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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.

Animefan007
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  • You are going to need the hypothesis: $\bigcap U_a\neq\emptyset $ otherwise the statement is false. –  Mar 18 '18 at 05:05
  • Is there any way that $U_a \subseteq U_b$ implies this? Since the index isn't explicitly stated, what can I infer from the ordering? – Animefan007 Mar 18 '18 at 05:13
  • No, I don't think $U_a \subseteq U_b\implies\bigcap_{a\in I} U_a\neq\emptyset$, check this https://math.stackexchange.com/questions/2000408/union-of-connected-sets-also-connected –  Mar 18 '18 at 05:22
  • @Isa. That the intersection is not empty is not needed as the sets are nested. – William Elliot Mar 18 '18 at 06:26
  • Yes, the sets are nested. – William Elliot Mar 18 '18 at 06:28
  • The connected sets (0,r), r > 0 are nested, their intersection is empty and their union is connected. – William Elliot Mar 18 '18 at 06:33
  • @WilliamElliot I though the nonempty intersection was a must have condition in order to prove that the union is connected, interesting to know that can be replaced with the nested condition. –  Mar 18 '18 at 06:57
  • @Isa. There are two cases: all the sets are empty; one is not empty. Proofs are simple. – William Elliot Mar 18 '18 at 09:41

2 Answers2

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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.

Henno Brandsma
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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.

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