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If $E$ and $F$ are connected subsets of $M$ with $E\cap F\ne\emptyset$, show that $E\cup F$ is connected.

My attempt:

Suppose $E\cup F$ is disconnected. $\boldsymbol{\Rightarrow \exists} \ open \ sets\ A, B\ne \emptyset$ such that $A\cap B=\emptyset$ and $A\cup B=E\cup F.$

Consider $E_1=A\cap E$ and $E_2=B\cap E$.

$A\cap B=\emptyset\ \boldsymbol{\Rightarrow} (A\cap E)\cap (B\cap E)=\emptyset.$

$A\cup B=E\cup F\ \boldsymbol{\Rightarrow}\ (A\cap E)\cup (B\cap E)=E$

claim: $A\cap E\ne \emptyset$ and $B\cap E\ne \emptyset$.

Suppose $A\cap E=\emptyset \boldsymbol{\Rightarrow} B=E \boldsymbol{\Rightarrow}A=F$. This is a contraction since $E\cap F\ne \emptyset$.

Maybe I should prove $E_1 $ and $E_2$ are open in E. I don't know how to do next.

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This hint should help you solve the problem: Since $E\cap F\neq \varnothing$, we have a point $x\in E\cap F$. In particular, $x\in E$ as well as $x\in F$. Now, consider the connected component containing $x$. Does it contain $E$? Does it contain $F$?

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  • $\begingroup$ I like this answer, but I don't think it addresses the OP's question, which is whether OP's attempted proof is valid. $\endgroup$ – John Hughes Oct 18 '16 at 18:13
  • $\begingroup$ @JohnHughes Fair enough. Thanks for posting a complementary answer. $\endgroup$ – Danu Oct 18 '16 at 18:18
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Your proof is not valid, for you need nonempty open sets that disjoint and cover $E$ to show that $E$ is disconnected. You've produced disjoint sets, but either might be empty, and in the case that $E$ and $F$ are the upper and lower semicircles of the unit circle within the plane, the intersection of $A$ with $E$ will be a subset of the upper semicircle, which will never be an open set in the plane unless it is the empty set.

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  • $\begingroup$ In your case, $E$ and $F$ can be open in the unit circle. @JohnHughes $\endgroup$ – User90 Oct 18 '16 at 18:36
  • $\begingroup$ I think it suffices to show that $E_1$ and $E_2$ are open in E not in M. Also, we know that a subset $X$ of $F$ is open in $F$ iff there exists an open subset $U$ of $M$ such that $X=F∩U$. So I can get that $E_1=A \cap E$ is open since $A$ is open in $E\cup F$. @JohnHughes $\endgroup$ – User90 Oct 18 '16 at 18:52
  • $\begingroup$ I agree, although this requires that OP know about the subspace topology, which I didn't want to assume, since at least when I teach this stuff, I'm likely to talk about connectedness pretty early. :) The real problem, of course, is the non-empty-ness. $\endgroup$ – John Hughes Oct 18 '16 at 19:41

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