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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.

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  • $\begingroup$ 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 $U\cup V=E\cup F$. $\endgroup$ – Tartaglia's Stutter Nov 2 '18 at 11:36
  • $\begingroup$ Both are the definitions of a disconnected set (in this case $E \cup F)$. I just took it from the book. $\endgroup$ – The Bosco Nov 2 '18 at 11:37
  • $\begingroup$ 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. $\endgroup$ – Tartaglia's Stutter Nov 2 '18 at 11:38
  • $\begingroup$ Then how do I proceed to the definition of disconected? $\endgroup$ – The Bosco Nov 2 '18 at 11:43
  • $\begingroup$ 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.) $E\cap F\neq \emptyset$. $\endgroup$ – Tartaglia's Stutter Nov 2 '18 at 11:48
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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 $E\cap V$.

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

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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.

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