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?



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".

  • $\begingroup$ 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

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