1
$\begingroup$

I want to prove the following statement.

If $X$ is a connected space then every open covering $\{U_j:j\in J\}$ has the following property: for each pair $U_{j_1},U_{j_n}$ there are finitely many $U_{j_2},...,U_{j_{n-1}}$ such that $U_{j_i}\cap U_{j_{i+1}}\neq\emptyset$ for every $i\in\{1,...,n-1\}$.

And I tried this:

Suppose the contrary. There exist an open covering $\{U_j:j\in J\}$ and $U_{j_1},U_{j_n}$ such that for every $U_{j_2},...,U_{j_{n-1}}$ exists $i\in\{1,...,n-1\}$ with $U_{j_i}\cap U_{j_{i+1}}=\emptyset$. I want to prove $X$ is not connected.

For example, if we consider an empty subfamily, we must get $U_{j_1}\cap U_{j_{n}}=\emptyset$. If $U$ is another element of the open covering, then either $U\cap U_{j_1}=\emptyset$ or $U\cap U_{j_n}=\emptyset$. Then I tried to prove something like this:

$$X=U_{j_1}\cup U_{j_n}\cup\bigcup_{j\in J\setminus\{j_1,j_n\}}U_j$$

But I don't really know if these sets are disjoint.

Can anyone give me a hint?

Thanks.

$\endgroup$
1
$\begingroup$

The common proof technique is a follows. Fix an open cover $\mathscr U$ and a point $x \in X$. Then define a subset $X' \subset X$ where we say $y \in X'$ if and only if there is some finite "chain" $\{U_1, U_2, \ldots , U_n\} \subset \mathscr U$ such that $x \in U_1$, each $U_m \cap U_{m+1} \ne \varnothing$, and $y \in U_n$. Show $X'$ is both open and closed (closed is harder). Thus we must have $X' = X$. Finally show this guarantees any pair of cover elements can be connected by a "chain".

$\endgroup$
  • $\begingroup$ Thanks @Daron Are you sure the closedness is hard? I don't know if I have a mistake in this. If $y\in X\setminus X'$, let $U_y$ in the cover such that $y\in U_y$. If $X\setminus X'$ were not open, $U_y$ can't be contained in $X\setminus X'$. Let $y_0\in U_y\cap X'$. Then there is a chain connecting $x$ and $y$. $\endgroup$ – Tanius Oct 15 '15 at 0:35
  • $\begingroup$ I made a mistake. I believe I had to say $y\in (X\setminus X')\setminus Int(X\setminus X')$. $\endgroup$ – Tanius Oct 15 '15 at 0:38
  • $\begingroup$ I didn't say it was hard, only harder than showing openness (which is at least shorter to write down once you know how). Your proof makes sense. I suppose you could combine the proof that $X'$ and $X-X'$ are clopen by showing their (shared) boundary is empty, since any boundary element lives inside a cover element that forms the last link in a chain that links a point of $X'$ to a point of $X-X'$. $\endgroup$ – Daron Oct 15 '15 at 9:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.