I understand the main principles of Cauchy sequences and metric spaces, but I have a particular question about determining whether or not a space is a complete space. If a space has all cauchy sequences converge to a point contained within that space, then it is a complete space.

My question is: How do you determine if there is a cauchy sequence that doesn't converge?

I realize this is a rather vague question, so let me clarify a bit using an example, and you can just provide an answer to this particular example. For the space containing all real numbers $\mathbb R^1$, if our space is defined as the interval $(0,1)$ then it is not a complete space since there is a Cauchy sequence $x_n = \frac{1}{n}$ for which the limit converges to $0$, which is not contained in the space.

  • How did they come up with this sequence $x_n = \frac{1}{n}$? Is it just from experience working in the field of complete spaces and you start recognizing things like this? Or is there a more formal way to find it?
  • For that sequence, if $n \le 1 $ then $x_n$ is not in the given space. Do we need to specify that $n \gt 1$ when defining the sequence?

Thanks! If you can answer those two bullet points I think I'd understand this a bit more thoroughly =).

  • $\begingroup$ (i) Yes, we do have to say $n\gt 1$, or else use the somewhat "uglier" $\frac{1}{n+1}$. (ii) Hard to answer, visualization, awareness of the geometry. Also, importantly, if you have "done it before" you can see whether a familiar idea can be recycled/modified. $\endgroup$ – André Nicolas Aug 26 '13 at 15:16

How did they come up with this sequence $x_n = 1/n$?

In my experience, coming up with counterexamples is something of an art, and sometimes much more difficult than proving a theorem! As you seem to agree, when proving theorems and solving problems, there is often some vague direction as to where you need to go; however, with these counterexamples, there does not usually seem like there is a well-defined route!

But this does not mean that counterexamples come out of nowhere. Instead, counterexamples tend to arise when you begin understanding the the object at hand a lot more. A complete space, for example, is an object in which sequences that clump together must eventually arrive at some destination. But in the set $(0, 1)$, no matter how close you get to the end points, you cannot get to the endpoint! So there is no way that $(0,1)$ can be a complete space. Formalizing this into the language of sequences gives us the sequence $x_n = 1/n$.

In short, the key is to really understand what you are dealing with. Also, when looking for counterexamples to certain theorems, you might want to examine the assumptions used in the proof, and exploit those weak spots when thinking up a counterexample.

For that sequence, if $n \leq 1$, then $x_n$ is not in the given space.

When talking about sequences, we usually only consider $n \geq 1$---in particular, the positive integers---as indices. So unless otherwise stated, you can assume that $n \in \mathbb{Z}^+$. Formally, sequences are usually defined as functions $\mathbf{x}: \mathbb{Z}^+ \to X$, where $X$ is some space. For convenience, however, we often write $\mathbf{x}(n) = x_n$, and denote the set of all $\mathbf{x}(n)$ by $\{x_n\}_{n=1}^\infty$, or something alike.

  • $\begingroup$ Ok, good answer =) thanks. Only thing is, for this sequence, $n = 1$ gives us a value that does not lie in the space. I mean...obviously all of the $x$'s must lie in our space, so would we just assume to start at $n = 2$? $\endgroup$ – Spaderdabomb Aug 26 '13 at 15:18
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    $\begingroup$ What you say about coming up with counterexamples must be a deep truth, because I agree with it but in this case I also agree with the opposite. ("It is the hallmark of any deep truth that its negation is also a deep truth." -- Niels Bohr) Namely, here I think the easiest way to proceed is to view this as an example of a theorem, rather than a counterexample. Theorem: if $X$ is a metric space and $Y$ is a subset of $X$ which is complete when endowed with the subspace metric, then $Y$ is closed in $X$. Proof: if not, there is a sequence in $Y$ converging to an element of $X \setminus Y$. $\endgroup$ – Pete L. Clark Aug 26 '13 at 15:24
  • $\begingroup$ I think that you’ve one of the most common ways of coming up with counterexamples: trying to prove that there is none, and seeing what makes the proof fail. $\endgroup$ – Brian M. Scott Aug 26 '13 at 16:40

