Proposition 1.12

If $$U_1$$ and $$U_2$$ are open, then $$U_1 \cap U_2$$ is open.

ian_mi
Let $$ x \in U_1 \cap U_2 $$. Then $$ x \in U_1 $$ and $$ U_1 $$ is open so $$ \exists \epsilon_1 > 0 $$ such that $$ B(x, \epsilon_1) \subseteq U_1 $$. Similarly, $$ x \in U_2 $$ and $$ U_2 $$ is open so $$ \exists \epsilon_2 > 0 $$ such that $$ B(x, \epsilon_2) \subseteq U_2 $$. Let $$ \epsilon = \min\{\epsilon_1, \epsilon_2\} $$. Then $$ B(x, \epsilon) \subseteq B(x, \epsilon_1) \subseteq U_1 $$ and $$ B(x, \epsilon) \subseteq B(x, \epsilon_2) \subseteq U_2 $$ so $$ B(x, \epsilon) \subseteq U_1 \cap U_2 $$. $$ \epsilon > 0 $$ hence $$ U_1 \cap U_2 $$ is open.

Ian mi 20:26, February 27, 2011 (UTC)ian_mi

'''Right to the point. I changed $$\subset$$ to $$\subseteq$$ as part of my personal mission to reserve the former for strict inclusion. :) --Steven.clontz 02:31, March 2, 2011 (UTC) '''

tagus
(What is wrong with the following counterexample? I don't see my mistake in reasoning, but I know that there's no way that I can be right):

Consider the two sets {$${ y : x = 0 }$$} and {$${ x : y = 0 }$$} in $$R^2$$ (visually, you can interpret this as a horizontal line and a vertical line intersecting at the origin). The intersection of these two open sets, then, is a single point, which is not open.

--208.54.4.55 21:24, March 7, 2011 (UTC)

'''A line is open in $$R^1$$, but not in $$R^2$$ because for any point on the line, any epsilon-neighborhood (in $$R^2$$) contains points (in$$R^2$$) not on the line. Olomana 21:58, March 7, 2011 (UTC)'''