# Topologically complete spaces

• Apr 28th 2012, 05:54 PM
ModusPonens
Topologically complete spaces
Hello.

I'm having trouble doing exercise 7.43.6.c of Munkres' Topology. It says:

Show that an open subspace of a topologically complete space is topologically complete.

Now, why isn't the open interval (0,1), considered as a subspace of R, a counterexample to what the question is afirming? (since the sequence 1/n is a Cauchy sequence that converges to 0, a point which isn't in (0,1))

Thanks.
• Apr 29th 2012, 04:58 AM
girdav
Re: Topologically complete spaces
Hello,
$(X,d)$ is topologically complete means that we can find a metric $d'$ which gives the same topology as $d$ such that $(X,d')$ is complete. Here, with $X=(0,1)$ it doesn't work with the usual metric but with $d'(x,y)=\left|\tan \frac{\pi x}2-\tan \frac{\pi y}2\right|$ it should.
• Apr 29th 2012, 10:27 AM
ModusPonens
Re: Topologically complete spaces
I think I get the idea. That metric is to avoid sequences that converge to the frontier of (0,1), right? By the way, if that's the idea, that metric doesn't avoid that sequences converge to 0. Shouldn't it be $d'(x,y)=|tan(\pi (x-1/2))-tan(\pi (y-1/2))|$
• Apr 29th 2012, 10:30 AM
girdav
Re: Topologically complete spaces
It's possible, I didn't make the computations.