1. ## Topology and metrics

find two metric functions (distance) $\displaystyle d1 , d2$ on the space $\displaystyle V=(0,1)$ (the interval 0,1).
$\displaystyle d1 , d2$ must support:
a. $\displaystyle V$ is complete with the metric $\displaystyle d1$ and incomplete with $\displaystyle d2$.
b. $\displaystyle d1 , d2$ induce the same topology on $\displaystyle V$ (same topological space).

I apologize for any spelling mistakes and appreciate your help

2. Originally Posted by aharonidan
find two metric functions (distance) $\displaystyle d1 , d2$ on the space $\displaystyle V=(0,1)$ (the interval 0,1).
$\displaystyle d1 , d2$ must support:
a. $\displaystyle V$ is complete with the metric $\displaystyle d1$ and incomplete with $\displaystyle d2$.
b. $\displaystyle d1 , d2$ induce the same topology on $\displaystyle V$ (same topological space).
With the usual metric d_1, where $\displaystyle d_1(x,y) = |x-y|$, V is incomplete. So the problem is to find another metric d_2, under which V is complete. Use the fact that that the interval (0,1) is homeomorphic to the whole real line. The homeomorphism can be used to transport the usual metric on $\displaystyle \mathbb{R}$ across to the unit interval, thereby giving it a complete metric.

Specifically, if $\displaystyle f0,1)\to \mathbb{R}$ is a homeomorphism, then you can take $\displaystyle d_2(x,y) = |f(x)-f(y)|$.

3. thanks for the help.
I wasn't clear enough or fully understood your soulution.
the functions (d1 , d2) must implement both a and b.

4. Originally Posted by aharonidan
thanks for the help.
I wasn't clear enough or fully understood your solution.
the functions (d1 , d2) must implement both a and b.
The functions d_1 and d_2 that I suggested do implement both a and b, apart from the fact that I had them the wrong way round. My d_1 makes the space incomplete and d_2 makes it complete. You wanted it the other way round, but I'm sure you can cope with that.

Take my d_1 first. It is the usual metric on (0,1), and it induces the usual topology. The space is not complete with this metric because for example the sequence (1/n) is Cauchy but does not converge to a point in the space.

The other metric is a bit more complicated. I suggested choosing a function f from (0,1) to $\displaystyle \mathbb{R}$ that is invertible and such that f and its inverse are both continuous. There are many such functions, for example $\displaystyle f(t) = \tan\bigl((t-\frac12)\pi\bigr)$. Now define $\displaystyle d_2(s,t) = |f(s)-f(t)|$. Check first that this is a metric on (0,1). Then notice that it induces the usual topology on the unit interval – that follows from the fact that f and its inverse are continuous.

Finally, the interval (0,1) is complete for the metric d_2. The reason for that is that if $\displaystyle (t_n)$ is Cauchy for the d_2 metric then $\displaystyle (f(t_n))$ will be Cauchy for the usual metric on $\displaystyle \mathbb{R}$. But $\displaystyle \mathbb{R}$ with its usual metric is complete and therefore the sequence $\displaystyle (f(t_n))$ converges to some point x. Thus $\displaystyle d_2\bigl(t_n,f^{-1}(x)\bigr) = |f(t_n) - x|\to0$, in other words $\displaystyle (t_n)$ converges in the d_2 metric.

I hope that makes things clearer.