As the topic, suppose $A,B$ are connected set with $A\cap \bar{B}\ne \emptyset$, does it implies $A\cup B$ is connected?
$A,B$ are connected set with $A\cap \bar{B}\ne \emptyset\implies A\cup B$ is connected??
1
$\begingroup$
general-topology
1 Answers
3
If $A\cup B$ is not connected but $A$ and $B$ are, the only possible separation of $A\cup B$ is into components $A$ and $B$. But if $\{A,B\}$ is a separation of $A\cup B$, there must be open sets $U$ and $V$ such that $U\cap(A\cup B)=A$ and $V\cap(A\cup B)=B$. Clearly, then, $A\subseteq U\subseteq X\setminus B$, where $X$ is the whole space; is this possible if $A\cap\operatorname{cl}B\ne\varnothing$?
-
0i don't quite see why you need to consider open set $U,V$? – 2012-11-21
-
0@Mathematics: What’s your definition of *not connected*? (I actually need only $U$, but the most common definition of *not connected* gives me both.) – 2012-11-21
-
0Y is not connected mean we can find open set $A,B$ both open in Y such that $A\cap B=\emptyset$ and $Y=A\cup B$ – 2012-11-21
-
0@Mathematics: You’ve misstated that a bit: you mean that you can find open sets $A$ and $B$ in $Y$ such that $A\cap V=\varnothing$ and $A\cup B=Y$. And I’m saying that here those open sets would have to be the given sets $A$ and $B$, since those are known to be connected. Thus, $A$ is relatively open in $A\cup B$, so there is an open $U$ in $X$ such that $U\cap(A\cup B)=A$, and **that** implies that $A\cap B=\varnothing$, contrary to hypothesis. – 2012-11-21
-
0do you mean $B$ instead of $V$ in the above comment? – 2012-11-21
-
0@Mathematics: Yes, I do. – 2012-11-21