# Cauchy sequences and complete spaces

I'm trying to fully understand the complete metric spaces so I came up with this reasoning: every metric space X can be extended to a complete metric space Y and every Cauchy sequence in X is still Cauchy in Y, but it's also convergent in Y. This means that Cauchy sequences are always convergent, but not always they converge inside the space.

Now, the closure of a set X is the set that contains all limits of all convergent sequences of elements of X, so it should also contain all limits of all Cauchy sequences.
But this is the definition of a complete metric space, so now I'm lost.

Is it true that every closed metric space is complete? I know it's true in Euclidean spaces, but I thought it wasn't true in general.

Edit: I just realised that we know that every closed subset of a complete metric space is complete, but since every metric space can be extended to a complete space, this means that every closed space is complete, right?

So is completeness less strong than closedness?

• I'll leave the answer to an analyst, but a quick comment: 'Cauchy sequences are always convergent' - no, most people would take issue with this statement, although your overall logic in that sentence is sound. If the sequence doesn't converge within the space in question, don't call it convergent. Welcome to MSE Emanuele, nice opening Q! Commented Sep 29, 2020 at 16:58

Every metric space is a closed subset of itself, and therefore, since there are non-complete metric spaces, the sentence “every closed metric space is complete” is false.

But, yes, if $$X$$ is a metric space, you can extend it to a complete metric space $$Y$$ (its completion), and the closure of $$X$$ in $$Y$$ is the whole $$Y$$.

• Dear Jose, thank you indeed for this enlightening example. Intuitively, I do believe that the closure of X is Y. On the other hand, I wish to enquire: closure -- with respect to what topology? Just any topology? Or the standard topology induced by the distance d ? Commented Sep 20, 2021 at 22:58
• @Michael_1812 The set $X$ is a subset of the metric space $Y$. I am talking about the closure of $X$ in $Y$. Commented Sep 20, 2021 at 23:09
• The closure of a set is the union of the set with all its limit points. A point p is a limit point of X iff every neighbourhood of p contains at least one point of X different from p. By saying `neighbourhood', we automatically imply the existence of some topology. Please correct me if I am wrong. [I suspect that you are implying the standard topology defined by open balls in the sense of the distance d of the metric space. Is that right?] Commented Sep 20, 2021 at 23:26
• Yes. The space $Y$ is a metric space with respect to some metric $d$, and its topology is obtained from that metric. Commented Sep 20, 2021 at 23:35
• This is what I thought. Thank you very much! Commented Sep 20, 2021 at 23:36
1. When you say : "so it should contain all limits of Cauchy sequences", it seems that you are concluding that the set of convergent sequences is the same as the set of Cauchy sequences. This happens when $$X$$ is complete, by definition.

2. In fact, a way to define the completion of $$(X,d)$$ is precisely to consider an equivalence relation on the set of all Cauchy sequences of $$X$$, where $$(a_n)$$ and $$(b_n)$$ are equivalent iff $$\lim d(a_n,b_n)=0$$. This will "fill up the holes in $$X$$". A well known result illustrates this idea : a metric space $$(X,d)$$ is complete iff any decreasing sequence of nonempty closed sets $$F_1\supset ...\supset F_n\supset ...$$ with $$diam(F_n)\rightarrow 0$$ is such that $$\bigcap F_n$$ is non empty (in fact, a singleton).

3. If $$F\subseteq X$$ is a closed subset of a complete metric space, consider any Cauchy sequence $$(x_n)$$ in $$F$$. Since the distance is the same in both spaces, the sequence is also Cauchy in $$X$$, where it converges to some $$x$$ due to completeness of $$X$$. But $$F$$ is closed, so $$x\in F$$.

• In item (3) of this proof, the closedness is understood in the sense of the so-called standard topology, the one induced by the distance $d$ and its open balls. Commented Sep 21, 2021 at 0:35