Finding an irrational not covered in standard proof that $\mu(\mathbb{Q} \cap [0,1]) = 0$ [duplicate]

Possible Duplicate:
How would one go about proving that the rationals are not the countable intersection of open sets?
Constructing a number not in $\bigcup\limits_{k=1}^{\infty} (q_k-\frac{\epsilon}{2^k},q_k+\frac{\epsilon}{2^k})$

When I was considering a comment on Conditions for integrability, I began to think of a few facts on Lebesgue measure. I often used to have to reconvince myself that the measure of the rationals was zero, and I thought about this in particular.

A standard way of showing that the measure of $\mathbb{Q} \cap [0,1]$ is $0$ is to come up with your favorite enumeration of the rationals in $[0,1]$, and then put an $\epsilon$-ball around the first, an $\frac{\epsilon}{2}$-ball around the second, an $\frac{\epsilon}{2^2}$-ball around the third, and so on, so that the overall length is bounded by (in this case) $4 \epsilon$. One might go on to show that the measure of $\mathbb{Q}$ is $0$ by showing that a countable union of null sets is null, by doing an analogous argument with $\epsilon$ covers.

Suppose for a moment that we look at the $\epsilon = 1/8$ covering, so that at most $1/2$ of $[0,1]$ is covered. Suppose we also that our enumeration of the rationals is the standard one of Cantor, i.e.

It might be easier if we don't remove duplicates (still a proof of denumerability) so that we know the general $n^\text{th}$ number easier.

Then my question is: can we explicitly find an irrational not covered by one of the $\epsilon$-balls? Perhaps it's deceptively easy. Can we find many?

Now, I'm not really set of the $\epsilon = 1/8$ case, nor on this particular enumeration. It's the spirit that counts, if you'll pardon the pun.

Solution 1:

The irrational number $\phi=(\sqrt5-1)/2$ has the property that $$|\phi-(p/q)|\gt1/(3q^2)$$ for all integers $p,q$ so you can certainly choose your enumeration and your epsilons in such a way as to avoid covering it.