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?

  • $\begingroup$ You haven't actually stated what you wanted to prove. It appears to be to show that $V$ is reachable by $U$ for any $U,V\in\mathcal U$. Is that correct? $\endgroup$ – Tim Raczkowski Oct 5 '15 at 0:20
  • $\begingroup$ yeah sorry, edited that, you are correct. $\endgroup$ – Jake Oct 5 '15 at 0:32
  • $\begingroup$ I don't know how to do this one ${}{}{}$ $\endgroup$ – D_S Jun 8 '16 at 2:38

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 $X$.

  • $\begingroup$ Ah yes I see. Do you know of a way to do it without a contradiction? I feel like this doesn't offer much insight to the problem somehow...I mean that I'm not very satisfied. $\endgroup$ – Jake Oct 5 '15 at 0:45
  • $\begingroup$ Sorry. Just came up with that argument on the fly. By the way, the argument is not a true proof by contradiction. Rather, the contrapositive is proved. $\endgroup$ – Tim Raczkowski Oct 5 '15 at 0:49
  • $\begingroup$ I imagine a direct proof of the statement would require a lot of work. Something else to keep in mind. Connectedness is defined as a negative. That is, there is not a separation. Because of that, proving the contrapositive is often a more efficient way to go. $\endgroup$ – Tim Raczkowski Oct 5 '15 at 1:01

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.
  • $\begingroup$ Nice idea Brian. When the OP was talking about a direct proof, I was thinking along the lines he/she had mentioned in the post. $\endgroup$ – Tim Raczkowski Oct 6 '15 at 14:34

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