Help Interpreting a Problem Concerning the Union of Connected Sets

Question

I have a problem where I am to prove that the union of connected sets is also connected, but the notation and set up is confusing me slightly.

Suppose $U_a$ is connected for all $a \in I$ where $I$ is an index set. Also, for all $a < b$ we have $U_a \subseteq U_b$. We are to prove that $$\bigcup_{a \in I} U_a$$ is also connected.

I'm not concerned as much with hints on proving the union is connected, but instead I'm confused on what I'm to infer for the $A < B$. Does this mean all of the $U_a$ are nested subsets?

Thanks for any help.

Answer 1

Your assumptions are somewhat unclear. The index set $A$ presumably has a (linear?) order and for $a < b$ we have $U_a \subseteq U_b$.

Suppose that $f: \bigcup_{a \in A} U_a \to \{0,1\}$ is continuous, where the image space has the discrete topology. By connectedness of each $U_a$ (I’ll assume they’re non-empty too WLOG) for each $a$ we have $i(a) \in \{0,1\}$ such that $f[U_a] = \{i(a)\}$. Now if $a \neq b$ we either have $a < b$ or $b < a$ (I am making the assumption that $A$ has a linear order) and in the first case we have $\{(i(a)\} = f[U_a] \subseteq f[U_b] = \{i(b)\}$ so that $(i(a) = i(b)$, the other case yields the same conclusion. So $f$ is constant and so $\bigcup_{a \in A} U_a$ is connected.

The linear order assumption is necessary as there are counterexamples for partially ordered index sets.

Answer 2

The OP does not like the way the problem is stated. How about this:

Let $(I, \le)$ be a totally ordered set that is also an index for a family of connected subspaces $(A_\alpha)_{ \, \alpha \in I} $ of a topological space $X$. Assume that whenever $\alpha_0, \alpha_1 \in I$,

$\tag 1 \alpha_0 \le \alpha_1 \text{ implies } A_{\alpha_0} \subset A_{\alpha_1}$

Then $\bigcup_{\alpha \in I} A_\alpha$ is connected.