Let $\{A_i\}$ a family of connected subsets on topological space $X$ and let $A_0 \in \{A_i\}$ such that $A_i \cap A_0 \neq \emptyset.$ Prove that $\bigcup A_i$ is connected.
My Attempt:
Call $S= \bigcup A_i$, obviously $A_0 \subset S$. We have $A_0 \cap A_i$ is clopen in $A_i$ and $A_i$ is connected then the only clopens in $A_i$ are $A_i$ and $\emptyset$, since $A_0 \cap A_i \neq \emptyset$ so $A_0 \cap Ai = A_i$, that is, $A_i \subset A_0$ for all $i$. Futhermore $S \subset A_0$ so $S = A_0.$
We conclude that the only set $A_0$ such that $A_0 \cap A_i \neq \emptyset$ is $S$, then $S$ is a disjointed union of non empty sets then $S$ isn't connected.
Where is my wrong??
There is a variation of this question and I need tips:
Let $\{A_i\}$ a family of path-connected subsets on topological space $X$ and let $A_0 \in \{A_i\}$ such that $A_i \cap A_0 \neq \emptyset.$ Prove that $\bigcup A_i$ is path-connected.
Best Answer
Suppose that $C$ is a clopen and non-empty subset of $A:= \bigcup_i A_i$ (i.e. clopen in its subspace topology). We must show $C=A$ for connectedness. Pick $p \in C$ and fix $i_1$ such that $p \in A_{i_1}$.
Because subspace topologies are transitive, $C \cap A_{i_1}$ is clopen in $A_{i_1}$ and non-empty as it still contains $p$. So by connectedness of $A_{i_1}$ we have
$$C\cap A_{i_1} = A_{i_1} \implies A_{i_1} \subseteq C\tag{1}$$
Also, by assumption, $A_0 \cap A_{i_1} \neq \emptyset$, so by $(1)$ we see that $C\cap A_0$ is clopen in $A_0$ and non-empty, as it contains $A_{i_1} \cap A_0$, so again by connectedness of $A_0$:
$$C\cap A_0 = A_0 \implies A_0 \subseteq C\tag{2}$$
Now for an arbitary $i$, $\emptyset \neq A_i \cap A_0 \subseteq A_i \cap C$, the latter set is clopen in $A_i$ so again $A_i \subseteq C$ and as this holds for all $i$, $A \subseteq C$ so $A=C$ and $A$ is connected.
For path connectedness we use the standard composition of paths: if $p_1:[0,1]\to X$ is a path from $x_0$ to $x_1$ and $p_2: [0,1]\to X$ is one from $x_1$ to $x_2$, the map $p_1 \ast p_2: [0,1]\to X$ can be defined as $(p_1 \ast p_2)(t)=p_1(2t)$ for $t \in [0,\frac12]$ and $(p_1 \ast p_2)(t)=p_2(2t-1)$ for $ t \in [\frac12,1]$ and by the pasting lemma $p_1 \ast p_2$ is continuous and a path from $x_0$ to $x_2$. This can be iterated via a third point as well, etc.
Apply this to $A$ for path-connected $A_i$: if $x,y \in A$ so $x \in A_{i_x}$ for some $i_x$ and $y \in A_{i_y}$ for some $i_y$, then fix $x' \in A_0 \cap A_{i_x}$ and $x'' \in A_0 \cap A_{i_y}$ and then combine a path $p_1$ from $x$ to $x'$ (both in the path-connected $A_{i_x}$), a path $p_2$ from $x'$ to $x''$ (both in $A_0$) and $p_3$ from $x''$ to $y$ (both in $A_{i_y}$) to get a path from $x$ to $y$ in $A_{i_x} \cup A_0 \cup A_{i_y} \subseteq A$.