# Simple Topology question

1. Oct 17, 2005

### 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!

2. Oct 17, 2005

### HallsofIvy

Staff Emeritus
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!

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