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?

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.

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.

If I'm right about the source of your doubts, here is the answer...

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.

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.