1

Let $\{A_\alpha\}$ be a collection of connected subspaces of $X$; let $A$ be connectted subspace of $X$. Show that if $A\cap A_\alpha \neq \emptyset$ for all $\alpha$, then $A\cup(\cup A_\alpha)$ is connected.

I know this theorem:enter image description here

And as every set in this union has a point in common with $A$, I guess I need to use this. If $\{U_\alpha \}$ was countable I could easily use this theorem with induction, but now I'm not sure if I can use such an argument.

I mean $A\cup A_1$ is connected, therefore $A\cup A_1 \cup A_2$ connected and so on.

Kasper
  • 14,040
  • 18
  • 71
  • 129

2 Answers2

3

If you assume that $A\cup(\cup A_{\alpha})$ is disconnected. Therefor it can be written as the union of two disjoint non-empty closed subsets $U_{1}$ and $U_{2}$. But because $A\cap A_{\alpha}\neq \emptyset$ for all $\alpha$ you get $A\cap U_{1}\neq \emptyset$ and $A\cap U_{2}\neq \emptyset$. But this would mean that $A$ is disconnected, this is a contradiction with the assumption that $A$ is connected .

Kaladin
  • 1,535
2

Note that $A \cup A_{\alpha}$ is connected for all $\alpha$ and therefore $A \cup (\bigcup A_{\alpha}) = \bigcup (A \cup A_{\alpha})$ is connected by your theorem because it is the union of connected sets that have $A$ in common.

dani_s
  • 1,660
  • 11
  • 12