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