Intuition: All Sets in an Open Cover Can be Finitely Connected to a Given Set

Question

Let $X$ be a connected topological space, and let $\mathcal{U}$ be an open cover of $X$. It's not hard to show that given any $U_0 \in \mathcal{U}$, for every $U \in \mathcal{U}$ there exists a finite sequence $U_0,\ldots,U_k$ of elements in $\mathcal{U}$ where $U_k = U$ and each $U_i \cap U_{i+1}$ is nonempty. But this is extremely surprising to me, since we're not assuming compactness or any other "finiteness" condition on the topological space, and the cover is entirely arbitrary. Before proving this result I would've had no trouble believing that there are "big" enough topological spaces, like the long line, where the result would not hold.

Is there a way I can look at this where the result seems more obvious? Does this result provide any deeper insight into limitations on the "size" of topological spaces?

Answer 1

It's as surprising or unsurprising as the fact that the transitive closure of a relation needs only to take points finitely many steps away.

Define a relation on $\mathcal{U}$, $U\sim U'$ iff. $U\cap U'\neq\emptyset$. The claim is then that the transitive closure of $\sim$ (it is already reflexive and symmetric) is the trivial relation. You see this by noting equivalence classes under this relation must have union a clopen subset of $X$, and the only clopen subsets are $X$ and $\emptyset$.

I'd say this is very unsurprising, and gives a strong justification for the definition of connectivity. That $X$ is connected means we can "build a bridge", or perhaps creating a series of links as in a bracelet, to go from any $U$ to any other $U'$. The converse holds, that is, we can use this property to prove $X$ is connected.

Answer 2

This answer is essentially the same as the equivalence relation in the original answer, but uses graph theory as an intuition. Probably not helpful if you don't know some basic graph theory.

On one level, you can think of this as being about paths in a graph (in the combinatorial sense of a graph.) You can take the bipartite graph with one part consisting of $X$ and the other part consisting of $\mathcal U,$ with an edge $(x,U)$ if $x\in U.$

There is a path from $U_0$ to $U$ in this graph if and only if there is a $U_1,\dots,U_k=U$ and an $x_i\in U_{i-1}\cap U_i$ for $i=1,2,\dots,k,$ which is what you wanted.

Then we can see each connected component of the graph will have nodes a subset $V\subseteq X$ and the set $\mathcal U'=\{U\in\mathcal U\mid U\subseteq V\}.$ In particular, in each component the $V$ will be a union of the elements in $\mathcal U'.$

This is all true when we are talking just about sets $X$ and a covering $\mathcal U,$ without any topology involved.

When $X$ is a topology, and the $\mathcal U$ is an open cover with non-empty sets, the associated $V$ in each component will be open, and they are pairwise disjoint, and thus, if $X$ is connected, there is only one component.

So this really about a discrete math concept, a graph, and thus we are restricted to finite paths. Graphs have disjoint connected components, so each $x\in X$ is only in one connected component.

It is only once we get to topology that we see that the connected components of the graph correspond to a partition of $X$ into open sets.