Let , be metric spaces and assume is compact. Show that if is a bijective and continious function, then is a homeomorphism.
And does this hold even if we dont assume that M is compact?
Yes the point I was trying to make is that being continuous is the same as being an open map or a closed map.
Open and closed maps - Wikipedia, the free encyclopedia
Since this is the cases we just need a few facts.
A map is closed if it maps closed sets to closed sets.
Let be closed. Since M is a compact space and A is a closed subset of M A is compact.
Now since is continuous by hypothesis is a compact set because the continuous image of a compact set is always compact.
Since compact sets are closed f must be a closed map.
I wasn't too familiar with open and closed maps, thats why i didn't get your point. But now it makes sense. thank you!
How about if you drop the assumption that M is compact. Could you prove that a continuous bijection not necessarily has a continuous inverse? Any examples?
Let M be the unit interval [0,1] with the discrete metric, and let N be the same set with the usual metric. The identity map is then a continuous bijection, but the inverse is not continuous.
The discrete metric is given by whenever . In this metric, every set is open. So every mapping from a space with the discrete metric is continuous.