$X$ connected space, $\mathcal{U}$ open cover, reachable sets in $\mathcal{U}$.

Question

I have this problem I cannot quite figure out. Let $X$ be connected, and $\mathcal{U}$ be an open cover of $X$. A nonempty set $V\in\mathcal{U}$ is reachable by a nonempty set $U\in\mathcal{U}$ if there exists a sequence $\{U_n\}_{n=1}^{N}$ of elements in $\mathcal{U}$ such that $U_i\cap U_{i+1}$ nonempty and $U_1=U$, $U_N=V$. I want to show every two nonempty elements of $\mathcal{U}$ are reachable.

Here's what I have: Let $U,V$ be nonempty elements of the open cover. Set $U=U_1$. Then since $U_1$ is open and $X$ is connected, $U_1$ isn't closed, so there exists a $u_1\in\overline{U_1}\backslash U_1$ and a set $U_2\in\mathcal{U}$ that contains $u_1$ since $\mathcal{U}$ is an open cover. So I kind of see how I can generate these elements in the open cover to make a sequence. What I cannot see is how I make these open sets "go towards" $V$. I know if I can show that there's some $y\in\overline{V}$, then I can set $U_{N-1}$ to be the open set in $\mathcal{U}$ that contains $y$ and I'll be done, but once again I can't figure out how to add some sort of "direction" to this, or why the finiteness comes into play.

Any suggestions?

Answer 1

You'll probably have better luck assuming that there is a pair $U,V\in\mathcal U$ for which $V$ is not reachable by $U$ and showing that $X$ is not connected.

Let $$A=\bigcup_{U'\text{ reachable by }U}U'$$ and $$B=\bigcup_{V'\text{ not reachable by }U}V'.$$

Now show that $A,B$ form a separation of $X$.

Answer 2

HINT: For $U,V\in\mathscr{U}$ write $U\sim V$ if and only if $V$ is reachable from $U$.

• Show that $\sim$ is an equivalence relation on $\mathscr{U}$.
• Show that if $\mathscr{C}\subseteq\mathscr{U}$ is an equivalence class of $\sim$, then $\bigcup\mathscr{C}$ is open.
• Conclude that $\left\{\bigcup\mathscr{C}:\mathscr{C}\text{ is an equivalence class of }\sim\right\}$ is a partition of $X$ into open sets.
• Now apply the hypothesis that $X$ is connected.