# Why is Completeness not a Topological Property?

I am trying to answer the question: Show why completeness is not a topological property.

My answer: $\mathbb{R}$ and the set $(0,1)$ are homeomorphic, but $\mathbb{R}$ is complete while $(0,1)$ is not.

My question to you all: Does this answer the question? I feel like I am not quite seeing what is going on with completeness and why it is not a topological property. Can someone give me another example?

• Completeness is a property of metric spaces. Not all topological spaces are metric spaces. – John Douma Dec 8 '15 at 5:09
• A property which is preserved under continuous maps is a topological property. Your counter example is enough to show that completeness isn't. – Non-Being Dec 8 '15 at 7:47
• @JohnDouma: Just because this is a property of metric spaces does not mean it cannot be invariant under homeomorphism. This is a good heuristic, but no proof at all. For instance, completeness PLUS total boundedness is a topological property (equivalent to compactness plus metrisability). Not all topological spaces are metric spaces, but metrisability is preserved by homeomorphisms, so it is not really an issue. – tomasz Dec 8 '15 at 14:05
• @Non-Being: Normality is a topological property that is not preserved by continuous maps. Topological properties are those that are preserved by homeomorphisms. – Brian M. Scott Dec 8 '15 at 18:36
• Oh, I don't know about normality. Thank you for the correction. – Non-Being Dec 10 '15 at 3:50

Another way to see it is: One can metrize the space $(0,1)$ in (at least) two different ways that generate the same topology. One such metric is the ordinary Euclidean metric on $(0,1)$; a second metric is what you get when you pull the Euclidean metric on $\mathbb{R}$ back to $(0,1)$ via a homeomorphism.

In one of those metrics, $(0,1)$ is complete, but in the other it is not -- again, despite the fact that the two metrics generate the same topology. That shows that completeness depends not on the topology per se but rather on the metric one associates with the topological space.

• why do you say that these two metrics generate the same topology? how is this claim related to the homeomorphism – Liu Sha Feb 21 '18 at 4:56

As the technical component of your question has already been addressed, I would like to tackle the intuitive component, implicit in:

I feel like I am not quite seeing what is going on with completeness and why it is not a topological property.

The source of this uncertainty is due, I suspect, to the fact that since topology adds sufficient structure to a set to cater for convergence (in particular, a topology determines whether or not any given sequence converges), surely it should be sufficient to accommodate completeness, which has convergence as its whole concern.

Intuitively, a complete space is one in which all of the sequences that are trying to converge actually do converge. It turns out that while the actually do part can be catered for by the topological structure, the trying to part can't be. Why?

Well, let's take as our example the sequence $$(\frac{1}{n})_{n\in\mathbb{N-\{1\}}}:=\frac{1}{2},\frac{1}{3},\frac{1}{4},...\subset(0,1)$$ This sequence certainly seems to be 'trying to' converge to $0$, but is it really? The reason our eyes say 'yes' is because our eyes add a metrical structure that the topology just doesn't 'see'. The topology doesn't 'see' these things as getting closer and closer to $0$ because it is distance-agnostic.

To visualize this, imagine the interval to be made of rubber. By pinching any two consecutive members of the sequence $\frac{1}{n}$ and $\frac{1}{n+1}$ and stretching them apart, you could make every separation one inch if you wanted, and you wouldn't destroy the topological structure of the thing. The result would be a sequence that no longer looks like it is trying to converge at all.

In short, to introduce the notion of 'trying to converge' you need to add metrical structure on top of the topology and there are infinitely many ways to do this. Some choices will lead to completeness, while others will lead to incompleteness.

• This is a superb explanation of the intutive aspects of the question. – mweiss Jun 20 '16 at 16:55
• This seems like a most excellent example, but I'm not quite understanding it yet. You're essentially saying that there's no notion of open neighborhood of $0$ in the topology of $(0,\,1)$, whereas we subconsciously add one, right? There's an infinite number of ways to define neighborhoods of $0$ when you add it to the interval to get $(0,\,1)\cup\{0\}$, right? I mean, I can apply the pinching and stretching argument to the sequence $1/2 + 1/3,\, 1/2 + 1/4, \,1/2 + 1/5,\, ...$ as well, the only difference here is that the limit belongs to the interval. – Selene Routley Jun 27 '16 at 0:09
• Otherwise put: it seems that this is simply a case of one's naturally wanting to think about the open interval embedded in $\mathbb{R}$ and its relationship with its complement in $\mathbb{R}$, whereas the topology of the interval doesn't know anything about any embedding, and there is a wide choice for possible embeddings. – Selene Routley Jun 27 '16 at 8:36
• @WetSavannaAnimalakaRodVance Hmmm... I'm pretty sure that's not what I'm saying, because even if we close the interval to $[0,1]$ to make it complete, neither 0 nor 1 are contained in any open neighborhood. Here are three claims to clarify my point: 1) If you assume the ordinary absolute-value metric, the sequence is trying to converge, but the limit point doesn't exist, and so the space is incomplete. 2) If we map $\mathbb{R}$ stereographically to $(0,1)$, and assign pairs of points in the latter the same distance their preimages had in $\mathbb{R}$, then suddenly the sequence diverges and... – Physics Footnotes Jun 29 '16 at 10:15
• @WetSavannaAnimalakaRodVance $(0,1)$ becomes complete by inheriting convergence from $\mathbb{R}$. 3) If $(0,1)$ does not ship with a metric, then it is neither complete, nor incomplete, as their is no concept of a Cauchy-sequence with respect to which we can declare certain limit-points to be missing. (I think I agree with your second statement about implied embeddings though). – Physics Footnotes Jun 29 '16 at 10:19

