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

. Then, consider Q

This set is clearly closed, and since

, the set must be a proper subset of R.

What set **exactly** are you talking about here? You had ....so? Apparently you added the lengths of all the intervals 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.