Is the completion of a set uncountable?

Question

We can get the completion of a set by including all the limit points of the set. If we take all the infinite Cauchy sequences and try to list them, then we can use Cantor's diagonalisation argument to show that the number of sequences is uncountable. Does this mean there are uncountably infinite limit points? or is it possible that infinitely many sequences converge to the same limit point, so that the completion of a set is countable?

Also how would you go about finding the completion of the natural numbers? Do you need to go through the process of constructing the integers and rationals or can it be done directly?

Thanks

Answer 1

So for a subset of the reals, any closed set is complete and vice-versa. Consider the set $E := \{\frac{1}{n} : n \ge 1\} \subseteq \mathbb{R}$. The closure, i.e. completion, of this set is $E\cup\{0\}$, which is countable.

In regards to your second question, the natural numbers are complete. Any Cauchy sequence of naturals is a sequence that is eventually constant and so converges to that constant value. The rationals are a different story because their metric is not the discrete one, and, further, they are dense in the reals.

Answer 2

Not in general, as the other answer exposes.

However, if your original set has no isolated point, then it is true that the completion is uncountable.

First, note that if the original set has no isolated points, then it is also true that the completion has no isolated points. To see this, suppose it had an isolated point $p$. Since the original set is dense on its completion, every neighbourhood of $p$ must intersect the original set. Being isolated, it follows that $p$ must be in the original set. But then $p$ is an isolated point also on the original set, a contradiction.

It follows then that every singleton has empty interior on the completion. Baire's category theorem then implies that the completion cannot be countable.