Your example is fine, as others have shown. There are generalisations of completeness to wider classes. E.g. there is a notion of uniform space, which is set with another type of structure, a uniformity, and every uniformity $\mathcal{U}$ on $X$ defines a topology $\mathcal{T}(\mathcal{U})$ on $X$. This is similar to how a metric also defines a topology. In fact, every metric defines a uniformity, and the topology generated by that uniformity is the same as the topology generated by the metric. Such a space has two structures, or even three (for metric spaces). The uniformity allows us to define notions like Cauchy sequence (or Cauchy net, even), uniform continuity, completeness, uniform connectedness, totally boundedness and others. But diffent uniformities, even if they generate the same topology (so from the most "coarse" standpoint they are the same) can be very different in completeness properties, e.g. There is a uniformity (or even metric) on $(0,1)$ that makes it complete, e.g. $d(x,y) = |x - y| + |\frac{1}{x} - \frac{1}{y}| + |\frac{1}{1-x} - \frac{1}{1-y}|$, and yields the same topology. If we also have a metric we can talk about Lipschitz mappings and Hausdorff dimension and other metric specific things, that cannot be defined in general uniform spaces, say. A uniformity can give a finer look at a certain topology, one could say.

A metric space that has an equivalent metric (same topology) but which is complete is called completely metrisable. This can be characterised using just topology, namely that $X$ is metrisable (this can be characterised using Bing-Nagata-Stone and other theorems) and $X$ is $G_\delta$ (countable intersection of open sets) in its Cech-Stone compactification. This latter property is called topologically complete. E.g. every locally compact Hausdorff space is topologically complete (as it is even open in any of its compactifications). In topologically complete spaces the Baire theorem holds, and this generalises the fact that it holds both in complete metric spaces (really completely metrisable ones) and in locally compact Hausdorff spaces.

Your answer is correct. Topological properties are preserved by homeomorphisms. $(0,1)$ is homeomorphic to $\mathbb{R}$, but yet $\mathbb{R}$ is complete, while $(0,1)$ is not under the Euclidean metric.

Also note that (as stated above) completeness is a property attributed to metric spaces, whereas other topological spaces exist.

The completenes property is dependent upon the metric. If you change the metric, the same Cauchy sequences which once converged may not under your new metric. Conceptually, size or distance are not properties preserved under homeomorphism. Topology is concerned with structure - limit point relations on the set - continuity, connectedness, compactness.

• What does this add to the existing answers? – Noah Schweber Jun 18 '16 at 21:44
• It answers the question. Completeness is not a topological invariant because it is dependent the metric. None of the other answers made that clear. – Rob Jun 18 '16 at 23:34
• That's simply false - read e.g. the first paragraph of the current highest-voted answer. – Noah Schweber Jun 19 '16 at 2:14
• Actually, none of the rated posts (with the possible exception of the last) answered the question. (0,1) is not complete for the simple reason that there are Cauchy sequences that approach either zero or one which, by definition, are excluded from the set. The change in the metric used to fold R1 into (0,1) has absolutely nothing to do with the completion property for these sets. I would appreciate no further comments from you, nor any more of your annoying downgrades. – Rob Jun 20 '16 at 11:50
• The question is not whether $(0, 1)$ is complete - the question is why completeness is not a topological property. One way to demonstrate this - as the top answer does - is to provide two metrics on $(0, 1)$ which induce the same topology, one of which is complete (the one coming from $\mathbb{R}$) but the other of which (the standard one) is not. So you are incorrect - the first answer is indeed quite relevant. (cont'd) – Noah Schweber Jun 21 '16 at 20:46