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Why do we care about the notion of a space being complete? Why don't just consider closed spaces? If the space is closed we know that the limits of a sequence exist and are in the set which is a property that is obviously desirable.

So what is the benefit in introducing this weaker notion of complete spaces and dealing with these weaker types of sequences called Cauchy sequences, as opposed to just using closed spaces and the stricter convergent sequences? What are we allowed to do with complete spaces that wouldn't be possible with closed spaces? It would also be interesting to hear the historical motivation for complete spaces if anybody is aware of it.

csss
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    Closed in what ambient space? Completeness is an intrinsic property, closedness isn't. – Daniel Fischer Dec 13 '16 at 19:46
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    Note that any topological space is closed (because the empty set is open). What is the definition of a closed space? How does this imply "that the limits of a sequence exist"? – hardmath Dec 13 '16 at 19:46
  • I can have a sequence of rational numbers that converges to an irrational number. It is a nice property when not only do we know that the limits exist but that they exist within that space. – Doug M Dec 13 '16 at 19:46
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    What do you mean by "closed", here? Typically "closed" only means something for a subset of an ambient space -- but if we take that definition, then it certainly is not true that limits need to exist. That's exactly why we have complete spaces. – Mees de Vries Dec 13 '16 at 19:46
  • @MeesdeVries I was under the impression that iff every convergent sequence in a set $A \in \mathbb{R}^n$ converges to a point in $A$, then $A$ is closed? So if the property of being closed gives us the existence of limits in the space, why would we care about Cauchy seqeunces which are just a weaker type of sequence? – csss Dec 14 '16 at 08:36
  • Well, there's the minor issue that ZF doesn't prove that all sequentially-closed subsets of $\mathbb{R}^n$ are closed. ​ Otherwise, it depends on whether you mean "convergent in $A$" or "convergent in $\mathbb{R}^n$". ​ ​ ​ ​ –  Dec 14 '16 at 10:37

3 Answers3

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First, a peripheral technical issue: Unlike completeness, closedness is not an absolute property; it is a relative property. A space $S$ is complete or not complete. But it makes no sense to say that $S$ is closed or not closed. All you can say is that it is closed relative to some larger space $T$. For example, consider the set $(0, \infty)$ of positive real numbers. This is not closed relative to $\Bbb R$, because it fails to contain its limit point $0$. But it is closed relative to $\Bbb R\setminus\{0\}$ because now it does contain all its limit points—zero is no longer a limit point because we removed it from the space. And of course it is closed relative to $(0,\infty)$ because every topological space is a closed subset of itself. So your suggestion does not even make sense, because there is no such thing as a "closed space".

(The issue can be seen more generally if you consider the definition of a closed set: $C$ is closed if $X\setminus C$ is open. But this depends on what $X$ is.)

But there is a larger issue. The Greeks noticed 2300 years ago that the rational numbers are incomplete. Putting it anachronistically, they noticed that there are Cauchy sequences that do not converge. A theory about such spaces is needed. You asked

what is the benefit in introducing this weaker notion of complete spaces and dealing with these weaker types of sequences called Cauchy sequences

This is like asking why physicists spend so much time trying to understand friction forces, when everyone knows the problems are much easier to solve if you simply assume that there is no friction. Why don't physicists take the easy way out? Because there is friction and the whole point of physics is to solve problems about how the world really is, friction and all.

Like physics, mathematics is not just made up from nothing. We want to solve certain kinds of problems and understand the way numbers and shapes work. The rational numbers seem to be a fundamental sort of object, one of the things that is important in the structure of the universe. The whole point of mathematics is to understand how things like the rational numbers work and how they are related to other things like the real numbers.

And the most important part of the answer to that question turns out to be: the rational numbers contain sequences where the elements get closer and closer together (cauchy sequences) and sort of converge, even though there is no rational number that they converge to. (A common example is $1, \frac32, \frac75, \frac{17}{12}, \frac{41}{29}\ldots$, with $\frac ab$ followed in each case by $\frac{a+2b}{a+b}$. We can show that if this sequence were to converge to some limit $L$, we would have $L^2 = 2$. But we have known for a long time that there is no such $L$.)

