Suppose that $V, W$ are closed in $X$ and that $V\cap W, V\cup W$ are connected. Show that $V$ and $W$ are connected.
My attempt:
Suppose that $V=U_1\cup U_2$ where $U_i$ are open in $V$, not empty and $U_1\cap U_2=\emptyset$. This induces a separation for $V\cap W$: $$ V\cap W = (U_1\cap W)\cup(U_2\cap W).$$
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$U_i\cap W$ is open in $V\cap W$
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$(U_1\cap W)\cup (U_2\cap W)=\emptyset$ as otherwise $U_1\cap U_2 \ne \emptyset$.
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Now, I need to show that both sets are not empty. We know that $V\cup W$ is connected and closed in $X$. So, for all $x\in V\cup W=U_1\cup U_2\cup W$ and all neighborhoods $Z$ (in $X$!) of $x$ we have that $Z\cap (V\cup W) \ne \emptyset$. We can choose $x\in U_1 \subseteq V\cup W$. But then I would have to find a neighborhood of $x$ in $X$. I can't find one… How can I complete this bullet point?
Thanks.
Best Answer
You certainly know that a space $Y$ is connected if and only if each continuous map $f : Y \to 2$ is constant. Here $2 = \{0,1\}$ with the discrete topology.
So let $f : V \to 2$ be continuous. Since $V \cap W$ is connected, $f$ is constant on $V \cap W$. W.l.o.g. we may assume $f(x) = 0$ for $x \in V \cap W$. Define $$F : V \cup W \to 2, F(x) = \begin{cases} f(x) & x \in V \\ 0 & x \in W \end{cases}$$ This map is well-defined and continuous because $V$ and $W$ are closed in $V \cup W$. Hence $F$ is constant which implies that $f$ is constant.