Prove that $\bigcup_{n=1}^\infty A_n $ is connected set

Question

Let $X$ be a topological space.

Question 1: Assume $(A_i)_{i \in I}$ be a family of connected subset and $A_i \cap A_j \neq \emptyset, \forall i, j\in I $. Prove that $\bigcup_{i\in I} A_i$ is connected.

My attempt:

As $\bigcap_{i \in I} A_i \subset A_i \cap A_j $ with fixed $i,j \in I$, so that $\bigcap_{i \in I} A_i \neq \emptyset$. For that reason with $A_i$ is connected forall $i \in I$ then according to my lemma I have studied $\bigcup_i A_i$ connected. Is it a true answer ? If not why it is wrong ?

Question 2: Assume that $(A_n)_{n \in \mathbb{N}}$ be a sequence of connected set and $A_n \cap A_{n+1} \neq \emptyset, \forall n $. Prove that $\bigcup_{n=1}^\infty A_n $ is connected set.

Can I apply the above proof ? If not what is the solution ? Thank you.

Answer 1

1) From $A_i\cap A_j\neq\varnothing$ for all $i,j\in I$ it cannot be concluded that $\bigcap_{i\in I}A_i\neq\varnothing$.

So your answer on the first question is not correct.

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Let $A:=\bigcup_{i\in I}A_i$ and suppose that $\{S,T\}$ is a separation of $A$ in the sense that $S$ and $T$ are non-empty subsets of $A$ that are both open in $A$ and satisfy $S\cap T=\varnothing$ and $S\cup T=A$.

Assuming the existence of a separation is actually the same thing as assuming that $A$ is not connected, and we will try to find a contradiction.

For every $i\in I$ the set $A_i$ must - because it is connected - be a subset of $S$ or a subset of $T$.

Then the condition that $A_i\cap A_j\neq\varnothing$ for every pair of indices $i,j\in J$ leads to the stronger conclusion that for all $i\in I$ the set $A_i$ is a subset of $S$ or for all $i\in I$ the set $A_i$ is a subset of $T$.

(Observe that $A_i\subseteq S$ and $A_j\subseteq T$ leads to $A_i\cap A_j=\varnothing$)

That however means that we have $A\subseteq S$ or $A\subseteq T$.

A contradiction is found now and we conclude that $A$ is connected.

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2) With induction on $n$ it can be proved that for every $n$ the set $B_n=\bigcup_{k=1}^nA_k$ is connected.

Further for every $n,m$ we have $B_n\cap B_m\neq\varnothing$ and now we can apply the former part.

The conclusion is that $\bigcup_{n=1}^{\infty}B_n$ is connected, and further it is evident that $\bigcup_{n=1}^{\infty}A_n=\bigcup_{n=1}^{\infty}B_n$.

Answer 2

Answer to Question 2: suppose $\cup_i A_i$ is the union of two disjoint non-empty open sets $C$ and $D$. By connectedness of $A_1$ we conclude that $A_1 \subset C$ or $A_1 \subset D$. Similarly $A_2 \subset C$ or $A_2 \subset D$. But $A_1 \subset C$ implies that $A_2 \subset C$ because $A_2 \subset D$ would imply that $A_1\cap A_2 =\emptyset$. Now use indiuction to show that each $A_i$ is contained in $C$ or each $A_i$ is contained in $D$. This leads to the contradiction that one of the sets $C,D$ must be empty.

Answer 3

This answers the first question where the number of A's may be uncountable.
Assuming there's more than one A, they all are not empty.
Well order I.
For each k in the order of I, let C(k) = $\cup${ A$_i$ : i < k }.
If the union of all the A's is not connected, then there is a
least j for which C(j) is not connected.
Show K = $\cup${ C(i) : i < j } is connected.
Show C(j) $\cap$ K is not empty.
Thus C(j) $\cup$ K = C(j+1) is connected, a contradiction.