is topologically complete means that we can find a metric which gives the same topology as such that is complete. Here, with it doesn't work with the usual metric but with it should.
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))
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