# Open sets on the unit circle?

1. Oct 9, 2009

### gmn

The question I had was to show that if a function is continuous, open and bijective then it is a homeomorphism. At first I said "no" because I thought of the example showing that [0,2pi) is not homeomorphic to the unit circle S. I knew that f(x)=(sinx,cosx) is a continuous bijection whose inverse fails to be continuous. I don't need help answering this question (I know that it is a homeomorphism) but I am confused because I keep reading that f fails to take open sets in [0,2pi) to open sets in the circle and intuitively (the reason I first thought the answer was no) I thought that f would take open sets to open sets. Specifically I read that f takes some open intervals in [0,2pi) to half-open intervals in S, and I couldn't figure out what half-open intervals in S means or what they would be. I also read that [a,b) is an open interval in R, which makes me confused because I would think that [a,b) is niether closed nor open in R. I thought I understood open sets but I don't think I do that well now.

I guess a concrete version of my question is how does f(x)=(sinx,cosx) fail to take open sets in [0,2pi) to open sets in S, specifically what are they? Is it because the circle is really in R2 ?

2. Oct 9, 2009

### Dick

You need to restrict the topology of the real line to the interval I=[0,2*pi). So the 'open sets' in I are intersections of open sets in the reals with I. E.g., (-pi,pi) intersect I is open in I. That means [0,pi) is an open set in the induced topology. 'I' is the whole space.