Let $\displaystyle A=Q \cap [0, 1] $ where A is a subset of $\displaystyle R$.

Give an example of an open covering of A which has no finite subcover.

My Answer:

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 {$\displaystyle U_i$} be the set of all $\displaystyle U_i$ for each $\displaystyle (p/q)_i$.

Then the set {$\displaystyle U_i$} 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??