But the real numbers behave differently: if elements of a sequence get closer and closer, there must be some single point $L$ that they get close to. The two kinds of numbers are fundamentally different in this way, and we are stuck with it, just like we are stuck with friction. We need to give that difference a name. The name is “completeness”.

MJD
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Note that completeness is a metric property whereas closedness is a topological property, and hence, depends on the topological space with which it is referred to. So, any set can be made closed by choosing the appropriate topology and as noted in the comments, every set is closed in itself. But, not all sets can be made complete by changing the metric maintaining the topology, an example being $\mathbb{Q}$, the set of rational numbers. Thus, completeness is crucial in discussion involving convergence, which is what mathematical analysis is most associated with. As pointed out in the comments, completeness is an intrisic property, and points must be added to make a non-topologically complete/non completely metrizable space complete. The historical importance of completeness comes noticing that all major branches of Analysis-Real, Fourier, Functional etc. have developed by completion of spaces that were hitherto not complete: The concept of irrational numbers was the first notion of completeness, though introduced without the exact notion of completeness, but nevertheless the idea was completion. The theory of Lebesgue integral is nothing but born out of completion of the space of Riemann integrable functions, The convergence theorem of Fourier series requires the completeness of $L^2(\mu)$ for its proof and the list goes on.

vidyarthi
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    "But, not all sets can be made complete by changing the metric." - What if I change the metric to the discrete metric? –  Dec 13 '16 at 20:08
  • @Bungo I think sets can be made closed/open with respect to the discrete metric, but not complete. – vidyarthi Dec 13 '16 at 20:14
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    Any metric space with the discrete metric is complete. This is because under that metric, a sequence is Cauchy if and only if it is eventually constant. –  Dec 13 '16 at 20:17
  • @Bungo But, arent $\mathbb{Q}$ non-completely metrizable? – vidyarthi Dec 13 '16 at 20:22
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    Yes, what you probably meant to say is that we can't make $\mathbb Q$ complete by changing the metric while maintaining the same topology. We can of course make it complete by changing to the discrete metric, but that changes the topology as well (all sets are now open). –  Dec 13 '16 at 20:24
  • @vidyarthi There is no complete metric on $\mathbb{Q}$ which induces the usual topology. But there is certainly a metric on the set $\mathbb{Q}$ which yields a complete metric space, namely the discrete one. Bungo is right - any set can be made a complete metric space via the discrete metric. If we demand that the metric be compatibile with some topology, however, that changes things. – Noah Schweber Dec 13 '16 at 20:25
  • @NoahSchweber thanks, edited the post. – vidyarthi Dec 13 '16 at 20:28
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If the space is closed we know that the limits of a sequence exist

That's... false. Try $0,1,0,1,\ldots$.

Did you mean that every sequence has a convergent subsequence? I hate to break this to you, but that's only true if the closed set is a subset of a complete space.

Or did you mean that every Cauchy sequence has a limit? Again... that's only true in a complete space.

Completeness is important, you just have only been using examples of closed sets that are in complete spaces. And it sounds like the properties you "like" are actually properties of complete spaces. Bluntly you are confusing a lot of related concepts.

  • "Closed" is a topological property. "Complete" is a property of metric spaces only. $[0,1]$ is closed in $\mathbb R$ and $[0,1] \cap \mathbb Q$ is closed in $[0,1] \cap \mathbb Q$.
  • "Closed" only makes sense relative to a containing topological space. "Complete" is an intrinsic property.
  • "Limit points" can be defined just in terms of open sets and topology. Closed sets contain all their limit points in any topological set.
  • "Completeness" means if a sequence tries to converge, then it hs something to actually converge to. "Convergence" only makes sense in metric spaces. This is not true for $\mathbb Q$ where sequences might try to converge to $\sqrt 2$ for instance. (1, 1.4, 1.41, 1.414... is such a sequence).
djechlin
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  • If limit points are defined in terms of open sets and topology, how are we using them when we talk of completeness? Are you saying there is a separate definition of limit points in terms of complete spaces that coincides with the definition in terms of open sets and topology? – csss Dec 14 '16 at 08:43
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    @csss : ​ We can use them for the equivalent definition "Every Cauchy filter has a limit point.". ​ ​ ​ ​ –  Dec 14 '16 at 10:41