230 CHAPTER 11. ABSTRACT MEASURE AND INTEGRATION
Proof: Let U be an open set and let y ∈U. Since U is open, B(y,r)⊆U for some r > 0and it can be assumed r/
√n ∈Q. Let
x ∈ B(
y,r
10√
n
)∩Qn
and consider the cube, Qx ∈B defined by
Qx ≡n
∏i=1
(xi−δ ,xi +δ )
where δ = r/4√
n. The following picture is roughly illustrative of what is taking place.
yx
Qx
B(y,r)
Then the diameter of Qx equals(n(
r2√
n
)2)1/2
=r2
and so, if z ∈ Qx, then
|z−y| ≤ |z−x|+ |x−y|
<r2+
r2= r.
Consequently, Qx ⊆U. Now also,(n
∑i=1
(xi− yi)2
)1/2
<r
10√
n
and so it follows that for each i,|xi− yi|<
r4√
n
since otherwise the above inequality would not hold. Therefore, y ∈ Qx ⊆U . Now let BUdenote those sets of B which are contained in U. Then ∪BU =U.