TITLE: Difference between topologically complete space and complete metric space
QUESTION [7 upvotes]: 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!

REPLY [2 votes]: 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.