco Nov 2 '18 at 11:43
• The answer below by hartkp should help lead you. One hint: you have two possilbe assumptions that you can contradict: 1.) both $E$ and $F$ are connected and 2.) $E\cap F\neq \emptyset$. – Tartaglia's Stutter Nov 2 '18 at 11:48

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 $$E\cap V$$.

• I saw something similar, but had trouble proving that if the first relation I wrote holds for $E \cup F$ it also holds for $E$ – The Bosco Nov 2 '18 at 11:47
• If $x$ is in the intersection for $E$ then it is also in the intersection for $E\cup F$. – hartkp Nov 2 '18 at 12:16

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.

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.