Proof: Openness of U∩V and U∪V in Rn

[stunner5000pt]

I am not well 'accustomed' to these kind of proofs so please bear with my stupidity
Suppose U and V are both open subsets in Rn. Prove that U intersection V and U union V are open as well.

Dfeinition of open is that you cna center a ball about a point a in a set such that that ball is completely contained in the set.

So let there be a ball with center a radius delta in U such that
B(a,\delta_{1}) = { x: ||x-a||< \delta_{1}} for U and
B(b,\delta_{2}) = { y: ||y-b||< \delta_{2}} for V

now the for the union
B(c, \delta) = {z: ||z-c|| < \delta}
and c belongs to the union of U and V. and picking delta = min (delta 1, delta 2) we can say that the ball is completely contained in U union V and the union is open?

If this correct we can move to the intersection...
do i do a similar procedure? Please help?

Thank you for you help and advice!

 

[HallsofIvy]

If B(a,\delta_{1}) = { x: ||x-a||< \delta_{1}} is in U and
B(b,\delta_{2}) = { y: ||y-b||< \delta_{2}} is in V, then
B(c, \delta) = {z: ||z-c|| < \delta}
with \delta= min(\delta_1, \delta_2)
is in both U and V and therefore is in both U union V and U intersect V- you don't have to do it two different ways!