Proposition 1.11

If $$U_i$$ is open for each $$i$$ in some indexing set $$I$$, then $$\bigcup_{i \in I} U_i$$ is open.

ian_mi
Let

$$\displaystyle x \in \bigcup_{i \in I} U_i $$

Then $$ \exists i_x \in I $$ such that $$ x \in U_{i_x} $$. $$ U_{i_x} $$ is open so $$ \exists \epsilon > 0 $$ such that

$$\displaystyle B(x, \epsilon) \subseteq U_{i_x} \subseteq \bigcup_{i \in I} U_i $$

Thus $$\bigcup_{i \in I} U_i $$ is open.

Ian mi 20:14, February 27, 2011 (UTC)ian_mi

'''Yep. I changed $$\subset$$ to $$\subseteq$$ to fit our notation. --Steven.clontz 03:37, March 2, 2011 (UTC)'''