That's a kind of a comment, if you'd like... I just recall how i gave an example of a metric space which is not complete, the lecturer asked me the question, and then i thought, hmmm, what if i take a complete space, for instance $[0,1]$ (i know it's complete since it is compact, a general fact) and then take away any point (any point there is a limit point, i.e. there is some sequence converging to it, hence this sequence will be Cauchy automatically), say $\frac{1}{2}$ (i don't know why i chose that particular one...). Then the sequence which converged to this point in $[0,1]$ will still be Cauchy, this property doesn't vanish anywhere, but it won't be convergent... In fact it's obvious that the original space (i chose it to be [0,1]) needn't be complete. You just have to take any metric space with a convergent sequence (hence you know it is Cauchy) and then take away its limit, and you'll certainly know at least one Cauchy sequence which is not convergent... That waqs my idea of building such an example. Sorry if this was kind of mess... Just a bit of my experience about studying completeness... :)

  • $\begingroup$ +1: Note that my answer can be viewed as a further development of the ideas here. $\endgroup$ – Pete L. Clark Aug 26 '13 at 15:38

In a case such as this, the intuition is often that you first identify something that is not in the space but is in its closure in a slightly bigger space: in this case $0$. Then you try to find a Cauchy sequence in the original space that converges to that point in the bigger space. There's nothing magical about $1/n$, it's just a sequence of positive numbers that converges to $0$.

Conventionally a sequence is usually taken as starting with $n=1$. If you wanted to start at some other $n$, you would have to specify that: e.g. if you're sequence was $x_n = \pi/2^n$, then you want $n \ge 2$ to make $x_n \in (0,1)$.

  • $\begingroup$ If we take $n = 1$ in my example, doesn't it also not lie in the space though? $\endgroup$ – Spaderdabomb Aug 26 '13 at 15:23

This is more of a sharpening of what other answers have already said, but here goes:

Theorem: Let $(X,d)$ be a metric space, and let $Y \subset X$. If $(Y,d)$ is complete, then $Y$ is closed in $X$.

Proof: If not, then there is $x \in X \setminus Y$ and a sequence $\{y_n\}$ in $Y$ with $y_n \rightarrow x$. Since $\{y_n\}$ is convergent in $X$, it is Cauchy in $X$, hence it is Cauchy in $Y$. Since limits in metric spaces are unique if they exist, $\{y_n\}$ is not convergent in $Y$.

This provides a nice approach to the present example, since $(0,1)$ is naturally given as a non-closed subset of the (complete) metric space $\mathbb{R}$. Note also that $(0,1)$ is homeomorphic to $\mathbb{R}$, but this is not good enough: here it is really the "extrinsic" rather than the "intrinsic" topology that matters.

If we add an additional hypothesis on $X$, we can get a better result:

Theorem: If $(X,d)$ is a complete metric space and $Y \subset X$, then $(Y,d)$ is complete if and only if $Y$ is closed in $X$.

Proof: Even without the completeness hypothesis on $X$ we saw that it is necessary for $Y$ to be closed. Conversely, if it is then every Cauchy sequence in $Y$ is convergent in the complete space $X$, and since $Y$ is closed, the limit is an element of $Y$. So $Y$ is complete.

[Corollary: Any compact metric space is complete.

Sneaky proof: Let $X$ be compact, and embed it in its completion $\tilde{X}$. Then $X$ is a compact subset of the Hausdorff space $\tilde{X}$, so it is closed. Apply the theorem.]

Here is the moral, I think: if you are trying to decide whether a metric space is complete, embed it in a complete metric space and see whether the image is closed. In theory this always succeeds since any metric space embeds in its completion. In practice, the last sentence is a tautology but nevertheless the strategy seems to work in many cases.


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