# Difference between topologically complete space and complete metric space

Definition: Topological space $$(X,\tau)$$ is called topologically complete if there is metric $$d$$ on $$X$$ which induces the topology $$\tau$$ of $$X$$ and $$(X,d)$$ is complete metric space.

Also the following fact is true: If $$f:X\to Y$$ where $$f$$ is homeomorphism and $$Y$$ is topologically complete then $$X$$ is also topologically complete.

The proof is not difficult because if $$d$$ is a metric on $$Y$$ which induces topology of $$Y$$ and $$(Y,d)$$ complete metric space then one can define the metric $$\rho$$ on $$X$$ as follows: $$\rho(x_1,x_2):=d(f(x_1),f(x_2))$$. One can show that $$\rho$$ induces the topology of $$X$$ and $$(X,\rho)$$ is complete metric space.

However, I was wondering about the following moment: the above reasoning shows that the notion of topologically complete is topological property. However completeness is not topological property. The standard example is $$(0,1)$$ and $$\mathbb{R}$$, they are homeomorphic, $$\mathbb{R}$$ is complete but $$(0,1)$$ is not since the sequence $$x_n=1-\frac{1}{n}$$ is Cauchy sequence but does not converge in $$(0,1)$$.

Can anyone explain to me why the above reasoning cannot be applied to the case of $$\mathbb{R}$$ and $$(0,1)$$? I guess that $$(0,1)$$ is not complete in the standard euclidean metric inherited from $$\mathbb{R}$$ but it maybe complete in the different metric which induces its subspace topology.

Anyway I would be very grateful for useful answer!

• What would it mean to "apply the above reasoning to the case of $\mathbb{R}$ and $(0,1)$"? Can you write out explicitly what argument you have in mind? Commented Nov 3, 2019 at 4:45
• I believe the reason why you can not apply the same reasoning comes from the definition of the metric $p(x_1,x_2)=d(f(x_1),f(x_2))$. For $(0,1)$ is an subspace of the real line you will have that $p$ is just a restriction of the usual metric thus the open set can not be complete with the metric $p$ and the construction fails. Commented Nov 3, 2019 at 4:52

The answer you suggested is correct.

"Topologically complete" means that there exists a metric with a certain property. But it might not be the metric you want.

For $$(0,1)$$ there is such a metric. But that metric is not the same as the usual one which it inherits from $$\Bbb R$$, although it does happen to induce the same topology.

In that metric, distances get bigger and bigger as you approach the ends of tne interval, so that (for example) the sequence $$10^{-1}, 10^{-2}, 10^{-3}, \ldots$$ is not a Cauchy sequence: the elements do not get closer together! So its non-convergence does not prevent $$(0,1)$$ from being metrically complete.

To see how this works, let's consider a slightly simpler example. Let $$\def\rplus{{\Bbb R^+}}\rplus$$ be the positive real numbers $$(0,\infty)$$. Certainly this space is incomplete with respect to the usual metric $$d$$, since it is missing $$0$$.

The mapping $$f:x\mapsto \log x$$ is continuous in both directions, and is a homeomorphism $$\rplus\leftrightarrow \Bbb R$$. Therefore, $$\rplus$$ is topologically complete.

But what about the sequence $$\big\langle\frac1i\big\rangle = \big\langle1, \frac12, \frac13, \frac14\ldots\big\rangle$$? Isn't it a Cauchy sequence in $$\rplus$$ that fails to converge to a point in $$\rplus$$? Let's look closely.

As you said:

one can define the metric $$\rho$$ on $$X$$ as follows: $$\rho(x_1,x_2):=d(f(x_1),f(x_2)).$$

So for $$x_1, x_2\in \rplus$$ take \begin{align}\rho(x_1,x_2) & = d(\log x_1, \log x_2) \\ & = \left| \log x_1 - \log x_2\right|.\end{align}

For $$\big\langle\frac1i\big\rangle$$ to be a Cauchy sequence in $$\rplus$$, we need to find, for any given $$\epsilon$$, some $$N$$ so that whenever $$m$$ and $$n$$ are larger than $$N$$, it is true that: \begin{align} \rho\bigg(\frac1m,\frac1n\bigg) & <\epsilon \\ \bigg|\log \frac1m - \log\frac1n\bigg| & <\epsilon \\ \bigg|\log n- \log m\bigg| & <\epsilon \end{align}

But this we obviously can't do, the logarithm function increases to infinity as $$m$$ and $$n$$ grow, and no matter how big $$m$$ is, there will always be a very much larger $$n$$ that will make $$\bigg|\log n- \log m\bigg|$$ as large as we like, and certainly one that will make it larger than $$\epsilon$$. So the sequence $$\big\langle\frac1i\big\rangle$$ is not a Cauchy sequence under the $$\rho$$ metric, and its failure to converge to a point in $$\rplus$$ is not a proof that $$\rplus$$ is metrically incomplete.

(This is what I meant when I said that the distances get bigger and bigger as you approach the end of the interval. In the usual metric, the sub-interval $$(0, e^{-9})$$ of $$\rplus$$ is very small. But in the $$\rho$$ metric, this same interval is enormous! It contains the two points $$e^{-10}$$ and $$e^{-1000}$$, which are at a distance of $$\left|\log e^{-10} - \log e^{-1000}\right| = 990$$ units apart. And it also contains the point $$e^{-10^{100}}$$ which is even farther away.)

The answer is, as you said:

I guess that $$(0,1)$$ is not complete in the standard euclidean metric inherited from $$\Bbb R$$ but it maybe complete in the different metric which induces its subspace topology.

Exactly so. $$\rplus$$ is not complete in the usual inherited euclidean metric, because in that metric $$\big\langle\frac1i\big\rangle$$ is a Cauchy sequence that does not converge to a point of $$\rplus$$. But the metric $$\rho$$ also induces the same topology, and under this metric, $$\rplus$$ is complete.

A preferred definition of topologically complete is
S is a topologically complete topological space
when S is homeomorphic to A complete metric space.

Clearly, complete metric spaces are topologically complete topological spaces. In particular, R with the usual metric is topologically complete, a topologically complete topological space.

(0,1) with the inherited subspace metric is not a complete metric space. As (0,1), however is homeomorphic to R, it is a topologically complete topological space.

• A preferred definition of "topologically complete" is being completely regular and a $G_\delta$ in its Cech-Stone compactification. Thm: if $X$ is topologically complete (in this sense) and also metrisable then $X$ has a complete compatible metric. Commented Nov 3, 2019 at 6:43