# How to prove that $X$ is a completion of $Y$, for some metric spcaes $(X,d)$ and $(Y,d)$?

In our Real Analysis class we did a proof for existence of completion of metric space. However, I do not understand how do we use it practically. i.e How do I prove that $$X$$ is a completion of $$Y$$, for some metric spcaes $$(X,d)$$ and $$(Y,d)$$?

For example : How do I prove that $$(\mathbb{R},d_1)$$ is a completeion of $$(\mathbb{Q},d_1)$$? Here, $$d_1$$ is the standard Euclidean metric

$$\underline{\text{My attempt}}$$ -

I believe we should prove that $$\mathbb{Q}$$ is subset of $$\mathbb{R}$$ and $$\mathbb{R}$$ is complete. This would mean that all Cauchy sequence in $$\mathbb{Q}$$ converges in $$\mathbb{R}$$.

However, consider this case -

Let $$(W,d)$$ be an incomplete Metric space and $$W \subset Y \subset X$$ and $$X,Y$$ are complete (with same metric).

Now, by my argument both $$X,Y$$ can be the completion of $$W$$. Is this alright? I read somewhere in Stack Exchange that completion is unique(I didn't really understand that answer)

$$\underline{\text{Approach to resolve this}}$$ - We choose the smallest, complete set which contains $$W$$ as its completion. We can prove that, since $$X$$ is complete any of it's closed subset is complete. We can also prove that $$\overline{W}$$ is the smallest closed set that contains $$W$$. Hence, $$\overline{W}$$ is the completion of $$W$$ (this would explain the notation in the completion proof).

Thus in the title - I just have to check if $$Y$$ is dense in $$X$$ to prove that $$X$$ is a completion of $$Y$$.

This would also solve my example question - We know that $$\mathbb{Q}$$ is dense in $$\mathbb{R}$$ and $$\mathbb{R}$$ is complete. Thus $$\mathbb{R}$$ is completion of $$\mathbb{Q}$$.

I am fairly confident that this is correct. However, I am a Physics Master's student this is my first "Real" Math course ;) So I would appreciate if someone let's me know if this right and if not the corrections.

• What is $d_1$? ${}$ – José Carlos Santos Sep 27 '19 at 11:13
• $d_1$ is the standard Euclidean metric – Indigo1729 Sep 27 '19 at 11:29
• This follows by the universal property of completion of metric spaces. – ε--δ Sep 27 '19 at 11:59

To show that $$\mathbb{R}$$ is the Cauchy completion of $$\mathbb{Q}$$, it is not sufficient to show that $$\mathbb{Q}$$ is contained in $$\mathbb{R}$$ and $$\mathbb{R}$$ is complete. In fact, this just means that the completion of $$\mathbb{Q}$$ is contained within $$\mathbb{R}$$.

For example, the completion of $$\mathbb{Q}$$ is contained in $$\mathbb{C}$$ and the complex numbers are complete, but the completion of the rational numbers is not the complex numbers.

You should also verify that every element of $$\mathbb{R}$$ is the limit of a sequence of rational numbers (which follows directly from the fact that $$\mathbb{Q}$$ is dense in $$\mathbb{R}$$ for example) in order to be sure that $$\mathbb{R}$$ is actually $$\mathbb{Q}$$'s completion. This resolves your worry about the uniqueness theorem for completions which you referred to.

One final point: one doesn't normally define the Cauchy completion in order to identify completions of metric spaces they already know (e.g. $$\mathbb{Q}$$ in your example). Instead, knowing that you can take the completion lets you construct new metric spaces which can be useful in their own right. Again for example, knowing that completions exist and taking the completion of $$\mathbb{Q}$$ is a way to construct the real numbers $$\mathbb{R}$$.

One nitpick: "I believe we should prove that $$\Bbb Q$$ is subset of $$\Bbb R$$ and $$\Bbb R$$ is complete. This would mean that all Cauchy sequence in $$\Bbb R$$ converge in $$\Bbb R$$."

To sum up, viewing $$\Bbb Q$$ as a subset of $$\Bbb R$$ (so we can avoid speaking of the embedding map and the image of $$\Bbb Q$$ under it), you need to show that

• the metric on $$\Bbb Q$$ is the restriction of the metric on $$\Bbb R$$
• $$\Bbb Q$$ is dense in $$\Bbb R$$ and of course
• $$\Bbb R$$ is complete under its metric

and that is essentially what you found out by yourself.

You're missing one... critical point!

It is not sufficient for $$X$$ to be a completion of $$W$$ that $$W \subseteq X$$ and $$X$$ is complete. You also need $$W$$ to be dense in $$X$$.

That being said, $$\mathbb R$$ is indeed a completion of $$\mathbb Q$$.