- #1

Buri

- 273

- 0

Proof:

Let {U_i} be an open cover for A. Therefore, there must exist a U_0 such that 0 is in U_0. Now since, U_0 is open and 1/n converges to 0, there must be infinite number of points of A in U_0. Now by the well ordering principle U_0 must contain a largest element (Note that {k, k+1,...} has a least element implies {1/k,1/(k+1),...} has a largest element <== this is probably what I'm not 100% sure of about the largest element)...So the points 1/(m-1), ..., 1 will only need AT MOST m-1 more open sets to cover. Therefore, a finite sub cover of {U_i} exists.

Is this proof correct?