Showing a topological space covered by connected subspaces is connected

Question

'Let $X$ be a topological space and let $(U_i)_{i \in I}$ be a cover of $X$ by connected subspaces $U_i$. Supposed for all $i,j \in I$ there exists some $n \geq 0$ and $k_0,...,k_n \in I$ such that $k_0 = i, k_n = j$ and $$U_{k_0} \cap U_{k_1} \neq \emptyset, U_{k_1} \cap U_{k_2} \neq \emptyset, ..., U_{k_{n-1}} \cap U_{n} \neq \emptyset$$ Show that $X$ is connected.

My intuition tells me the proof is along the lines of since each $U_i$ is connected, there are no subsets $A,B \subseteq U_i$ for $i \in I$ such that $A$ and $B$ are disjoint (unless they are the empty set.) Also, since the intersection of $U_{k_i} \cap X/U_{k_i}$ is non empty (as there always exists at least some other subcover it intersects with) there are no two nonempty subsets $A,B \subset X$ such that $A \cap B = X$. Thus $X$ is connected.

I was looking for some help formulating this into a nice mathematical proof. Thanks.

Answer 1

You've got the right rough picture. Here's an outline of a proof that you should be able to fill in. Proceed by contradiction:

1) Start by assuming there are disjoint open sets $A$ and $B$ whose union is $X$.

2) Pick a point in $A$ and a point in $B$ and construct the $U_i$ connecting them.

3) Show that there is some $U_i$ that has nonempty intersection with both $A$ and $B$.

4) Show that this contradicts the assumptions on $A$ and $B$ and the fact that $U_i$ is connected.

Answer 2

I assume you know the following basic lemma: if $A \subset X, B \subset X$ are connected, and $A \cap B \neq \emptyset$, then $A \cup B$ is connected as well. (If not, say so in the comments, and I'll add its proof).

Suppose $X = A \cup B$ where $A,B$ are open, non-empty and disjoint (so striving for a contradiction). Pick $x \in A, y \in B$, and then pick $i \in I $ such that $x \in U_i$, $y \in U_j$, using that they form a cover.

The assumption tells us that there is $n\ge 0$ and there are $k_0,\ldots,k_n \in I$ such that $k_0 = i, k_n = j$ and for all $l \in \{0,\ldots,n-1\}$, $U_{k_l} \cap U_{k_{l+1}} \neq \emptyset$.

Now, for all $j \le n$, $V_j = \cup_{l = 0}^{l = j} U_{k_l}$ is connected. This is shown by induction, using the first lemma: $j = 0$ is just the asumption that $U_{k_0} = U_i$ is connected. If the statement holds for $j = j_0$, $j_0 < n$, it holds for $j=j_0 + 1$ as well, as $V_{j_0}$ is connected (inductive assumption), and $U_{k_{j_0}}$ (a subset of it) intersects $U_{k_{j_0+1}}$ by the chaining property, and $V_{j_0 + 1}$ is a union of these two intersecting connected subsets. This shows the inductive step as well, and so in particular, $V_n = \cup_{l=0}^n U_{k_l}$ is connected, but now $A \cap V_n$ and $B \cap V_n$ are both open in $V_n$, still disjoint, and both non-empty as $x \in A \cap U_{k_0}$ and $y \in B \cap U_{k_n}$. This contradicts the connectedness of $V_n$, and we are done.