# Thread: example of open covering with no finite subcovering

1. ## example of open covering with no finite subcovering

Let $A=Q \cap [0, 1]$ where A is a subset of $R$.
Give an example of an open covering of A which has no finite subcover.

Let p/q be any element of Q. Let I be the closes irrational number to p/q.
Let U= ( (p/q)-I, (p/q)+I ) be an open set containing p/q.
Let { $U_i$} be the set of all $U_i$ for each $(p/q)_i$.
Then the set { $U_i$} for all i, is an infinite open covering of A which has no finite subcovering of A.

Is that right? I'm not sure there IS such an I because, for any rational p/q, aren't there an infinite number of irrationals infinitely close to p/q??

2. Originally Posted by 234578
Is that right? I'm not sure there IS such an I because, for any rational p/q, aren't there an infinite number of irrationals infinitely close to p/q??
Indeed, there is no "closest" irrational number to p/q, because for any irrational number i , you can find another one j , which is closer to p/q, that means |j-p/q|<|i-p/q|.
This is because the irrational numbers are dense in R.

Hint:
First try to find an open covering of an open interval (a,b) , which has no finite subcovering.
Then pick out two irrational numbers a<b in [0,1] and consider the open covering (-1,a), (a,b), (b,2) of A.