There is nothing to prove about union.
so the union is .
If then
But
For every n∈N, let An=[0,1/n], closed interval of the real line. Find the union and the intersection of the indexed family {An}n∈N. Justify your answer.
Well, I got the union:
∪F=[0,1] (where F=the above indexed family)
and the intersection:
∩F={0}
I am having troubles proving this.
For the union I choose an arbitrary x∈∪F. x is an element of ∪F ⇔ ∃n∈N(x∈An). Clearly, if x∈An for some n∈N ⇒ x∈[0,1] for some n∈N. So, 0<x<1/n≤1.
Now, for the second inclusion we choose some x∈[0,1]....?
I don't really know how to prove either. My teacher says the archimedean principle should be in the proof. I have my final tomorrow and wouldl appreciate some help with this. It's one of the concepts I never really got in the class.
Thanks.