Proving $\cup_{j\in J} A_j$ is connected.

Question

Let $\{A_j:j\in J\}$ be a family of connected subspaces of a topological space $(X,\tau)$. If $\bigcap_\limits{j\in J}^{}A_j\neq\emptyset$, show that $\bigcup_{j\in J} A_j$ is connected.

Assuming $\cup_{j\in J} A_j$ is disconnected. Then there exists $U,V$ open in the subspace topology such that $U\cup V=\bigcup_{j\in J} A_j$ and $U\cap V=\emptyset$ As $A_j$ is connected $\forall j$ then $A_j$ for any $j$ is a subset of, let's say $U$ otherwise $A_j=A_j\cap U\cup A_j\cap V$ which would imply $A_j$ to be disconnected. If we pick $A_{j+1}$ then $A_{j+1}\subset U$ since $A_j\cap A_{j+1}\neq\emptyset$ and $A_j$ is connected. By induction $A_j\subset U\:\forall j\in J$ which implies that $V=\emptyset$ contradicting the fact $\cup_{j\in J} A_j$ forms a disconnected subset.

Question:

Is the proof right? If not. Why?

Thanks in advance!

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What does $j+1$ mean? Are you assuming that $J\subset\mathbb N$? — José Carlos Santos, Nov 07 '18 at 15:24
@JoséCarlosSantos It was a way to refer to the next element. But since $J$ is not necessarily a subset of $\mathbb{N}$. How should I refer to the next element in the set sequence? — Pedro Gomes, Nov 07 '18 at 15:27
The index set $J$ could possibly not possess any kind of ordering: it is not necessarily a directed set. — Daniele Tampieri, Nov 07 '18 at 15:29
Even when a set is ordered, it can happen that there is a next element. The rationals form an ordered set. What is the next element after $0$? — José Carlos Santos, Nov 07 '18 at 15:32
@JoséCarlosSantos Yes I understood due to the comments the problem with my approach lied with the fact $J$ is not necessarily a subset of the natural numbers and may not even be ordered. So what should I do in such a case? — Pedro Gomes, Nov 07 '18 at 15:34

Henno Brandsma · Accepted Answer · 2018-11-07T22:11:12.707

Suppose $p\in \bigcap_{j \in J} A_j$. We can assume nothing about $J$ (no order etc.)

Suppose $\bigcup_{j \in J} A_j = U_1 \cup U_2$, where both $U_1$ and $U_2$ are open in the union and disjoint.

Fix any $j$: $A_j \cap U_1$ and $A_j \cap U_2$ form a disjoint relatively open partition of the connected set $A_j$, so one of the sets must equal $A_j$ and the other must be empty. Then so $U_{i(j)} \supseteq A_j$ for $i(j)\in \{1,2\}$.

Now if $j \neq j'$ then $p \in A_j \cap A_{j'} \subseteq U_{i(j)}\cap U_{i(j')}$ which shows that $i(j) = i(j')$, so there is one $U_1$, say, such that $U_1 \supseteq A_j$ for all $j$ and then $U_1= \bigcup_{j \in J}A_j $ as well, and the other open set must be empty and this shows that the union is connected.

Proving $\cup_{j\in J} A_j$ is connected.

1 Answers1