Originally Posted by

**davismj** I was asked to verify whether or not the closure of Q is R.

I said, it sure is, and gave what I thought was sound reasoning.

Then I was presented with a counterexample:

Since Q is countable, denote each element of Q as $\displaystyle x_n$. Then, consider Q $\displaystyle \cup [x_n - 2^{-n},x_n +2^{-n}].$ This set is clearly closed, and since $\displaystyle \sum_{n=0}^{\infty} \frac{2}{2^n} = 2$, the set must be a proper subset of R.

What set **exactly** are you talking about here? You had $\displaystyle \mathbb{Q}\cup [x_n-2^{-n},x_n+2^{-n}]$....so? Apparently you added the lengths of all the

intervals $\displaystyle [x_n-2^{-n},x_n+2^{-n}]^\infty_{n=0}$ and you got, wrongly by the way, 2 (it is 4, but it never minds)...so, again? What is your point?

Tonio

I cannot refute this, but it doesn't seem right to me.